Solutions Manual For Foundations of College Chemistry, 16th Edition by Morris Hein, Susan, Cary Will by ACADEMIAMILL

CHAPTER 1

AN INTRODUCTION TO CHEMISTRY SOLUTIONS TO REVIEW QUESTIONS 1. (a)

(b)

A hypothesis is a tentative explanation of certain facts to provide a basis for further experimentation. A theory is an explanation of the general principles of certain phenomena with considerable evidence to support it. A theory is an explanation of the general principles of certain phenomena with considerable evidence to support it. A scientific law is a simple statement of natural phenomena to which no exceptions are known under the given conditions.

2. (a)

hypothesis

(b)

hypothesis

(c)

observation

(d)

theory

(e)

observation

(f)

scientific law

3. (a)

A liquid has a definite volume but not a definite shape.

(b)

A gas has an indefinite volume and high compressibility.

(c)

A solid has a definite shape.

(d)

A liquid has an indefinite shape and slight compressibility.

4. A crystalline solid has a regular, repeating, three-dimensional, geometric pattern. An amorphous solid does not. (a)

A solid that has a regular, repeating pattern is a crystalline solid.

(b)

A plastic solid is amorphous.

(c)

A solid that has no regular repeating pattern is amorphous.

(d)

Glass is an amorphous solid.

(e)

Gold is a crystalline solid.

5. A phase is a homogeneous part of a system separated from other parts by a physical boundary. 6. There are six phases present. 7. Another name for a homogeneous mixture is solution. 8. Alcohol, mercury, and water are the only liquids in the table which are not mixtures. Mercury is an element; alcohol and water are compounds.

– Chapter 1 –

9. Air is the only gas mixture found in the table. The other gases are elements or compounds. 10. Three phases are present within the bottle; solid and liquid are observed visually, while gas is detected by the immediate odor. 11. The system is heterogeneous as three phases are present. 12. A system containing only one substance is not necessarily homogeneous. Two phases may be present. Example: ice in water. 13. A system containing two or more substances is not necessarily heterogeneous. In a solution only one phase is present. Examples: sugar dissolved in water, dilute sulfuric acid. 14. Homogeneous mixtures contain only one phase, while heterogeneous mixtures contain two or more phases. 15. (a) sugar, a compound and (c) gold, an element 16. Using the steps of the scientific method to help determine why your cell phone has stopped working. (a)

Observation:

My cell phone has stopped working.

(b)

Hypothesis:

I think that the battery needs to be recharged.

(c)

Experiment:

Plug in the phone to recharge the battery and allow sufficient time for the battery to fully recharge. Turn the phone back on. The phone now works again.

(d)

Theory:

The battery in the phone has a limited charge time and needs to be recharged on a regular basis in order to keep it in working order.

– Chapter 1 – SOLUTIONS TO EXERCISES 1. Two states are present; solid and gas. 2. Two states are present; solid and liquid. 3. The photo represents a heterogeneous mixture. 4. The maple leaf represents a heterogeneous mixture. 5. (a)

homogeneous

(b)

heterogeneous

(c)

heterogeneous

(d)

heterogeneous

6. (a)

homogeneous

(b)

homogeneous

(c)

heterogeneous

(d)

heterogeneous

7. Typical answers could be Substance chocolate syrup margarine nondairy creamer beef bouillon toothpaste antibacterial soap first aid spray sunblock lotion

Main or Active Ingredient high fructose corn syrup vegetable oil blend corn syrup solids salt sodium fluoride triclocarban lidocaine HCl ethylhexyl p-methoxycinnamate

8. The steps of the scientific method can be used to predict the outcome of the semester in the following way: (a)

Collect the facts or data. They include the number of classes you have enrolled in, the amount of time you will give to each class, the number of hours required for your off campus job, and the amount of time for social activities.

(b)

Formulate a hypotheses. You predict that the amount of time you have allocated each week for class work will be sufficient to result in good grades at the end of the semester.

(c)

Plan and do additional experiments to test the hypothesis. In the first two weeks of the semester keep a record of your performance in each class, including grades on homework, quizzes, and exams.

(d)

Modify the hypothesis. If your grades in any class are not good, your hypothesis was incorrect. It will be necessary to increase the amount of time allocated to that class.

– Chapter 1 – 9. (a)

water, a pure substance

(b)

chicken stock, a mixture

(c)

salt, a pure substance

(d)

mustard flour, a mixture

10. Hypothesis – The soup is not a good source of nutrition because it contains 12 grams of fat per serving. Test – Search for the number of grams of fat per day recommended for an adult. Compare this number to the grams per fat on the label. 11. Observations – Calcite crystals have the ability to reflect light around an object rendering it “unseeable”. “Invisibility cloaks” work only with laser light aimed directly at the crystal. Hypotheses – If larger crystals were used, larger objects could be hidden. These “invisibility cloaks” will improve in the future. 12. (a), (b), Picture (2) best represents a homogeneous mixture. Pictures (1) and (c) (3) show heterogeneous mixtures, and picture (4) does not show a mixture, as only one species is present. Picture (1) likely shows a compound, as one of the components of the mixture is made up of more than one type of “ball”. Picture (2) shows a component with more than one part, but the parts seem identical, and therefore it could be representing a diatomic molecule. 13. (a) two phases, solid and gas (b) two phases, liquid and gas (c)

two phases, solid and liquid

CHAPTER 2

STANDARDS FOR MEASUREMENT SOLUTIONS TO REVIEW QUESTIONS 1.

The exponent will be positive for a large number and negative for a small number.

The exponent will decrease.

The last digit in a measurement is uncertain because if the quantity were to be measured multiple times, the last digit would vary.

It must be written in scientific notation as 6.420 × 105 g.

Zeroes are significant when they are between nonzero digits or at the end of a number that includes a decimal point.

Rule 1. When the first digit after those you want to retain is 4 or less, that digit and all others to its right are dropped. The last digit retained is not changed. Rule 2. When the first digit after those you want to retain is 5 or greater, that digit and all others to the right of it are dropped and the last digit retained is increased by one.

No, the number of significant digits in the calculated value may not be more than the number of significant figures in any of the measurements.

Yes, the number of significant digits depends on the precision of each of the individual measurements and the calculated value may have more or fewer significant figures than the original measurements as long as the precision is no greater than the measurement with the lowest precision. An example of a calculation with an increase in significant figures is in Example 2.9.

100 cm = 1 m

1,000,000,000 nm = 1 m

 1 m   1,000,000,000 nm  (1 cm)    = 10,000,000 nm m   100 cm  

10. 1000 mg = 1 g

1000 g = 1 kg

 1000 g   1000 mg    = 1,000,000 mg g  kg   

(1 kg ) 

11. Weight is a measure of how much attraction the earth’s gravity has for an object (or person). In this case, the farther the astronaut is from the Earth the less gravitational force is pulling on him or her. Less gravitational attraction means the astronaut will weigh less. The mass of the astronaut is the amount of matter that makes up him or her. This does not change as the astronaut moves away from the Earth. 12. They are equivalent units. 13.

(3.5 in.) 

2.54 cm   = 8.9 cm  1 in. 

14. Heat is a form of energy, while temperature is a measure of the intensity of heat (how hot the system is). 1

– Chapter 2 –

15. The number of degrees between the freezing and boiling point of water are Fahrenheit

180°F

Celsius

100°C

Kelvin

100 K

16. The three materials would sort out according to their densities with the most dense (mercury) at the bottom and the least dense (glycerin) at the top. In the cylinder, the solid magnesium would sink in the glycerin and float on the liquid mercury. 17. Order of increasing density: ethyl alcohol, vegetable oil, salt, lead 18. The density of ice must be less than 0.91 g/mL and greater than 0.789 g/mL. 19. The density of water is 1.0 g/mL at approximately 4°C. However, when water changes from a liquid to a solid at 0°C there is actually an increase in volume. The density of ice at 0°C is 0.917 g/mL. Therefore, ice floats in water because solid water is less dense than liquid water. 20. If you collect a container of oxygen gas, you should store it with the mouth up. Oxygen gas is denser than air. 21. density of gold = 19.3 g/mL

density of silver = 10.5 g/mL

 1 mL   = 1.3 mL  19.3 g 

( 25 g gold ) 

25 g of silver has the greater volume.  1 mL   = 2.4 mL  10.5 g 

( 25 g silver ) 

– Chapter 2 – SOLUTIONS TO EXERCISES

(a) kilogram = 1000 grams (b) centimetre = 1/100 of a meter (0.01 m) (c) microliter = 1/1,000,000 of a liter (0.000001 L) (d) millimetre = 1/1000 of a meter (0.001 m) (e) decilitre = 1/10 of a liter (0.1 L)

(a) 1000 meters = 1 kilometer

(d)

0.01 meter = 1 centimeter

(b) 0.1 gram = 1 decigram

(e)

0.001 liter = 1 milliliter

(a) gram = g

(d)

micrometer = μm

(b) microgram = μg

(e)

millilitre = mL

(f)

decilitre = dL

(a) milligram = mg

(e)

angstrom= Å

(b) kilogram = kg

(f)

microlitre = μL

(a) 2050

the first zero is significant; the last zero is not significant

(b) 9.00 × 102

zeros are significant

the first two zeros are not significant; the last zero is significant

(d) 0.075

zeros are not significant

(e) 300.

zeros are significant

(f) 285.00

zeros are significant

(a) 0.005

zeros are not significant

(b) 1500

zeros are not significant

zero is significant

(d) 10.000

zeros are significant

(e) 6.070 × 104

zeros are significant

(f) 0.2300

the first zero is not significant; the last two zeros are significant 3

– Chapter 2 –

Significant figures (a) 0.025

(2 sig. fig.)

(3 sig. fig.)

(b) 22.4

(3 sig. fig.)

(d) 5.50 × 103

(3 sig. fig.)

Significant figures (a) 40.0 (b) 0.081

(6 sig. fig.) −3

(2 sig. fig.)

(d) 4.090 × 10

Round each of the following numbers to four significant figures: (a) 47.70

(c)

100.2

(b) 1.026

(d)

16.50

10. Round each of the following numbers to four significant figures: (a) 0.07794

(c)

5.005

(b) 21.63

(d)

74.29

(a) 6.7 × 104

(c)

7.8 × 10−3

(b) 6.54 × 10−2

(d)

4.11 × 105

(a) 4.56 × 10−2

(c)

4.030 × 101

(b) 4.0822 × 103

(d)

1.2 × 107

11. Exponential notation

12. Exponential notation

13. (a)

12.62 1.5 0.25 14.37 = 14.4

(b) (2.25 × 103)(4.80 × 104) = 10.8 × 107 = 1.08 × 108 (c)

( 452 )( 6.2 ) = 195.97 = 2.0×102 14.3

(d) (0.0394) (12.8) = 0.504 (e)

0.4278 = 0.00718 = 7.18×10 −3 59.6

(f) 10.4 + (3.75)(1.5 × 104) = 5.6 × 104

(4 sig. fig.)

– Chapter 2 –

14. (a)

15.2 −2.75 15.67 28.1

(b) (4.68)(12.5) = 58.5 (c)

182.6 = 40. 4.6

1986 (d)

23.84 0.012 2009.852 = 2010. = 2.010 × 103

(e)

29.3 = 2.49 ×10 −4 ( 284 )( 415 )

(f) (2.92 × 10−3)(6.14 × 105) = 1.79 × 103 15. Fractions to decimals (3 significant figures) (a)

5 = 0.833 6

(c)

12 = 0.750 16

(b)

3 = 0.429 7

(d)

9 = 0.500 18

(c)

1.67 = 1

2 5 or 3 3

(d)

0.888 =

8 9

(c)

0.525 = 0.25 x

16. Decimals to fractions

17.

1 4

(a)

0.25 =

(b)

0.625 =

(a)

3.42x = 6.5 6.5 = 1.9 x= 3.42

(b)

5 8

x = 7.05 12.3 x = ( 7.05 ) (12.3 ) = 86.7

0.525 = 0.25x 0.525 x= = 2.1 0.25

– Chapter 2 –

18.

(a)

212 − 32 1.8

72 = 1.8x + 32

(c)

72 − 32 = 1.8x 40. = 1.8x 40. =x 1.8 22 = x

x = 1.0 × 10 2

(b) 8.9

g 40.90 g = mL x

g    8.9  x = 40.90 g mL   40.90 g = 4.6 mL x= g 8.9 mL

19. (a)

( 28.0 cm ) 

1m   = 0.280 m  100 cm 

(b)

(1000. m ) 

(c)

( 9.28 cm ) 

(d)

(10.68 g ) 

(e)

1 g   1 kg  ( 6.8 ×10 mg )  1000   = 6.8×10 kg mg 1000 g

(f)

(8.54 g ) 

(g)

( 25.0 mL ) 

(h)

( 22.4 L ) 

20. (a)

( 4.5 cm ) 

(b)

(12 nm ) 

1 km   = 1.000 km  1000 m  10 mm   = 92.8 mm  1 cm 

 1000 mg  4  = 1.068 × 10 mg  1g  −2







 1 kg   = 0.0854 kg  1000 g 

1L  −2  = 2.50 × 10 L 1000 mL  

 106 μL  7  = 2.24 × 10 μL 1 L   1 m  1 Å  8   −10  = 4.5 × 10 Å  100 cm   10 m 

 10 −9 m   100 cm  −6   = 1.2 × 10 cm  1 nm   1 m  6

– Chapter 2 –

(c)

(8.0 km ) 

1000 m   1000 mm  6   = 8.0 × 10 mm 1 km 1 m   

(d)

(164 mg ) 

(e)

( 0.65 kg ) 

(f)

(5.5 kg ) 

(g)

( 0.468 L ) 

(h)

( 9.0 μL ) 

21. (a)

( 42.2 in.) 

(b)

( 0.64 mi ) 

(c)

cm  2 ( 2.00 in.2 )  2.54  = 12.9 cm 1 in. 

(d)

( 42.8 kg ) 

(e)

(3.5 qt ) 

(f)

( 20.0 L ) 



1g   = 0.164 g  1000 mg   1000 g   1000 mg  5   = 6.5 × 10 mg  1 kg   1 g 

 1000 g  3  = 5.5 ×10 g  1 kg  1000 mL   = 468 mL  1L 

 1 L   1000 mL  −3  = 9.0 ×10 mL  6 μ 10 L 1 L    

2.54 cm   = 107 cm  1 in.  5280 ft   12 in.  4   = 4.1×10 in. 1 mi 1 ft    2

 2.205 lb   = 94.4 lb  kg 

 946 mL  3  = 3.3 × 10 mL  1 qt  1 qt   1 gal   = 5.29 gal   0.946 L   4 qt 

22. (a) The conversation is: m → cm → in. → ft

(35.6 m ) 

100 cm   1 in.   1 ft     = 117 ft  1 m   2.54 cm   12 in. 

(b)

(16.5 km ) 

1 mi   = 10.3 mi  1.609 km  3

 2.54 cm   10 mm  4 3 (c) ( 4.5 in. )     = 7.4 ×10 mm  1 in.   1 cm  3

(d)

( 95 lb ) 

453.6 g  4  = 4.3 ×10 g 1 lb  

– Chapter 2 –

 4 qt   0.946 L  (e) (20.0 gal)    = 75.7 L  1 gal   1 qt  (f)

The conversation is: ft 3 → in.3 → cm 3 → m3 3

cm   1 m  3 3 ( 4.5 ×10 ft )  121 ftin.   2.54    = 1.3 ×10 m 1 in.   100 cm  4

23. First convert 6 ft 6 in. into inches and then convert to centimeters and meters.

( 6 ft ) 

12 in.   + 6 in. = 78 in.  ft 

( 78 in.) 

2.54 cm  2  = ( 2.0 × 10 cm ) in.  

( 2.0 × 10 cm )  1001 mcm  = 2.0 m ( 2 significant figures) 2

24. We must convert ounces to mL.

( 46.0 oz ) 

29.6 mL  3  = 1.36 ×10 mL ( 3 significant figures )  1 oz 

25. The conversion is: L → mL → mg → g

1000 mL  500. mg   1 g   = 5.00 g    1 L  100. mL   1000 mg 

(1 L ) 

26. The conversion is: tablet → g → mg → grains

0.500 g   1000 mg  1 grain    = 8.33 grains   1 tablet   1 g  60 mg 

(1 tablet ) 

27. The conversion is: hr → min → s → m → km → mi

(5 hours ) 

60 min   60 s   0.11 m   1 km   1 mi       = 1.2 miles  1 hr   1 min   1 s   1000 m   1.61 km 

28. The conversion is: cm → in. → ft → yr → day

1 in.  1 ft  1 yr   365 day   = 3.54 days     2.54 cm  12 in.  3.38 ft   1 yr 

(1 cm ) 

29. The conversion is: lb → g → mg → ant

454 g ant   1000 mg ant  1 ant  4   = 3.6 ×10 ant   1 lb ant   1 g ant  85 mg ant 

( 6.7 lb ant ) 

– Chapter 2 –

30. The conversion is: lb → g → egg

 454 g egg  1 lg egg    = 5.8 lg eggs  1 lb egg  58.5 g egg 

( 0.75 lb egg ) 

31. The conversion is: lb body → lb fat → kg fat

 11.2 lb fat   1 kg    = 11.4 kg fat  100 lb body   2.205 lb 

( 225 lb body ) 

32. The conversion is: carats → mg → g → lb

200 mg   1 g  1 lb  −3   = 2.54 ×10 lb   1 carat   1000 mg  453.6 g 

(5.75 carats ) 

33. The conversion is:

yd mi km m m → → → → s s s s min

 100. yd   1 mi   1.609 km  1000 m  60 s  2 m       = 1.1×10 min  52 s   1760 yd   1 mi  1 km  1 min  34. The conversion is:

mi km m cm cm cm → → → → → hr hr hr hr min s

 133 mi   1.609 km   1000 m   100 cm   1 hr   1 min  3 cm        = 5.94 × 10 s  1 hr   1 mi   1 km   1 m   60 min   60 s 

35.

(a)

29,035 ft − 21,002 ft = 8033 ( total feet climbed )

(16 hr ) 

60 min  3  = 960 min 960 min + 42 min = 1.0 ×10 min  hr 

 8003 ft  1 mi  −3 mi    = 1.5 ×10 3 min  1.0 × 10 min  5280 ft 

36.

(b)

 8003 ft   1 mi  1.609 km  1000 m  1 min  −2 m       = 4.1×10 3 mi  km  60 s  s  1.0×10 min   5280 ft 

(a)

The conversion is:

m cm in. ft ft → → → → hr hr hr hr min

ft  4500 m   100 cm   1 in.   1 ft   1 hr        = 50 min  5 hr   m   2.54 cm   12 in.   60 min 

(b)

The conversion is:

m km km km → → → hr hr min s

 4500 m   1 km   1 hr   1 min  −4 km      = 3 × 10 s  5 hr   1000 m   60 min   60 s 

– Chapter 2 –

37. The conversion is:

mi ft in. cm cm cm → → → → → hr hr hr hr min s

 155 mi   5280 ft   12 in   2.54 cm   1 hr   1 min  6930 cm       = sec  hr   1 mi   1 ft   1 in   60 min   60 sec 

38. The conversion is:

mi ft in cm cm cm → → → → → hr hr hr hr min s

 125 mi   5280 ft   12 in   2.54 cm   1 hr   1 min  5590 cm       = sec  hr   1 mi   1 ft   1 in   60 min   60 sec 

39. The conversion is: gal → qt → L → mL → cup

 4 qt  1 L   1000 mL  1 cup  4     = 1.60 ×10 cups coffee  1 gal  1.06 qt   1 L  473 mL 

( 2010 gal ) 

40. The conversion is: lb → g → number of tilapia

g   1 tilapia  ( 475 ×10 lb )  454  = 4.03 ×10 tilapia  1 lb  535 g 6





41. The conversion is: gal → qt → mL → drops

 4 qt  946 mL   20. drops  4    = 7.6 ×10 drops gal qt mL    

(1.0 gal ) 

42. The conversion is: gal → qt → L

 4 qt  0.946 L    = 160 L  gal  qt 

( 42 gal ) 

43. The conversion is: ft3 → in.3 → cm3 → mL 3

cm   1 mL  = 2.83 × 10 4 mL (1.00 ft3 )  121 ftin.   2.54   3  1 in.   1 cm 

44. V = A × h A = area h = height V = volume cm 3 cm 3 → → m2 nm m 3 V  200 cm 3   1 nm  1 m  5 2 A = =  −9   = 4 ×10 m h  0.5 nm   10 m  100 cm 

The conversion is:

45. (a) ( 27 cm) ( 21 cm)( 4.4 cm) = 2.5 ×10 cm 3

 1L  3 3 3 (b) 2.5 ×10 cm is 2.5 ×10 mL   = 2.5 L  1000 mL  3

(c)

1 in.  2 3 ( 2.5 ×103 cm3 )  2.54  = 1.5×10 in. cm 

– Chapter 2 –

2.54 cm   1 mL   1 L   1 qt   1 gal  46. (16 in.)( 8 in.)(10 in.)   = 6 gal     3   1 in.   1 cm   1000 mL   0.946 L   4 qt  3

 73.14 gC   = 183 g C  100 g ketone 

47.

( 250. g ketone ) 

48.

( 475. g nectaryl ) 



10.98 g H   = 52.2 g H  100 g nectaryl 

  283.4 g potassium dichromate  mass part  49. (a)  ×100  ×100 =   mass whole   283.4 g potassium dichromate+ 650.0 g water 

 283.4 g potassium dichromate  =  ×100 = 30.36% potassium dichromate 933.4 g soln    30.36 g potassium dichromate   = 83.5 g potassium dichromate 100 g solution  

(b)

( 275. g solution ) 

(c)

(15.0 g potassium dichromate ) 



 100 g soln  = 49.4 g soln  30.36 g potassium dichromate 

  95.4 g sodium nitrate  mass part  50. (a)   ×100 ×100 =   mass whole   95.4 g sodium nitrate + 753 g water   95.4 g sodium nitrate  =  ×100 = 11.3% sodium nitrate 848 g soln    11.3 g sodium nitrate   = 39.6 g sodium nitrate 100 g solution  

(b)

(350. g solution ) 

(c)

(50.0 g sodium nitrate ) 



 100 g soln  = 442 g soln  11.3 g sodium nitrate 

51. (a) °F=1.8°C + 32 = (1.8) (38.8) + 32 = 101.8°F (b) Yes, the child has a fever since 101.8°F > 98.6°F 52. °F = 1.8°C +32 53. (a)

(1.8) ( 45) +32 =113°F Summer!

Remember to express the answer to the same 162 −32 = 72.2°C precision as the original measurement. 1.8

(b) °C + 273 = K

0.0 − 32 + 273 = 255.2 K 1.8

−8.6 − 32 = −22.6°C 1.8

– Chapter 2 –

273 + 273 = 546 K

(c)

°C = 100 − 273 = −173°C

(d)

( −173)(1.8) + 32 = −279°F = −300°F (1 significant figure in 100 K )

55. °C =

( °F − 32 ) = ( 425°F − 32°F ) = 218°C 1.8

1.8°F/°C

K = °C + 273 = 218 + 273 = 491 K

56. °C =

( °F − 32 ) = ( 247°F − 32°F ) = 119°C 1.8

1.8°F/°C

K = °C + 273 = 119 + 273 = 392 K

°F = °C

57.

°F = 1.8 ( °C ) + 32 °F = 1.8 ( °F ) + 32

substitute °F for °C

−32 = 0.8 ( °F ) −32 = °F 0.8 −40 = °F −40°F = −40°C

°F = −°C

58.

°F = 1.8 ( °C ) + 32

substitute − °C for °F

−°C = 1.8 ( °C ) + 32

2.8 ( °C ) = −32

−32 2.8 °C = −11.4 −11.4°C = 11.4°F °C =

59. °C =

−60° − 32 −92° = = −51°C 1.8 1.8

60. °F = (1.8 × 801°) + 32 =1470°F 61. d =

m 59.82 g g = = 0.920 v 65.0 mL mL

62. d =

m 20.41 g g = = 0.810 v 25.2 mL mL

63. 50.92 g − 25.23 g = 25.69 g ( mass of liquid ) d=

m 25.69 g g = = 1.03 v 25.0 mL mL

– Chapter 2 –

64. 54.6 mL − 50.0 mL = 4.6 mL ( volume of zinc ) d=

m 32.95 g g = = 7.2 v 4.6 mL mL

65. The conversion is g → mL

 1 mL   = 16 mL  0.929 g 

(15 g ) 

66. The conversion is g → mL

 1 mL   = 63 mL  1.20 g 

( 75 g ) 

67. The conversion is kg cayenne → kg salsa → lb salsa

 100 kg salsa  2.20 lb salsa    = 85.3 lb  8.41 kg cayenne  1 kg salsa 

(3.26 kg cayenne ) 

68. The conversion is kg pecan → NCB ( nutty candy bars ) → lb NCB

 100 kg NCB  2.20 lb NCB    = 70.lb NCB  22.0 kg pecan  1 kg NCB 

( 7.0 kg pecan )  69. (a)

91.67 g Cu 8.33 g Ni and 100 g quarter 100 g quarter

(b) The conversion is ton Cu →g Cu →g quarters →quarters 1 quarters       1 ton Cu  1 lb Cu   91.67 g Cu  5.670 g quarters 

( 3.0 ton Cu ) 

2000 lb Cu  454 g Cu   100 g quarters 

= 5.2 × 105 quarters

70. (a)

92.5 g Ag 7.50 g Cu and 100 g sterling 100 g sterling

(b) The conversion is lb Ag → g Ag → g sterling → silver hearts

 454 g Ag  100 g sterling  1 silver heart     = 701 silver hearts  1 lb Ag  9.5 g Ag  35.0 g sterling 

(50.0 lb Ag )  71. a) 1055 kg 1)

1055 is the numerical value

kg is the unit

4 significant figures

105 is known, 5 is estimated

1.1 × 103 kg

– Chapter 2 –

b) 1,650,763 wavelengths 1) 1,650,763 is the numerical value 2) wavelengths is the unit 3) 7 significant figures 4) 165076 is known, 3 is estimated 5) 1.7 × 106 wavelengths c)

1.077 g/cm3 1) 1.077 is the numerical value 2) g/cm3 is the unit 3) 4 significant figures 4) 1.07 is known, 7 is estimated 5) 1.1 × 100 g/cm3

d) 0.00845 L 1) 0.00845 is the numerical value 2) L is the unit 3) 3 significant figures 4) 84 is known, 5 is estimated 5) 8.5 × 10−3L e)

776,000 g 1) 776,000 is the numerical value 2) g is the unit 3) 3 significant figures 4) 77 is known, 6 is estimated 5) 7.8 × 105 g

72. (a) report 10.01 grams (b) report 10.012 grams (c) report 10.0124 grams 73. (a)

(175 Skittles ) 

1.134 g Skittle   =198 g Skittles  1 Skittle 

 5.3 mL Skittle   1 L Skittles  (b) (175 Skittles )    = 0.15 L Skittles  6 Skittle   1000 mL Skittles  (c)



1 Skittles   = 286.6 Skittles  1.134 g Skittles 

(325.0 Skittles ) 

 1000 mL Skittle   6 Skittles  (d) ( 0.550 L Skittles )    = 620 Skittles  1 L Skittle   5.3 mL Skittles  14

– Chapter 2 –

(e) The mass measurement is more precise. It is also more accurate. Using calculations similar to those above there should be 309 Skittles in 350 grams and the average value is 310 Skittles. There should be 367 Skittles in 0.325 L of Skittles and the average value is 384 Skittles. 74. A graduated cylinder would be the best choice for adding 100 mL of solvent to a reaction. While the volumetric flask is also labeled 100 mL, volumetric flasks are typically used for doing dilutions. The other three pieces of glassware could also be used, but they hold smaller volumes so it would take a longer time to measure out 100 mL. Also, because you would have to repeat the measurement many times using the other glassware, there is a greater chance for error. 75. The conversion is: g → mL

 1 mL   = 14.5 mL  1.484 g 

( 21.5 g ) 

76. The conversion is: g → mL

 1 mL   = 26 mL  0.97 g 

( 25.27 g ) 

77. The conversion is: day → cups → mg → g → lb

 4.00 ×108 cups  160 mg  1 g  1 lb  5     = 1×10 lb day cup 1000 mg 453.6 g     

(1 day ) 

78. Mass of gear carried by paladin after drinking strength potion = 115 lb + 50.0 lb = 165 lb Amount of mass potion paladin can carry = 165 lb – 92 lb = 73 lb

The conversion is: lb → g → mL → vials potion 454 g   1 ml   1 vial potion     = 3.43 vials potion  1 lb   93 g   50.0 mL 

( 73 lb ) 

You would only be able to collect 3 vials, because 4 would put you over your mass limit. 79. The conversion is : sequins → cm3 → g → kg

 0.0241 cm 3   41.6 g sequins   1 kg  4560 sequins ( )   = 4.57 kg sequins  1 cm3   1000 g   1 sequins   The conversion is: kg → lb  2.20 lb   = 10.1 lb sequins  1 kg 

( 4.57 kg sequins ) 

– Chapter 2 –

80. V = side3 = ( 0.50 m ) = 0.13 m 3 3

( 0.13 m )  100.mcm   10001 Lcm  = 130 L 3

( volume of the cube ) ( volume of cube )

Yes, the cube will hold the solution. 130 L − 8.5 L = 120 L additional solution is necessary to fill the container. 81. The conversion is:

μg m

→

μg L

→

μg day

 180 μ g   1 m   4 L  = 4000 μ g ingested/day    2 × 10 3  1 m 1000 L day     3

(1 significant figure )

Yes, the nurse is at risk. This is well over the toxic limit. 82. (a)

The conversion is ac ft → L → kL

 1.233×106 L   1 kL  6   = 1.70 ×10 kL 1 ac ft 1000 L   

(1380 ac ft )  (b)

The conversion is day → kL → L → qt → gal  1.70×106 kL   1000 L   1.06 qt   1 gal  10 (31 day )    = 1.40 ×10 gal    1 day   1 kL   1 L   4 qt 

83. (a)

Convert 20.27 K to °C K − 273.15 = °C 20.27 K − 273.15 = −252.88°C

(b)

Convert 20.27 K to °F °F = (1.8× °C ) + 32

°F = (1.8× −252.88 ) + 32 = −455.18 + 32

°F = −423.18°F 84. °F = 1.8°C + 32 Convert 36°C to °F

Convert 38°C to °F

(1.8)(36 ) + 32 = 97°F (1.8)( 38) + 32 = 100°F

Sauropods have a body temperature close to the body temperature of other warmblooded mammals such as dogs, humans, and cows. 85.

The conversion is : L → dL → mg → g 10 dL   130 mg   1 g   = 6.1 g    1 L   1 dL   1000 mg 

( 4.7 L ) 

86. A sample of gold will sink to the bottom of the mercury and a sample of iron pyrite will float.

– Chapter 2 –

87.

The conversion is : El → ton → lb

( 5.3 El ) 

6.00 ton   2000 lb  4   = 6.4 × 10 lb  1 El   1 ton 

88. The conversion is: hands → in. → cm → m

(14.2 hands ) 

4 in.   2.54 cm  1 m     = 1.44 m  1 hand   1 in.  100 cm 

89. The conversion is: days → hr → gal → qt → L

 24 hr   22.5 gal   4 qt  0.946 L  3    = 5.1×10 L   1 day   12 hr   1 gal  1 qt 

(30 days )  90. d =

m The cube with the largest volume has the lowest density. Use Table 2.6. V

Cube A - lowest density Cube B

1.74 g/mL - magnesium 2.7 g/mL - aluminium

Cube C - highest density 10.5 g/mL - silver

91. The conversion is: ft → in. → cm → m → nm → nanotubes 9  12 in.  2.54 cm  1 m   10 nm   1 nanotube  9 40.0 ft ( )      = 9.4 ×10 nanotubes  1 ft  1 in.  100 cm   1 m   1.3 nm  92.

 0.25 c peanuts   = 1.4 c peanuts  67 mg Mg 

(380 mg Mg ) 

2 2 2 93. The conversion is: acres → ft → mi → km

 43560 ft 2   1 mi   1.609 km  2     = 0.506 km 1 acre 5280 ft 1 mi     

(125 acres ) 

94. The volume of the aluminum cube is m 500. g V= = = 185 mL d 2.70 g mL

Density of Al is 2.70 g/mL. This is the same volume as the gold cube thus

m = dV = (185 mL)(19.3 g/mL) = 3.57 ×103 g of gold Density of Au is 19.3 g/mL. m 24.12 g 0.965 g = = V 25.0 mL mL 96. 150.50 g − 88.25 g = 62.25 g

95. d =

( density of water at 90°C )

m m 62.25 g thus V = = = 49.8 mL V d 1.25 g mL The container must hold at least 50 mL. d=

( mass of liquid ) ( volume of liquid )

– Chapter 2 –

97. H2 O

50 g = 50 mL g 1.0 mL alcohol 50 g = 60 mL g 0.789 mL

Ethyl alcohol has the greater volume due to its lower density. 98. Volume of sulfuric acid

 1 mL    (100. g ) = 54.3 mL  1.84 g  99.

The conversion is: cup → oz → qt → L → mg 











31.4 mg NMP oz coffee 1 qt coffee 1 L coffee     ( 2.00 cup coffee )  10 1 cup coffee   32 oz coffee   1.057 qt coffee   1 L coffee  = 18.6 mg NMP

100.

m , as the volume increases, the density decreases. As solids are heated V the density decreases due to an increase in the volume of the solid.

Since d =

1 mL  101. V = ( 2.00 cm )(15.0 cm )( 6.00 cm )  = 180.mL ( volume of bar ) 3   1 cm  m d = = 3300 g/180.mL = 18.3 g/mL V

The density of pure gold is 19.3 g/mL (from Table 2.5), therefore, the gold bar is not pure gold, since its density is only 18.3 g/mL, or it is hollow inside. 102. m = dV = ( 0.789 g/mL ) ( 35.0 mL ) = 27.6 g ethyl alcohol 27.6 g + 49.28 g = 76.9 g ( mass of cylinder and alcohol )

103. The conversion is g → oz → gr → scruples

 12 oz   480 gr  1 scruple    = 537 scruples   373 g   oz   20 gr 

( 695 g ) 

– Chapter 2 –

104. The conversion is: days → teaspoons → mL

 2 teaspoons× 4   5 mL    = 400 mL day    teaspoon 

(10 days ) 

Since you will need a total of 400 mL for the 10 days and the bottle contains 500 mL, you have purchased enough. 105. (a)

The conversion is kg → g → mL → L

 1000 g brine  1 mL brine   1 L brine     = 6.36 L brine  1 kg brine  1.28 g brine   1000 mL brine 

(8.14 kg brine )  (b)

The conversion is g NaCl → g brine  100 g brine   = 1990 g brine  7.55 g NaCl 

(150.0 g NaCl ) 

106. The density of lead is 11.34 g/mL. The density of aluminum is 2.70 g/mL. The density of silver is 10.5 g/mL. The density of the unknown piece of metal can be calculated from the mass (20.25 g) and the volume (57.5 mL − 50 mL = 7.5 mL) of the metal. Density of the unknown metal = 20.25 g/7.5 mL = 2.7 g/mL. The metal must be aluminum. 107.

Volume of slug

30.7 mL − 25.0 mL = 5.7 mL

Density of slug

Mass of liquid, cylinder, and slug

m 15.454 g = = 2.7 g/mL V 5.7 mL 125.934 g

Mass of slug ( subtract )

Mass of cylinder ( subtract )

−89.450 g

Mass of the liquid

21.030 g

Density of liquid d =

108.

−15.454 g

m 21.030 g = = 0.841 g/mL V 25.0 mL

7.8 g bronze needed   100 g bronze melted     1 mL bronze needed   90.0 g bronze needed 

( 225 mL bronze needed ) 

= 2.0 ×103 g bronze melted 109. The conversion is: km → m → cm → in. → ft → mi → hr → min → s → ns 











m 100 cm 1 in. 1 ft 1 mi 1 hr 60 min 60 s         (730.0 km )  1000 1 km 1 m  2.54 cm  12 in.   5280 ft  1.86 ×108 mi  1 hr  1 min  109 ns    = 8.78 ×106 ns  1s   

8,780,000 ns is good to only 3 significant figures. You would need at least 6 significant digits to detect the difference between 8,780,000 ns and (8,780,000 + 60) ns. Sometimes experimenters do not see differences due to the precision of their measuring techniques.

– Chapter 2 –

110.

(1.5 m ) 

100 cm   1 in.   1 ft     = 4.9 ft  1 m   2.54 cm   12 in.  100 cm  1 in.  1 ft  ( 4 m )     = 13 ft  1 m  2.54 cm  12 in. 

( 27°C ×1.8) + 32 = 81°F

CHAPTER 3

ELEMENTS AND COMPOUNDS SOLUTIONS TO REVIEW QUESTIONS 1.

Silicon

25.7%

Hydrogen 0.9%

25.7 g Si = 30 g Si/1 g H (1 sig. fig.) 0.9 g H

In 100 g

Si is 28 times heavier than H, thus since 30 > 28, there are more Si atoms than H atoms. 2. 3.

(a)

(c)

(e)

(g)

(b) F

(d)

(f)

(h)

(a)

Iron

(c)

Carbon

(e)

Beryllium

(g)

Argon

(b) Magnesium

(d)

Phosphorus

(f)

Cobalt

(h)

Mercury

The symbol of an element represents the element itself. It may stand for a single atom or a given quantity of the element.

sodium

silver

potassium

tungsten

iron

gold

antimony

mercury

tin

lead

hydrogen

sulfur

boron

potassium

carbon

vanadium

nitrogen

yttrium

oxygen

iodine

fluorine

tungsten

phosphorus

uranium

0 metalloids

5 nonmetals

1 metal

Hydrogen

Chlorine

Cl2

Nitrogen

Bromine

Br2

Oxygen

Iodine

Fluorine

– Chapter 3 –

(a) CO – 1 atom of carbon and 1 atom of oxygen (b) H2 – 1 molecule of hydrogen (made of 2 atoms of hydrogen) (c) S8 – 1 molecule of sulfur (made of 8 atoms of sulfur) (d) CS – 1 atom of carbon and 1 atom of sulfur

Co – 1 atom of cobalt 2 H – 2 atoms of hydrogen 8 S – 8 atoms of sulfur Cs – 1 atom of cesium

10. In an element all atoms are alike, while a compound contains two or more elements (different atoms) which are chemically combined. Compounds can be decomposed into simpler substances while elements cannot. 11. 92 metals

7 metalloids

19 nonmetals (based on 118 elements)

12. 7 metals

1 metalloid

2 nonmetals

13. (a) iodine

(b) bromine

14. A compound is composed of two or more elements which are chemically combined in a definite proportion by mass. Its properties differ from those of its components. A mixture is the physical combining of two or more substances (not necessarily elements). The composition may vary, the substances retain their properties, and they may generally be separated by physical means. 15. Molecular compounds exist as molecules formed from two or more atoms of elements bonded together. Ionic compounds exist as cations and anions held together by electrical attractions. 16. Compounds are distinguished from one another by their characteristic physical and chemical properties. 17. Cations are positively charged, while anions are charged negatively.

– Chapter 3 –

SOLUTIONS TO EXERCISES 1.

Diatomic molecules: (a) HCl, (b) O2 (h) NO

Diatomic molecules: (a) I2 (d) HF (f) Cl2

(a) Potassium, iodine

(d)

Calcium, bromine

(b) Sodium, carbon, oxygen

(e)

Hydrogen, carbon, oxygen

(a) Magnesium, bromine

(d)

Barium, sulfur, oxygen

(b) Carbon, chlorine

(e)

Aluminum, phosphorus, oxygen

(a) ZnO

(c)

NaOH

(b) KClO3

(d)

C2H6O

(a) AlBr3

(c)

PbCrO4

(b) CaF2

(d)

C6H6

(a) C6H10OS2 (b) C18H27NO3 (c) C10H16

(a) C55H86NO24 (b) C15H14O6 (c) C30H48O3

(a) 2 atoms iron, 3 atoms oxygen (b) 2 atoms calcium, 2 atoms nitrogen, 6 atoms oxygen (c) 1 atom cobalt, 4 atoms carbon, 6 atoms hydrogen, 4 atoms oxygen (d) 3 atoms carbon, 6 atoms hydrogen, 1 atom oxygen (e) 2 atoms potassium, 1 atom carbon, 3 atoms oxygen (f) 3 atoms copper, 2 atoms phosphorus, 8 atoms oxygen (g) 2 atoms carbon, 6 atoms hydrogen, 1 atom oxygen

(h) 2 atoms sodium, 2 atoms chromium, 7 atoms oxygen 10. (a) 4 atoms hydrogen, 2 atoms carbon, 2 atoms oxygen (b) 3 atoms nitrogen, 12 atoms hydrogen, 1 atom phosphorus, 4 atoms oxygen (c) 1 atom magnesium, 2 atoms hydrogen, 2 atoms sulfur, 6 atoms oxygen (d) 1 atom zinc, 2 atoms chlorine (e) 1 atom nickel, 1 atom carbon, 3 atoms oxygen (f) 1 atom potassium, 1 atom manganese, 4 atoms oxygen (g) 4 atoms carbon, 10 atoms hydrogen (h) 1 atom lead, 1 atom chromium, 4 atoms oxygen

– Chapter 3 –

11. (a) 9 atoms

(b) 14 atoms

(d)

45 atoms

12. (a) 9 atoms

(b) 12 atoms

(d)

12 atoms

13. (a) 6 atoms C (b) 3 atoms C

(d)

8 atoms C

14. (a) 6 atoms O (b) 4 atoms O

(d)

21 atoms O

15. (a) mixture

(d)

mixture

(b) mixture

(e)

pure substance

(f)

mixture

16. (a) mixture

(d)

pure substance

(b) mixture

(e)

mixture

(f)

mixture

17. (c) compound

(e)

element

18. (c) element

(d)

compound

19. (a) mixture

(c)

element

(c)

mixture

(b) compound 20. (a) compound (b) compound 21. Yes. The gaseous elements are all found on the extreme right of the periodic table. They are the entire last column and in the upper right corner of the table. Hydrogen is the exception and located at the upper left of the table. 22. No. The only common liquid elements (at room temperature) are mercury and bromine. 23.

18 metals ×100 = 50% metals 36 elements

24.

26 solids ×100 = 72% solids 36 elements

25. The formula for water is H2O. There is one atom of oxygen for every two atoms of hydrogen. The molar mass of oxygen is 16.00 g and the molar mass of hydrogen is 1.008 g. For H2O the mass of two hydrogen atoms is 2.016 g and the mass of one oxygen atom is 16.00 g. The ratio of hydrogen to oxygen is approximately 2:16 or 1:8. Therefore, there is 1 gram of hydrogen for every 8 grams of oxygen. 26. The formula for hydrogen peroxide is H2O2. There are two atoms of oxygen for every two atoms of hydrogen. The molar mass of oxygen is 16.00 g and the molar mass of hydrogen is 1.008 g. For hydrogen peroxide the total mass of hydrogen is 2.016 g and the total mass of oxygen is 32.00 g for a ratio of hydrogen to oxygen of approximately 2:32 or 1:16. Therefore, there is 1 gram of hydrogen for every 16 grams of oxygen.

– Chapter 3 –

27. (a) No (b) Compound; molecular (c) No (d) Compound; molecular (e) Compound; molecular 28. (a) Compound; molecular (b) No (c) Compound; ionic (d) No (e) Compound; molecular 29. To a small sample of the mixture, add water and observe that the salt dissolves but the pepper does not. After the small trial, add water to the entire mixture to dissolve the salt. Separate the undissolved pepper from the salt solution by filtering the mixture. Coffee filters or strong paper towels would work well for this process. 30. The atoms that make up each ionic compound are on opposite ends of the periodic table from one another. An ionic compound is made up of a metal-nonmetal combination. 31. (a) 1 carbon atom and 1 oxygen atom, total number of atoms = 2 (b) 1 boron atom and 3 fluorine atoms, total number of atoms = 4 (c) 1 hydrogen atom, 1 nitrogen atom, 3 oxygen atoms, total number of atoms = 5 (d) 1 potassium atom, 1 manganese atom, 4 oxygen atoms, total number of atoms = 6 (e) 1 calcium atom, 2 nitrogen atoms, 6 oxygen atoms, total number of atoms = 9 (f)

3 iron atoms, 2 phosphorus atoms, 8 oxygen atoms, total number of atoms = 13

32. (a) 181 atoms/module

63 C 88 C 1 Co 14 N 14 O 1P 181 atoms (b)

63 C ×100 = 35% C atoms 181 atoms

(c)

1 Co 1 = metals 181 atoms 181

– Chapter 3 –

33. The conversion is: cm3 → L, → mg, → g, → $ −4 1L   4 × 10 mg   1 g   $19.40  15 3  × 1 10 cm ( )  1000 cm3   L   1000 mg   g  = $8 ×106    

34. Ca(H2PO4)2

(10 formula units ) 

4 atoms H   = 40 atoms H  formula unit 

35. C145H293O168

145 C 293 H 168 O 606 atoms/molecule 36. (a)

magnesium, manganese, molybdenum, mendeleevium, mercury, meitnerium

(b)

carbon, phosphorus, sulfur, selenium, iodine, astatine, boron

(c)

sodium, potassium, iron, silver, tin, antimony

37. Add water to the mixture to dissolve the sugar. Filter the mixture to separate the sugar solution from the insoluble sand. Add another small amount of water to remove last traces of sugar. Filter. Allow the water to evaporate from the sugar solution to obtain crystals of sugar. Sand is the insoluble residue. 38. HNO3 has 5 atoms/molecule 7 dozen = 84

(84 molecules )(5 atoms/molecule ) = 420 atoms  12 molecules   5 atoms  or ( 7 dz )    = 420 atoms dz    molecule  39.

(a) As temperatures decreases, density increases. (b) approximately 1.28 g/L at 5°C approximately 1.18 g/L at 25°C approximately 1.09 g/L at 70°C

– Chapter 3 –

40. Each represents eight units of sulfur. In 8 S the atoms are separate and distinct. In S8 the atoms are joined as a unit (molecule). 41. (a) NaCl

(d)

Fe2S3

(g) C6H12O6

(b) H2SO4

(e)

K3PO4

(h) C2H5OH

(f)

Ca(CN)2

(i)

Cr(NO3)3

42. Let X = grams sea water

 5.0 ×10 −8 %I 2    ( X ) = 1.0 g I 2 100   (1.0 g I2 )(100 ) = 2.0 ×109 g sea water X= (5.0 ×10−8%I2 ) 1 kg  ( 2.0 ×10 g )  1000  = 2.0 × 10 kg sea water g 9





43. (a) 12 carbons, 22 hydrogens, 11 oxygens (b) 7 carbons, 5 hydrogens, 3 oxygens, 1 nitrogen, 1 sulfur (c) 14 carbons, 18 hydrogens, 5 oxygens, 2 nitrogens (d) 4 carbons, 4 hydrogens, 3 oxygens, 1 nitrogen, 1 sulfur, 1 potassium (e) 12 carbons, 19 hydrogens, 8 oxygens, 3 chlorines 44. Cobalt should be written Co as CO is the formula for carbon monoxide. 45. C8N4O2H10 46. (a) Picture (3) because fluorine gas exists as a diatomic molecule. (b) Other elements that exist as diatomic molecules are oxygen, nitrogen, chlorine, hydrogen, bromine, and iodine. (c) Picture (2) could represent SO3 gas. 47. (a) noble gases

(d)

alkaline earth metals

(b) metalloids

(e)

transition metals

(f)

halogens

48. (a) 6

(b) 1

(c)

(d) 2

(e) 6

49. (a) Bromine is the only nonmetal element that is a liquid at room temperature. (b) Mercury is the only metal element that is a liquid at room temperature. (c) Iodine (d) (b) diatomic, (c) noble gas 50. (a) NH4Cl

(d)

FeF2

(b) H2SO4

(e)

Pb3(PO4)2

(f)

Al2O3

– Chapter 3 –

51. Group 1A oxides: Li2O, Na2O, K2O, Rb2O, Cs2O Group 2A oxides: BeO, MgO, CaO, SrO, BaO 52. (a) Arachidic acid Arachidonic acid

20 carbons, 40 hydrogens, and 2 oxygens 20 carbons, 32 hydrogens, and 2 oxygens

(b) Stearic acid

18 carbons, 36 hydrogens, and 2 oxygens

Linoleic acid

18 carbons, 32 hydrogens, and 2 oxygens

40 H 2H = 20 C C 36 H 2 H Stearic acid = 18 C C

32 H 1.6 H = 20 C C 32 H 1.8 H Linoleic acid = 18 C C

Arachidonic acid

(d) Saturated molecules have more H’s per C than unsaturated molecules. Saturated molecules must have more hydrogen atoms. 53. (a) Let x = RDA of iron 60% of x = 11 mg Fe x=

11 mg Fe ×100% = 18 mg Fe 60%

(b) density of iron = 7.86 g/mL

 1 g  1 mL  m = (11 mg Fe )    d  1000 mg  7.86 g 

= 1.4 ×10−3 mL Fe 54. If Alfred inspects the bottles carefully, he should be able to see whether the contents are solid (silver) or liquid (mercury). Alternatively, since mercury is more dense than silver, the bottle of mercury should weigh more than the bottle of silver (the question indicated that both bottles were of similar size and both were full). Density is mass/volume.

CHAPTER 4

PROPERTIES OF MATTER SOLUTIONS TO REVIEW QUESTIONS 1.

Gaseous state: 393 K = 120.0°C; boiling point of acetic acid is 118.0°C.

Liquid state: melting point of chlorine is −101.6°C and boiling point of chlorine is −34.6°C.

(a) 118.0°C+273.15=391.2 K (b) (118.0°C)(1.8)+32=244.4°F

(a) (16.7°C)(1.8)+32=62.1°F (b) 16.7°C+273.15=289.9 K

Small bubbles appear at each electrode, and a gas collects above each electrode. The system now contains water and two different gases.

A new substance is always formed during a chemical change, but never formed during physical changes.

Reading the problem carefully and writing down all of the important information including units is the critical first step in solving any chemistry problem.

No, the last step is always a check of the answer to make sure it makes sense.

Potential energy is the energy of position. By the position of an object, it has the potential of movement to a lower energy state. Kinetic energy is the energy matter possesses due to its motion.

10. A food Calorie is equal to 1000 cal or 1 kcal. 11. Iron requires more energy to heat because it has a higher specific heat. 12. A molecule of octane would produce more carbon dioxide because it contains more carbon atoms. 13. All hydrocarbons are composed only of the elements carbon and hydrogen. 14. Carbon 15. Fossil fuels produce carbon dioxide, a greenhouse gas. The supply of fossil fuels is also decreasing and mining or drilling for these fuels is harmful to the environment. Other sources of energy which may become important are wind and solar.

– Chapter 4 –

SOLUTIONS TO EXERCISES

(a) physical

(e) chemical

(b) chemical

(f)

(c)

physical

(g) chemical

(d) physical

(h) chemical

(a) physical

(e) physical

(b) physical

(f)

(c)

chemical

(g) chemical

(d) physical

(h) chemical

physical

Although the appearance of the platinum wire changed during the heating, the original appearance was restored when the wire cooled. No change in the composition of the platinum could be detected.

A copper wire, like the platinum wire, changes to a glowing red color when heated (physical change). Upon cooling, the original appearance of the copper wire is not restored, but a new substance, black copper(II) oxide appears (chemical change).

Reactants:

copper, oxygen

Product:

copper(II) oxide

Reactant:

water

Product:

hydrogen, oxygen

6. 7.

(a) chemical

(d)

chemical

(b) physical

(e)

chemical

(f)

physical

(a) physical

(d)

physical

(b) chemical

(e)

chemical

(f)

chemical

(a) kinetic energy

(d)

potential energy

(b) kinetic energy

(e)

kinetic energy

(d)

kinetic energy

(e)

kinetic energy

The kinetic energy is converted to thermal energy (heat), chiefly in the brake system, and eventually dissipated into the atmosphere.

12.

The transformation of kinetic energy to thermal energy (heat) is responsible for the fiery re-entry of a space vehicle.

– Chapter 4 –

13. (a) +

(d)

(b) +

(e)

−

14. (a) +

(d)

−

(b) −

(e)

−

heat = ( m ) ( specific heat )( Δt )

= (125 g ) ( 0.900 J/g°C )( 95.5°C −19.0°C ) = 8.61× 103 J

16.

heat = ( m ) ( specific heat )( Δt )

= ( 65 g ) ( 0.128 J/g°C )( 98.5°C − 22.0°C ) = 6.4 ×10 2 J

17.

heat = ( m ) ( specific heat )( Δt )

= ( 25.0 g ) ( 2.138 J/g°C )( 78.5°C − 22.5°C ) = 2.99 × 103 J

18.

heat = ( m ) ( specific heat )( Δt )

= ( 35.0 g ) ( 2.604 J/g°C )(82.4°C − 21.2°C ) = 5.58 × 103 J

19.

heat = ( m )( specific heat )( Δt ) ; change kJ to J

specific heat = 20.

heat 2.50×103 J = = 0.230 J/g°C m ( Δt ) (135 g )(100.0°C −19.5°C )

heat = ( m )( specific heat )( Δt ) ; change kJ to J

specific heat =

heat 1.075 ×10 4 J = = 0.128 J/g°C m ( Δt ) ( 275 g )( 327.5°C − 21.2°C )

21. heat lost by copper = heat gained by water x = final temperature ( m ) ( specific heat )( Δt ) = ( m ) ( specific heat )( Δt )

(155.0 g ) ( 0.385 J/g°C )(150.0°C − x ) = ( 250.0g )( 4.184 J/g°C )( x − 19.8°C ) 8950 J − 59.675x J/°C = 1046 x J/°C − 20710 J 8950 J + 20,710 J = 1046 x J/°C + 59.675x J/°C 29,660 J = 1106 x J/°C 26.8°C = x

– Chapter 4 –

22. heat lost by copper = heat gained by water x = final temperature

( m ) ( specific heat )( Δt ) = ( m ) ( specific heat )( Δt ) ( 225.0 g ) ( 0.900 J/g°C )(125.5°C − x ) = (500.0g )( 4.184 J/g°C )( x − 22.5°C ) 25, 410 J − 202.5x J/°C = 2090 x J/°C − 47,070 J 25,410 J + 47,070 J = 2090 x J/°C + 202.5x J/°C 72,480 J = 2290 x J/°C 31.7°C = x 23. (a) 1 (b) 2 (c) 1 (d) 2 (e) 4 24. (a) 3 (b) 4 (c) 2 (d) 1 (e) 2 25. Physical properties of zeolites Crystalline solids, low density, porous, absorb water Chemical properties of zeolites Composed of silicon, aluminum, and oxygen, formed from reaction of volcanic rocks and ash with alkaline water 26. (a)

heat lost by metal = heat gained by water

( m )( specific heat )( Δt ) = ( m )( specific heat )( Δt ) (110.0 g )( specific heat )( 92.0°C − 24.2°C ) = ( 75.0g )( 4.184 J/g°C )( 24.2°C − 21.0°C ) ( specific heat ) 7458 g°C = 1004.16 J specific heat = 0.13 J/g°C (b) Iron has a specific heat of 0.47 J/g°C and lead has a specific heat of 0.13 J/g°C. The metal could possibly be lead, but further tests would be needed to determine this. The metal cannot be iron. 27. (a) heat = 262 Cal or 262 kcal or 262,000 cal  453.6 g  4 Mass of man = (165 lb )   = 7.48 × 10 g 1 lb    3.47 J   1 cal  0.829 cal specific heat =  =  g °C  g °C   4.184 J  heat = ( m )( specific heat )( Δt ) Δt =

( 26,2000 cal ) heat = = 4.22°C ( m )( specific heat ) ( 7.48 × 10 4 g ) ( 0.829 cal/g°C )

– Chapter 4 –

(b)

heat = 250 Cal or 250 kcal or 250,000 cal

 453.6 g  4 Mass of man = (165 lb )   = 7.48 ×10 g 1 lb    3.47 J   1 cal 0.829 cal  specific heat =  =    g °C   g°C   4.184 J heat = ( m )( specific heat )( Δt ) Δt =

( 250, 000 cal ) heat = = 4.0°C ( m )( specific heat ) ( 7.48 × 10 4 g ) ( 0.829 cal/g°C )

28. heat = ( m ) ( specific heat )( Δt )

mass =

heat 1.69×10 4 J = = 52.5 g H2 O ( specific heat )( Δt ) ( 4.184 J/g°C )(100.0°C − 23.0°C )

29. heat = ( m ) ( specific heat )( Δt ) mass =

heat ; convert grams to pounds ( specific heat )( Δt )

3.25 × 10 4 J = 238 g gold ( 0.131 J/g°C )(1064.4°C − 23.2°C )  1 lb   = 0.525 lb gold  453.6 g 

( 238 g ) 

30. heat = (m)(specific heat)(Δt) change kJ to J x = final temperature 4.00 × 10 4 J = ( 500.0 g )( 4.184 J/g°C )( x − 10.0°C ) 4.00 × 10 4 J = 2092 x J/°C − 20,920 J 60,920 J = 2092 x J/°C 60,920°C = 2092 x 29.1°C = x

31. heat lost by iron = heat gained by water x = mass of iron in grams (m)(specific heat)(Δt) = (m)(specific heat)(Δt) x ( 0.473 J/g°C )(125.0°C − 25.6°C ) = ( 375 g )( 4.184 J/g°C )( 25.6°C − 19.8°C ) 47.0x J/g = 9.1× 103 J x=

9.1× 103 J = 190 g Fe 47.0 J/g

– Chapter 4 –

32. heat lost by copper = heat gained by water x = mass of copper in grams (m)(specific heat)(Δt) = (m)(specific heat)(Δt)

x ( 0.385 J/g°C )( 275.1°C − 29.7°C ) = ( 272 g )( 4.184 J/g°C )( 29.7°C − 21.0°C ) 94.5x J/g = 9.9 ×103 J x=

9.9 ×103 J = 1.0 × 102 g Cu 94.5 J/g

33. (a) heat = (m)(specific heat)(Δt) (100.0 g)(0.0921 cal/g°C)(100.0°C−10.0°C) = 829 cal to heat Cu (b) let x = temperature of Al after adding 829 cal 829 cal = (100.0 g)(0.215 cal/g°C)(x – 10.0°C) 829 cal = (21.5 cal/°C)x – 215 cal x = 48.6°C (final temperature for aluminum) Therefore the copper gets hotter since it ended up at 100.0°C. Note: You can figure this out without calculation if you consider the specific heats of the metals. Since the specific heat of copper is much less than aluminum the copper heats more easily. 34. heat lost by iron = heat gained by water

x = initial temperature of iron

( m )( specific heat )( Δt ) = ( m )( specific heat )( Δt ) (500.0 g )( 0.473 J/g°C )( x − 90.0°C ) = ( 400.0 g )( 4.184 J/g°C )( 90.0°C − 10.0°C ) J   4 5  237  x − 2.13 ×10 J = 1.34 ×10 J °C   J   5  237  x = 1.55 ×10 J°C °C   1.55 × 105 J°C 237 J x = 654°C x=

35. heat lost by metal = heat gained by water

x = specific heat of metal

( m )( specific heat )( Δt ) = ( m )( specific heat )( Δt ) ( 20.0 g )( x )( 203.°C − 29.0°C ) = (100.0 g )( 4.184 J/g°C )( 29.0°C − 25.0°C ) ( 3480 g°C ) x = 1674 J x = 0.48 J/g°C 36. heat lost = heat gained x = final temperature

( m )( specific heat )( Δt ) = ( m )( specific heat )( Δt ) ( specific heats are the same )

– Chapter 4 –

Let x = final temperature

 J   J  (50.0°C − x ) = (50.0 g )    ( x − 10.0°C )  g°C   g°C 

(10.0 g ) 

500 g°C − 10.0 gx = 50.0 gx − 500 g°C 1.00 × 103 g°C = 60.0 gx x=

1.00 × 103 g°C = 16.7°C 60.0 g

37. Specific heats for the metals are Fe: 0.473 J/g°C; Cu: 0.385 J/g°C; Al: 0.900 J/g°C. The metal with the lowest specific heat will warm most quickly, therefore, the copper pan heats fastest, and fries the egg fastest. 38. In order for the water to boil, both the pan and water must reach 100.0°C. Specific heat for copper is 0.385 J/g°C.



(300.0 g )  0.385

 J  J  (100°C − 25°C ) + (800.0 g )  4.184 (100°C − 25°C )  g°C  g°C  

 = 8.7 × 103 J + 2.5 × 105 J

=2.6 × 105 J needed to heat the pan and water to 100°C 1s  ( 2.6 × 10 J )  628  = 414 s = 6.9 min = 6 min + 54 s J 5

The water will boil at 6:06 and 54 s p.m. 39. Heat is transferred from the molecules of water on the surface to the air above them. As you blow you move the warmed air molecules away from the surface replacing them with cooler ones which are warmed by the coffee and cool it in a repeating cycle. Inserting a spoon into hot coffee cools the coffee by heat transfer as well. Heat is transferred from the coffee to the spoon lowering the temperature of the coffee and raising the temperature of the spoon. 40. The potatoes will cook at the same rate whether the water boils vigorously or slowly. Once the boiling point is reached the water temperature remains constant. The energy available is the same so the cooking time should be equal. 41.

( 250 mL ) ( 0.04 ) = 10 mL fat (10 mL ) ( 0.8 g/mL ) = 8 g fat in a glass of milk

42. (a) heat = (m)(specific heat)(Δt) m = 99.3 g

1.33 J g°C 100°C Δt = 15.3°F × = 8.50°C 180°F  1.33 J  heat = ( 99.3 g )   ( 8.50°C ) = 1120 J or 1.12 kJ  g°C 

specific heat =

– Chapter 4 –

(b)

heat = ( m )( specific heat )( Δt )

m = 86.2 g 100°C = 14.4°C 180°F heat = 1.71 kJ = 1710 J Δt = 25.9°F ×

specific heat =

( heat ) = (1710 J ) 1.38 J = ( m )( Δt ) ( 86.2 g )(14.4°C ) g°C

The glove appears to be made of wool felt. (c)

heat = 1.65 kJ or 1650 J

m = 50.0 g heat = ( m )( specific heat )( Δt ) Δt =

heat ( m )( specific heat )

For wool felt specific heat = Δt =

1.38 J g°C

(1650 cal ) heat = = 23.9°C or 43.0°F ( m )( specific heat ) ( 50.0 g )(1.38 cal/g°C )

For cotton or paper specific heat = Δt =

(1650 cal ) heat = = 24.8°C or 44.7°F ( m )( specific heat ) ( 50.0 g )(1.33 cal/g°C )

For rubber specific heat = Δt =

3.65 J g°C

(1650 cal ) heat = = 9.04°C or 16.3°F ( m )( specific heat ) ( 50.0 g )(3.65 cal/g°C )

For silicon specific heat = Δt =

1.33 J g°C

1.46 J g°C

(1650 cal ) heat = = 22.6°C or 40.7°F ( m )( specific heat ) ( 50.0 g )(1.46 cal/g°C )

(d) Rubber (e) The higher the specific heat the less the glove will increase in temperature as it absorbs heat. This means a high specific heat is good for an oven mitt. 43. Note that 100 MW =

100 MJ s

 7.50 hr   60 min   60 s   100 MJ   106 J   1 GJ   1 ton coal   9       s   1 MJ   10 J   26.6 GJ   1 day   1 hr   1 min  

(1 day ) 

= 102 tons of coal

– Chapter 4 –

44. (a) Container holds a mixture of sulfur and oxygen. (b) No. If the container were sealed, the total mass would remain the same whether a reaction took place or not. The mass of the reactants must equal the mass of the products. (c) No. Density is mass/volume. The volume is the container volume, which does not change. Since the total mass remains constant even if a reaction has taken place, the density of the container, including its contents, remains constant. The density of each individual component within the container may have changed, but the total density of the container is constant. 45. Knowns:

initial temp = 5 °C; final temp = 61°C Δt = 61− 5 = 56°C specific heat of water = 4.184 J/g°C

heat in J = 1.0 ×103 J solving for: mass equation: heat in J = mass × specific heat × Δt rearrange:

mass =

heat in J ( specific heat ) ( Δt )

solve:

mass =

1.0 ×103 J = 4.267 g = 4.3 g ( to 2 significant figures ) ( 4.184J/g°C ) (56°C )

46. (2) 43 kJ. We know the mass of water (180 g) and can determine the temperature difference by subtraction. The unknown quantity is the number of kilojoules of heat energy required. Specific heat has units involving joules so the heat energy is first found in joules and then divided by 1000 to give the answer in kilojoules (and two significant figures). joules = m × specific heat × ΔT = (180 g )( 4.184 J/g°C )( 82 − 25 ) °C = 42928 J = 43 kJ

47. According to the law of conservation of energy the amount of potential energy the ball has initially should equal the amount of kinetic energy it has at the bottom of the hill. If the hill that the ball rolled down was frictionless then the ball should role up the other side until it has reached the same level as where it started. However, no hill is really frictionless. Therefore, some of the ball’s potential energy is converted to kinetic energy of motion and some is lost by way of heat friction between the ball and the surface of the hill. 48. A chemical change has occurred. Hydrogen molecules and oxygen molecules have combined to form water molecules. 49. (a) As the water is heated, molecules that exist in the liquid state are changed into molecules in the gaseous state. (b) A physical change has occurred. No new substance was formed during the heating process. Only a change in state occurred.

– Chapter 4 –

50. Upon heating the substance, the appearance of the bright blue solid changes to black and a brownish orange gas is released. A chemical change has occurred. 51. (a) Tim’s bowl should require less energy. Both bowls hold the same volume, but since snow is less dense than a solid block of ice, the mass of water in Tim’s bowl is less than the mass of water in Sue’s bowl. (Both bowls contain ice at 12°F.) (b)

12°F − 32 = −11°C 1.8 Temperature change: −11°C to 25°C = 36°C

(c) temperature change: −11°C to 0°C = 11°C specific heat of ice = 2.059 J/g°C  1000 mL  vol. of H2 O = 1 qt = ( 0.946 L )   L   = 946 mL  1g  mass of ice = mass of water = 0.946 mL    1 mL  = 946 g

(

)

heat required = ( m )( sp.ht.)( Δt )

 2.059 J   1 kJ  = ( 946 g )   (11°C )   = 21 kJ  1000 J   g°C 

(d) Physical changes 52.

(300 kJ ) 

1000 J   75 J transferred to food  5   = 2.25 × 10 J transferred to food 100 J available  1 kJ   

heat = ( m )( specific heat )( Δt ) mass =

( heat )

( 2.25 ×10 J ) 5

( specific heat )( Δt ) ( 4.184 J/g°C )(100.0°C − 15.0°C )

= 633 g water

 1 mL water   1 oz water    = 21.4 oz water or between 2 and 3 cups  1.00 g water   29.6 mL water 

( 633 g water ) 

CHAPTER 5

EARLY ATOMIC THEORY AND STRUCTURE SOLUTIONS TO REVIEW QUESTIONS 1.

•

Elements are composed of indivisable particles called atoms.

•

Atoms of the same element have the same properties; atoms of different elements have different properties.

•

Compounds are composed of atoms joined together to form compounds.

•

Atoms combine in whole number ratios to form compounds.

•

Atoms may combine in different ratios to form more than one compound.

Dalton used Democritus’ idea that all matter was composed of tiny indivisible particles or atoms when formulating his theory.

Dalton said that compounds could form only by combining whole atoms, not parts of atoms. Thus chemical formulas will always show whole numbers of atoms combining.

An atom is electrically neutral, containing equal numbers of protons and electrons. An ion has a charge resulting from an imbalance between the numbers of protons and electrons.

The force of attraction increases as the distance between the charged particles decreases.

Cations are ions with a positive charge and anions are ions with a negative charge.

The neutron is about 1840 times heavier than an electron.

Particle

Charge

Mass

proton

1 amu

neutron

1 amu

electron

−1

Element

Atomic number

(a) copper

(b) nitrogen

(d) radium

(e) zinc

10. Isotopic notation ZA E Z represents the atomic number A represents the mass number

– Chapter 5 –

11. Isotopes contain the same number of protons and the same number of electrons. Isotopes have different numbers of neutrons and thus different atomic masses. 12. The mass number is equal to the sum of the number of protons and the number of neutrons in an atom. It is not possible to have a partial proton or neutron in an atom, thus the total number of nuclear particles will always be a whole number.

– Chapter 5 –

SOLUTIONS TO EXERCISES 1.

N3−, O2−, and Te3−

Mg2+, Cr3+, Ba2+, Ca2+, and Y3+

Gold nuclei are very massive (compared to an alpha particle) and have a large positive charge. As the positive alpha particles approach the atom, some are deflected by this positive charge. Alpha particles approaching a gold nucleus directly are deflected backwards by the massive positive nucleus.

(a) The nucleus of the atom contains most of the mass since only a collision with a very dense, massive object would cause an alpha particle to be deflected back towards the source. (b) The deflection of the positive alpha particles from their initial flight indicates the nucleus of the atom is also positively charged. (c) Most alpha particles pass through the gold foil undeflected leading to the conclusion that the atom is mostly empty space.

In the atom, protons and neutrons are found within the nucleus. Electrons occupy the remaining space within the atom outside the nucleus.

The nucleus of an atom contains nearly all of its mass.

(a) Dalton contributed the concept that each element is composed of atoms which are unique, and can combine in ratios of small whole numbers. (b) Thomson discovered the electron, determined its properties, and found that the mass of a proton is 1840 times the mass of the electron. He developed the Thomson model of the atom. (c) Rutherford devised the model of a nuclear atom with the positive charge and mass concentrated in the nucleus. Most of the atom is empty space.

Electrons:

Dalton – electrons are not part of his model Thomson – electrons are scattered throughout the positive mass of matter in the atom Rutherford – electrons are located out in space away from the central positive nucleus

Positive matter:

Dalton – no positive matter in his model Thomson – positive matter is distributed throughout the atom Rutherford – positive matter is concentrated in a small central nucleus

Atomic masses are not whole numbers because: (a) the neutron and proton do not have identical masses and neither is exactly 1 amu. (b) most elements exist in nature as a mixture of isotopes with different atomic masses due to different numbers of neutrons. The atomic mass given in the periodic table is the average mass of all these isotopes.

– Chapter 5 –

10. The isotope of 126 C with a mass of 12 is an exact number by definition. The mass of 63 Cu , will not be an exact number for reasons given in other isotopes, such as 29 Exercise 9.

11. The isotopes of hydrogen are protium, deuterium, and tritium. 12. All three isotopes of hydrogen have the same number of protons (1) and electrons (1). They differ in the number of neutrons (0, 1, and 2). 13. All five isotopes have nuclei that contain 32 protons. The numbers of neutrons are: Isotope mass number

Neutrons

14. All five isotopes have nuclei that contain 30 protons. The numbers of neutrons are: Isotope mass number

Neutrons

15. This element has 9 protons and 10 neutrons so it is an atom of fluorine-19. The isotopic notation for fluorine-19 is 199F . 16. This element has 5 protons and 5 neutrons so it is an atom of boron-10. The isotopic notation for boron-10 is 105 B . 17.

This atom contains 6 protons, 6 neutrons, and 6 electrons.

– Chapter 5 –

18.

This atom contains 7 protons, 7 neutrons, and 7 electrons. 19. (a)

65 29

(b)

45 20

(c)

84 36

20. (a)

109 47

(b)

18 8

(c)

57 26

21. (a) 29 protons, 29 electrons, and 34 neutrons (b) 16 protons, 16 electrons, and 16 neutrons (c) 25 protons, 25 electrons, and 30 neutrons (d) 19 protons, 19 electrons, and 20 neutrons 22. (a) 26 protons, 26 electrons, and 28 neutrons (b) 11 protons, 11 electrons, and 12 neutrons (c) 35 protons, 35 electrons, and 44 neutrons (d) 15 protons, 15 electrons, and 16 neutrons 23. (a) 33 (b) Arsenic, As (c) 43 (d) The charge is −3, this is an anion. (e)

76 33

24. (a) 56 (b) Barium, Ba (c) 79 (d) The charge is +2, this is a cation. (e)

135 56

– Chapter 5 –

25. (a) A nitride anion has a charge of −3. (b) 14.01 u (c)

 9.110 ×10 −28 g    1 amu   = 0.001646 u −24  electron   1.6606 ×10 g 

(3 electrons ) 

(d) 14.01 u + 0.001646 u = 14.01 u. Using the rules for significant figures, the additional mass added by the extra electrons does not significantly affect the mass of the nitrogen atom. 26. (a) A calcium cation has a charge of +2. (b) 40.08 u (c)

 9.110 ×10 −28 g    1 amu 2 electrons ( )   = 0.001097 u −24  electron   1.6606 ×10 g 

(d) 40.08 u – 0.001097 u = 40.08 u. Using the rules for significant figures, the mass lost does not significantly affect the mass of the calcium atom. 27. For each isotope: (%)(amu) = that portion of the average atomic mass for that isotope. Add together to obtain the average atomic mass. (0.5145)(89.905 amu) + (0.1122)(90.906 amu) + (0.1715)(91.905 amu) + (0.1738)(93.906 amu) + (0.0280)(95.908 amu) 46.26 amu + 10.20 amu + 15.76 amu + 16.32 amu + 2.69 amu =91.23 amu = average atomic mass of Zr 28. For each isotope: (%)(amu) = that portion of the average atomic mass for that isotope. The sum of the portions = the average atomic mass. (0.080)(45.953) + (0.073)(46.952) + (0.738)(47.948) + (0.055)(48.948) + (1.000 − 0.946)x amu = 47.9 amu = 3.7 amu + 3.4 amu + 35.4 amu + 2.7 amu + 0.054x amu = 47.9 amu = 45.2 amu + 0.054x = 47.9 amu x amu =

2.7 amu 0.054

0.054x = 47.9 amu − 45.2 amu

x = 50 = mass of the fifth isotope of titanium

29. (0.6917)(62.9296 amu) + (1.0000 − 0.6917)(64.9278 amu) = 43.53 amu + 20.02 amu = 63.55 amu = average atomic mass The element is copper (see periodic table). 30. (0.7577)(34.9689 amu) + (1.0000 − 0.7577)(36.9659 amu) = 26.50 amu + 8.96 amu = 35.46 amu = average atomic mass The element is chlorine (see periodic table).

– Chapter 5 –

4 31. Vsphere = π r 3 3

rA = radius of atom, rN = radius of nucleus

4 3 π rA r3 (1.0 ×10 −8 )3 Vatom 1.0 ×1015 = =3 = A3 = Vnucleus 4 π r3 rN (1.0 ×10 −13 )3 1.0 N 3 32.

3.0×10−8 cm 1.5 ×105 = 2.0 ×10 −13 cm 1.0

(The ratio of atomic volume to nuclear volume is 1.0×1015 :1.0.)

(The ratio of the diameter of an Al atom to its nucleus diameter is 1.5×105 :1.0.)

33. (a) Boron-10 has 5 protons and 5 neutrons, boron-11 has 5 protons and 6 neutrons. (b) Boron-10, 105 B ; boron-11, 115B (c)

Mass protons in both isotopes 5 (1.673 ×10 −24 g ) = 8.365 × 10 −24 g Mass neutrons in boron 10 Mass neutrons in boron 11

5 (1.675 ×10 −24 g ) = 8.375 × 10 −24 g 6 (1.675 ×10 −24 g ) = 1.005 × 10 −23 g

  8.365 ×10 −24 g % p in 105 B =   × 100 = 49.97% proton −24 −24  8.365 × 10 g + 8.375 × 10 g    8.365 × 10 −24 g % p in B =   × 100 = 45.42% proton −24 −23  8.365 × 10 g + 1.005 × 10 g  11 5

34. (a) Iridium-191 has 77 protons and 114 neutrons, iridium-193 has 77 protons and 116 neutrons. r 193 r (b) Iridium-191, 191 77 I ; iridium-193, 77 I

(c)

Mass protons in both isotopes 77 (1.673 ×10−24 g ) = 1.288 ×10 −22 g Mass neutrons in iridium-193 Mass neutrons in iridium-193

114(1.675 ×10−24 g ) = 1.910 ×10 −22 g 116(1.675 ×10−24 g ) = 1.943 ×10 −22 g

  1.288 × 10 −22 g % p in 191 Ir = × 100 = 40.28% proton  77 −22 −22   1.288 × 10 g + 1.910 × 10 g    1.288 × 10 −22 g % p in 193 Ir = × 100 = 39.86% proton  77 −22 −22   1.288 × 10 g + 1.943 × 10 g 

35. (1) (b), (2) (d), (3) (a), (4) (c) 36. (a) In Rutherford’s experiment the majority of alpha particles passed through the gold foil without deflection. This shows that the atom is mostly empty space and the nucleus is very small. (b) In Thomson’s experiments with the cathode ray tube, rays were observed coming from both the anode and the cathode.

– Chapter 5 –

(c) In Rutherford’s experiment an alpha particle was occasionally dramatically deflected by the nucleus of a gold atom. The direction of deflection showed the nucleus to be positive. Positive charges repel each other. 37. Elements (a) and (c) are isotopes of phosphorus. 38. hydrogen 11H 39.

deuterium 21H

tritium 31H

(8.5 in.) 

2.54 cm   1 atom Si  8   = 9.2 ×10 atoms Si −8 in. 2.34 cm ×10   

40. The properties of an element are related to the number of protons and electrons. If the number of neutrons differs, isotopes result. Isotopes of an element are still the same element even though the nuclear composition of the atoms are different. 41.

156

Dy has 90 neutrons; 160Gd has 96 neutrons; 162Er has 94 neutrons; 165Ho has 98 neutrons. In order of increasing number of neutrons: Dy<Er<Gd<Ho On the periodic table, the order is based on increasing number of protons, so the order is Gd<Dy<Ho<Er.

42. Na = 12 Zn = 35 Cd = 67 Ar = 22 Cl = 20 Mg = 12 43. Because the average atomic mass is very close to the atomic mass of lithium-7, this must be the more common isotope. 44. Percent of sample 60 Q = x Percent of sample 63Q = 1 − x

( x )( 60 amu ) + (1 − x )( 63 amu ) = 61.5 amu 60x amu + 63 amu − 63x amu = 61.5 amu 63 amu − 61.5 amu = 63x amu − 60x amu 1.5 = 3x 0.50 = x 60

Q = 50%

45. (a) Compare the mass of the unknown element to the mass of a carbon-12 atom.

(3.27 ×10

−19

 1 g  12.0 amu  mg unknown element )  = 197 amu  −23   1000 mg  1.9927 ×10 g 

The atomic mass of the unknown element is 197 amu. (b) The unknown element is Au, gold (see periodic table). 46. Li = 7

Fe = 56

Al = 27

H=1

Br = 81

O = 18

– Chapter 5 –

47. (1) Isotope A – the average amu is closer to the mass of the lighter isotope; with only two isotopes, the weighted average amu will be closer in mass to the more abundant isotope. (2) Isotope B (3) Isotope A (4) Isotope B 48.

453.6 g   1 atom Ag  24  = 1.3 ×10 atoms Ag  −22  lb   1.79×10 g Ag 

( 0.52 lb Ag ) 

49. These are the elements that have the same number of protons, neutrons, and electrons. Protons

Neutrons Electrons

50. C+ O+ O2+

6 protons, 5 electrons 8 protons, 7 electrons 8 protons, 6 electrons

51. (a) An ion is a charged atom or group of atoms. The charge can be either positive or negative. (b) Electrons have negligible mass compared with the mass of protons and neutrons. The only difference between Ca and Ca2+ is two electrons. The mass of those two electrons is insignificant compared with the mass of the protons and neutrons present (and whose numbers do not change). 52. Number Number Mineral Supplement Mineral use Ion provided protons electrons Calcium carbonate Bones and Teeth Ca2+ 20 18 2+ Iron(II) sulfate Hemoglobin Fe 26 24 3+ Chromium(III) nitrate Insulin Cr 24 21 Magnesium sulfate Bones Mg2+ 12 10 2+ Zinc sulfate Cellular metabolism Zn 30 28 − Potassium iodide Thyroid function I 53 54 9

– Chapter 5 –

53. Element

Symbol

Atomic #

Mass #

Protons Neutrons Electrons

Chlorine

Gold

197

118

Barium

135

Argon

Nickel

Element

Symbol

Atomic #

Mass #

Protons

Neutrons Electrons

Xenon

134

Silver

107

Fluorine

Uranium

235

143

Potassium

Cl Au

54.

55.

56. (a) 246.82 (b) The heavier isotope. Since the isotopes differ by only one mass number, the nuclides must weigh 246 and 247. If there were equal amounts of each isotope, the average mass would be 246.5. Since the average mass is greater than 246.5, the heavier isotope must be more abundant. 57. Cl2 O7Cl : 17p × 2 = 34 protons

O : 8p × 7 = 56 protons 90 protons in Cl2 O7 Since the molecule is electrically neutral, the number of electrons is equal to the number of protons, so Cl2O7 has 90 electrons. The number of neutrons cannot be precisely determined unless it is known which isotopes of Cl and O are in this particular molecule.

– Chapter 5 –

58. Electrons are not in the nucleus. An ion of Ca (Ca2+) and an atom of Ar have the same number of electrons. 59. The mass of one electron is 9.110 × 10−28 grams.

 (13 ) ( 9.110 ×10 −28 g )   (100 ) = 0.02644% electrons (a) Aluminum has 13 electrons.   4.480 ×10 −23 g     (15 ) ( 9.110 ×10 −28 g )  (b) Phosphorus has 15 electrons.   (100 ) = 0.02657% electrons  5.143× 10 −23 g   

 ( 36 ) ( 9.110 ×10 −28 g )   (100 ) = 0.02356% electrons (c) Krypton has 36 electrons.   1.392 ×10 −22 g     ( 78 ) ( 9.110 ×10 −28 g )   (100 ) = 0.02193% electrons (d) Platinum has 78 electrons.   3.240 ×10 −22 g    60. The mass of one proton is 1.673 × 10−24 grams.

 ( 34 ) (1.673×10 −24 g )   (100 ) = 43.39% protons (a) Selenium has 34 protons.   1.311×10 −22 g     ( 54 ) (1.673×10 −24 g )   (100 ) = 41.44% protons (b) Xenon has 54 protons.   2.180 ×10 −22 g     (17 ) (1.673×10 −24 g )   (100 ) = 48.31% protons (c) Chlorine has 17 protons.   5.887 ×10 −23 g     ( 56 ) (1.673×10 −24 g )   (100 ) = 41.09% protons (d) Barium has 56 protons.   2.280 ×10 −22 g    61. The electron region is the area around the nucleus where electrons are most likely to be located. 62.

– Chapter 5 –

Average atomic mass: ( 0.1081)( 269.14 amu ) = 29.09 amu

( 0.3407 )( 270.51 amu ) = 92.16 amu ( 0.5512 )( 271.23 amu ) = 149.50 amu Total = 270.75 amu

An atomic mass of 270.75 amu would come somewhere after Bohrium (mass = 264 amu). So, the atomic number of this new element would be greater than 107,most likely either meitnerium(109) or darmstadtium (110). 63. (a) Let x = abundance of heavier isotope

303.9303 ( x ) + 300.9326 (1 − x ) = 303.001

303.9303 ( x ) − 300.9326x = 303.001 − 300.9326 2.9977x = 2.068

x = 0.6899 1 − x = 0.3101 % Abundance of heavier isotope = 68.99% % Abundance of lighter isotope = 31.01% (b) (c)

304 120

Wz, 301 120Wz

Mass number − atomic number = 303 −120 = 183 neutrons

64. (a) 12 amu × 7.18 = 86.16 amu (b) The atom is most likely Rb or Sr. Other remote possibilities are Kr or Y. (c) Because of the possible presence of isotopes, the atom cannot be positively identified. The periodic table gives average masses. (d) M forms a +1 cation and is most likely in Group 1A. The unknown atom is most likely 86 37 Rb . 65. The presence of isotopes contradicts Dalton’s theory that all atoms of the same element are identical. Also, the discovery of protons, neutrons, and electrons suggests that there are particles smaller than the atom and that the atom is not indivisible. Thomson proposed a model of an atom with no clearly defined nucleus. Rutherford passed alpha particles through gold foil and inspected the angles at which the alpha particles were deflected. From his results, he proposed the idea of an atom having a small dense nucleus.

CHAPTER 6

NOMENCLATURE OF INORGANIC COMPOUNDS SOLUTIONS TO REVIEW QUESTIONS 1.

Water (H2O)–dihydrogen monoxide Ammonia (NH3)–nitrogen trihydride

An atom must gain one or more electrons to form an anion. An atom must lose one or more electrons to form a cation.

No, if elements combine in a one-to-one ratio the charges on their ions must be equal and opposite in sign. They could be +1, −1, or +2, −2 or +3, −3, etc.

The metal or cation is always named first.

(a) K2S

(d) H3PO4

(b) Co(BrO3)2

(e) Fe2O3

(f)

Mg(OH)2

The system for naming binary compounds composed of two nonmetals uses the name of the first element and the stem of the second element plus the suffix –ide. A prefix is attached to each element indicating the number of atoms of that element in the formula. Thus, N2O5 is dinitrogen pentoxide. Aluminum forms only one series of compounds in which the cation is always Al3+. Thus the name for Al2O3, aluminum oxide, does not need to be distinguished from any other compound.

Barium forms only one series of compounds in which the cation is always Ba2+. Thus the name for BaCl2 (barium chloride) does not need to be distinguished from any other compound. Iron forms two series of compounds in which the iron ion is Fe2+ or Fe3+. Thus the name iron chloride does not indicate which compound is in question. Therefore, FeCl2 is called iron(II) chloride to indicate that the compound contains the Fe2+ ion. Neither uses prefixes because prefixes are used to name compounds composed of two nonmetals.

The mono prefix is generally not used for the first element.

The Stock System clearly indicates cation charges. The classical system works, but requires a chemist to memorize the common charges on all cations. It is much easier to use the Stock System and much more error free.

10. Nickel(II) compounds (a) NiSO4

(d) Ni(OH)2

(g) NiCr2O7

(b) Ni3P2

(e) Ni(IO2)2

(h) NiBr2

(f) Ni(C2H3O2)2

(i) Ni(NO3)2

(j) Ni(ClO)2

– Chapter 6 –

11. (a) HBrO

hypobromous acid

(b) HIO

hypoiodous acid

HBrO2

bromous acid

HIO2

iodous acid

HbrO3

bromic acid

HIO3

iodic acid

HbrO4

perbromic acid

HIO4

periodic acid

12. Include hydrogen or dihydrogen in the name to tell how many hydrogens are left behind. For example Phosphate ion PO43− Hydrogen phosphate ion HPO42− Dihydrogen phosphate ion H2PO4−

– Chapter 6 –

SOLUTIONS TO EXERCISES 1.

Formulas of compounds: (a) Ba and S

BaS

(d) Mg and N

Mg3N2

(b) Cs and P

Cs3P

(e) Ca and I

CaI2

LiBr

(f) H and Cl

HCl

Formulas of compounds: (a) Al and S

Al2S3

(d) Sr and O

SrO

(b) H and F

(e) Cs and P

Cs3P

K3N

(f) Al and Cl

AlCl3

(a) potassium

K+

(b) ammonium

NH+4

Cu+

(d) titanium(IV)

Ti4+

(e) nickel(III)

Ni3+

(f)

cesium

Cs+

(g) mercury(II)

Hg2+

(a) fluoride

F−

(b) acetate

C 2 H 3 O 2−

I−

(d) carbonate

CO 32 −

(e) sulfide

S2−

(f)

nitrate

NO 3−

(g) phosphide

p3−

(a) sodium hydrogen carbonate

(d)

acetic acid

(b) mercury

(e)

magnesium sulfate heptahydrate

(f)

sodium hydroxide

– Chapter 6 –

(a) sodium thiosulfate

(d)

sodium chloride

(b) dinitrogen monoxide

(e)

magnesium hydroxide

(f)

lead(II) sulphide

7. Br−

O2−

NO 3−

PO34−

CO 32 −

KBr

K2O

KNO3

K3PO4

K2CO3

MgBr2

MgO

Mg(NO3)2

Mg3(PO4)2

MgCO3

Al3+

AlBr3

Al2O3

Al(NO3)3

AlPO4

Al2(CO3)3

Zn2+

ZnBr2

ZnO

Zn(NO3)2

Zn3(PO4)2

ZnCO3

H+

HBr

H2O

HNO3

H3PO4

H2CO3

Ion K+ 2+

Ion

SO 24 −

OH−

AsO 34−

C 2 H 3 O 2−

CrO 24 −

NH +4

( NH4 )2 SO4

NH 4 OH

( NH4 )3 AsO4

NH4C2H3O2

(NH4)2CrO4

Ca2+

CaSO4

Ca(OH)2

Ca3(AsO4)2

Ca(C2H3O2)2

CaCrO4

Fe3+

Fe2(SO4)3

Fe(OH)3

FeAsO4

Fe(C2H3O2)3

Fe2(CrO4)3

Ag+

Ag2SO4

AgOH

Ag3AsO4

AgC2H3O2

Ag2CrO4

Cu2+

CuSO4

Cu(OH)2

Cu3(AsO4)2

Cu(C2H3O2)2

CuCrO4

K+ compounds: potassium bromide, potassium oxide, potassium nitrate, potassium phosphate, potassium carbonate Mg2+ compounds: magnesium bromide, magnesium oxide, magnesium nitrate, magnesium phosphate, magnesium carbonate Al3+ compounds: aluminum bromide, aluminum oxide, aluminum nitrate, aluminum phosphate, aluminum carbonate Zn2+ compounds: zinc bromide, zinc oxide, zinc nitrate, zinc phosphate, zinc carbonate H+ compounds: hydrogen bromide (or hydrobromic acid), water, nitric acid, phosphoric acid, carbonic acid

10. NH +4 compounds: ammonium sulfate, ammonium hydroxide, ammonium arsenate, ammonium acetate, ammonium chromate Ca2+ compounds: calcium sulfate, calcium hydroxide, calcium arsenate, calcium acetate, calcium chromate Fe3+ compounds: iron(III) sulfate, iron(III) hydroxide, iron(III) arsenate, iron(III) acetate, iron(III) chromate

– Chapter 6 –

Ag+ compounds: silver sulfate, silver hydroxide, silver arsenate, silver acetate, silver chromate Cu2+ compounds: copper(II) sulfate, copper(II) hydroxide, copper(II) arsenate, copper(II) acetate, copper (II) chromate 11. Nonmetal binary compound formulas: (a) diphosphorus pentoxide, P2O5

(e) carbon tetrachloride, CCl4

(b) carbon dioxide, CO2

(f)

(g) boron trifluoride, BF3

(d) sulfur hexachloride, SCl6

(h) tetranitrogen hexasulfide, N4S6

dichlorine heptoxide, Cl2O7

12. Metal-nonmetal binary compound formulas: (a) potassium nitride, K3N

(e) calcium nitride, Ca3N2

(b) barium oxide, BaO

(f) cesium bromide, CsBr

(g) manganese(III) iodide, MnI3

(d) strontium phosphide, Sr3P2

(h) sodium selenide, Na2Se

13. Naming binary metal-nonmetal compounds: (a) BaO, barium oxide

(e) Al2O3, aluminum oxide

(b) K2S, potassium sulfide

(f)

(g) SrI2, strontium iodide

(d) Cs2S, cesium sulphide

(h) Mg3N2, magnesium nitride

CaBr2, calcium bromide

14. Naming binary nonmetal compounds: (a) PBr5, phosphorus pentabromide

(e) SiCl4, silicon tetrachloride

(b) I4O9, tetraiodine nonoxide

(f) ClO2, chlorine dioxide

(g) P4S7, tetraphosphorus heptasulfide

(d) S2F10, disulfur decafluoride

(h) IF6, iodine hexafluoride

copper(II) chloride

(d) FeCl3

iron(III) chloride

(b) FeCl2

iron(II) chloride

(e) SnF2

tin(II) fluoride

iron(II) nitrate

(f)

vanadium(III) phosphate

15. (a) CuCl2

VPO4

16. Formulas: (a) tin(IV) bromide

SnBr4

(d) mercury(II) nitrite

(b) copper(I) sulfate

Cu2SO4

(e) cobalt(III) carbonate Co2(CO3)3

Ni3(BO3)2

(f)

iron(II) acetate

Hg(NO2)2 Fe(C2H3O2)2

17. Formulas of acids: (a) hydrochloric acid, HCl

(d) carbonic acid, H2CO3

(b) chloric acid, HClO3

(e)

sulfurous acid, H2SO3

(f)

phosphoric acid, H3PO4

– Chapter 6 –

18. Formulas of acids: (a) acetic acid, HC2H3O2

(d) boric acid, H3BO3

(b) hydrofluoric acid, HF

(e) nitrous acid, HNO2

(f) hypochlorous acid, HClO

19. Naming acids: (a) HNO2, nitrous acid

(f)

(b) H2SO4, sulfuric acid

(g) HF, hydrofluoric acid

(h) HBrO3, bromic acid

(d) HBr, hydrobromic acid

(i)

HIO4, periodic acid

(a) H3PO4, phosphoric acid

(f)

HNO3, nitric acid

(b) H2CO3 carbonic acid

(g) HI, hydroiodic acid

(h) HClO4 perchloric acid

(d) HCl, hydrochloric acid

(i)

HC2H3O2, acetic acid

(e) H3PO3, phosphorous acid 20. Naming acids:

H2SO3, sulfurous acid

(e) HClO, hypochlorous acid 21. Formulas for: (a) silver sulphite

Ag2SO3

(b) cobalt(II) bromide

CoBr2

Sn(OH)2

(d) aluminum sulphite

Al2(SO4)3

(e) lead(II) chloride

PbCl2

(f)

(NH4)2CO3

ammonium carbonate

(g) chromium(III) oxide

Cr2O3

(h) cupric chloride

CuCl2

22. Formulas for: (a) sodium chromate

Na2CrO4

(b) magnesium hydride

MgH2

Ni(C2H3O2)2

(d) calcium chlorate

Ca(ClO3)2

(e) magnesium bromate

Mg(BrO3)2

(f)

KH2PO4

potassium dihydrogen phosphate

(g) manganese(II) hydroxide

Mn(OH)2

(h) cobalt(II) hydrogen carbonate

Co(HCO3)2

– Chapter 6 –

23. Names for: (a) ZnSO4

zinc sulfite

(b) Hg2S

mercury(I) sulfite

copper(II) carbonate

(d) Cd(NO3)2

cadmium nitrate

(e) Al(C2H3O2)3

aluminum acetate

(f)

cobalt(II) fluoride

CoF2

(g) Cr(ClO3)3

chromium(III) chlorate

(h) Ag3PO4

silver phosphate

(i)

MnS

manganese(II) sulfite

(j)

BaCrO4

barium chromate

24. Names for: (a) Ca(HSO4)2

calcium hydrogen sulfate

(b) As2(SO3)3

arsenic(III) sulfite

tin(II) nitrite

(d) CuI

copper(I) iodide

(e) KHCO3

potassium hydrogen carbonate

(f)

BiAsO4

bismuth(III) arsenate

(g) (NH4)2CO3

ammonium carbonate

(h) (NH4)2HPO4

ammonium monohydrogen phosphate

(i)

NaClO

sodium hypochlorite

(j)

KMnO4

potassium permanganate

25. Lead commonly forms two cations, lead(II) and lead(IV). This means that the likely compounds found in the mines were lead(II) sulfide, PbS or lead(IV) sulfide, PbS2. Lead(II) oxide is given the common name galena. 26. (a) Lead(II) carbonate PbCO3 Lead(II) hydroxide Pb(OH)2 (b) Lead(II) oxide PbO Lead(II) acetate Pb(C2H3O2)2 27. (a)

K → K+ + e−

(b) I + e − → I − (c)

Br + e − → Br −

(d)

Fe → Fe 2+ + 2e −

(e)

Ca → Ca 2+ + 2e −

(f)

O + 2e − → O2−

– Chapter 6 –

28. (a) sulfate (b) phosphate (c) nitrate (d) chlorate (e) hydroxide (f) carbonate (d)

Ca(ClO3)2

(b) Ca3(PO4)2

(e)

Ca(OH)2

(f)

CaCO3

30. (a) K2SO4

(d)

KClO3

(b) K3PO4

(e)

KOH

(f)

K2CO3

29. (a) CaSO4

31. (a) CBr4 carbon tetrabromide

(c)

PCl5 phosphorus pentachloride

(b) BF3 boron trifluoride 32. Formula: KCl 33. (1) NaHCO3 (4) NH4HS

Name: potassium chloride (2) MgNH4PO4

(3)

NaKSO4

(5) KHSO4

(6)

KAl(SO4)2

34. (a) sulfide ion (d) carbonate ion 35. (a) The second part of the name must include a prefix, even if it is only for one atom of that element. So although we know it is a C and an O in the compound, we don’t know how many O’s. We assume there is only one C, as ‘‘mono’’ is not needed for the first element. (b) There is more than one possible cobalt ion. Is the cobalt in this compound made up of Co2+or Co3+. The appropriate Roman numeral after the cobalt is needed. (c) Since barium is a metal, it is assumed the compound is ionic. The second half of the name should be the name of the polyatomic ion, not named as though it is a binary nonmetallic compound.

– Chapter 6 –

36.

ide:

suffix is used to indicate a binary compound except for hydroxides, cyanides, and ammonium compounds.

ous:

used as a suffix to name an acid that has a lower oxygen content than the -ic acid (e.g., HNO2, nitrous acid, and HNO3, nitric acid). Also used as a suffix to name the lower ionic charge of a metal with more than one possible charge (e.g., Fe2+, ferrous, and Fe3+, ferric).

ic:

used as a suffix to name an acid that has a higher oxygen content than the -ous acid (e.g., HNO3 nitric acid, and HNO2 nitrous acid). Also used as a suffix to name the higher ionic charge of a metal with more than one possible charge (e.g., Fe2+, ferrous, and Fe3+, ferric).

hypo:

used as a prefix in naming an acid that has a lower oxygen content than the -ous acid when there are more than two oxyacids with the same elements (e.g., HClO, hypochlorous acid, and HClO2, chlorous acid).

per:

used as a prefix in naming an acid that has a higher oxygen content than the -ic acid when there are more than two oxyacids with the same elements (e.g., HClO4, perchloric acid, and HClO3, chloric acid).

ite:

the suffix used in naming a salt derived from an -ous acid. For example, HNO2 (nitrous acid); NaNO2 (sodium nitrite).

ate:

the suffix used in naming a salt derived from an -ic acid. For example, H2SO4 (sulfuric acid); CaSO4 (calcium sulfate).

Roman numerals: In the Stock System, Roman numerals are used in naming compounds that contain metals that may exist as more than one type of cation. The charge of a metal is indicated by a Roman numeral written in parenthesis immediately after the name of the metal. For example, FeCl2 [iron (II) chloride]. 37. (a) Calcium carbonate or CaCO3 is the principal ingredient in lime, chalk, and marble. It is harmless, though it does leave a residue on plumbing. There should be no concern regarding this spill. (b) Acetic acid or HC2H3O2 is found in vinegar. This is not a particularly hazardous spill, though it will be widely apparent due to the characteristic odor of vinegar. (Perhaps it could be cleaned up with lots of lettuce?) (c) Dihydrogen oxide or H2O is commonly known as water and it may be safely dumped into the sewer. 38. (a) 50 e−, 50 p

(b) 48e−, 50 p

39. The formula for a compound must be electrically neutral. Therefore X=+3 and Y=−2 since in X2Y3 this would give 2(+3) + 3(−2) = 0. 40. Li3Fe(CN)6 AlFe(CN)6 Zn3[Fe(CN)6]2

– Chapter 6 –

41. (a) N3− nitride

One has oxygen the other does not, charges on the ions differ.

NO2− nitrite (b) NO2− nitrite

The number of oxygens differ, but the charge is the same.

NO3− nitrate (c) HNO2 nitrous acid

The number of oxygens in the compounds differ but they both have only one hydrogen and one nitrogen atom.

HNO3 nitric acid 42. (a) Iron can form cations with different charges (e.g., Fe2+ or Fe3+). The Roman numeral indicating which cation of iron is involved is missing. This name cannot be fixed unless the particular cation of iron is specified. (b) K2Cr2O7. Potassium is generally involved in ionic compounds. The naming system used was for covalent compounds. The name should be potassium dichromate. (Dichromate is the name of the Cr2 O 72 − anion.) (c) Sulfur and oxygen are both nonmetals and form a covalent compound. The number of each atom involved needs to be specified for covalent compounds. There are two oxides of sulfur—SO2 and SO3. Both elements are nonmetals, so the names should be sulfur dioxide and sulfur trioxide, respectively. 43. No. Each compound, SO2 and SO3, has a definite composition of sulfur and oxygen by mass. The law of multiple proportions says that two elements may combine in different ratios to form more than one compound. 44. (a) and (c) don’t need any further information. Although silver and zinc are transition metals, they form only one ion each, Ag+ and Zn2+. Therefore their ions are unambiguous. (b) and (d), mercury and copper, have more than one choice, and therefore the charge on the cation must be specified with the Roman numeral in the name of the compound. 45. Acids: HCl, H2CO3, H2SO4 Nonmetallic binary compounds: CO, PBr3, P2O5, N2O3 46. Phosphate has a −3 charge; therefore, the formula for the ionic compound is M3(PO4)2. P has 15 protons; therefore, M3(PO4)2 has 30 phosphorus protons. 30 × 6 5 = 36 protons in 3 M

3 ( number of protons in M ) =

number of protons in M =

36 = 12 protons 3

from the periodic table, M is Mg.

– Chapter 6 –

47. (a) Potassium chloride

KCl

(b) Calcium carbonate

CaCO3

FeSO4

(d) Zinc oxide

ZnO

(e) Manganese sulfate

MnSO4 and Mn2(SO4)3 would both be described by this name. Better names would be manganese(II) sulfate for MnSO4 and manganese(III) sulfate for Mn2(SO4)3

(f) Copper sulfate

CuSO4 and Cu2SO4 would both be described by this name. Better names would be copper(II) sulfate and cupric sulfate for CuSO4 or copper(I) sulfate or cuprous sulfate for Cu2SO4

(g) Manganous oxide

MnO

(h) Potassium iodide

(i) Cobalt carbonate

CoCO3 and Co2(CO3)3 would both be described by this name. Better names would be cobalt(II) carbonate for CoCO3 and cobalt(III) carbonate for Co2(CO3)3

(j) Sodium chloride

NaCl

CHAPTER 7

QUANTITATIVE COMPOSITION OF COMPOUNDS SOLUTIONS TO REVIEW QUESTIONS 1.

A mole is an amount of substance containing the same number of atoms as there are atoms in exactly 12 g of carbon-12. It is Avogadro’s number (6.022 × 1023) of anything (atoms, molecules, ping-pong balls, etc.).

A mole of gold (197.0 g) has a higher mass than a mole of potassium (39.10 g).

Both samples (Au and K) contain the same number of atoms. (6.022 × 1023).

A mole of gold atoms contains more electrons than a mole of potassium atoms, as each Au atom has 79 e−, while each K atom has only 19 e−.

6.022 × 1023.

There are Avogadro’s number of particles in one mole of substance.

1 mole of ozone has the greater number of oxygen atoms.

The molar mass of an element is the mass of one mole (or 6.022 × 1023 atoms) of that element.

No. Avogadro’s number is a constant. The mole is defined as Avogadro’s number of C-12 atoms. Changing the atomic mass to 50 amu would change only the size of the atomic mass unit, not Avogadro’s number.

10. (a)

A mole of oxygen atoms (O) contains 6.022 × 1023 atoms.

(b) A mole of oxygen molecules (O2) contains 6.022 × 1023 molecules. (c) A mole of oxygen molecules (O2) contains 1.204 × 1024 atoms. (d) A mole of oxygen atoms (O) has a mass of 16.00 grams. (e) A mole of oxygen molecules (O2) has a mass of 32.00 grams. 11. 6.022 × 1023 molecules in one molar mass of H2SO4. 4.215 × 1024 atoms in one molar mass of H2SO4. 12. Either the chemical formula or experimental data giving the mass of the component elements in a sample. 13. There is 56.2% oxygen. (Remember the total percentage of all components must equal 100%.)

 total mass of the element  14.   (100 ) = mass percentage of the element  molar mass of the element  15. Choosing 100.0 g of a compound allows us to simply drop the % sign and use grams instead of percent. 16. C4H6 and C8H12 both have the empirical formula C2H3.

– Chapter 7 –

17. The molecular formula represents the total number of atoms of each element in a molecule. The empirical formula represents the lowest number ratio of atoms of each element in a molecule. 18. The molar mass is the most useful additional information that can be used to determine the molecular formula of a compound from its empirical formula. 19. This formula tells us the number of units of the empirical formula found in the molecule.

– Chapter 7 –

SOLUTIONS TO EXERCISES

Molar masses (a)

KBr

39.10 g

79.90 g 119.0

(b)

Na2SO4

45.98 g

32.07 g

64.00 g 142.1 g

(c)

Pb(NO3)2

207.2 g

28.02 g

96.00 g 331.2 g

(d)

C2H5OH

24.02 g

6.048 g

16.00 g 46.1 g

(e)

HC2H3O2

4.032 g

24.02 g

32.00 g 60.1 g

(f)

Fe3O4

167.6 g

64.00 g 231.6 g

(g)

C12H22O11

144.1 g

22.18 g

176.0 g 342.3 g

(h)

Al2(SO4)3

53.96 g

96.21 g

192.0 g 342.2 g

– Chapter 7 –

(i)

(NH4)2HPO4

9.072 g

28.02 g

30.97 g

64.00 g 132.1 g

Molar masses (a)

NaOH

22.99 g

16.00 g

1.008 g 40.00 g

(b)

Ag2CO3

215.8 g

12.01 g

48.00 g 275.8 g

(c)

Cr2O3

104.0 g

48.00 g 152.0 g

(d)

(NH4)2CO3

28.02 g

8.064 g

12.01 g

48.00 g 96.09 g

(e)

Mg(HCO3)2

24.31 g

2.016 g

24.02 g

96.00 g 146.3 g

(f)

C6H5COOH

84.07 g

6.048 g

32.00 g 122.1 g

(g)

C6H12O6

72.06 g

12.10 g

96.00 g 180.2 g

– Chapter 7 –

(h)

K4Fe(CN)6

156.4 g

55.85 g

72.06 g

84.06 g 368.4 g

(i)

BaCl2 • 2 H2O

137.3 g

70.90 g

4.032 g

32.00 g 244.2 g

Moles of atoms

 1 mol Zn   = 1.24 mol Zn  65.39 g Zn 

(a)

(81.4 g Zn ) 

(b)

( 0.688 g Mg ) 

(c)

1 mol Cu  (3.9×10 atoms Cu )  6.022 ×10  = 6.5×10 mol Cu atoms Cu 

(d)

(382 g Co ) 

(e)

( 0.072 g Sn ) 

(f)

2 atoms N  1 mol N atoms  (8.5×10 molecules N )  1 molecule   N  6.022 ×10 atoms N 

 1 mol Mg  −2  = 2.83×10 mol Mg 24.31 g Mg   −2

 1 mol Co   = 6.48 mol Co  58.93 g Co   1 mol Sn  −4  = 6.1×10 mol Sn 118.7 g Sn   







= 28 mol N atoms

Number of moles

 1 mol NaOH   = 0.625 mol NaOH  40.00 g NaOH 

(a)

( 25.0 g NaOH ) 

(b)

( 44.0 g Br2 ) 

(c)

( 0.684 g MgCl2 ) 

(d)

(14.8 g CH3OH ) 

 1 mol Br2   = 0.275 mol Br2  159.8 g Br2 

 1 mol MgCl2  −3  = 7.18 ×10 mol MgCl2  95.21 g MgCl2 

 1 mol CH3OH   = 0.462 mol CH3OH  32.04 g CH3OH 

– Chapter 7 –

( 2.88 g Na 2SO4 ) 

(f)

( 4.20 lb ZnI2 ) 

453.6 g   1 mol ZnI 2   = 5.97 mol ZnI2   1 lb   319.2 g ZnI 2 

Number of grams (a)

( 0.550 mol Au ) 

(b)

(37.2 mol H2 O ) 

(c)

( 47.5 mol Cl2 ) 

(d)

(3.15 mol NH4 NO3 ) 

197.0 g Au   = 108 g Au  1 mol Au   18.02 g H2 O   = 670 g H2 O  mol H2 O 

 70.90 g Cl2   = 3370 g Cl2  mol Cl2   80.05 g NH 4 NO3   = 252 g NH4 NO3  mol NH 4 NO3 

Number of grams (a)

(b)

 1 mol Na 2SO 4  −2  = 2.03×10 mol Na 2SO4  142.1 g Na 2SO 4 

(e)





g H SO ( 4.25 ×10 mol H SO )  98.09  = 0.0417 g H SO mol H SO −4









g CCI ( 4.5×10 molecules CCI )  6.022 ×101 molmolecules   153.8  = 11 g CCI mol CCI 22





47.87 g Ti   = 0.122 g Ti  mol Ti 

(c)

( 0.00255 mol Ti ) 

(d)

gS  (1.5×10 atom S )  6.02232.07  = 8.0 ×10 g S ×10 atoms S  −7

Number of molecules  6.022 ×10 23 molecules  24  = 1.5 ×10 molecules S8 mol  

(a)

( 2.5 mol S8 ) 

(b)

 6.022 ×10 23 molecules  25 (81.4 mol NH3 )   = 4.9 ×10 molecules NH3 mol  

(c)

(39.6 g C 2H5OH ) 

(d)

( 225 g Cl2 ) 

 6.022 ×10 23 molecules  23  = 5.18 ×10 molecules C 2 H5 OH 46.07 g C H OH 2 5  

 6.022 ×10 23 molecules  24  = 1.91×10 molecules Cl2 70.90 g Cl 2  

– Chapter 7 –

Number of molecules  6.022 ×10 23 molecules  24  = 5.8 ×10 molecules C 2H4 mol  

(a)

( 9.6 mol C 2H4 ) 

(b)

( 2.76 mol N 2 O ) 

(c)

( 23.2 g CH3OH ) 

(d)

 6.022 ×10 23 molecules  23 32.7 g CCI (  = 1.28 ×10 molecules CCI 4 4 ) 153.8 g CCI 4  

 6.022 ×10 23 molecules  24  = 1.66 ×10 molecules N 2 O mol    6.022 ×10 23 molecules  23  = 4.36 ×10 molecules CH3OH 32.04 g CH OH 3  

( 99 atoms Fe ) 

1 mol Fe   55.85 g Fe  −21   = 9.182 ×10 g Fe 23 6.022×10 atoms Fe 1 mol Fe   

10.

( 69 molecules CO ) 

1 mol CO  28.01 g CO  −21   = 3.209×10 g CO 23  6.022 ×10 atoms CO   1 mol CO 

11. Number of atoms  7 atoms 2  = 1.8 ×10 atoms 1 molecule P O 2 5   

(a)

( 25 molecules P2O5 ) 

(b)

(3.62 mol O2 ) 

(c)

(19.4 mol CS2 ) 

(d)

(1.25 g Na ) 

(e)

( 2.7 g CO2 ) 

(f)

( 0.72 g CH4 ) 

 6.022 ×10 23 molecules   2 atoms  24   = 4.36 ×10 atoms mol O2   1 molecule    6.022 ×10 23 molecules   3 atoms  25   = 3.50 ×10 atoms mol CS 1 molecule   2  

 6.022 ×10 23 atoms  22  = 3.27 ×10 atoms 22.99 g Na    6.022 ×10 23 molecules   3 atoms  23   = 1.1× 10 atoms 44.01 g CO 2   1 molecule    6.022 ×10 23 molecules   5 atoms  23   = 1.4 ×10 atoms 16.04 g CH 1 molecule   4  

12. Number of atoms (a)

( 2 molecules CH3COOH ) 

(b)

( 0.55 mol C2H6 ) 

8 atoms   = 16 atoms  1 molecule 

 6.022×1023 molecules CH3COOH   8 atoms    mol C 2H6    1 molecule 

= 2.6 ×1024 atoms

– Chapter 7 –

(c)

13.

  6.022 ×10 23 molecules  3 atoms 25  = 4.5×10 atoms  mol   1 molecule H2 O 

( 25 mol H2 O ) 

 6.022 ×10 23 atoms  23  = 2.30 ×10 atoms 197.0 g Au  

(d)

( 75.1 g Au ) 

(e)

 6.022 ×10 23 molecules   4 atoms  24 75 g PCl (  3 )  = 1.3 ×10 atoms 137.3 g PCl3    1 molecule 

(f)

(15 g C6H12 O6 ) 

 6.022 ×10 23 molecules   24 atoms  24   = 1.2 ×10 atoms  180.2 g C 6H12 O6   1 molecule  

1 g C16H12FN 3O3  1 mol C16H12FN 3O3    3  10 mg C16H12FN 3O3  313.3 g C16H12FN 3O3 

(80.0 mg C16H12 FN3 O3 ) 

 6.022 ×1023 molecules C16 H12 FN3 O3  20   = 1.54 ×10 molec C16 H12FN 3O3 1 mol C16H12FN 3O3  

 1 g C18H27 NO3  1 mol C18H27 NO3    6  10 μg C18H27 NO3  305.4 g C18H27 NO3 

( 20.0 μg C18H27 NO3 )  14.

 6.022 ×1023 molecules C18H27 NO3  16   = 3.94 ×10 molec C18H27 NO3 1 mol C18H27 NO3  

15. Number of grams (a)

(1 atom He ) 

4.003 g  −24  = 6.647 ×10 g He 23 6.022 ×10 atoms  

(b)

( 25 atoms C ) 

(c)

( 4 molecules N 2 O5 ) 

(d)

(144 molecules C6H5 NH2 ) 

12.01 g  −22  = 4.986 ×10 g C 23  6.022 ×10 atoms  108.0 g  −22  = 7.175×10 g N 2 O5 23 6.022 ×10 molecules   93.13 g  −20  = 2.227×10 g C6 H5 NH2 23  6.022 ×10 molecules 

16. Number of grams (a)

(1 atom Xe ) 

131.3 g  −22  = 2.180 ×10 g Xe 23  6.022 ×10 atoms 

(b)

(51 atoms Cl ) 

(c)

( 9 molecules CH3COOH ) 

35.45 g  −21  = 3.002 ×10 g Cl 23  6.022 ×10 atoms  60.05 g  −22  = 8.975 ×10 g CH3COOH 23 6.022 ×10 molecules  

– Chapter 7 –

116.1 g   15 molecules C H O ( NH ) )  (  (d)  6.022 ×10 molecules  4

2 2

= 2.892 ×10 −21 g C 4H4 O2 ( NH2 )2

 1000 g   1 mol CO 2  2  = 8.4 ×10 mol CO 2   kg  44.01 g CO2 

17. (a)

(37 kg CO2 ) 

(b)

(5 atoms Pb ) 

(c)

 6.022 ×10 23 molecules O2  2 atoms O  24 6 mol O (   = 7 ×10 atoms O 2 ) mol O2  1 molecule O2  

(d)

(37 molecules P4 ) 

1 mol Pb  −24  = 8 ×10 mol Pb 23  6.022 ×10 atoms Pb 



 123.9 g P4 −21  = 7.6 ×10 g P4 23  6.022 ×10 molecules P4 

1 mol W  −22  = 4.57 ×10 mol W 23  6.022 ×10 atoms W 

18. (a)

( 275 atoms W ) 

(b)

( 45 mol H2 O ) 

(c)

(12 molecules SO2 ) 

(d)

( 65 mol Cl2 ) 

 18.02 g H2 O   1 kg    = 0.81 kg H2 O  mol H2 O   1000 g  

 64.07 g SO2 −21  = 1.277 ×10 g SO2 23 6.022 ×10 molecules SO  2 

 6.022 ×1023 molecules Cl2  2 atoms Cl  25   = 7.8 ×10 atoms Cl mol Cl 1 molecule Cl   2 2 

19. One molecule of tetraphosphorus decoxide (P4O10) contains: (a)



 1 mol P4 O10 −24  = 1.661× 10 mol P4 O10 23 6.022 ×10 molecules P O 4 10  

(1 molecule P4 O10 ) 

(1.661×10 (b)

−24

 283.9 g P4 O10  −22 mol P4 O10 )   = 4.716 × 10 g P4 O10  mol P4 O10  

 4 P atoms  = 4 atoms P  1 molecule P4 O10 

(c)

(1 molecule P4 O10 ) 

(d)

(1 molecule P4 O10 ) 

(e)

(1 molecule P4 O10 ) 



10 atoms O   = 10 atoms O  1 molecule P4 O10 



 14 atoms  = 14 total atoms  1 molecule P4 O10 

– Chapter 7 –

20. 125 grams of disulfur decofluoride (S2F10) contains:  1 mol S2F10   = 0.492 mol S2F10  254.1 g S2F10 

(a)

(125 g S2F10 ) 

(b)

( 0.492 mol S2 F10 ) 

(c)

( 0.492 mol S2F10 ) 

(d)

( 0.492 mol S2F10 ) 

(e)

( 0.492 mol S2F10 ) 

 6.022 ×10 23 molecules  23  = 2.96 ×10 molecules S2 F10 mol  

  6.022 ×1023 molecules   12 atoms 24  = 3.56 ×10 total atoms  mol   1 molecule S2F10 

 6.022 ×10 23 molecules   2 S atoms  23  = 5.93 ×10 atoms S  F mol 1 molecule S   2 10 

 6.022×1023 molecules  10 F atoms  24  = 2.96 ×10 atoms F  mol molecule S F   2 10 

21. Atoms of carbon in: 

 7 atoms C 2  = 1.8 ×10 atoms H molecule C H CH 6 5 3  

(a)

( 25 molecules C6H5CH3 ) 

(b)

(3.5 mol H2CO3 ) 

(c)

(52 g CH3CH2OH ) 

  6.022 ×10 23 molecules   1 atom C 24  = 2.1×10 atoms H  mol molecule H CO   2 3 

  6.022 ×10 23 molecules   2 atoms C   46.07 g    molecule CH3CH 2 OH 

= 1.4 ×10 24 atoms H

22. Atoms of hydrogen in:  6 atoms H  = 138 atoms H  1 molecule CH3CH2 COOH  

(a)

( 23 molecules CH3CH2COOH ) 

(b)

( 7.4 mol H3PO4 ) 

(c)

(57 g C 6H5ONH2 ) 

  6.022 ×10 23 molecules   3 atoms H 25  = 1.3 ×10 atoms H  mol    1 molecule H3 PO 4 

  6.022 ×10 23 molecules   7 atoms H   109.1 g    1 molecule C 6H5 ONH2 

= 2.2 ×10 24 atoms H

23. (a)

molar mass = 15 (12.01 g/mol ) + 26 (1.008 g/mol ) +1(16.00 g/mol )

= 180.2 + 26.21 + 16.00 = 222.4 g/mol (b)

 222.4 g C15H26 O  3  = 1.32 ×10 g C15H26 O 1 mol C H O 15 26  

(5.95 mol C15H26 O ) 

– Chapter 7 –

(c)





C H O ( 7.61×10 mol C H O )  6.022 ×101 molmolecules  = 4.58 ×10 molecules C H O C H O −4





15 mol C   = 107 mol C  1 mol C15H26 O 



(d)

( 7.11 mol C15H26 O ) 

(e)

( 4.29 g C15H26 O ) 

(f)

mol H  1 mol C H O  222.4 g C H O (8.64 ×10 atoms H )  6.022 1×10    atom H  26 mol H  1 mol C H O

 1 mol C15H26 O     16.00 g O  1 mol O    = 0.309 g O  222.4 g C15H26 O   1 mol C15H26 O   1 mol O 

= 1.23 ×10

−3









g C15H26 O



(g)

  222.4 g C15H26 O  1 mol C15H26 O   23  6.022 ×10 molecules C15H26 O   1 mol C15H26 O 

(1 molecule C15H26 O ) 

= 3.69 ×10 −22 g C15H26 O  1 mol C15H26 O  15 mol C  6.022 ×1023 atom C  8.00 g C H O ( )    15 26 (h) 1 mol C   222.4 g C15H26 O  1 mol C15H26 O  = 3.25×1023 atom C or  1 mol C15H26 O   6.022 ×10 23 molecules C15H26 O    1 mol C15H26 O   222.4 g C15H26 O 

(8.00 g C15H26 O ) 

  15 atom C 23   = 3.25 ×10 atom C 1 molecule C H O 15 26   24. (a)

molar mass = 11(12.01 g/mol ) + 20 (1.008 g/mol ) + 2 (16.00 g/mol )

= 132.1+ 20.16 + 32.00 = 184.3 g/mol  184.3 g C11H20 O2  3  = 1.50 ×10 g C11H20 O2  1 mol C11H20 O2 

(b)

(8.13 mol C11H20 O2 ) 

(c)

(5.77 ×10

(d)

(35.7 mol C11H20 O2 ) 

(e)

(85.1 g C11H20 O2 ) 

 6.022 ×10 23 molecules C11H20 O 2  mol C11H20 O 2 )   1 mol C11H20 O 2   = 3.47 ×1012 molecules C11H20 O2 −12

20 mol H   = 714 mol H  1 mol C11H20 O2 



 1 mol C11H20 O2  11 mol C   12.01 g C     = 61.0 g C  184.3 g C11H20 O2  1 mol C11H20 O2   1 mol C 

– Chapter 7 –

(f)







mol O  1 mol C H O  184.3 g C H O ( 9.47 ×10 atoms O )  6.022 1×10    atom O  2 mol O  1 mol C H O 11

= 1.45 ×10

−2



g C11H20 O2

  184.3 g C11H20 O2  1 mol C11H20 O2   23  6.022 ×10 molecules C11H20 O2  1 mol C11H20 O2 



(g)

(1 molecule C11H20 O2 ) 

= 3.06 ×10 −22 g C11H20 O2

 1 mol C11H20 O2  11 mol C  6.022 ×1023 atom C  4.50 g C H O (    11 20 2 )  (h) 1 mol C   184.3 g C11H20 O2  1 mol C11H20 O2  = 1.62 ×10 23 atom C or   1 mol C11H20 O2  6.022 ×1023 molecules C11H20 O2  11 atom C    184.3 g C H O 1 mol C H O 1 molecule C H O 11 20 2  11 20 2 11 20 2   

( 4.50 g C11H20O2 ) 

= 1.62 ×1023 atom C

25. The number of grams of: (a) Silver in 25.0 g AgBr

 107.9 g Ag   = 14.4 g Ag  187.8 g AgBr 

( 25.0 g AgBr ) 

(b) Nitrogen in 6.34 mol (NH4)3PO4





gN  = 266 g N ( 6.34 mol ( NH ) PO )  mol42.03 ( NH ) PO  4 3



4 3



→ mol SO3 ⎯⎯ → gO The conversion is : molecules SO3 ⎯⎯ 



gO (8.45 ×10 molecules SO )  6.022 ×101 molmolecules   48.00  = 6.74 g O mol SO 22

26. The number of grams of: (a) Chlorine in 5.00 g PbCl2  70.90 g Cl   = 1.27 g Cl  278.1 g PbCl2 

(5.00 g PbCl2 ) 

(b) Hydrogen in 4.50 mol H2SO4  2.016 g H   = 9.07 g H  1 mol H2SO 4 

( 4.50 mol H2SO4 ) 





– Chapter 7 –



gH (5.45 ×10 molecules NH )  6.022 ×101 molmolecules   3.024  = 0.274 g H mol NH 22





27. Percent composition (a) NaBr

22.99 g

79.90 g 102.9 g

(b) KHCO3

39.10 g

1.008 g

48.00 g

12.01 g

 12.01 g    (100 ) = 12.00% C  100.1 g   48.00 g    (100 ) = 47.95% O  100.1 g 

55.85 g

3 Cl

106.54 g

 55.85 g    (100 ) = 34.41% Fe  162.3 g   106.4 g    (100 ) = 65.56% Cl  162.3 g 

28.09 g

4 Cl

141.8 g 169.9 g

(e) Al2(SO4)3

 39.10 g    (100 ) = 39.06% K  100.1 g   1.008 g    (100 ) = 1.007% H  100.1 g 

100.1 g

162.3 g (d) SiCl4

 22.99 g    (100 ) = 22.34% Na  102.9 g   79.90 g    (100 ) = 77.65% Br  102.9 g 

2Al

53.96 g

96.21 g

12 O

192.0 g 342.2 g

 28.09 g    (100 ) = 16.53% Si  169.9 g   141.8 g    (100 ) = 83.46% Cl  169.9 g   53.96 g    (100 ) = 15.77% Al  342.2 g   96.21 g    (100 ) = 28.12% S  342.2 g   192.0 g    (100 ) = 56.11% O  342.2 g 

– Chapter 7 –

(f)

AgNO3

107.9 g

14.01 g

48.00 g 169.9 g

28.

 107.9 g    (100 ) = 63.51% Ag  169.9 g   14.01 g    (100 ) = 8.246% N  169.9 g   48.00 g    (100 ) = 28.25% O  169.9 g 

Percent composition (a) ZnCl2

65.39 g

2 Cl

70.90 g 136.3 g

(b) NH4C2H3O2

14.01 g

7.056 g

24.02 g

32.00 g

 14.01 g    (100 ) = 18.17% N  77.09 g 

 7.056 g    (100 ) = 9.153% H  77.09 g 

77.09 g

 24.02 g    (100 ) = 31.16% C  77.09 g   32.00 g    (100 ) = 41.51% O  77.09 g 

24.31 g

61.94 g

 24.31 g    (100 ) = 12.26% Mg  198.3 g 

112.0 g 198.3 g

(d) (NH4)2SO4

 65.39 g    (100 ) = 47.98% Zn  136.3 g   70.90 g    (100 ) = 52.02% Cl  136.3 g 

28.02 g

8.064 g

32.07 g

64.00 g 132.2 g

 61.94 g    (100 ) = 31.24% P  198.3 g   112.0 g    (100 ) = 56.48% O  198.3 g   28.02 g    (100 ) = 21.20% N  132.2 g   8.064 g    (100 ) = 6.100% H  132.2 g 

 32.07 g    (100 ) = 24.26% S  132.2 g   64.00 g    (100 ) = 48.41% O  132.2 g 

– Chapter 7 –

Fe 3N 9O

55.85 g 42.03 g 144.0 g 241.9 g

 55.85 g    (100 ) = 23.09% Fe  241.9 g 

I 3 Cl

126.9 g 106.4 g 233.3 g

 126.9 g    (100 ) = 54.39% I  233.3 g   106.4 g    (100 ) = 45.61% Cl  233.3 g 

Fe O

55.85 g 16.00 g

 55.85 g    (100 ) = 77.73% Fe  71.85 g 

(e) Fe(NO3)3

(f)

29.

ICl3

Percent of iron (a) FeO

(b) Fe2O3

2Fe 3O

3 Fe 4O

(d) K4Fe(CN)6

Fe 4K 6C 6N

30. Percent of chlorine (a) KCl

(b) BaCl2

71.85 g 111.7 g 48.00 g 159.7 g 167.6 g 64.00 g 231.6 g 55.85 g 156.4 g 72.06 g 84.06 g 368.4 g

 42.03 g    (100 ) = 17.37% N  241.9 g   144.0 g    (100 ) = 59.53% O  241.9 g 

 111.7 g    (100 ) = 69.94% Fe  159.7 g   167.6 g    (100 ) = 72.37% Fe  231.6 g   55.85 g    (100 ) = 15.16% Fe  368.4 g 

K Cl

39.10 g 35.45 g

 35.45 g    (100 ) = 47.55% Cl  74.55 g 

Ba 2 Cl

74.55 g 137.3 g 70.90 g

 70.90 g    (100 ) = 34.05% Cl  208.2 g 

Si 4 Cl

208.2 g 28.09 g 141.8 g

 141.8g    (100 ) = 83.46% Cl  169.9 g 

169.9 g 6.941 g Li (d) LiCl 35.45 g Cl 42.39 g Highest % Cl is LiCl; lowest % Cl is in BaCl2.

 35.45 g    (100 ) = 83.63% Cl  42.39 g 

– Chapter 7 –

31. Percent composition 73.16 g barium silicide −33.62 g Ba

 39.54 g Si    (100 ) = 54.05% Si  73.16 g   33.62 g Ba    (100 ) = 45.95% Ba  73.16 g 

39.54 g Si

32. Percent composition 7.52 g ajoene −3.09 g S −0.453 g H −0.513 g O 3.46 g C

 3.09 g S    (100 ) = 41.1% S  7.52 g   0.453 g H    (100 ) = 6.02% H  7.52 g   0.513 g O    (100 ) = 6.82% O  7.52 g   3.46 g C    (100 ) = 46.0% C  7.52 g 

33. (a) H2O has the higher percent Hydrogen (b) N2O3 has the lower percent Nitrogen (c) Both have the same percent Oxygen 34. (a) KClO3 is lower.

(Because a K atom has more mass than a Na atom.)

(b) KHSO4 is higher.

(Because a H atom has less mass than a K atom.)

(Because only one Cr atom is present.)

35. Empirical formulas from percent composition. (a) Step 1. Express each element as grams/100 g material. 63.6% N = 63.6 g N/100 g material 36.4% O = 36.4 g O/100 g material Step 2. Calculate the relative moles of each element.

 1 mol N   = 4.54 mol N  14.01 g N 

( 63.6 g N ) 

 1 mol O   = 2.28 mol O  16.00 g O 

(36.4 g O ) 

Step 3. Change these moles to whole numbers by dividing each by the smaller number. 4.54 mol N = 1.99 mol N 2.28 2.28 mol O = 1.00 mol O 2.28 The simplest ratio of N:O is 2:1. The empirical formula, therefore, is N2O.

– Chapter 7 –

(b) 46.7% N, 53.3% O

 1 mol N   = 3.33 mol N  14.01 g N 

3.33 mol N = 1.00 mol N 3.33

 1 mol O   = 3.33 mol O  16.00 g O 

3.33 mol O = 1.00 mol O 3.33

( 46.7 g N )  (53.3 g N ) 

The empirical formula is NO. (c) 25.9 % N, 74.1% O  1 mol N  1.85 mol N = 1.00 mol N  = 1.85 mol N 1.85  14.01 g N 

( 25.9 g N ) 

 1 mol O   = 4.63 mol O  16.00 g O 

( 74.1 g O ) 

4.63 mol O = 2.50 mol O 1.85

Since these values are not whole numbers, multiply each by 2 to change them to whole numbers. (1.00 mol N)(2) = 2.00 mol N; (2.5 mol O)(2) = 5.00 mol O The empirical formula is N2O5. (d) 43.4% Na, 11.3% C, 45.3% O  1 mol Na  1.89 mol Na = 2.01 mol Na  = 1.89 mol Na 0.941  22.99 g Na 

( 43.4 g Na ) 

 1 mol C   = 0.941 mol C  12.01 g C 

0.941 mol C = 1.00 mol C 0.941

 1 mol O   = 2.83 mol O  16.00 g O 

2.83 mol O = 3.00 mol O 0.941

(11.3 g C ) 

( 45.3 g O ) 

The empirical formula is Na2CO3. (e) 18.8% Na, 29.0% Cl, 52.3% O  1 mol Na   = 0.818 mol Na  22.99 g Na 

0.818 mol Na = 1.00 mol Na 0.818

 1 mol Cl   = 0.818 mol Cl  35.45 g Cl 

0.818 mol Cl = 1.00 mol Cl 0.818

 1 mol O   = 3.27 mol O  16.00 g O 

3.27 mol O = 4.00 mol O 0.818

(18.8 g Na )  ( 29.0 g Cl )  (52.3 g O ) 

The empirical formula is NaClO4. (f)

72.02% Mn, 27.98% O  1 mol Mn  1.311 mol Mn = 1.000 mol Mn  = 1.311 mol Mn 1.311  54.94 g Mn 

( 72.02 g Mn ) 

 1 mol O   = 1.749 mol O  16.00 g O 

( 27.98 g O ) 

1.749 mol O = 1.334 mol O 1.311

– Chapter 7 –

Multiply both values by 3 to give whole numbers. (1.000 mol Mn) (3) = 3.000 mol Mn; (1.334 mol O) (3) = 4.002 mol O The empirical formula is Mn3O4. 36. Empirical formulas from percent composition (a) 64.1% Cu, 35.9% Cl  1 mol Cu  1.01 mol Cu = 1.00 mol Cu  = 1.01 mol Cu 1.01  63.55 g Cu 

( 64.1 g Cu ) 

 1 mol Cl   = 1.01 mol Cl  35.45 g Cl 

1.01 mol Cl = 1.00 mol Cl 1.01

(35.9 g Cl ) 

The empirical formula is CuCl. (b) 47.2% Cu, 52.8% Cl  1 mol Cu   = 0.743 mol Cu  63.55 g Cu 

0.743 mol Cu = 1.00 mol Cu 0.743

 1 mol Cl   = 1.49 mol Cl  35.45 g Cl 

1.49 mol Cl = 2.01 mol Cl 0.743

( 47.2 g Cu )  (52.8 g Cl ) 

The empirical formula is CuCl2. (c) 51.9% Cr, 48.1% S  1 mol Cr   = 0.998 mol Cr  52.00 g Cr 

0.998 mol Cr = 1.00 mol Cr 0.998

 1 mol S   = 1.50 mol S  32.07 g S 

1.50 mol S = 1.50 mol S 0.998

(51.9 g Cr )  ( 48.1 g S ) 

Multiply both values by 2 to give whole numbers. (1.00 mol Cr) (2) = 2.00 mol Cr; (1.50 mol S) (2) = 3.00 mol S The empirical formula is Cr2S3. (d) 55.3% K, 14.6% P, 30.1% O  1 mol K  1.41 mol K = 2.99 mol K  = 1.41 mol K 0.471  39.10 g K 

(55.3 g K ) 

 1 mol P   = 0.471 mol P  30.97 g P 

0.471 mol P = 1.00 mol P 0.471

 1 mol O   = 1.88 mol O  16.00 g O 

1.88 mol O = 3.99 mol O 0.471

(14.6 g P ) 

(30.1 g O ) 

The empirical formula is K3PO4.

– Chapter 7 –

(e) 38.9% Ba, 29.4% Cr, 31.7% O  1 mol Ba   = 0.283 mol Ba  137.3 g Ba 

0.283 mol Ba = 1.00 mol Ba 0.283

 1 mol Cr   = 0.565 mol Cr  52.00 g Cr 

0.565 mol Cr = 2.00 mol Cr 0.283

 1 mol O   = 1.98 mol O  16.00 g O 

1.98 mol O = 7.00 mol O 0.283

(38.9 g Ba )  ( 29.4 g Cr )  (31.7 g O ) 

The empirical formula is BaCr2O7. (f)

3.99% P, 82.3% Br, 13.7% Cl  1 mol P   = 0.129 mol P  30.97 g P 

0.129 mol P = 1.00 mol P 0.129

 1 mol Br   = 1.03 mol Br  79.90 g Br 

1.03 mol Br = 7.98 mol Br 0.129

 1 mol Cl   = 0.386 mol Cl  35.45 g Cl 

0.386 mol Cl = 2.99 mol Cl 0.129

(3.99 g P ) 

(82.3 g Br )  (13.7 g Cl ) 

The empirical formula is PBr8Cl3. 37. Empirical formula  1 mol Zn   = 0.3988 mol Zn  65.39 g 

( 26.08 g Zn )  (a)

0.3988 mol Zn = 1.00 mol Zn 0.3988

 1 mol C   = 0.399 mol C  12.01 g 

0.399 mol C = 1.00 mol C 0.3988

 1 mol O   = 1.196 mol O  16.00 g 

1.196 mol O = 2.999 mol O 0.3988

( 4.79 g C ) 

(19.14 g O ) 

The empirical formula is ZnCO3. (b) 150.0 g compound −57.66 g C −7.26 g H 85.1

g Cl  1 mol C   = 4.801 mol C  12.01 g C 

4.801 mol C = 2.000 mol C 2.40

 1 mol H   = 7.20 mol H  1.008 g H 

7.20 mol H = 3.00 mol H 2.40

 1 mol Cl   = 2.40 mol Cl  35.45 g 

2.40 mol Cl = 1.00 mol Cl 2.40

(57.66 g C )  ( 7.26 g H ) 

(85.1 g Cl ) 

The empirical formula is C2H3Cl.

– Chapter 7 –

0.824 mol V = 1.00 mol V 0.824

 1 mol O   = 2.06 mol O  16.00 g O 

2.06 mol O = 2.50 mol O 0.824

( 42.0 g V )  (33.0 g O ) 

Multiplying both values by 2 gives the empirical formula V2O5. (d)

 1 mol Ni  1.148 mol Ni = 1.501 mol Ni  = 1.148 mol Ni 0.7649  58.69 g Ni 

( 67.35 g Ni ) 

 1 mol O  3.060 mol O = 4.001 mol O  = 3.060 mol O 0.7649  16.00 g O 

( 48.96 g O ) 

 1 mol P   = 0.7649 mol P  30.97 g P 

( 23.69 g P ) 

0.7649 mol P = 1.00 mol P 0.7649

Multiplying all values by 2 gives the empirical formula Ni3O8P2. 38. Empirical formula (a)

 1 mol C   = 4.586 mol C  12.01 g C 

4.586 mol C = 6.000 mol C 0.7643 mol

 1 mol H   = 3.82 mol H  1.008 g H 

3.82 mol H = 5.00 mol H 0.7643

(55.08 g C )  (3.85 g H ) 

 1 mol Br   = 0.7643 mol Br  79.90 g Br 

( 61.07 g Br ) 

0.7643 mol Br = 1.000 mol Br 0.7643

The empirical formula is C6H5Br. (b) 65.2 g compound − 36.8 g Ag − 12.1 g Cl = 16.3 g O  1 mol Ag   = 0.341 mol Ag  107.9 g Ag 

0.341 mol Ag = 1.00 mol Ag 0.341

 1 mol Cl   = 0.341 mol Cl  35.45 g Cl 

0.341 mol Cl = 1.00 mol Cl 0.341

 1 mol O   = 1.02 mol O  16.00 g O 

1.02 mol O = 2.99 mol O 0.341

(36.8 g Ag )  (12.1 g Cl )  (16.3 g O ) 

The empirical formula is AgClO3. (c) 25.25 g sulfide − 12.99 gV = 12.26 g S  1 mol V   = 0.2550 mol V  50.94 g V 

0.2550 mol V = 1.000 mol V 0.2550

 1 mol S   = 0.3823 mol S  32.07 g S 

0.3823 mol S = 1.499 mol S 0.2550

(12.99 g V )  (12.26 g S ) 

Multiplying both values by 2 gives the empirical formula V2S3.

– Chapter 7 –

(d)

 1 mol Zn   = 0.581 mol Zn  65.39 g 

0.581 mol Zn = 1.50 mol Zn 0.387

 1 mol P   = 0.387 mol P  30.97 g 

0.387 mol P = 1.00 mol P 0.387

(38.0 g Zn )  (12.0 g P ) 

Multiplying both values by 2 gives the empirical formula Zn3P2. 39. 15.267 g sulfide − 12.272 g Au = 2.995 g S  1 mol Au   = 0.06229 mol Au  197.0 g Au 

(12.272 g Au ) 

 1 mol S   = 0.09339 mol S  32.07 g S 

0.06229 mol Au = 1.000 mol Au 0.06229 0.09339 mol S = 1.499 mol S 0.06229

( 2.995 g S ) 

Multiplying both values by 2 gives the empirical formula Au2S3. 40. 10.724 g oxide − 7.143 g Ti = 3.581 g O  1 mol Ti   = 0.1492 mol Ti  47.88 g Ti 

0.1492 mol Ti = 1.000 mol Ti 0.1492

 1 mol O   = 0.2238 mol O  16.00 g O 

0.2238 mol O = 1.500 mol O 0.1492

( 7.143 g Ti )  ( 3.581 g O ) 

Multiplying both values by 2 gives the empirical formula Ti2O3. 41. 5.000 g compound – (0.6375 g C + 0.1070 g H) = 4.256 g S  1 mol C   0.05308 mol C   = 0.05308 mol C   = 1.000 mol C  0.05308 mol   12.01 g C 

( 0.6375 g C ) 

 1 mol H   = 0.1062 mol H  1.008 g H 

( 0.1070 g H ) 

 1 mol S   = 0.1327 mol S  32.07 g S 

( 4.256 g S ) 

 0.1062 mol H    = 2.001 mol H  0.05308 mol   0.1327 mol S    = 2.500 mol S  0.05308 mol 

Multiplying all values by 2 gives the empirical formula C2H4S5. This is also the molecular formula of lenthionine. 42. Empirical formula 5.276 g compound − 3.898 g Hg = 1.378 g Cl  1 mol Hg   = 0.01943 mol Hg  200.6 g Hg 

0.01943 mol Hg = 1.000 mol Hg 0.01943

 1 mol Cl   = 0.03887 mol Cl  35.45 g Cl 

0.03887 mol Cl = 2.001 mol Cl 0.01943

(3.898 g Hg )  (1.378 g Cl ) 

The empirical formula is HgCl2.

– Chapter 7 –

43. Empirical and molecular formula of traumatic acid 63.13% C, 8.830% H, 28.03% O; molar mass = 228 g  1 mol C   = 5.256 mol C  12.01 g C 

 5.256 mol C    = 3.000 mol C 1.752  

( 63.13 g C ) 

 1 mol H   8.760 mol H   = 8.760 mol H   = 5.000 mol H 1.752    1.008 g H 

(8.830 g H ) 

 1 mol O   = 1.752 mol O  16.00 g O 

( 28.03 g O ) 

 1.752 mol O    = 1.000 mol O  1.752 mol 

The empirical formula for traumatic acid is C3H5O. The empirical formula mass is 57 g.

molar mass 228 = =4 empirical formula mass 57 The molecular formula is four times that of the empirical formula. Molecular formula is (C3H5O)4 = C12H20O4. 44. Empirical and molecular formulas of dixanthogen 29.73% C, 4.16% H, 13.20% O, 52.91% S; molar mass = 242.4 g

 1 mol C   = 2.475 mol C  12.01 g C 

2.475 mol C = 3.000 mol C 0.8250

 1 mol H   = 4.13 mol H  1.008 g H 

4.13 mol H = 5.01 mol H 0.8250

 1 mol O   = 0.8250 mol O  16.00 g O 

0.8250 mol O = 1.000 mol O 0.8250

 1 mol S   = 1.650 mol S  32.07 g S 

1.650 mol S = 2.000 mol S 0.8250

( 29.73 g C )  ( 4.16 g H ) 

(13.20 g O )  (52.91 g S ) 

The empirical formula is C3H5OS2. The empirical formula mass is 121.2 g.

molar mass 242.4 g = =2 empirical formula mass 121.2 g The molecular formula is twice that of the empirical formula. Molecular formula is (C3H5OS2)2 = C6H10O2S4. 45. 86.65 g C × 1 mol C = 7.215 mol C 12.01 g C 1 mol H 4.48 g H × = 4.44 mol H 1.008 g H 1 mol O 8.88 g O × = 0.555 mol O 16.00 g O

7.215 mol C = 13 mol C 0.555 4.44 mol H = 8 mol H 0.555 0.555 mol O = 1 mol O 0.555

The empirical formula is C13H8O. 22

– Chapter 7 –

1 mol C 6.746 mol C = 6.746 mol C = 15 mol C 12.01 g C 0.450 1 mol H 11.70 mol H 11.79 g H × = 11.70 mol H = 26 mol H 1.008 g H 0.450 1 mol O 0.450 mol O 7.20 g O × = 0.450 mol O = 1 mol O 16.00 g O 0.450

46. 81.02 g C ×

The empirical formula is C15H26O. 47. Molecular formula of oxalic acid (ethanedioic acid) 26.7% C, 2.24% H, 71.1% O; molar mass = 90.04  1 mol C  26.7 g C   = 2.22 mol C  12.01 g C   1 mol H  2.2 g H   = 2.2 mol H  1.008 g H 

2.22 mol C = 1.0 mol C 2.2

 1 mol O  71.1 g O   = 4.44 mol O  16.00 g O 

4.44 mol O = 2.0 mol O 2.2

2.2 mol H = 1.0 mol H 2.2

The empirical formula is CHO2, making the empirical formula mass 45.02 g.

molar mass 90.04 g = =2 empirical formula mass 45.02 g The molecular formula is twice that of the empirical formula. Molecular formula = (CHO2)2 = C2H2O4 48. Molecular formula of butyric acid 54.5% C, 9.2% H, 36.3% O; molar mass = 88.11  1 mol C   = 4.54 mol C  12.01 g C 

4.54 mol C = 2.00 mol C 2.27

 1 mol H   = 9.1 mol H  1.008 g H 

9.1 mol H = 4.0 mol H 2.27

 1 mol O   = 2.27 mol O  16.00 g O 

2.27 mol O = 1.0 mol O 2.27

(54.5 g C )  ( 9.2 g H ) 

(36.3 g O ) 

The empirical formula is C2H4O, making the empirical formula mass 44.05 g.

molar mass 88.11 g = =2 empirical formula mass 44.05 g The molecular formula is twice that of the empirical formula. Molecular formula = (C2H4O)2 = C4H8O2

– Chapter 7 –

12.04 g (100 ) = 30.45% 39.54 g 39.54 g −12.04 g % oxygen = (100 ) = 69.55% 39.54 g

49. % nitrogen =

12.04 g N = 0.8594 mol N 14.01 g/mol 27.50 g O moles of oxygen = = 1.719 mol O 16.00 g/mol O

empirical formula: moles of nitrogen =

0.8594 mol = 1.000 0.8594 mol 1.719 mol = 2.000 relative number of oxygen atoms = 0.8594 mol

relative number of nitrogen atoms =

empirical formula = NO2 molecular formula: (molar mass of NO2) x = 92.02 g, 46.01 x = 92.02, x = 2 The molecular formula is twice the empirical formula. molecular formula = N2O4 50. Total mass of C + H + O = 30.21 g + 5.08 g + 40.24 g + 75.53 g 30.21 g % carbon = (100 ) = 40.0% 75.53 g 5.08 g % hydrogen = (100 ) = 6.73% 75.53 g 40.24 g % oxygen = (100 ) = 53.3% 75.53 g 30.21 g C empirical formula: moles of carbon = = 2.515 mol C 12.01 g/mol 5.080 g H moles of hydrogen = = 5.03 mol H 1.008 g/mol 40.24 g O moles of oxygen = = 2.515 mol O 16.00 g/mol 2.515 mol relative number of carbon atoms = = 1.000 2.515 mol 5.03 mol relative number of hydrogen atoms = = 2.00 2.515 mol 2.515 mol relative number of oxygen atoms = = 1.000 2.515 mol empirical formula = CH2O molecular formula: (molar mass of CH2O) x = 180.18 g/mol, (30.03 g/mol)x = 180.18 g/mol,

– Chapter 7 –

180.18 g/mol =6 30.03 g/mol

The molecular formula is six times the empirical formula. molecular formula = C6H12O6 51. What is compound XYZ3?

( 0.4004 )(100.09 g ) = 40.08 g ( calcium ) Y: ( 0.1200 )(100.09 g ) = 12.01 g ( carbon ) X:

( 0.4796 )(100.09 g ) = 48.00 g;

48.00 g = 16.00 g ( oxygen ) 3

Elements determined from atomic masses in the periodic table. XYZ3 = CaCO3 52. What is compound X2(YZ3)3?

53.96 g = 26.98 g ( aluminum ) 2 84.27 g Y: ( 0.2986 )( 282.23 g ) = = 28.09 g ( silicon ) 3 143.99 g Z: ( 0.5102 )( 282.23 g ) = = 16.00 g ( oxygen ) 9 X:

( 0.1912 )( 282.23 g ) =

Elements determined from atomic masses in the periodic table. X2(YZ3)3=Al2(SiO3)3  6.022 ×10 23 molecules  4 atoms P  23  = 8.43 ×10 atoms P  mol   molecule P4 

53.

( 0.350 mol P4 ) 

54.

(10.0 g K ) 

 1 mol K   1 mol Na  22.99 g Na     = 5.88 g Na  39.10 g K   1 mol K  mol Na 

 8.64 ×10 −22 g   6.022 ×10 23 molecules  520 g 55.   = 1 mol  1 molecule    mol

453.6 g   6.022 ×10 23 molecules  24  = 4 × 10 molecules C12 H22 O11  1 lb 342.3 g   

56.

(5 lb C12H22 O11 ) 

57.

 1 acre ( 6.022 × 10 apple trees )  472 apple  =1.27×10 acres of apple trees trees 23





 6.022 ×10 23 dollars  13 58.   = 8.6 ×10 dollars/person 9  7.0 ×10 people  → ft3 ⎯⎯ → in.3 ⎯⎯ → cm3 ⎯⎯ → drops 59. The conversion is: mi3 ⎯⎯ 3

(a)

ft   12.0 in.   2.54 cm   20 drops  = 8 ×1016 drops (1 mi3 )  5280       3  mile   ft   inch   1.0 cm 

– Chapter 7 –

(b)





( 6.022 ×10 drops )  8×101 midrops  = 8 ×10 mi 23





60. 1 mol Ag = 107.9 g Ag  1 cm 3  3  = 10.3 cm ( volume of cube )  10.5 g 

(a)

(107.9 Ag ) 

(b)

10.3 cm 3 = volume of cube = ( side )

side = 3 10.3 cm 3 = 2.18 cm 61. (a) Determine the molar mass of each compound. CO2, 44.01 g; O2, 32.00 g; H2O, 18.02 g; CH3OH, 32.04 g. The 1.00 gram sample with the lowest molar mass will contain the most molecules. Thus, H2O will contain the most molecules. 23  1 mol   ( 3 ) ( 6.022 ×10 atoms )   = 1.00 ×1023 atoms (b) (1.00 g H2 O )     18.02 g mol    23  1 mol   ( 6 ) ( 6.022 ×10 atoms )   = 1.13×10 23 atoms (1.00 g CH3OH )     32.04 g mol    23  1 mol   ( 3 ) ( 6.022 × 10 atoms )   = 4.10 ×1022 atoms (1.00 g CO2 )     mol  44.01 g    23  1 mol   ( 2 ) ( 6.022 ×10 atoms )    = 3.76 ×10 22 atoms (1.00 g O2 )    mol 32.00 g   

The 1.00 g sample of CH3OH contains the most atoms. 62. 1 mol Fe2S3 = 207.9 g Fe2S3 = 6.022 × 1023 formula units 207.9 g Fe S unit    ( 6.022 ×10 atoms )  1 formula   = 41.58 g Fe S 5 atoms 6.022 ×10 formula units   23

2 3

→ mol P ⎯⎯ → mol Ca ⎯⎯ → g Ca 63. The conversion is g P ⎯⎯

 1 mol P   3 mol Ca  40.08 g Ca     = 1.94 g Ca  30.97 g P   2 mol P  1 mol Ca 

(1.00 g P ) 

1.94 g Ca combines with 1.00 g P. 64. Grams of Fe per ton of ore that contains 5% FeSO4. The conversion is: ton ⎯⎯ → lb ⎯⎯ → g ⎯⎯ → gFeSO4 ⎯⎯ → g Fe  55.85 g Fe  2000 lb   453.6 g  4  = 2 × 10 g Fe   ( 0.05 FeSO4 )  151.9 g FeSO  ton   lb   4 

(1.0 ton ) 

1.0 ton of iron ore contains 2 × 104 g Fe.

2 3

– Chapter 7 –

65. From the formula, 2 Li (13.88 g) combine with 1 S (32.07 g).

 13.88 g Li    ( 20.0 g S ) = 8.66 g Li  32.07 g S  66. (a) HgCO3

200.6 g

12.01 g

48.00 g

 200.6 g Hg    (100 ) = 76.98% Hg  260.6 g 

260.6 g (b) Ca(ClO3)2

96.00 g

2 Cl

70.90 g

40.08 g

 96.00 g O    (100 ) = 46.38% O  207.0 g 

207.0 g (c) C10H14N2

28.02 g

10 C

120.1 g

14 H

14.11 g

 28.02 g N    (100 ) = 17.27% N  162.6 g 

162.2 g (d) C55H72MgN4O5 Mg

24.31 g

55 C

660.55 g

72 H

72.58 g

56.04 g

80.00 g

 24.31 g Mg    (100 ) = 2.721% Mg  893.5 g 

893.5 g 67. According to the formula, 1 mol (65.39 g) Zn combines with 1 mol (32.07 g) S.

 32.07 g S   = 9.56 g S  65.39 g Zn 

(19.5 g Zn ) 

19.5 g Zn require 9.56 g S for complete reaction. Therefore, there is not sufficient S present (9.40 g) to react with the Zn. 68. Percent composition of C15H20O6 15 C

180.2 g

20 H

20.16 g

96.00 g

180.2 g C    (100 ) = 60.80% C  296.4 g   20.16 g H    (100 ) = 6.802 % H  296.4 g   96.00 g O    (100 ) = 32.39 % O  296.4 g 

296.4 g

– Chapter 7 –

69.

Percent composition of C17H21NO•HCl 17 C

204.2 g

22 H

22.18 g

14.01 g

16.00 g

35.45 g

 204.2 g C    (100 ) = 69.98% C  291.8 g   22.18 g H    (100 ) = 7.60% H  291.8 g   14.01 g N    (100 ) = 4.80% N  291.8 g   16.00 g O    (100 ) = 5.48% O  291.8 g 

291.8 g

 35.45 g Cl    (100 ) = 12.15% Cl  291.8 g  70. Percent composition of sucrose 144.1 g 12 C 22 H

22.18 g

11 O

176.0 g

144.1 g C    (100 ) = 42.10% C  342.3 g   22.18 g H    (100 ) = 6.480% H  342.3 g 

342.3 g

 176.0 g O    (100 ) = 51.42% O  342.3 g 

71. Molecular formula of aspirin 60.0% C, 4.48% H, 35.5% O; molar mass of aspirin = 180.2  1 mol C   = 5.00 mol C  12.01 g C 

5.00 mol C = 2.25 mol C 2.22

 1 mol H   = 4.44 mol H  1.008 g H 

4.44 mol H = 2.00 mol H 2.22

 1 mol O   = 2.22 mol O  16.00 g O 

2.22 mol O = 1.00 mol O 2.22

( 60.0 g C ) 

( 4.48 g H )  (35.5 g O ) 

Multiplying each by 4 gives the empirical formula C9H8O4. The empirical formula mass is 180.2 g. Since the empirical formula mass equals the molar mass, the molecular formula is the same as the empirical formula C9H8O4. 72. First calculate the percent oxygen in Al2(SO4)3. 2 Al 53.96 g  192.0 g    (100 ) = 56.11% O 3S 96.21 g  342.2 g  12 O 192.0 g 342.2 g Second calculate grams of oxygen in 8.50 g of Al2(SO4)3. Now take 56.11% of 8.50 g. (8.50 g Al2(SO4)3)(0.5611) = 4.77 g O

– Chapter 7 –

73. (a) 7.79% C, 92.21% Cl  1 mol C   = 0.649 mol C  12.01 g C 

0.649 mol C = 1.00 mol C 0.649

 1 mol Cl   = 2.601 mol Cl  35.45 g Cl 

2.601 mol Cl = 4.01 mol Cl 0.649

( 7.79 g C ) 

( 92.21 g Cl ) 

The empirical formula is CCl4. The empirical formula mass is 153.8 which equals the molar mass, therefore the molecular formula is CCl4. (b) 10.13% C, 89.87% Cl  1 mol C   = 0.8435 mol C  12.01 g C 

0.8435 mol C = 1.000 mol C 0.8435

 1 mol Cl   = 2.535 mol Cl  35.45 g Cl 

2.535 mol Cl = 3.005 mol Cl 0.8435

(10.13 g C ) 

(89.87 g Cl ) 

The empirical formula is CCl3. The empirical formula mass is 118.4 g. molar mass 236.7 g = = 1.999 empirical formula mass 118.4 g The molecular formula is twice that of the empirical formula. Molecular formula = C2Cl6 (c) 25.26% C, 74.74% Cl  1 mol C   = 2.103 mol C  12.01 g C 

2.103 mol C = 1.000 mol C 2.103

 1 mol Cl   = 2.108 mol Cl  35.45 g Cl 

2.108 mol Cl = 1.002 mol Cl 2.103

( 25.26 g C ) 

( 74.74 g Cl ) 

The empirical formula is CCl. The empirical formula mass is 47.46 g.

molar mass 284.8 g = = 6.000 empirical formula mass 47.46 g The molecular formula is six times that of the empirical formula. Molecular formula = C6Cl6 (d) 11.25% C, 88.75% Cl  1 mol C   = 0.9367 mol C  12.01 g C 

0.9367 mol C = 1.000 mol C 0.9367

 1 mol Cl   = 2.504 mol Cl  35.45 g Cl 

2.504 mol Cl = 2.673 mol Cl 0.9367

(11.25 g C ) 

(88.75 g Cl ) 

Multiplying each by 3 gives the empirical formula C3Cl8. The empirical formula mass is 319.6. Since the molar mass is also 319.6 the molecular formula is C3Cl8.

– Chapter 7 –

74. The conversion is: s → min → hr → day → yr

hr  1 day   1 year  ( 6.022 ×10 s )  160mins   601 min  = 1.910 ×10 years    24 hr  365 days 23





75. The conversion is: g → mol → atom  1 mol Cu   6.022 ×10 23 atoms  22 2.5 g Cu ( )  = 2.4 ×10 atoms Cu  mol  63.55 g Cu   

76. The conversion is: molecules → mol → g 1 trillion = 1012





gC HO (1000. ×10 molecules C H O )  6.022 ×101 molmolecules   92.09  mol C H O 12

= 1.529×10 77.

−7





g C3H8 O3

1 mol people  ( 7.0 ×10 people )  6.022  = 1.2 ×10 ×10 people 9



−14



mol of people

78. Empirical formula and molecular formula are the same, BrI. The percent composition is 38.6% Br2 and 61.4% I2. Number of moles of Br2 =

5.40 g = 0.0676 mol 79.90 g/mol

Number of moles of I 2 =

8.58 g 126.9 g/mol

= 0.0676 mol

The number of moles of each element are in a ratio of 1:1, so the empirical formula is BrI. The mass of one mole of BrI is 206.8 g, which matches the value given in the problem, so the molecular formula is also BrI. Percent composition is calculated as follows:

%Br =

5.40 g mass Br2 = ×100 = 38.6% total mass ( 5.40 + 8.58 ) g

Likewise

%I 2 =

8.58 g ×100 = 61.4% 13.98 g

79. Empirical formula 23.3% Co, 25.3% Mo, 51.4% Cl  1 mol Co   = 0.935 mol Co  58.93 g Co 

0.395 mol Co = 1.50 mol Co 0.264

 1 mol Mo   = 0.264 mol Mo  95.94 g Mo 

0.264 mol Mo = 1.00 mol Mo 0.264

 1 mol Cl   = 1.45 mol Cl  35.45 g Cl 

1.45 mol Cl = 5.49 mol Cl 0.264

( 23.3 g Co ) 

( 25.3 g Mo )  (51.4 g Cl ) 

Multiplying all values by 2 gives the empirical formula Co3Mo2Cl11.

– Chapter 7 –

80. The conversion is: g Al ⎯⎯ → mol Al ⎯⎯ → mol Mg ⎯⎯ → g Mg

 1 mol Al   2 mol Mg   24.31 g Mg    = 32 g Mg   26.98 g Al   1 mol Al   mol Mg 

(18 g Al ) 

81. (10.0 g compound) (0.177) = 1.77 g N  1 mol N   = 0.126 mol N  14.01 g N 

(1.77 g N ) 

1 mol  (3.8 ×10 atoms H )  6.022 ×10  = 0.63 mol H atoms  23

To determine the mol C, first find grams of H and subtract the grams of H and N from the grams of the sample.

( 0.63 mol H ) 

1.008 g H   = 0.64 g H  mol H 

10.0 g sample −1.77 g N −0.64 g H 7.6 g C

 1 mol C   = 0.63 mol C  12.01 g C 

( 7.6 g C ) 

Now determine the empirical formula from the moles of C, H, and N.

N H C

0.126 mol N = 1.00 mol N 0.126 0.63 mol H = 5.0 mol H 0.126 0.63 mol C = 5.0 mol C 0.126

The empirical formula is C5H5N. 82. Let x = molar mass of A2O 0.400 x = 16.00 g O ( since A 2 O has only one mol of O atoms ) x = 40.0 g O/mol A 2 O 40.0 = 16.00 + 2y

y = molar mass of A

40.0 −16.00 = 2y g 12.0 =y mol Look in the periodic table for the element that has 12.0 g/mol. The element is carbon. The mystery element is carbon.

– Chapter 7 –

83. (a) CH2O

(divide the molecular formula by 6)

(b) C4H9

(divide the molecular formula by 2)

(divide the molecular formula by 3)

(d) C25H52

(divide the molecular formula by 1)

(e) C6H2Cl2O

(divide the molecular formula by 2)

84.

 9.0 μ g Cu 2+ ions  1 g  1 mol Cu 2+ ions  6.022 ×1023 Cu2 + ions     6  2+ 2+ 1 L water   10 μ g  63.55 g Cu ions  1 mol Cu ions 

(1 L water ) 

= 8.5×1016 Cu 2+ ions 85.

 1 mol H2   6.022 ×10 23 molecules H2    1 molecule O2    6 1 mol H2  2.016 g H2     1.0 ×10 molecule H2 

(3.0 g H2 ) 

= 9.0 ×1017 molecules O2

86.

 10.75 hr   60 min   60 s  250.0 kg H2 O   1000 g H2 O   1 mol H2 O        1s   1 kg H2 O   18.02 g H2 O   1 day   1 hr   1 min 

(1 day ) 

= 5.369×108 mol H2 O 87. 115 g Fe. We can start by finding out how many moles of arsenic are represented by 154 g.

154 g = 2.06 mol As 74.92 g/mol 2.06 moles of As contains the same number of atoms as 2.06 moles of Fe. To convert this value into grams of Fe, we need only to multiply the molar by the number of moles. Fe  ( 2.06 mol )  55.85 gmol  = 115 g Fe ( 3 significant figures ) 

88. Empirical formula of 38.65% C, 9.74% H, 51.61% S  1 mol C   3.218 mol C   = 3.218 mol C   = 2.000 mol C 1.609    12.01 g C 

(38.65 g C ) 

 1 mol H   = 9.66 mol H  1.008 g H 

 9.66 mol H    = 600 mol H  1.609 

 1 mol S   = 1.609 mol S  32.07 g S 

 1.609 mol S    = 1.000 mol S  1.609 

( 9.74 g H ) 

(51.61 g S ) 

The empirical formula is C2H6S.

– Chapter 7 –

89. First determine the elements in compound A(BC)3: A:

( 0.3459 )( 78.01 g ) = 26.98 g ( aluminum )

48.00 g = 16.00 g ( oxygen ) 3 3.03 g C: ( 0.0388 )( 78.01 g ) = = 1.01 g ( hydrogen ) 3 B:

( 0.6153)( 78.01 g ) =

Elements determined from atomic masses in the periodic table. A(BC)3 = Al(OH)3 Then compound A2B3 = Al2O3 with a molar mass of 2 ( 26.98 g ) + 3 (16.00 g ) = 102.0 g

% Al = % O= 90. (a)

2 ( 26.98 g ) (100 ) = 52.90% 102.0 g

3 (16.00 ) (100 ) 102.0

= 47.06%

Percent composition of the original unknown compound Convert g CO2 to g C and g H2O to g H.  12.01 g C   = 1.303 g C  44.01 g CO2 

( 4.776 g CO2 ) 

 2.016 g H   = 0.3282 g H  18.02 g H2 O 

( 2.934 g H2 O ) 

2.500 g compound −1.303 g C − 0.3282 g H = 0.8688 g O  1.303 g C    (100 ) = 52.12% C  2.500 g   0.3282 g H    (100 ) = 13.13% H  2.500 g   0.8688 g O    (100 ) = 34.75% O  2.500 g  (b) Empirical formula of unknown compound: 52.12% C, 13.13% H, 34.76% O.  1 mol C   = 4.340  12.01 g 

(52.12 g C ) 

4.340 mol C = 1.997 mol C 2.173

 1 mol H  13.03 mol H = 5.996 mol H  = 13.03 2.173  1.008 g H 

(13.13 g H ) 

 1 mol C   = 2.173  16.00 g 

(34.76 g O ) 

2.173 mol O = 1.000 mol O 2.173

The empirical formula is C2H6O.

– Chapter 7 –

 1 g   1 mol C10H16  6.022 ×10 23 molecules  91. 1.00 pg C10 H16  12    1 mol  10 pg   136.2 g C10 H16      1 photon 1s −9    = 1.67 ×10 s 18 1 molecule C H 2.64 ×10 photons  10 16   

92. (a)

molar mass = 1( Sr ) + 4 ( Ca ) + 8 ( Cr ) + 8 ( C ) + 69 ( O ) +1( S ) + 66 ( H )

= 1( 87.62 g/mol ) + 4 ( 40.08 g/mol ) + 8 ( 52.00 g/mol ) + 8 (12.01 g/mol ) +69 (16.00 g/mol ) +1( 32.07 g/mol ) + 66 (1.008 g/mol ) = 87.62 +160.3 + 416.0 + 96.08 +1104 + 32.07 + 66.53 = 1963 g/mol

(1.00 mol SrCa Cr ( CO ) (SO )( OH ) 25H O ) 4

(b)

3 8

  8 mol C    1 mol SrCa 4Cr8 ( CO3 )8 ( SO4 ) ( OH )16 25H2 O   6.022 ×10 23 atom C  24   = 4.82 ×10 atom C 1 mol C  

(1.00 mol SrCa Cr ( CO ) (SO )( OH ) 25H O ) 4

(c)

3 8

  69 mol O    1 mol SrCa 4Cr8 ( CO3 )8 ( SO 4 ) ( OH )16 25H2 O 

 6.022 ×10 23 atom O  25   = 4.16 ×10 atom O 1 mol O  

(d)

 1 mol SrCa 4Cr8 ( CO3 )8 ( SO4 ) ( OH )16 25H2 O   4 mol Ca  

(1.00 mol Ca ) 

 1963 g SrCa 4Cr8 ( CO3 )8 ( SO 4 ) ( OH )16 25H2 O    = 491 g SrCa 4Cr8 ( CO3 )8 ( SO 4 ) ( OH )16 25H2 O  1 mol SrCa 4Cr8 ( CO3 )8 ( SO 4 ) ( OH )16 25H2 O 

(e)

(1.00 kg SrCa Cr ( CO ) (SO ) ( OH ) 25H O ) 4

3 8

 1000 g SrCa 4Cr8 ( CO3 )8 ( SO 4 ) ( OH )16 25H2 O     1 kg SrCa 4 Cr8 ( CO3 )8 ( SO 4 ) ( OH )16 25H2 O   1 mol SrCa 4 Cr8 ( CO3 )8 ( SO 4 ) ( OH )16 25H2 O     1963 g SrCa 4Cr8 ( CO3 )8 ( SO 4 ) ( OH )16 25H2 O    1 mol S    1 mol SrCa 4Cr8 ( CO3 )8 ( SO 4 ) ( OH )16 25H2 O   32.07 g S    = 16.3 g S  1 mol S 

– Chapter 7 –

93. (a)

125 g flour   32 g starch  1 mol starch     5  1 c flour   100 g flour  1.08 ×10 g starch 

( 2.0 c flour ) 

 6.022 × 10 23 molec starch  20   = 2.7 ×10 molecules starch 1 mol starch  

 11 g NaHCO3   1 mol NaHCO3     1 tsp baking soda   84.01 g NaHCO3 

(1.0 tsp baking soda )  (b)

 6.022 ×1023 units NaHCO3  22   = 7.9 ×10 units NaHCO3 1 mol NaHCO 3    220.g C12H22 O11  1 mol C12H22 O11    1 c sugar   342.3 g C12H22 O11 

(1.3 c sugar )  (c)

(d)

 6.022 ×1023 molecules C12H22 O11  23   = 5.0 ×10 molecules C12H22 O11 1 mol C H O  12 22 11   6.3 g protein   1 mol protein   6.022 × 10 23 molecules protein  2 eggs ( )    1 egg 1 mol protein    4500 g protein    = 1.7 ×1021 molecules protein

 23 mg C8H8O3   1 g C8H8O3  1 mol C8H8O3      1 tsp vanilla   1000 mg C8H8O3  152.1 g C 8H8O3 

(1.0 tsp vanilla )  (e)

(f)

 6.022 ×10 23 molecules C8H8O3  19   = 9.1×10 molecules C8H8O3 1 mol C H O 8 8 3   227 g butter   81 g fat  1 mol fat  6.022 ×10 23 molecules fat      1 mol fat  1 c butter   100 g butter  800 g fat  

(1.0 c butter ) 

= 1.4 ×10 23 molecules fat

94. (a)

 1 mol  104 g O2 = (104 g O2 )   = 3.25 mol O2  32.00 g  X + O 2 → CO 2 + 3.25 mol

2 mol

H2 O 2.5 mol

( multiply moles bt 4 )

4X +13 O2 → 8 CO 2 +10 H2 O

Oxygen is balanced. By inspection, X must have 8/4 C atoms and 20/4 atoms (2 C and 5 H). Empirical formula is C2H5. (b) Additional information needed is the molar mass of X. 95. Distributing one mole of pennies among the entire population would give each person more money. The problem can be solved by comparing either the moles of pennies distributed in both cases, or the amount of money each person receives.

– Chapter 7 –

Comparing moles of pennies distributed: if everyone was given $999,900, then 99,990,000 pennies would be distributed per person, so total pennies given: population … 7.016 × 109 persons 99,990,000 pennies per person 1 mol pennies = 6.022 × 1023 pennies solution map:

# people  # pennies  mol pennies

7.833×109 persons ×

99,990,000 pennies 1 mol × 1 person 6.022 ×10 23 pennies

= 1.301×10 −6 moles of pennies distributed Less than 1 mole of pennies is distributed in this scenario. OR … comparing money each person receives: population … 7.833 × 109 persons 1 dollar = 100 pennies 1mol pennies = 6.022 × 1023 pennies solution map: 1 mol pennies  dollars  dollars per person 1 mol of pennies ×

1 dollar 1 6.022 ×10 23 pennies × × = $7.688 ×1011 9 1 mol 100 pennies 7.833 ×10 persons

In other words, 1 mol of pennies distributed equally would give over 760 billion dollars to each person!

CHAPTER 8

CHEMICAL EQUATIONS SOLUTIONS TO REVIEW QUESTIONS 1.

The coefficients in a balanced chemical equation represent the number of moles (or molecules or formula units) of each of the chemical species in the reaction.

The physical state of a substance may be a solid, a liquid, or a gas. The symbols indicate whether a substance is a solid, a liquid, a gas, or is in an aqueous solution. A solid is indicated by (s), a liquid by (l), a gas by (g), and an aqueous solution by (aq).

The purpose of balancing chemical equations is to conform to the Law of Conservation of Mass. Ratios of reactants and products can then be easily determined.

(a) Yes. It is necessary to conserve atoms to follow the Law of Conservation of Mass. (b) No. Molecules can be taken apart and rearranged to form different molecules in reactions. (c) Moles of molecules are not conserved (b). Moles of atoms are conserved (a).

Chemical equations can only be balanced by changing the number of each substance reacted or produced. If the subscripts are changed then the identity of the reactants and products is also changed. Equations must be balanced using the actual reactants and products.

6. 2 Al(OH)3

+ 3 H2SO4

Coefficients

Reactants

Al(OH)3

H2SO4

Subscripts (shown in bold) Products

Al2(SO4)3

→ Al2(SO4)3 + 1

2 Al(OH)3 +

3 H2SO4

6 H2O 6

→ Al2(SO4)3+6 H2O

H2O

A combustion reaction is an exothermic process (usually burning) done in the presence of oxygen.

The activity series given in Table 8.2 shows the relative activity of certain metals and halogens. As you move up the table starting with gold (Au) and ending with potassium (K) the activity increases. The same is true as you move up from iodine (I2) to fluorine (F2). The table is useful for predicting the products of some reactions because an element in the series will replace any element given below it. For example, hydrogen can replace copper, silver, mercury, or gold in a chemical reaction.

The major types of chemical reactions are combination reactions, decomposition reactions, single displacement reactions, and double displacement reactions.

– Chapter 8 –

10. Indicators that a chemical reaction has occurred are •

Formation of a precipitate

•

Formation of a gas

•

A temperature change

11. A chemical change that absorbs heat energy is said to be an endothermic reaction. The products are at a higher energy level than the reactants. A chemical change that liberates heat energy is said to be an exothermic reaction. The products are at a lower energy level than the reactants. 12. Although an exothermic reaction will liberate more heat than it absorbs, some heat is still necessary to get the reaction started. The energy required to start a reaction is the activation energy. 13. Methane, carbon dioxide, and water 14. Carbon dioxide levels fall in the spring as plants grow and incorporate carbon dioxide into their cells. Carbon dioxide levels rise in the fall as plants begin to decay back into the soil releasing carbon dioxide back into the air.

– Chapter 8 –

SOLUTIONS TO EXERCISES 1.

(a) exothermic

(d) exothermic

(b) endothermic

(e) endothermic

(f) endothermic

(a) endothermic

(d) exothermic

(b) exothermic

(e) exothermic

(f) exothermic

(a) DD

(b) C

(d) D

(a) SD

(b) C

(d) D

(a) 2H2 + O2 →2 H2O combination (b) 3 N2H4(l) → 4 NH3 (g) + N2(g)

decomposition

double displacement

Δ → Al2O3 + 3CO2 (d) Al2 ( CO3 )3 ⎯⎯

decomposition

(e) 2 NH4I + Cl2 → 2 NH4Cl + I2

single displacement

(a) H2 + Br2 → 2 HBr

combination

6. (b) BaO2(s) + H2SO4(aq) → BaSO4(s) + H2O2(aq) double displacement

Δ (c) Ba(ClO3)2 ⎯⎯ → BaCl2 + 3 O2

decomposition

(d) CrCl3 + 3AgNO3 → Cr(NO3)3 + 3 AgCl

double displacement

(e) 2 H2O2 → 2H2O + O2

decomposition

(a) Na2CrO4 + CaCl2 → CaCrO4 + 2 NaCl Double displacement reaction (b) 2 Al + 3 CuCl2 → 3 Cu + 2 AlCl3 Single replacement reaction (c)

(d)

– Chapter 8 –

(e)

(a) 2 FeCl3 + 3Li2CO3 → 6 LiCl + Fe2 (CO3)3 Double displacement reaction (b) 3 Cd + 2 TiCl3 → 2 Ti + 3 CdCl2 Single replacement reaction (c)

(d)

(e)

(a) C3H8(g) + 5 O2(g) →3 CO2(g) + 4H2O(g) (b) 2 C10H22(l) + 3l O2(g) → 20 CO2(g) + 22 H2O(g) (c) 2 C6H14(l) + 19 O2(g) → 12 CO2(g) + 14 H2O(g) (d) C4H9OH(l) + 6 O2(g) → 4 CO2(g) + 5 H2O(g)

10. (a) 2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(g) (b) C34H64(l) + 50 O2(g) → 34 CO2(g) + 32 H2O(g) (c) 2 C2H2(l) + 5 O2(g) → 4 CO2(g) + 2 H2O(g) (d) 2 CH3OH(l) + 3 O2(g) → 2 CO2(g) + 4 H2O(g) 11. (a) 2 MnO2 + CO → Mn2O3 + CO2 (b) Cu2O(s) + C(s) → 2 Cu(s) + CO(g) (c) 4 C3H5(NO3)3 → 12 CO2 + 10 H2O + 6 N2 + O2 (d) 4 FeS + 7 O2 → 2 Fe2O3 + 4 SO2 (e) 2 Cu(NO3)2 → 2 CuO + 4 NO2 +O2 (f)

3 NO2 + H2O → 2 HNO3 + NO

(g) 2 Fe(s) + 3 S(l) → Fe2S3(s) (h) 4 HCN +5 O2 → 2 N2 + 4 CO2 + 2 H2O (i)

2 B5H9 + 12 O2 → 5 B2O3 + 9 H2O 4

– Chapter 8 –

12. (a) 2 SO2 + O2 → 2 SO3 (b) Li2O(s) + H2O(l) → 2 LiOH(aq) (c) 2 Na + 2H2O → 2 NaOH + H2 (d) 2 AgNO3 + Ni → Ni(NO3)2 + 2 Ag (e) Bi2S3 + 6 HCl → 2 BiCl3 + 3 H2S (f)

2 PbO2 ⎯Δ⎯ → 2 PbO + O2

(g) Hg2(C2H3O2)2(aq) + 2 KCl(aq) → Hg2Cl2(s) + 2KC2H3O2(aq) (h) 2 KI + Br2 →2 KBr + I2 (i)

2 K3PO4 +3 BaCl2 → 6 KCl + Ba3(PO4)2

13. (a) Mg(s) + 2 HBr(aq) → H2(g) + MgBr2(aq) (b) Ca(ClO3)2(s) ⎯Δ⎯ → CaCl2(s) + 3 O2(g) (c) 4 Li(s) + O2(g) → 2 Li2O(s) (d) 3 Ba(BrO3)2(aq) +2 Na3PO4(aq) → Ba3(PO4)2(s) + 6 NaBrO3(aq) (e) 2 HC2H3O2(aq) + Na2CO3(aq) → 2 NaC2H3O2(aq) + CO2(g) + H2O(l) (f)

3 AgNO3(aq) + AlI3(aq) → 3 AgI(s) + Al(NO3)3(aq)

14. (a) MgCO3(s) ⎯Δ⎯ → MgO(s) + CO2(g) (b) Ca(OH)2(s) + 2 HClO3(aq) → Ca(ClO3)2(aq) + 2 H2 O(l) (c) Fe2(SO4)3(aq) + 6 NaOH(aq) → 2 Fe(OH)3(s) + 3 Na2SO4(aq) (d) Zn(s) + 2 HC2H3O2(aq) → H2(g) + Zn(C2H3O2)2(aq) (e) SO3(g) + H2O(l) → H2SO4(aq) (f)

Na2CO3(aq) + CoCl2(aq) → CoCO3(s) + 2 NaCl(aq)

15. (a) 2 H2O2 → 2 H2O + O2 (b) formula : KNO3 → KNO2 + O2 balanced equation: 2 KNO3 → 2 KNO2 + O2 (c) Br2 +H2S → 2 HBr + S (d) 2 K + 2 H2O → 2 KOH + H2 (e) 3 Fe + 4 H2O → Fe3O4 + 4 H2 (f)

2 NaHCO3 + H2SO4 → Na2SO4 + 2 H2O + 2 CO2

16. (a) NH4NO2 → N2 + 2 H2O (b) formula: CaO + HCl → CaCl2 + H2O balanced equation: CaO + 2 HCl → CaCl2 + H2O (c) Cu + 2 H2SO4 → CuSO4 + 2 H2O + SO2 (d) N2 + 3 H2 → 2 NH3

– Chapter 8 –

(e) 2 C6H14 + 19 O2 → 12 CO2 + 14 H2O (f)

2 ZnS + 3 O2 → 2 ZnO + 2 SO2

17. (a) H2SO4(aq) + 2 NaOH(aq) → 2 H2O(l) + Na2SO4(aq) + heat (b) Pb(NO3)2(aq) + 2 KBr(aq) → PbBr2(s) + 2 KNO3(aq) (c) NH4Cl(aq) + AgNO3(aq) → AgCl(s) + NH4NO3(aq) (d) CaCO3(s) + 2 HC2H3O2(aq) → Ca(C2H3O2)2(aq) + H2O(l) + CO2(g) 18. (a) CuSO4(aq) + 2 KOH(aq) → Cu(OH)2(s) + K2SO4(aq) (b) H3PO4(aq) + 3 NaOH(aq) → Na3PO4(aq) + 3 H2O(l) + heat (c) 3 NaHCO3(s) + H3PO4(aq) → Na3PO4(aq) + 3 H2O(l) + 3CO2(g) (d) 2 AlCl3(aq) + 3 Pb(NO3)2 →3 PbCl2(s) +2 Al(NO3)3(aq) 19. (a) 2 Ca(s) + 2H2O(l) → Ca(OH)2(aq) + H2(g) (b) Br2(l) + 2 KI(aq) → I2(s) + 2 KBr(aq) (c) Cu(s) + HCl(aq) → no reaction (d) 2 Al(s) + 3 H2SO4(aq) → 3 H2(g) + Al2(SO4)3(aq) 20. (a) Cu(s) + NiCl2(aq) → no reaction (b) 2 Rb(s) + 2 H2O(l) → H2(g) + 2 RbOH(aq) (c) I2(s) + CaCl2(aq) → no reaction (d) 3 Mg(s) + 2 Al(NO3)3(aq) → 2 Al(s) + 3 Mg(NO3)2(aq) 21. (a) PbCl2(aq) + Mn(s) → MnCl2(aq) + Pb(s) (b) Mn (c) Ba (d) No, Because Pb<Mn and Mn<Ba so Pb<Ba meaning that lead is less active than barium and cannot replace it as an ion. 22. (a) 3 NiCl2(aq) + 2 Al(s) → 2 AlCl3(aq) + 3 Ni(s) (b) Al (c) Ni (d) Yes, because Al>Ni and Ni<Hg so Al>Hg and a reaction should occur. Ni(s) + HgCl2(aq) → Hg(l) + NiCl2(aq) The nickel should begin to dissolve and a pool of mercury should begin to appear. 23. (a) Sr(s) + 2 H2O(l) → H2(g) + Sr(OH)2(s) (b) BaCl2(aq) + 2 AgNO3(aq) → Ba(NO3)2(aq) + 2 AgCl(s) (c) Mg(s) + ZnBr2(aq) → Zn(s) + MgBr2(aq) (d) 2 K(s) + Cl2(g) → 2 KCl(s)

– Chapter 8 –

24. (a) Li2O(s) + H2O → 2 LiOH(aq) (b) Na2SO4(aq) + Pb(NO3)2(aq) → 2 NaNO3(aq) + PbSO4(s) (c) Zn(s) + CuSO4(aq) → Cu(s) + ZnSO4(aq) (d) 4 Al(s) + 3 O2(g) → 2 Al2O3(s) 25. (a) 2 Ba + O2 → 2 BaO Δ (b) 2 NaHCO3 ⎯⎯ → Na2CO3 + H2O + CO2

(c) Ni + CuSO4 → NiSO4 + Cu (d) MgO + 2 HCl → MgCl2 + H2O (e) H3PO4 + 3 KOH → K3PO4 + 3 H2O 26. (a) C + O2 → CO2 Δ (b) 2 Al(ClO3)3 ⎯⎯ → 9 O2 + 2 AlCl3

(c) CuBr2 + Cl2 → CuCl2 + Br2 (d) 2 SbCl3 + 3(NH4)2S→Sb2S3 + 6 NH4Cl (e) 2 NaNO3 ⎯Δ⎯ → 2 NaNO2 + O2 27. (a) One mole of MgBr2 reacts with two moles of AgNO3 to yield one mole of Mg(NO3)2 and two moles of AgBr. (b) One mole of N2 reacts with three moles of H2 to produce two moles of NH3. (c) Two moles of C3H7OH react with nine moles of O2 to form six moles of CO2 and eight moles of H2O. 28. (a) Two moles of Na react with one mole of Cl2 to produce two moles of NaCl and release 822 kJ of energy. The reaction is exothermic. (b) One mole of PCI5 absorbs 92.9 kJ of energy to produce one mole of PC13 and one mole of Cl2. The reaction is endothermic. (c) 1 mol of solid sulfur will react with 2 moles of gaseous carbon monoxide to produce 1 mol of gaseous sulfur dioxide, 2 moles of solid carbon, and 76 kilojoules of energy in this exothermic reaction. 29. (a) 2 HgO(s) + 182 kJ → 2 Hg(l) + O2(g) (b) 2 H2(g) + O2(g) → 2 H2O(l) + 571.6 kJ 30. (a) Ca(s) + 2 H2O(l) → Ca(OH)2(aq) + H2(g) + 635.1 kJ (b) 2 BrF3 + 601.6 kJ → Br2 + 3 F2 Δ 31. (a) decomposition reaction, 2 AgClO3(s) ⎯⎯ → 2 AgCl(s) + 3 O2(g)

(b) single-displacement, Fe(s) + H2SO4(aq) → H2(g) + FeSO4(aq) (c) combination reaction, Zn(s) + Cl2(g) → ZnCl2(s) (d) double-displacement, HBr(aq) + KOH(aq) → KBr(aq) + H2O(l)

– Chapter 8 –

32. (a) single-displacement, Ni(s) + Pb(NO3)2(aq) → Pb(s) + Ni(NO3)2(aq) (b) combination, MgO(s) + H2O(l) → Mg(OH)2(s) (c) decomposition, 2 HgO(s) → 2 Hg(l) + O2(g) (d) double-displacement, PbCl2(aq) + (NH4)2CO3(aq) → PbCO3(s) + 2 NH4Cl(aq) 33. Combinations that form a precipitate: Ca(NO3)2(aq) + (NH4)2SO4(aq) → CaSO4(s) + 2 NH4NO3(aq) Ca(NO3)2(aq) + (NH4)2CO3(aq) → CaCO3(s) + 2 NH4NO3(aq) AgNO3(aq) + NH4Cl(aq) → AgCl(s) + NH4NO3(aq) 2 AgNO3(aq) + (NH4)2SO4(aq) → Ag2SO4(s) + 2 NH4NO3(aq) 2 AgNO3(aq) + (NH4)2CO3(aq) → Ag2CO3(s) + 2 NH4NO3(aq) 34. P4O10 + 12 HClO4

→ 6 Cl2O7 + 4 H3PO4

10 O + 12(4 O)

6(7 O) + 4(4 O)

10 O + 48 O

42 O + 16 O

58 O

35. In 5 Ni3(PO4)2 there are: (a) 15 atoms of N

(b) 10 atoms of P

(d) 65 total atoms

36. CaCO3(s) + 2 HC2H3O2(aq) → Ca(C2H3O2)2(aq) + CO2(g) + H2O(l) 37.

38. (a) C(s) + O2(g) → CO2(g) (b) CO2(g) + C(s) → 2 CO(g) (c) Fe2O3(s) + 3 CO (g) ⎯Δ⎯ → 2 Fe(l) + 3 CO2(g) Fe3O4(s) + 4 CO(g) ⎯Δ⎯ → 3 Fe(l) + 4 CO2(g) 39. 2 B + 3 G2 → 2 BG3 40. The metals that should be chosen are (b) zinc, (c) aluminum, and (e) calcium. These metals are more active than nickel, therefore will react in a solution of nickel(II) chloride; (a) copper and (d) lead are less active and will not react with nickel(II) chloride solution (see Table 8.2).

– Chapter 8 –

Equations Zn + NiCl2 → Ni + ZnCl2 2 Al + 3 NiCl2 → 3 Ni + 2 AlCl3 Ca + NiCl2 → Ni + CaCl2 41. Ti + Ni(NO3)2 → Reaction occurs Ti + Pb(NO3)2 → Reaction occurs Ti + Mg(NO3)2→ No reaction Ti is above Ni and Pb in the activity series since both react. Ti is below Mg in the series since it will not replace Mg. From the printed activity series in the chapter Ni lies above Pb so the order is: Mg Ti Ni Pb 42. (a) 4 Cs + O2 → 2 Cs2O (b) 2 Al + 3 S → Al2S3 (c) SO3 + H2O → H2SO4 (d) Na2O + H2O → 2 NaOH 43. (a) 2 H2(g) + O2(g) → 2 H2O(g) (b) CH4(g) + H2O(g) → 3 H2(g) + CO(g) CO(g) + H2O(g) → H2(g) + CO2(g) (c) CH4(g) + 2 O2(g) → 2 H2O(g) + CO2(g) (d) One molecule of carbon dioxide is produced from one molecule of methane regardless of whether it is converted into hydrogen gas or burned in oxygen. If the goal is to reduce carbon dioxide emission, there is no advantage in using hydrogen synthesized from methane as a fuel. 44. (a) Pb(NO3)2(aq) + K2CrO4(aq) → 2 KNO3(aq) + PbCrO4(s) (b) CdCl2(aq) + Li2S(aq) → CdS(s) + 2 LiCl(aq)

Lead(II) chromate Cadmium sulfide

(c) Pb(C2H3O2)2(aq) + H2CO3(aq) → PbCO3(s) + 2 HC2H3O2(aq) Lead(II) carbonate (d) Hg(NO3)2(aq) + Na2S(aq) →HgS(s) + 2 NaNO3(aq)

Mercury(II) sulfide

(e) 4 Ti(s) + 3 O2(g) → 2 Ti2O3(s)

Titanium(III) oxide

(f)

4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)

Iron(III) oxide

45. (a) 2 ZnO ⎯ ⎯ → 2 Zn + O2 (b) SnO2 ⎯Δ⎯ → Sn + O2 or 2 SnO2 ⎯Δ⎯ → 2 SnO + O2 (c) Na2CO3 ⎯Δ⎯ → Na2O + CO2 (d) Mg(ClO3)2 ⎯Δ⎯ → MgCl2 + 3 O2

– Chapter 8 –

46. (a) Mg(s) + 2 HCl(aq) → H2(g) + MgCl2(aq) (b) 2 NaBr(aq) + Cl2 (g) → 2 NaCl(aq) + Br2(l) (c) 3 Zn(s) + 2 Fe(NO3)3(aq) → 2 Fe(s) + 3 Zn(NO3)2(aq) (d) 2 Al(s) + 3 Cu(NO3)2(aq) → 3 Cu(s) + 2 Al(NO3)3(aq) 47. (a) 2 (NH4)3PO4(aq) + 3 Ba(NO3)2(aq) → Ba3(PO4)2(s) + 6 NH4NO3(aq) (b) Na2S(aq) + Pb(C2H3O2)2(aq) → PbS(s) + 2 NaC2H3O2(aq) (c) CuSO4(aq) + Ca(ClO3)2(aq) → CaSO4(s) + Cu(ClO3)2(aq) (d) Ba(OH)2(aq) + H2C2O4(aq) → BaC2O2(aq) + 2 H2O(l) (e) H3PO4(aq) + 3 KOH(aq) → K3PO4(aq) + 3 H2O(l) (f)

H2SO4(aq) + Na2CO3(aq) → Na2SO4(aq) + H2O(l) + CO2(g)

48. (a) K2SO4(aq) + Ba(C2H3O)2(aq) → 2 KC2H3O2(aq) + BaSO4(s) (b) H2SO4(aq) + 2 LiOH(aq) → Li2SO4(aq) + 2 H2O(l) (c) (NH4)3PO4(aq) + NaBr(aq) → No reaction (d) CaI2(aq) + 2 AgNO3(aq) → Ca(NO3)2(aq) + 2 AgI(s) (e) 2 HNO3(aq) + Sr(OH)2(aq) → Sr(NO3)2(aq) + 2H2O(l) (f)

CsNO3(aq) + Ca(OH)2(aq) → No reaction

49. (a) CH4 + 2 O2 →CO2 + 2 H2O (b) 2 H2 + O2 → 2 H2O 50. (a) CH4 + 2 O2 → CO2 +2 H2O (b) 2 C3H6 + 9 O2 → 6 CO2 + 6 H2O (c) C6H5CH3 + 9 O2 → 7 CO2 + 4 H2O 51. 1.

combustion of fossil fuels

destruction of the rainforests by burning

increased population

52. Carbon dioxide, methane, and water are all considered to be greenhouse gases. They each act to trap the heat near the surface of the earth in the same manner in which a greenhouse is warmed. 53. The effects of global warming can be reduced by: 1.

developing new energy sources (not dependent on fossil fuels)

conservation of energy resources

recycling

decreased destruction of the rainforests and other forests

54. About half the carbon dioxide released into the atmosphere remains in the air. The rest is absorbed by plants and used in photosynthesis or is dissolved in the oceans.

– Chapter 8 –

55. In the Northern Hemisphere, the concentration of CO2 peaks once in May, dropping as plants use the CO2 to produce growth, until October when the second peak occurs as a result of the fallen leaves decaying. 56. The reaction profile shows an exothermic reaction that means heat is given off during the reaction and therefore the beaker should feel hotter after the reaction than it did before the reaction. In exothermic reactions heat can be written as a product. 57. (a) endothermic

(b) reactant (net absorption of energy)

(c)

58. (a)

(b) All the anionic salts are sodium salts because you need to use soluble reagents and all sodium salts are soluble. 59. (a) Compound A must have a lower activation energy than compound B because B requires heat to overcome the activation energy for the reaction.

(b) (i) 2 NaHCO3 → Na2CO3 + H2O + CO2 Decomposition of 0.500 mol NaHCO3 requires 85.5 kJ of heat. If 24.0 g CO2 is produced, then

– Chapter 8 –

 1 mol CO2   1 mol H2 O   18.02 g      44.01 g   1 mol CO2   1 mol H2 O 

( 24.0 g CO2 ) 

= 9.83 g H2 O produced

0.500 mol NaHCO3 produces 

1 mol CO2   44.01 g CO2     2 mol NaHCO3   1 mol CO2 

( 0.500 mol NaHCO3 ) 

= 11.0 g CO2

Producing 11.0 g CO2 required 85.5 KJ  24.0 g  (85.5 kJ ) Producing 24:0 g CO2 requires  11.0 g  = 187 kJ

(ii) NaHCO3 could be compound B. Since heat was absorbed for the decomposition of NaHCO3, the reaction was endothermic. Decomposition of A was exothermic. 60. (a) double decomposition (precipitation) (b) mercury(II) iodide (HgI2) 61. (a) double displacement

2 Al(OH)3 + 3 H2SO4 → Al2(SO4)3 + 6 H2O

(b) single displacement

2 Na + 2 H2O → 2 NaOH + H2(g)

2 HCl + K2CO3 → 2 KCl + H2O + CO2(g)

(d) decomposition

Δ CaCO3 ⎯⎯ → CaO + CO2 ( g )

(e) single displacement

Zn + Pb(NO3)2 → Zn(NO3)2 + Pb

(f)

Cl2 + 2 KBr → Br2 + 2 KCl

single displacement

(g) double displacement

Ba(NO3)2 + Na2SO4 → BaSO4 + 2 NaNO3

(h) combination

4 K + O2 → 2 K2O

(i)

double displacement

HNO3 + NaOH → NaNO3 + H2O

(j)

double displacement

AgNO3 + NaCl → AgCl + NaNO3

CHAPTER 9

CALCULATIONS FROM CHEMICAL EQUATIONS SOLUTIONS TO REVIEW QUESTIONS 1.

A mole ratio is the ratio between the mole amounts of two atoms and/or molecules involved in a chemical reaction.

In order to convert grams to moles the molar mass of the compound under consideration needs to be determined.

The balanced equation is Ca3P2 + 6 H2O → 3 Ca(OH)2 + 2 PH3  2 mol PH3  (a) Correct: (1 mol Ca 3P2 )   = 2 mol PH3  1 mol Ca 3P2 

(b) Incorrect: 1 g Ca3P2 would produce 0.4g PH3  1 mol   2 mol PH3   33.99 g     = 0.4 g PH3  182.2 g   1 mol Ca 3P2   mol 

(1 g Ca3P2 ) 

(c) Correct: see equation (d) Correct: see equation (e) Incorrect: 2 mol Ca3P2 requires 12 mol H2O to produce 4.0 mol PH3.  6 mol H 2 O   = 12 mol H 2 O  1 mol Ca 3P2 

( 2 mol Ca3P2 )  (f)

Correct: 2 mol Ca3P2 will react with 12 mol H2O (3 mol H2O are present in excess) and 6 mol Ca(OH)2 will be formed.  3 mol Ca(OH)2   = 6 mol Ca ( OH )2  1 mol Ca 3P2 

( 2 mol Ca3P2 ) 

 1 mol   6 mol H 2 O   18.02 g  (g) Incorrect: ( 200 g Ca 3 P2 )     = 119 g H 2 O  182.2 g   1 mol Ca 3 3P2   mol 

The amount of water present (100 g) is less than needed to react with 200 g Ca3P2 Ca3P2•H2O is the limiting reactant. (h) Incorrect: water is the limiting reactant.  1 mol   2 mol PH3   33.99 g     = 62.9 g PH3 ( theoretical )  18.02 g   6 mol H 2 O   mol 

(100 g H2 O ) 

– Chapter 9 –

The balanced equation is 2 CH4 + 3 O2 + 2 NH3 → 2 HCN + 6 H2O (a) Correct  2 mol HCN  (b) Incorrect: (16 mol O 2 )   = 10.7 mol HCN ( not 12 mol HCN )  3 mol O 2 

 6 mol H2 O  (d) Incorrect: (12 mol HCN )   = 36 mol H2 O ( not 4 mol H2 O )  2 mol HCN  (e) Correct (f)

Incorrect: O2 is the limiting reactant  2 mol HCN   = 2 mol HCN ( not 3 mol HCN )  3 mol O 2 

(3 mol O2 )  5.

The molar mass of the product is also needed to convert moles of product into grams.

(a) 2

g H2 → mol H2 → mol NH3 → g NH3

The theoretical yield of a chemical reaction is the maximum amount of product that can be produced based on a balanced equation. The actual yield of a reaction is the actual amount of product obtained.

You can calculate the percent yield of a chemical reaction by dividing the actual yield by the theoretical yield and multiplying by 100.

(b) 4

– Chapter 9 –

SOLUTIONS TO EXERCISES

 1 mol   = 0.247 mol KNO3  101.1 g 

(a)

( 25.0 g KNO3 ) 

(b)

(37 mmol NaOH ) 

(c)

1 mol  (5.4 ×10 g ( NH ) C O )  124.1  = 4.4 mol ( NH ) C O g

1 mol   = 0.037 mol NaOH  1000 mmol 

4 2



4 2



(d) The conversion is mL sol → g sol → g H2SO4 → mol H2SO4 1.727 g   0.800 g H2SO 4  1 mol    = 0.396 mol H2SO 4   mL   g solution  98.09 g 

( 28.1 mL solution )  2.

 1000 g   1 mol    = 25.0 mol NaHCO3  kg   84.01 g 

(a)

( 2.10 kg NaHCO3 ) 

(b)

(525 mg ZnCl2 ) 

(c)

( 9.8 ×10 molecules CO )  6.022 ×101 molmolecules  = 16 mol CO

(d)

( 250 mL C 2H5OH ) 

(a)

g (1.84 mol Fe ( OH) )  106.9  = 197 g Fe ( OH ) mol 

(b)

(125 kg CaCO3 ) 

(c)

(13.7 mol NH3 ) 

(d)

( 72 mmol HCl ) 

(e)

( 400.0 mL Br2 ) 

(a)

( 0.00844 mol NiSO4 ) 

(b)

( 0.0600 mol HC2H3O2 ) 



1 g   1 mol  −3   = 3.85 ×10 mol ZnCl2 1000 mg 136.3 g   

0.789 g   1 mol   = 4.3 mol C 2 H5 OH   mL   46.07 g 

 1000 g  5  = 1.25 × 10 g CaCO3  kg 

17.03 g   = 233 g NH3  mol 

1 mol   36.46 g    = 2.6 g HCl  1000 mmol   mol 

3.119 g  3  = 1.248 ×10 g Br2  mL  154.8 g   = 1.31 g NiSO4  mol  60.05 g   = 3.60 g HC2 H3O2  mol 

– Chapter 9 –

(c)

( 0.725 mol Bi2S3 ) 

514.2 g   = 373 g Bi2S3  mol 

(d)

g ( 4.50 × 10 molecules C H O )  6.022 × 101 molmolecules   180.2  = 1.35 g C H O mol

(e)

( 75 mL solution ) 







1.175 g   0.200 g K 2 CrO 4   = 18 g K 2 CrO4  g solution  mL   

Larger number of molecules: 10.0 g H2O or 10.0 g H2O2 Water has a lower molar mass than hydrogen peroxide. 10.0 grams of water has a lower molar mass, contains more moles, and therefore more molecules than 10.0 g of H2O2.

Larger number of molecules: 25.0 g HCl or 85 g C6H12O6

 1 mol   6.022 × 1023 molecules  23  = 4.13 ×10 molecules HCl  mol  36.46 g  

( 25.0 g HCl ) 

 1 mol  1 mol   6.022 ×1023 molecules     mol  180.2 g  180.2 g  

(85.0 g C6H12 O6 ) 

= 2.84 ×1023 molecules C6 H12 O6 HCl contains more molecules. 7.

Mole ratios 12 CO2 + 11 H2O → C12H22O11 + 12 O2

(a) 12 mol CO 2 11 mol H2 O

(d) 1 mol C12 H 22 O11 12 mol CO 2

(b)

11 mol H2 O 1 mol C12 H 22 O11

(e) 11 mol H2 O 12 mol O 2

(c)

12 mol O 2 12 mol CO 2

(d)

12 mol O 2 1 mol C12 H 22 O11

Mole ratios C4H9OH + 6 O2 → 4 CO2 + 5 H2O (a)

6 mol O2 1 mol C 4H9 OH

(d) 1 mol C 4H9 OH 4 mol CO2

(b) 5 mol H2 O 6 mol O2

(e)

5 mol H2 O 1 mol C 4H9 OH

(f)

4 mol CO2 6 mol O 2

– Chapter 9 –

The balanced equation is CO2 + 4 H2 → CH4 + 2 H2O

 2 mol H2 O   = 30. mol H2 O  1 mol CO2 

(a)

(15 mol CO2 ) 

(b)

(18 mol H2 O ) 

 1 mol CH4   = 9.0 mol CH4  2 mol H2 O 

10. The balanced equation is H2SO4 + 2 NaOH → Na2SO4 + 2 H2O

 2 mol NaOH   = 34 mol NaOH  1 mol H2SO4 

(a)

(17 mol H2SO4 ) 

(b)

( 21 mol NaOH ) 

1 mol Na 2SO 4   = 11 mol Na 2SO 4  2 mol NaOH 

11. The balanced equation is Mn O2(s) + 4 HCl(aq) → Cl2(g) + MnCl2(aq) + 2 H2O(l)

 4 mol HCl   = 5.32 mol HCl  1 mol MnO2 

(a)

(1.33 mol MnO2 ) 

(b)

(1.25 mol H2 O ) 

(c)

( 2.63 mol MnO2 ) 

(d)

( 20.0 kg MnCl2 ) 

 1 mol MnCl2   = 0.625 mol MnCl2  2 mol H2 O   1 mol Cl2   70.90 g    = 186 g Cl2  1 mol MnO2   1 mol 

 1000 g   1 mol  4 mol HCl   = 636 mol HCl    1 kg   125.8 g  1 mol MnCl2 

12. Al4C3 + 12 H2O → 4 Al(OH)3 + 3 CH4

 1 mol   12 mol H2 O   = 8.33 mol H2 O   144.0 g   1 mol Al 4C3 

(a)

(100. g Al4C3 ) 

(b)

( 0.600 mol CH4 ) 

(c)

( 275 g Al4C3 ) 

(d)

H O   18.02 g  ( 4.22 mol Al ( OH ) )  12 mol   = 228 g H O 1 mol   1 mol 

 4 mol Al ( OH )3   = 0.800 mol Al ( OH )3  3 mol CH4 

 1 mol   3 mol CH4   = 5.73 mol CH4   144.0 g   1 mol Al 4C3  2

– Chapter 9 –

13. Grams of CaCl2 CaCO3 + 2 HCl → CaCl2 + H2O + CO2 The conversion is g CaCO3 → mol CaCO3 → mol CaCl2 → g CaCl2

 1 mol   1 mol CaCl2   111.0 g     = 66.5 g CaCl2  100.1 g   1 mol CaCO3   mol 

( 60.0 g CaCO3 )  14. Grams of AlBr3

2 Al + 6 HBr → 2 AlBr3 + 3 H2 The conversion is g Al → mol Al → mol AlBr3 → g AlBr3  1 mol   2 mol AlBr3  266.7 g    = 249 g AlBr3   26.98 g   2 mol Al   mol 

( 25.2 g Al ) 

15. The balanced equation is Fe2O3 + 3 C → 2 Fe + 3 CO The conversion is kg Fe2O3 → kmol Fe2O3 → kmol Fe → kg Fe

 1 kmol   2 kmol Fe   55.85 kg     = 122 kg Fe  159.7 kg   1 kmol Fe2 O3   kmol 

(175 kg Fe2 O3 ) 

16. The balanced equation is 3 Fe + 4 H2O → Fe3O4 + 4 H2 Calculate the grams of both H2O and Fe to produce 375 g Fe3O4

 1 mol   4 mol H2 O   18.02 g     = 117 g H2 O  231.6 g   1 mol Fe3O4   mol 

(375 g Fe3O4 ) 

 1 mol   3 mol Fe   55.85 g     = 271 g Fe  231.6 g   1 mol Fe3O4   mol 

(375 g Fe3O4 ) 

17. The balanced equation is 2 C12H4Cl6 + 23 O2 + 2 H2O → 24 CO2 + 12 HCl

 2 mol H2 O   = 0.870 mol H2 O  23 mol O2 

(a)

(10.0 mol O2 ) 

(b)

( 21.6 mol H2 O ) 

(c)

( 76.5 g HCl ) 

(d)

(115.27 g CO2 ) 

(e)

( 2.5 kg C12 H4 Cl6 ) 

 12 mol HCl   36.46 g  3   = 4.73 × 10 g HCl  2 mol H2 O   mol 

 1 mol   24 mol CO2    = 4.20 mol CO2  36.46 g   12 mol HCl 

 1 mol   2 mol C12H4Cl6   360.9 g     = 78.77 g C12 H4Cl6  44.01 g  24 mol CO2   mol   1000 g   1 mol   12 mol HCl   36.46 g  3  360.9 g   2 mol C H Cl  mol  = 1.5 × 10 g HCl 1 kg      12 4 6 

– Chapter 9 –

18. 4 HgS + 4 CaO → 4 Hg + 3 CaS + CaSO4 (a)

( 2.5 mol CaO ) 

3 mol CaS   = 1.9 mol CaS  4 mol CaO 

(b)

( 9.75 mol CaSO4 ) 

(c)

( 97.25 g HgS ) 

(d)

(87.6 g HgS ) 

(e)

( 9.25 kg Hg ) 

 4 mol Hg   200.6 g  3   = 7.82 × 10 g Hg mol 1 mol CaSO    4 

 1 mol   4 mol CaO    = 0.4179 mol CaO  232.7 g   4 mol HgS 

 1 mol   4 mol Hg   200.6 g     = 75.5 Hg  232.7 g   4 mol HgS   mol   1000 g   1 mol   3 mol CaS   72.15 g  3     = 2.50 × 10 g CaS 1 kg 200.6 g 4 mol Hg 1 mol      

19.

20.

Note that because chlorine gas exists as a diatomic molecule, only a single atom of chlorine would remain if all of the silver were to react. In actual reactions we count in terms of moles and a ½ mole of chlorine gas would remain if 3 moles of silver reacted with 2 moles of chlorine gas.

– Chapter 9 –

21.

22.

23. (a)

KOH + HNO3 15.0 g 11.0 g

→ KNO3

+ H2 O

Choose one of the products and calculate its mass that would be produced from each given reactant. Using KNO3 as the product:  1 mol   1 mol KNO3   101.1 g    = 27.0 g KNO3   56.10 g   1 mol KOH   mol 

(15.0 g KOH ) 

 1 mol   1 mol KNO3   101.1 g    = 17.6 g KNO3   63.02 g   1 mol KOH   mol 

(11.0 g HNO3 ) 

Since HNO3 produces less KNO3, it is the limiting reactant and KOH is in excess. (b)

2 NaOH + H2SO4 10.0 g 10.0 g

→ Na 2SO 4

+ 2H2 O

Choose one of the products and calculate its mass that would be produced from each given reactant.

– Chapter 9 –

Using H2O as the product:

 1 mol   2 mol H2 O  18.02 g     = 4.51 H2 O  40.00 g   2 mol NaOH   mol 

(10.0 g NaOH ) 

 1 mol   2 mol H2 O   18.02 g     = 3.67 H2 O  98.09 g  1 mol H2SO4   mol 

(10.0 g H2SO4 ) 

Since H2SO4 produces less H2O, it is the limiting reactant and NaOH is in excess. 24. (a)

2 Bi ( NO3 )3

+ 3 H2S → Bi2S3

+ 6 HNO3

6.00 g

50.0 g

Choose one of the products and calculate its mass that would be produced from each given reactant. Using Bi2S3 as the product: 



1 mol Bi S 1 mol   514.2 g    (50.0 g Bi ( NO ) )  395.0    = 32.5 g Bi S g 2 mol Bi ( NO )  mol  3 3



2 3



3 3

2 3



 1 mol   1 mol Bi2S3   514.2 g     = 30.2 g Bi2S3  34.09 g  3 mol H2S   mol 

( 6.00 g H2S) 

Since H2S produces less Bi2S3, it is the limiting reactant and Bi(NO3)3 is in excess. (b)

3 Fe 40.0 g

+ 4H2 O → Fe3O 4

+ 4H2

16.0 g

Choose one of the products and calculate its mass that would be produced from each given reactant. Using H2 as the product:

 1 mol   4 mol H2   2.016 g     = 1.93 g H2  55.85 g   3 mol Fe   mol 

( 40.0 g Fe ) 

 1 mol   4 mol H2   2.016 g     = 1.79 g H2  18.02 g  4 mol H2 O   mol 

(16.0 g H2 O ) 

Since H2O produces less H2, it is the limiting reactant and Fe is in excess. 25. Limiting reactant calculations 2 Al(OH)3 + 3 H2SO4 → Al2(SO4)3 + 6 H2O (a) Reaction between 45.0 g Al(OH)3 and 45.0 g H2SO4. Convert each amount to moles of Al2(SO4)3.  1 mol Al 2 ( SO 4 )3 

1 mol   = 0.288 mol Al ( SO ) ( 45.0 g Al ( OH ) )  78.00  g  2 mol Al ( OH )  3







4 3

 1 mol   1 mol Al 2 ( SO 4 )3   = 0.153 mol Al 2 ( SO 4 )3   98.09 g   3 mol H2SO 4 

( 45.0 g H2SO 4 ) 

H2SO4 is the limiting reactant. The yield is 0.153 mol Al2(SO4)3.

– Chapter 9 –

(b) Reaction between 50.0 g H2SO4 and 30.0 g Al(OH)3. Calculate the grams of Al2(SO4)3 from each reactant.  1 mol   1 mol Al 2 ( SO 4 )3   342.2 g   = 58.1 g Al 2 ( SO 4 )3    98.09 g   3 mol H 2SO 4   mol 

( 50.0 g H 2SO 4 ) 

 1 mol Al 2 ( SO 4 )3   342.2 g  = 65.7 mol Al 2 ( SO 4 )3  ( )3   mol  

(30.0 g Al ( OH ) )  78.09 g   2 mol Al OH 1 mol



H2SO4 is the limiting reactant. The yield is 58.1 g Al2 (SO4)3. Al(OH)3 is the excess reactant. (c) Reaction between 2.5 mol Al(OH)3 and 5.5 mol H2SO4. Convert each moment to moles of product.  1 mol Al 2 ( SO 4 )3 

( 2.5 mol Al ( OH ) )  2 mol Al ( OH )  = 1.3 mol Al (SO ) 2

4 3

3    1 mol Al 2 ( SO 4 )3  ( 5.5 mol H2SO 4 )   = 1.8 mol Al2 ( SO 4 )3  3 mol H2SO 4 

Al(OH)3 is the limiting reactant. 1.3 mol Al2 (SO4)3 is produced.





6 mol H O  = 7.5 mol H O produced ( 2.5 mol Al ( OH) )  2 mol Al ( OH )  2

3    mol H2SO4  = 3.8 mol H2SO 4 produced ( 2.5 mol Al ( OH)3 )  23mol  Al OH ( ) 3  

5.5 mol H2SO4 – 3.8 mol H2SO4 = 1.7 mol H2SO4 unreacted When the reaction is complete, 1.3 mol Al2(SO4)3, 7.5 mol H2O, and 1.7 mol H2SO4 will be in the container. 26. The balanced equation is P4 + 6 Cl2 → 4 PCl3 (a) Reaction between 20.5 g P4 and 20.5 g Cl2. Convert each amount to moles of PCl3.

 1 mol   4 mol PCl3   = 0.662 mol PCl3   123.9 g  1 mol P4 

( 20.5 g P4 ) 

 1 mol   4 mol PCl3   = 0.193 mol PCl3   70.90 g  6 mol Cl2 

( 20.5 g Cl2 ) 

Cl2 is the limiting reactant. The yield is 0.193 mol PCl3.

– Chapter 9 –

(b) Reaction between 55 g Cl2 and 25 g P4. Calculate the grams of PCl3 from each reactant.

 1 mol   4 mol PCl3   137.3 g     = 71 g PCl3  70.90 g  6 mol Cl2   mol 

(55 g Cl2 ) 

 1 mol   4 mol PCl3   137.3 g  2    = 1.1×10 g PCl3  123.9 g  6 mol P4   mol 

( 25 g P4 ) 

Cl2 is the limiting reactant. The yield is 71 g PCl3. P4 is the excess reactant. (c) Reaction between 15 mol P4 and 35 mol Cl2. Convert each amount to moles of product.

 4 mol PCl3   = 60. mol PCl3  1 mol P4 

(15 mol P4 ) 

 4 mol PCl3   = 23 mol PCl3 produced  6 mol Cl2 

(35 mol g Cl2 ) 

Cl2 is the limiting reactant.

 1 mol P4   = 5.8 mol P4 reacted  6 mol Cl2 

(35 mol Cl2 ) 

15 mol P4 − 5.8 mol P4 = 9 mol P4 left over When the reaction is complete, 23 mol PCl3 and 9 mol P4 will be in the container. 27. X 8 + 12 O2 → 8 XO3 The conversion is g O2 → mol O2 → mol X8  1 mol   1 mol X 8   = 0.3125 ol X 8   32.00 g   12 mol O2 

(120.0 g O2 ) 

80.0 g X 8 = 0.3125 mol X 8

80.0 g = 256 g/ mol X 8 0.3125 mol g 256 mol = 32.0 g molar mass X = 8 mol Using the periodic table we find the element with 32.0 g/mol is sulfur.

28. X + 2 HCl → XCl2 + H2 The conversion is g H2 → mol H2 → mol X

 1 mol   1 mol X   = 1.20 mol X   2.016 g  1 mol H2 

( 2.42 g H2 ) 

78.5 g X = 1.20 mol X

78.5 g = 65.4 g/mol 1.20 mol Using the periodic table we find that the element with atomic mass 65.4 is zinc.

– Chapter 9 –

29. Limiting reactant calculation and percent yield SiO2 + 2 C → Si + 2 CO kg SiO2 → g SiO2 → mol SiO2 → mol Si → g Si → kg Si  1000 g   1 mol   1 mol Si   28.09 g   1 kg      = 18.7 kg Si   1 kg   60.09 g   1 mol SiO2   1 mol   1000 g 

( 40.0 kg SiO2 ) 

kg C → g C → mol C → mol Si → g Si → kg Si  1000 g   1 mol   1 mol Si   28.09 g   1 kg     = 29.6 kg Si    1 kg   12.01 g   2 mol C   1 mol   1000 g 

( 25.3 kg C ) 

The limiting reactant is SiO2. 18.7 kg Si is the theoretical yield.  actual yield   14.4 kg  Percent yield =   (100 ) =   (100 ) = 77.0% yield of Si  theoretical yield   18.7 kg  30. Limiting reactant calculation and percent yield 2 Cr2 O3 + 3 Si → 4 Cr + 3 SiO2  1 mol   4 mol Cr   52.00 g     = 239.5 g Cr  152.0 g   2 mol Cr2 O3   1 mol 

(350.0 g Cr2 O3 ) 

 1 mol   4 mol Cr   52.00 g     = 580.0 g Cr  28.09 g   3 mol Si   1 mol 

( 235.0 g Si ) 

The limiting reactant is Cr2O3. 239.5 g Cr is the theoretical yield.  actual yield   213.2 g  Percent yield =   (100 ) =   (100 ) = 89.02% yield  theoretical yield   239.5 g  31. 3 Cu + 8 HNO3 → 3 Cu(NO3)2 + 4 H2O + 2 NO (a) Limiting reactant problem

 1 mol   3 mol Cu ( NO3 )2   187.6 g     = 81.2 g Cu ( NO3 )2 3 mol Cu  63.55 g     1 mol 

( 27.5 g Cu ) 

 1 mol   3 mol Cu ( NO3 )2   187.6 g     = 140 g Cu ( NO3 )2  63.02 g   3 mol HNO3   1 mol 

(125 g HNO3 ) 

Copper is the limiting reactant. 81.2 g of Cu(NO3)2 is the theoretical yield. (b) The excess reactant is HNO3. (c) The percent yield is 87.3%. The actual yield is (0.873)(81.2 g) = 70.9 g Cu(NO3)2.

– Chapter 9 –

32. Fe2O3(s) + 6 HCl(aq) → 2 FeCl3(aq) + 3 H2O(l) (a) Limiting reactant problem

 1 mol   2 mol FeCl3   = 0.44 mol FeCl3   159.7 g   1 mol Fe2 O3 

(35 g Fe2 O3 ) 

 1 mol   2 mol FeCl3    = 0.32 mol FeCl3  36.46 g   6 mol HCl 

(35 g HCl ) 

HCl is the limiting reactant.  1 mol   3 mol H2 O    = 0.48 mol H2 O  36.46 g   6 mol HCl 

(35 g HCl ) 

0.32 mol FeCl3 and 0.48 mol H2O are the theoretical yields. (b) Fe2O3 is the excess reactant.

 1 mol   1 mol Fe2 O3  159.7 g    = 26 g Fe 2 O3 reacted   36.46 g   6 mol HCl   mol 

(35 g HCl ) 

35 g Fe2 O3 − 26 g Fe2 O3 = 9 g Fe2 O3 excess (c) The percent yield is 92.5%. The actual yield is 162.2 g  ( 0.925 )( 0.32 mol FeCl3 )   = 48 g FeCl3  m ol  33. No. There are not enough screwdrivers, wrenches, or pliers. 2400 screwdrivers, 3600 wrenches, and 1200 pliers are needed for 600 tool sets. 34. The balanced equation is C6H12 O6 → 2 C2H5OH+2 CO2

( 575 lb C 6 H12 O6 )  

453.6 g   1 mol   2 mol C 2 H5 OH   46.07 g   1 mL   lb

1L            180.2 g   1 mol C 6 H12 O6   mol   0.789 g   1000 mL 

= 169 L C 2 H5 OH

35. Consider the reaction A→2B and assume that you have 1 gram of A. This does not guarantee that you will produce 1 gram of B because A and B have different molar masses. One gram of A does not contain the same number of molecules as 1 gram of B. However, 1 mole of A does have the same number of molecules as one mole of B. (Remember, 1 mole = 6.022 × 1023 molecules always.) If you determine the number of moles in one gram of A and multiply by 2 to get the number of moles of B . . . then from that you can determine the grams of B using its molar mass. Equations are written in terms of moles not grams. 36. 4 KO2 + 2 H2O + 4 CO2 → 4 KHCO3 + 3 O2

(a)

 0.85 g CO 2   1 mol   4 mol KO 2  0.019 mol KO2 =    min min    44.01 g   4 mol CO 2   0.019 mol KO 2    (10.0 min ) = 0.19 mol KO 2 min  

– Chapter 9 –

(b) The conversion is

g CO2 mol CO 2 mol O 2 g O2 g O2 min → → → → → min g CO 2 mol CO 2 mol O 2 hr hr

 0.85 g CO2   1 mol   3 mol O2   32.00 g  60 min  28 g O2    =   min hr    44.01 g  4 mol CO2   1.0 mol  1.0 hr  H2SO 4 37. C12 H 22 O11 ⎯⎯⎯ →12 C + 11 H 2 O

(a)

  12.01 g  453.6 g   1 mol   12 mol C 2     = 3.8 ×10 g C  lb   342.3 g  1 mol C12H22 O11   mol 

( 2.0 lb C12H22 O11 ) 

453.6 g   1 mol   11 mol H2 O   18.02 g  2    = 5.3 ×10 g H2 O  lb 342.3 g 1 mol C H O mol     12 22 11  

( 2.0 lb C12H22 O11 ) 

From 2.0 lb C12H22O11, 3.8 × 102 g C and 5.3 × 102 g H2O are yielded. (b)

 1 mol C12H22 O11  11 mol H2 O   18.02 g   1 mL      342.3 g    1 mol C12H22 O11   mol   0.994 g 

( 25.2 g C12H22 O11 )  = 14.7 mL H2 O

38. 4 C3H5 ( NO3 )3 ( l ) → 12 CO2 ( g ) + 10 H2 O ( g ) + 6 N2 ( g ) + O2 ( g ) 



 1 mol  12 mol CO  44.01 g  65.0 g C H ( NO ) )    (    = 37.8 g CO (a) 227.1 g 4 mol C H ( NO )  1 mol  3

3 3







3 3





1 mol  10 mol H O  18.02 g    ( 65.0 g C H ( NO ) )  227.1    = 12.9 g H O g 4 mol C H ( NO )  1 mol  3

3 3





   28.02 g  1 mol   6 mol N 2  ( 65.0 g C3H5 ( NO3 )3 )  227.1    = 12.0 g N 2 g   4 mol C3H5 ( NO3 )3   1 mol   3



3 3



1 mol O 1 mol   32.00 g    ( 65.0 g C H ( NO ) )  227.1    = 2.29 g O g 4 mol C H ( NO )  1 mol  3

3 3





3 3



From 65.0 g C5H5(NO3)3, 37.8 g CO2, 12.9 g H2O, 12.0 g N2, and 2.29 g O2 are yielded. 



 1 mol  29 mol gas 65.0 g C H ( NO ) )   = 2.08 mol gas   (b) ( 227.1 g 4 mol C H ( NO )  3

3 3





3 3



39. 2 CH3OH + 3 O2 → 2 CO2 + 4 H2O The conversion is mL CH3OH → g CH3OH → mol CH3OH → mol O2 → g O2

0.787 g   1 mol   3 mol O2   32.00 g      = 70.7 g O2  mL   32.04 g   2 mol CH3OH   mol 

( 60.0 mL CH3OH ) 

– Chapter 9 –

40. The balanced equation is 7 H2O2 + N2H4 → 2 HNO3 + 8 H2O The conversion is  1000 g   1 mol   2 mol HNO3   63.02 g  5  = 2.9 × 10 g HNO3       1 kg   32.05 g   1 mol N 2 H 4   mol 

(a)

( 75 kg N 2H 4 ) 

(b)

( 250 L H2 O2 ) 

1000 mL   1.41 g   1 mol   8 mol H2 O   18.02 g        1 L   1 L   34.02 g   7 mol H2 O2   mol 

= 2.1× 105 g H2 O

(c)

 1 mol  1 mol N 2H4   32.05 g     = 97.6 g N 2H4  34.02 g  7 mol H2 O2   mol 

( 725 g H2O2 ) 

(d) Reaction between 750 g of N2H4 and 125 g of H2O2. Convert each amount to grams of H2O.

 1 mol   8 mol H2 O   18.02 g  3    = 3.4 × 10 g H2 O 32.05 g 1 mol N H mol    2 4 

( 750 g N 2 H4 ) 

 1 mol   8 mol H2 O   18.02 g     = 75.7 g H2 O  34.02 g   7 mol H2 O2   mol 

(125 g H2 O2 ) 

75.7 g H2O can be produced. (e) Since H2O2 is the limiting reactant, N2H4 is in excess.

 1 mol   1 mol N 2H4   32.05 g     = 16.8 g N 2H4 reacted  34.02 g  7 mol H2 O2   mol 

(125g H2 O2 ) 

750 g N2H4 given – 16.8 g N2H4 used = 730 g N2H4 remaining 41. The balanced equation is 16 HCl + 2 KMnO4 → 5 Cl2 + 2 KCl + 2 MnCl2 + 8 H2O (a) Reaction between 25 g KMnO4 and 85 g HCl. Convert each to moles of MnCl2.

 1 mol KMnO 4  2 mol MnCl2    = 0.16 mol MnCl2  158.04 g KMnO 4  2 mol KMnO 4 

( 25 g KMnO4 ) 

 1 mol   2 mol MnCl2    = 0.29 mol MnCl2  36.46 g   16 mol HCl 

(85 g HCl ) 

KMO4 is the limiting reactant; 0.16 mol MnCl2 is produced.  1 mol   8 mol H2 O   18.02 g    = 73 g H2 O   74.55 g   2 mol KCl   mol 

(b)

( 75 g KCl ) 

(c)

(150 g HCl ) 

 1 mol   5 mol Cl2  70.90 g    = 91 g Cl2   36.46 g   16 mol HCl   mol 

 75 g  Theoretical yield is 91 g Cl2; Percent yield =   (100 ) = 82% yield  91 g  15

– Chapter 9 –

(d) Reaction between 25 g HCl and 25 g KMnO4. Convert each amount to grams of Cl2.

 1 mol   5 mol Cl2   70.90 g    = 15 g Cl2   36.46 g   16 mol HCl   mol 

( 25 g HCl ) 

 1 mol   5 mol Cl2   70.90 g     = 28 g Cl2  158.04 g  2 mol KMnO4   mol 

( 25 g KMnO4 ) 

HCl is the limiting reactant; KMnO4 is in excess; 15 g Cl2 will be produced. (e) Calculate the mass of unreacted KMnO4  1 mol   2 mol KMnO 4   158.0 g    = 14 g KMnO 4 will react.   36.46 g   16 mol HCl   mol 

( 25 g HCl ) 

Unreacted KMnO4 = 25 g – 14 g = 11 g KMnO4 remain unreacted. 42. The balance equation is 4 Ag + 2 H2S + O2 → 2 Ag2S + 2 H2O (a)

 1 mol   2 mol Ag 2S   247.9 g     = 1.3 g Ag 2S  107.9 g   4 mol Ag   mol 

(1.1 g Ag ) 

 1 mol   2 mol Ag 2S   247.9 g     = 1.0 g Ag 2S  34.09 g  2 mol H2S   mol 

( 0.14 g H2S) 

 1 mol   2 mol Ag 2S   247.9 g     = 1.2 g Ag 2S  32.00 g  1 mol O2   mol 

( 0.080 g O2 ) 

H2S is limiting reactant; 1.0 g Ag2S forms. (b)

 1 mol  2 mol H2S   34.09 g     = 0.17 g H2S reacts  107.9 g  4 mol Ag   mol 

(1.1 g Ag ) 

0.17 g H2S – 0.14 g H2S = 0.03 grams more H2S needed to completely react with Ag. 43. Limiting reactant calculation and percent yield C3H5 ( C15H31CO2 )3 + 3 KOH → C 3H5 ( OH )3 + 3 C15H31CO2 K 



3 mol C H CO K 1 mol   294.5 g    ( 600.0 g C H ( C H CO ) )  807.3    g 1 mol C H ( C H CO )  1 mol  3

2 3





2 3



= 656.6 g C15H31CO2 K  1 mol   3 mol C15H31CO2 K   294.5 g    = 3149 g C15H31CO2 K  3 mol KOH   1 mol   56.11 g  

( 600.0 g KOH ) 

The limiting reactant is the triglyceride (C3H5(C15H31CO2)3). 656.6 g soap (C15H31CO2K) is the theoretical yield.  actual yield   361.7 g  Percent yield =   (100 ) =   (100 ) = 55.09% yield of soap  theoretical yield   656.6 g 

– Chapter 9 –

44. Equation: CaO + H2O → Ca(OH)2

 1 mol   1 mol Ca ( OH )2   74.10 g     = 46.97 g Ca ( OH )2  56.08 g   1 mol CaO   1 mol 

(35.55 g CaO ) 

45. (a) Pb(NO3)2(aq) + Na2CrO4(aq) → 2 NaNO3(aq) + PbCrO4(s) (b)





1 mol  1 mol PbCrO  323.2 g    ( 26.41 g Pb ( NO ) )  331.2    = 36.57 g PbCrO g 1 mol Pb ( NO )  1 mol  3 2





3 2



 1 mol   1 mol PbCrO 4   323.2 g     = 36.57 g PbCrO 4  162.0 g   1 mol Na 2 CrO 4   1 mol 

(18.33 g Na 2CrO 4 ) 

Theoretical yield is 25.77 g PbCrO4.

 actual yield   21.23 g  Percent yield =   (100 ) =   (100 )  theoretical yield   25.77 g  = 82.38% yield of PbCrO4 46. C2H5OH + 3 O2 → 2 CO2 + 3 H2O (a)



2 mol CO 2   = 5.0 mol CO 2  1 mol C 2 H5 OH 

( 2.5 mol C 2H5OH ) 

 2 mol CO 2   = 5.0 mol CO 2  3 mol O 2 

( 7.5 mol O 2 ) 

Neither reactant is limiting. 

3 mol H 2 O   = 7.5 mol H 2 O  1 mol C 2 H5 OH 

( 2.5 mol C 2H5OH ) 

When the reaction is complete, there will be 5.0 mol CO2 and 7.5 mol H2O. (b)

 1 mol   2 mol CO 2   44.01 g     = 430.g CO 2  46.07 g   1 mol C 2 H5 OH   mol 

( 225 g C 2H5OH ) 

 1 mol   3 mol H 2 O   18.02 g     = 264 g H 2 O  46.07 g   1 mol C 2 H5 OH   mol 

( 225 g C 2H5OH ) 

47. The balanced equation is Zn + 2 HCl → ZnCl2 + H2 180.0 g Zn − 35 g Zn = 145 g Zn reacted with HCl (a)

 1 mol   1 mol H2   2.016 g    = 4.47 g H2 produced   65.39 g   1 mol Zn   mol 

(145 g Zn ) 

– Chapter 9 –

 1 mol   2 mol HCl   36.46 g     = 162 g HCl reacted  65.39 g   1 mol Zn   mol 

(b)

(145 g Zn ) 

(c)

(180.0 g Zn ) 

 1 mol   2 mol HCl   36.46 g     = 201 g HCl reacts  65.39 g   1 mol Zn   mol 

201 g – 162 g = 39 g more HCl needed to react with the 180.0 g Zn 48. Fe ( s )

+ CuSO 4 ( aq ) → Cu ( s ) + FeSO 4 ( aq )

2.0 mol

3.0 mol

(a) 2.0 mol Fe react with 2.0 mol CuSO4 to yield 2.0 mol Cu and 2.0 mol FeSO4. 1.0 mol CuSO4 is unreacted. At the completion of the reaction, there will be 2.0 mol Cu, 2.0 mol FeSO4, and 1.0 mol CuSO4. (b) Determine which reactant is limiting and then calculate the g FeSO4 produced from that reactant.

 1 mol   1 mol Cu  63.55 g     = 22.8 g Cu  55.85 g   1 mol Fe  mol 

( 20.0 g Fe ) 

 1 mol   1 mol Cu   63.55 g     = 15.9 g Cu  159.6 g  1 mol CuSO 4   mol 

( 40.0 g CuSO4 ) 

Since CuSO4 produces less Cu, it is the limiting reactant. Determine the mass of FeSO4 produced from 40.0 g CuSO4.

 1 mol   1 mol FeSO4   151.9 g     = 38.1 g FeSO4 produced  159.6 g  1 mol CuSO4   mol 

( 40.0 g CuSO4 ) 

Calculate the mass of unreacted Fe

 1 mol   1 mol Fe   55.85 g     = 14.0 g Fe will react  159.6 g  1 mol FeSO4   mol 

( 40.0 g CuSO4 ) 

Unreacted Fe = 20.0 g – 14.0 g = 6.0 g. Therefore, at the completion of the reaction, 15.9 g Cu, 38.1 g FeSO4, 6.0 g Fe, and no CuSO4 remain. 49. Limiting reactant calculation CO(g) + 2 H2(g) → CH3OH(l) Reaction between 40.0 g CO and 10.0 g H2: determine the limiting reactant by calculating the amount of CH3OH that would be formed from each reactant.

 1 mol   1 mol CH3OH  32.04 g    = 45.8 g CH3OH   28.01 g   1 mol CO   mol 

( 40.0 g CO ) 

 1 mol   1 mol CH3OH   32.04 g     = 79.5 g CH3OH  2.016 g  2 mol H2   mol 

(10.0 g H2 ) 

CO is limiting; H2 is in excess; 45.8 g CH3OH will be produced.

– Chapter 9 –

Calculate the mass of unreacted H2  1 mol   2 mol H2   2.016 g    = 5.76 g H2 react   28.01 g   1 mol CO   mol 

( 40.0 g CO ) 

10.0 g H2 – 5.76 g H2 = 4.2 g H2 remain unreacted 50. The balanced equation is C6H12O6 → 2 C2H5OH + 2 CO2 (a) First calculate the theoretical yield.  1 mol   2 mol C 2 H5 OH   46.07 g      180.2 g   1 mol C 6H12 O6   mol 

( 750 g C 6H12 O6 ) 

= 3.8 × 10 2 g C 2H5 OH ( theoretical yield )

Then take 84.6% of the theoretical yield to obtain the actual yield. 2 theoretical yield )( 84.6 ) ( 3.8 ×10 g C 2 H5 OH ) ( 84.6 ) ( actual yield = =

100 = 3.2 ×10 g C 2 H5 OH

100

(b) 475 g C2H5OH represents 84.6% of the theoretical yield. Calculate the theoretical yield. theoretical yield =

475 g = 561 g C 2 H5 OH 0.846

Now calculate the g C6H12O6 needed to produce 561 g C2H5OH.

 1 mol   1 mol C 6H12 O6   180.2 g  3    = 1.10 × 10 g C 6H12 O6 46.07 g 2 mol C H O H m o l    2 5 

( 561 g C 2H5OH )  51. (a) MX2

(b) The mole ratio X2:MX2 is 1:2 so 0.342 moles of X2 requires a minimum 0.684 moles of MX2. (c) 8 MX2 theoretically yields 8 MX3 but only 6 MX3 were obtained at the end of the reaction. The actual yield is therefore 75%. 52. The balanced equations are

CaCl2 ( aq ) + 2 AgNO3 ( aq ) → Ca ( NO3 )2 ( aq ) + 2 AgCl ( s ) MgCl2 ( aq ) + 2 AgNO3 ( aq ) → Mg ( NO3 )2 ( aq ) + 2 AgCl ( s ) 1 mol of each salt will produce the same amount (2 mol) of AgCl. MgCl2 has a higher percentage of Cl than CaCl2 because Mg has a lower atomic mass than Ca. Therefore, on an equal mass basis, MgCl2 will produce more AgCl than will CaCl2. Calculations show that 1.00 g MgCl2 produces 3.01 g AgCl, and 1.00 g CaCl2 produces 2.56 g AgCl.

– Chapter 9 –

53. The balanced equation is Li2O + H2O → 2 LiOH The conversion is g H2O → mol H2O → mol LiO2 → g Li2O → kg Li2O

 2500 g H2 O   1 mol   1 mol Li2 O   29.88 g   1 kg  4.1 kg Li2 O     =   astronaut day   18.02 g   1 mol H2 O   mol   1000 g  astronaut day  4.1 kg Li2 O  2   ( 30 days )( 3 astronauts ) = 3.7 ×10 kg Li2 O  astronaut day  54. The balanced equation is H2SO4 + 2 NaCl → Na2SO4 + 2 HCl First calculate the g HCl to be produced

( 20.0 L HCl solution ) 

1000 mL   1.20 g  4   ( 0.420 ) = 1.01× 10 g HCl  1 L   1.00 mL 

Then calculate the g H2SO4 required to produce the HCl 1 mol   1 mol H SO  (1.01×10 g HCl )  36.46   = 139 g H SO g  2 mol HCl  4





Finally, calculate the kg H2SO4 (96%)

 1.00 g H2SO 4 solution   1 kg    = 0.14 kg concentrated H2SO4 0.96 g H2SO4    1000 g 

(1.39 g H2SO4 ) 

55. SiO2 + 3 C → SiC + 2 CO The conversion is kg SiO2 → g SiO2 → mol SiO2 → mol SiC → g SiC → kg SiC  1000 g  1 mol   1 mol SiC   40.10 g   1 kg        1 kg  60.09 g   1 mol SiO2   1 mol   1000 kg 

( 28.1 kg SiO2 )  = 18.75 kg SiC

56. Percent yield of H2SO4

 1 mol   = 3.118 mol S to start with  32.07 g 

(100.0 g S ) 

3.118 mol S → 3.118 mol SO2 − 10% = 2.806 mol SO2

 3.118     −0.3118   2.8062    2.806 mol SO2→2.806 mol SO3 – 10% = 2.525 mol SO3

 2.806     −0.2806   2.5254   

– Chapter 9 –

2.525 mol SO3 → 2.525 mol H2SO4 – 10% = 2.273 mol H2SO4  2.525     −0.2525   2.2725   

( 2.273 mol H2SO 4 ) 

98.09 g   = 223.0 g H2SO 4 formed  mol 

(3.118 mol S ) 

1 mol H2SO 4   = 3.118 mol H 2SO 4 ( theoretical yield )  1 mol S 

 2.273 mol H2SO 4    (100 ) = 72.90% yield  3.118 mol H2SO 4 

Alternate Solution:

Calculation of yield. There are three chemical steps to the formation of H2SO4. Each step has a 10% loss of yield. Step 1:

100% yield – 10% = 90.00% yield

Step 2:

90.00% yield – 10% = 81.00% yield

Step 3:

81.00% yield – 10% = 72.90% yield

Now calculate the grams of product. One mole of sulfur will yield a maximum of 1 mol H2SO4. Therefore 3.118 mol S will give a maximum of 3.118 mol H2SO4.

(3.118 mol S ) 

1 mol H2SO 4   98.09 g   ( 0.7290 ) = 223.0 g H 2SO 4 yield  1 mol    mol 

57. According to the equations, the moles of CO2 come from both reactions and the moles H2O come from only the first reaction. So the mol NaHCO3 = 2 × mol H2O = 2 × 0.0357 mol = 0.0714 mol NaHCO3

( 0.0714 mol NaHCO3 ) 

84.01 g   = 6.00 g NaHCO3 in the sample  mol 

(10.00 g NaHCO3 + Na 2CO3 ) − 6.00 g NaHCO3 = 4.00 g Na 2CO3 in the sample  6.00 g NaHCO3    (100 ) = 60.0% NaHCO3 10.00 g    4.00 g Na 2CO3    (100 ) = 40.0% Na 2 CO3 10.00 g  

58.

– Chapter 9 –

(a) C6H12O6 → 2 C2H5OH + 2 CO2(g) Calculate the grams of C6H12O6 that produced 11.2 g C2H5OH.

 1 mol   1 mol C 6H12 O6   180.1 g      46.07 g   2 mol C 2 H5 OH   1 mol 

(11.2 g C 2H5OH ) 

= 21.9 g C 6H12 O6 25.0 g – 21.9 g = 3.1 g C6H12O6 left unreacted Volume of CO2 produced

 1 mol   2 mol CO2   24.0 L      46.07 g   2 mol C 2H5OH   mol 

(11.2 g C2H5OH ) 

= 5.83 L The assumptions made are that the conditions before and after the reaction are the same and that all reactants went to products. (b) The theoretical yield is

 1 mol   2 mol C 2H5 OH   46.07 g     = 12.8 g C 2 H5 OH  180.1 g   1 mol C 6H12 O6   mol 

( 25.0 g C6H12 O6 ) 

 11.2 g  % yield =   (100 ) = 87.5%  12.8 g  (c) decomposition reaction 59. (a) double decomposition (precipitation) (b) lead(II) iodide (PbI2) (c) Pb(NO3)2(aq) + 2 KI(aq) → 2 KNO3(aq) + PbI2(s) If Pb(NO3)2 is limiting, the theoretical yield is





1 mol  1 mol PbI  461.0 g    ( 25 g Pb ( NO ) )  331.2    = 35 g PbI g 1 mol Pb ( NO )  mol  3 2





3 2



If KI is limiting, the theoretical yield is  1 mol   1 mol PbI 2   461.0 g    = 35 g PbI 2   166.0 g   2 mol KI   mol 

( 25 g KI ) 

 7.66 g  percent yield =   (100 ) = 22%  35 g 

60. (a) Balance the equation 2 XNO3 + CaCl2 → 2 XCl + Ca(NO3)2

 1 mol   2 mol XCl   = 0.555 mol XCl   111.0 g  1 mol CaCl2 

(30.8 g CaCl2 ) 

– Chapter 9 –

Therefore, the molar mass of the XCl is 79.6 g = 143 g/mol 0.555 mol mass of ( X + Cl ) = mass of XCl mass of X = 143 − 35.45 = 107.6 X = Ag ( from periodic table ) (b) No. Ag is below H in the activity series. 61. The balanced equation is 2 KClO3 → 2 KCl + 3 O2 12.82 g mixture – 9.45 g residue = 3.37 g O2 lost by heating Because the O2 lost came only from KClO3, we can use it to calculate the amount of KClO3 in the mixture. The conversion is g O2 → mol O2 → mol KClO3 → g KClO3

 1 mol   2 mol KClO3   122.6 g     = 861 g KClO3 in then mixture  32.00 g  3 mol O 2   mol 

(3.37 g O2 ) 

 8.61 g KClO3    (100 ) = 67.2% KClO3  12.82 g sample  62. The balanced equation is Al(OH)3(s) + 3 HCl(aq) → AlCl3(aq) + 3 H2O(l) The conversion is L HCl → g HCl → mol HCl → mol Al(OH)3 →g Al(OH)3

 2.5 L   3.0 g HCl   1 mol   1 mol Al ( OH )3   78.00 g        = 5.3 g Al ( OH )3 /day L   36.46 g   3 mol HCl   mol   day   Now calculate the number of 400. mg tablets that can be made from 5.3 g A1(OH)3

  5.3 g Al ( OH )3   1000 mg   1 tablet  = 13 tablets/day     day g   400. mg Al ( OH )3    64. 4 P + 5 O2 → P4O10 P4O10 + 6 H2O → 4 H3PO4 In the first reaction

 1 mol   = 0.646 mol P  30.97 g 

( 20.0 g P ) 

 1 mol   = 0.938 mol O2  32.00 g 

(30.0 g O2 ) 

This is a ratio of

0.646 mol P 3.44 mol P = 0.938 mol O2 5.00 mol O2

– Chapter 9 –

Therefore, P is the limiting reactant and the P4O10 produced is

( 0.646 mol P ) 

1 mol P4 O10   = 0.162 mol P4 O10  4 mol P 

In the second reaction  1 mol   = 0.832 mol H2 O  18.02 g 

(15.0 g H2 O ) 

and we have 0.162 mol P4O10. The ratio of

H2 O 0.832 mol 5.14 mol is = P4 O10 0.162 mol 1.00 mol

Therefore, H2O is the limiting reactant and the H3PO4 produced is

 4 mol H3PO4   97.99 g    = 54.4 g H3PO4  6 mol H2 O   mol 

( 0.832 mol H2 O ) 

65. 2.31 g 2.000 gallons is equal to 7568 mL which has a density of 1.018 g/mL, The mass of the solution therefore is

(1.018 g/ mL ) ( 7568 mL ) = 7704 g Of this mass, 7704 g, 300. ppm are silver ions. To calculate how many grams of silver might be reclaimed with no loss, we set up the following equation.   1   300. ppm ( 7704 g ) = 2.31 g ( 3 significant figures ) 6  1×10 ppm   

(

)

66. Limiting reactant-like calculation Calculate the number of bags of supplement that can be made from each starting material. 25 lb oats + 17.5 lb molasses + 30.0 lb alfalfa + 28.5 lb apples → 1 bag supplement

( 4.20 tons oats ) 

2000 lb  1 bag    = 336 bags supplement  1 ton  25 lb oats   3.785 L   1000 mL  1.46 g   1 lb   1 bag       1 gal 1 L 1 mL 453.6 g 17.5 lb molasses      

( 2490 gal molasses ) 

= 1730 bags supplement   12.5 kg apple   1 lb 1 bag    1 crate    0.4536 kg  28.5 lb apple 

( 250.0 crates apple ) 

= 242 bags supplement  62.8 kg alfalfa   1 lb 1 bag     1 bale    0.4536 kg   30.0 lb alfalfa 

(350.0 bales ) 

= 1620 bags supplement The limiting reactant is apples. 242 bags of feed supplement is the theoretical yield.  $2.25  Income = ( 242 bag )   = $545  1 bag  24

CHAPTER 10

MODERN ATOMIC THEORY AND THE PERIODIC TABLE SOLUTIONS TO REVIEW QUESTIONS 1.

Wavelength is defined as the distance between consecutive peaks in a wave. It is generally symbolized by the Greek letter lambda, λ. Frequency is a measure of the number of waves that pass a specific point every second. It is generally symbolized by the Greek letter nu, ν.

Visible light ranges in wavelength from about 4 × 10−7 m to 7 × 10−7 m. Red light has a longer wavelength than blue light.

Photon

An electron orbital is a region in space around the nucleus of an atom where an electron is most probably found.

The electrons in the atom are located in the orbitals with the lowest energies.

(a) In the Bohr model, when an electron absorbs energy, it makes a quantized jump to a higher energy level. (b) When an electron loses energy and makes the quantized jump to a lower energy level, the excess energy is given off as light energy.

The main difference is that the Bohr orbit has an electron traveling a specific path around the nucleus while an orbital is a region in space where the electron is most probably found.

Bohr’s model was inadequate since it could not account for atoms more complex than hydrogen. It was modified by Schrödinger into the modern concept of the atom in which electrons exhibit wave and particle properties. The motion of electrons is determined only by probability functions as a region in space, or a cloud surrounding the nucleus.

Both 1s and 2s orbitals are spherical in shape and located symmetrically around the nucleus. The sizes of the spheres are different—the radius of the 2s orbital is larger than the 1s. The electrons in 2s orbitals are located further from the nucleus.

10. The letters used to designate the energy sublevels are s, p, d, and f. 11. 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p 12. s–2 electrons per shell p–6 electrons per shell after the first energy level d–10 electrons per shell after the second energy level 13.

– Chapter 10 –

14. A second electron may enter an orbital already occupied by an electron if its spin is opposite that of the electron already in the orbital and all other orbitals of the same sublevel contain an electron. 15. The valence shell is the outermost energy level of an atom. 16. Valence electrons are the electrons located in the outermost energy level of an atom. Valence electrons are involved in bonding. They are important because ion formation involves the gain or loss of valence electrons. Covalent bonding involves sharing valence electrons. 17. 4 is the fourth principal energy level f indicates the energy sublevel 3 indicates the number of electrons in the f sublevel 18. Ir, Zr, and Ag are not representative elements; they are transition elements. 19. Elements in the p-block all have one to six electrons in the p sublevel. 20. Atomic #

Symbol

Elements with atomic numbers 7, 15, and 33 are all in the same group on the periodic table. They have an outermost electron structure of s2p3. 21. The first three elements that have six electrons in their outermost energy level are O, S, and Se. 22. The greatest number of elements in any period is 32. The 6th period has this number of elements. 23. The elements in Group A always have their last electrons in the outermost energy level, while the last electrons in Group B lie in an inner level. 24. Pairs of elements which are out of sequence with respect to atomic masses are: Ar and K; Co and Ni; Te and I; Th and Pa; U and Np; Pu and Am; Lr and Rf; Sg and Bh. 25. Dimitri Mendeleev, of Russia and Lothar Meyer, of Germany both independently published results that led to the current periodic table. 26. Dimitri Mendeleev is credited with being the father of the modern periodic table.

– Chapter 10 –

SOLUTIONS TO EXERCISES 1. Element

Total Electrons

Valence Electrons

(a) Li

(b) Mg

(d) F

Total Electrons

Valence Electrons

(a) Na

(b) As

(d) Al

2. Element

Electron configurations (a)

1s22s22p63s23p64s23d1

(b)

1s22s22p63s23p64s23d104p65s1

(c)

1s22s22p63s23p64s23d104p5

(d)

1s22s22p63s23p4

(a) Mn

1s22s22p63s23p64s23d5

(b)

1s22s22p63s23p64s23d104p6

(c)

1s22s22p63s23p64s23d104p1

(d)

1s22s22p1

The spectral lines of hydrogen are produced by energy emitted when the electron from a hydrogen atom, which has absorbed energy, falls from a higher energy level to a lower energy level (closer to the nucleus).

Bohr said that a number of orbits were available for electrons, each corresponding to an energy level. When an electron falls from a higher energy orbit to a lower energy orbit, energy is given off as a specific wavelength of light. Only those energies in the visible range are seen in the hydrogen spectrum. Each line corresponds to a change from one orbit to another.

16 orbitals in the 4th principal energy level; 1 in s, 3 in p, 5 in d, and 7 in f. The s and p orbitals are in the 4th period, the d orbitals are in the 5th period, and the f orbitals are in the 6th period.

18 electrons in the third energy level; 2 in s, 6 in p, 10 in d

– Chapter 10 –

(a) (b) (c) (d)

(e)

10. (a) (b) (c) (d) (e) 11. (a) O

1s22s22p4

(b) Ca 1s22s22p63s23p64s2 (c) Ar 1s22s22p63s23p6 (d) Br 1s22s22p63s23p64s23d104p5 (e) Fe 1s22s22p63s23p64s23d6 12. (a) Li 1s22s1 (b) P

1s22s22p63s23p3

1s22s22p63s23p64s1

13. (a) Incorrect – the 2p sublevel should be completely filled before the 3s sublevel is populated. (b) Correct (c) Correct (d) Incorrect – electrons in the 3d sublevel should not be paired until all 3d orbitals are populated.

– Chapter 10 –

14. (a) Correct (b) Incorrect – the 3d sublevel should be populated before the 4p sublevel. (c) Incorrect – the second electron in the 4s orbital must be represented by a down arrow (d) Correct 15. (a) Neon

(c)

Gallium

(d)

Manganese

16. (a) Nitrogen

(c)

Calcium

(b) Nickel

(d)

Sulfur

19. (a) oxygen, O

(c)

chlorine, Cl

(b) sodium, Na

(d)

chromium, Cr

20. (a) nitrogen, N

(c)

radon, Rn

(d)

antimony, Sb

(b) Phosphorus

17.

18.

(b) aluminum, Al 21.

22.

– Chapter 10 –

23.

24.

25.

32 16

(b)

60 28

26. (a) (b)

27 13

48 22

27. The eleventh electron of sodium is located in the third energy level because the first and second levels are filled. Also the properties of sodium are similar to the other elements in Group 1A. 28. The last electron in potassium is located in the fourth energy level because the 4s orbital is at a lower energy level than the 3d orbital. Also the properties of potassium are similar to the other elements in Group 1A. 29. Noble gases all have filled s and p orbitals in the outermost energy level. 30. Noble gases each have filled s and p orbitals in the outermost energy level.

– Chapter 10 –

31. Moving from left to right in any period of elements, the atomic number increases by one from one element to the next and the atomic radius generally decreases. Each period (except period 1) begins with an alkali metal and ends with a noble gas. There is a trend in properties of the elements changing from metallic to nonmetallic from the beginning to the end of the period. 32. The elements in a group have the same number of outer energy level electrons. They are located vertically on the periodic table. 33. (a)

(b)

(c)

(d)

(e)

(a)

(b)

(c)

(d)

(e)

34. 35. The outermost energy level contains one electron in an s orbital. 36. All of these elements have a s2d10 electron configuration in their outermost energy levels. 37. (a) and (g) (b) and (d) 38. (a) and (f) (e) and (h) 39. 12, 38 since they are in the same group or family of elements. 40. 7, 33 since they are in the same group or family of elements. 41. (a) K, metal

(b) Pu, metal

(d) Sb, metalloid

(a) I, non-metal

(b) W, metal

(d) Ge, metalloid

42.

43. (a) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p4 (b) [Kr] 5s2 4d10 5p4 (c) 6 (d) [Kr] 5s2 4d10 5p4 44. (a) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3 (b) [Ar] 4s2 3d10 4p4 (c) 5 (d) [Ar] 4s2 3d10 4p3 45. Period 6, lanthanide series, contains the first element with an electron in an f orbital. 46. Period 4, Group 3B, contains the first element with an electron in a d orbital. 7

– Chapter 10 –

47. Group 7A contains 7 valence electrons. Group 7B contains 2 electrons in the outermost level and 5 electrons in an inner d orbital. Group A elements are representative while Group B elements are transition elements. 48. Group 3A contains 3 valence electrons. Group 3B contains 2 electrons in the outermost level and one electron in an inner d orbital. Group A elements are representative while Group B elements are transition elements. 49. (a) arsenic

(c)

lithium

(b) cobalt

(d)

chlorine

(a) lead

(c)

gallium

(b) samarium

(d)

iridium

50.

51. The valence energy level of an atom can be determined by looking at what period the element is in. Period 1 corresponds to valence energy level 1, period 2 to valence energy level 2 and so on. The number of valence electrons for elements 1–18 can be determined by looking at the group number. For example, boron is under Group 3A, therefore it has three valence shell electrons. 52. (a) Mg (b) P (c) K (d) F (e) Se (f)

53. (a) Na+, (d) F−, and (e) Ne all have 8 valence electrons. 54. (a) 7A, Halogens

(d)

8A, Noble Gases

(b) 2A, Alkaline Earth Metals

(e)

8A, Noble Gases

(f)

1A, Alkali Metals

– Chapter 10 –

55. a.

c. d. e.

f. 56. a.

57. (a) No, the electronic configuration predicted by the periodic table is 1s2 2s2 2p6 3s2 3p6 4s2 3d4. (b)

23 3  7.19 g   1 mol   6.022 ×10 atoms  23 5.00 cm ( )  1 cm 3   52.00 g    = 4.16 × 10 atoms Cr 1 mol   

3 4 4 (c) V = π r 3 = π (1.40 ×10 −8 ) = 1.15 ×10 −23 cm 3 3 3

(d)

atom  (5.00 cm )  1.5 ×1 10  = 4.35 ×10 atoms Cr cm  3

−23

58. Each of the different elements has a characteristic emission spectra which will be observed as different colors in the fireworks. 59. Sb 1s22s22p63s23p64s23d104p65s24d105p3 or [Kr] 5s24d105p3 60. Bi

1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p3 or [Xe] 6s24f145d106p3

61. (a) The four most abundant elements in the earth’s crust, seawater, and air are: O: 1s22s22p4

Si: 1s22s22p63s23p2

Fe: 1s22s22p63s23p64s23d6

Al: 1s22s22p63s23p1

– Chapter 10 –

(b) The five most abundant elements in the human body are: O: 1s22s22p4

C: 1s22s22p4

H: 1s1 N: 1s22s22p3

Ca: 1s22s22p63s23p64s2 62. Maximum number of electrons (a) Any orbital can hold a maximum of two electrons. (b) A d sublevel can hold a maximum of 10 electrons. (c) The third principal energy level can hold two electrons in 3s, six electrons in 3p, and 10 electrons in 3d for a total of 18 electrons. (d) Any orbital can hold a maximum of two electrons. (e) An f sublevel can hold a maximum of 14 electrons. 63. Name of elements (a) Magnesium

(b)

Phosphorus

64. Nitrogen has more valence electrons on more energy levels than hydrogen. More varied electron transitions are possible. 65. (a) Ne (b) Ge (c) F (d) N 66. The outermost electron structure for the elements in 7A is s2p5. 67. (a) Only 2 electrons are allowed in an orbital. (b) Electrons will fill all orbitals before they pair. 68. When an electron drops from a higher energy level (excited state) to a lower energy level it releases energy in the form of light. The line spectra represent these light emissions. 69. The ground state is represented by the electron configuration of an atom with all electrons in the lowest possible energy states. Ground state of carbon 1s2 2s2 2p2 Possible excited state of carbon 1s2 2s2 2p1 3s1 70. Transition elements are found in Groups 1B–8B, lanthanides and actinides. 71. In transition elements the last electron added is in a d or f orbital. The last electron added in a representative element is in an s or p orbital. 72. Using the periodic table in Chapter 10 of the textbook, find the following items of information about each element. Symbol

Atomic Number

Atomic Mass

No. of Valence Electrons

(1) Phosphorus

30.97

(2) Fluorine

19.00

(3) Mercury

200.6

(4) Cesium

132.9

73. Elements 7, 15, 33, 51, and 83 all have 5 electrons in their valence shell.

– Chapter 10 –

74. Family names (a) Alkali Metals

(b)

Alkaline Earth Metals

(c)

Halogens

75. Sublevels (a) sublevel p

(b)

sublevel d

(c)

sublevel f

76. (a) Na

representative element

Metal

(b) N

representative element

Non-metal

Metal

(d) Ra

representative element

Metal

(e) As

representative element

metalloid

(f)

Noble gas

nonmetal

77. If element 36 is a noble gas, 35 would be in periodic Group 7A and 37 would be in periodic Group 1A. 78. Answers will vary but should at least include a statement about: (1) Numbering of the elements and their relationship to atomic structure; (2) division of the elements into periods and groups; (3) division of the elements into metals, nonmetals, and metalloids; (4) identification and location of the representative and transition elements. 79. (a) The two elements are isotopes. (b) The two elements are adjacent to each other in the same period. 80. Most gases are located in the upper right part of the periodic table (H is an exception). They are nonmetals. Liquids show no pattern. Neither do solids, except the vast majority of solids are metals. 81. excited sulfur atom: electron configuration: 1s22s22p63s13p5 orbital diagram: 82. Electrons are located in seven principal energy levels. The outermost energy level has one electron residing in a 7s orbital. 83. Metals are located on the left side of the periodic table. The elements in Group 1A have only one valence electron and those in Group 2A have only two valence electrons. All metals easily lose their valence electrons to obtain a Noble Gas configuration. Nonmetals are located on the right side of the periodic table where they are only a few electrons short of a noble gas configuration. Nonmetals gain valence electrons to obtain a noble gas configuration.

– Chapter 10 –

84. On the periodic table, the period number corresponds to the principal energy level in which the s and p sublevels are filling. The group number of the Main Representative elements corresponds to the number of electrons filling in the principal energy level. Groups 1A and 2A are known as the s-block elements and Groups 3A through 8A are known as the p-block elements. 85. The compound will be ionic because there is a very large difference in electronegativity between elements in Group 2A and those in Group 7A of the periodic table. The Lewis structure is

86. Having an even atomic number has no bearing on electrons being paired. An even atomic number means only that there is an even number of electrons. For example, carbon is atomic number six, and it has two unpaired p electrons: 1s 2 2s 2 2 p1x 2 p1y .

CHAPTER 11

CHEMICAL BONDS: THE FORMATION OF COMPOUNDS FROM ATOMS SOLUTIONS TO REVIEW QUESTIONS 1.

smallest Cl, Mg, Na, K, Rb largest

More energy is required for neon because it has a very stable outer electron structure consisting of an octet of electrons in filled orbitals (noble gas electron structure). Sodium, an alkali metal, has a relatively unstable outer electron structure with a single electron in an unfilled orbital. The sodium electron is also farther away from the nucleus and is shielded by more inner electrons than are neon outer electrons.

When a third electron is removed from beryllium, it must come from a very stable electron structure corresponding to that of the noble gas, helium. In addition, the third electron must be removed from a +2 beryllium ion, which increases the difficulty of removing it.

The first ionization energy decreases from top to bottom because in the successive alkali metals, the outermost electron is farther away from the nucleus and is more shielded from the positive nucleus by additional electron energy levels.

The first ionization energy decreases from top to bottom because the outermost electrons in the successive noble gases are farther away from the nucleus and are more shielded by additional inner electron energy levels.

Helium has a much higher first ionization energy than does hydrogen because helium has a very stable outer electron structure, consisting of a filled principal energy level.

After the first electron is removed from an atom of lithium, much more energy would be required to remove a second electron since that one would come from a noble gas electron configuration, a filled principal energy level.

(a)

Li > Be

(d)

Cl > F

(b)

Rb > K

(e)

Se > 0

(c)

Al > P

(f)

As > Kr

The first element in each group has the smallest radius.

10. Atomic size increases down a column since each successive element has an additional energy level which contains electrons located farther from the nucleus. 11. By losing one electron, a potassium atom acquires a noble gas structure and becomes a K+ ion. To become a K2+ ion requires the loss of a second electron and breaking into the noble gas structure of the K+ ion. This requires too much energy to generally occur. 12. An aluminum ion has a +3 charge because it has lost 3 electrons in acquiring a noble gas electron structure.

– Chapter 11 –

13. A Lewis structure is a representation of the bonding in a compound. Valence electrons are responsible for bonding, therefore they are the only electrons that need to be shown in a Lewis structure. 14. Valence electrons are the electrons found in the outermost s and p energy levels of an atom. 15. Metals are less electronegative than nonmetals. Therefore, metals lose electrons more easily than nonmetals. So, metals will transfer electrons to nonmetals leaving the metals with a positive charge and the nonmetals with a negative charge. 16. The noble gases are the most stable of all the elements because they have a complete octet (8 electrons) in their valence level. When the elements in Groups 1A(1), 2A(2), 6A(16), and 7A(17) form ions, they do so to establish a stable electron structure. (a) The elements in Group 1A(1) lose an electron (obtain a positive charge) in order to achieve a noble gas electron configuration. (b) The elements in Group 2A(2) lose electrons (obtain a positive charge) in order to achieve a noble gas electron configuration. (c) The elements in Group 6A(16) gain electrons (obtain a negative charge) in order to achieve a noble gas electron configuration. (d) The elements in Group 7A(17) gain an electron (obtain a negative charge) in order to achieve a noble gas electron configuration. 17. All compounds must be neutrally charged. This means the overall charge on an ionic compound must be zero. 18. The alkaline earth elements. 19. Mg3(PO4)2, Be3(PO4)2, Sr3(PO4)2, and Ba3(PO4)2. Note that the basic formula is the same for all of the elements in the same family when they form an ionic compound with a phosphate ion. 20. When magnesium loses two electrons it will achieve the same electron configuration as the noble gas neon. Elements tend to gain or lose electrons to achieve a noble gas electron configuration. 21. The term molecule is used to describe covalent compounds that have a distinct set of atoms in their structure. Ionic compounds are composed of collections of ions which come together in ratios that balance their charges, but there are no independent molecules formed. 22. A single covalent bond is composed of two electrons. A maximum of three covalent bonds may be formed between any two atoms. 23. A covalent bond is formed by the overlap of the orbitals on individual atoms. There is a sharing of electrons in the region where the orbitals overlap. 24. In a Lewis structure the dash represents a two electron covalent bond. 25. Not all molecules that contain polar bonds are polar due to dipoles that cancel each other by acting in equal and opposite direction.

– Chapter 11 –

26. The more electronegative atom in the bond between two atoms will more strongly attract electrons so it will have a partial negative charge (δ−). The less electronegative atom will have a partial positive charge (δ+) because the bonding pair of electrons has been pulled away by the more electronegative atom. 27. (a) Elements with the highest electronegativities are found in the upper right hand corner of the periodic table. (b) Elements with the lowest electronegativities are found in the lower left of the periodic table. 28. A Lewis structure is a visual representation of the arrangement of atoms and electrons in a molecule or an ion. It shows how the atoms in a molecule are bonded together. 29. The dots in a Lewis structure represent electrons. The lines represent bonding pairs of electrons. 30. There are times when two or more Lewis structures can be drawn for a single skeleton structure. These different structures all exist and are called resonance structures. 31. Resonance structures 32. The Lewis structure for an ion should be drawn inside of a set of square brackets and the charge of the entire ion specified at the upper right-hand outer corner of the brackets. An example is [Na+]. 33. Na2SO4 and CaCO3 are two of many possible examples of compounds with both ionic and covalent bonds. 34. The electron pair arrangement is the arrangement of both bonding and nonbonding electrons in a Lewis structure. The molecular shape of a molecule is the three dimensional arrangement of its atoms in space.

– Chapter 11 –

SOLUTIONS TO EXERCISES 1.

Drawing 1

K+ ion

Drawing 2

K atom

Drawing 1

Cl atom

Drawing 2

Cl− ion

Both particles have the same number of protons, so K+ with one fewer electron is smaller.

Both particles have the same number of protons, so Cl− with one more electron is larger.

(a) A calcium atom is larger because it has electrons in the 4th shell, while a calcium ion does not. In addition, Ca2+ ion has 20 protons and 18 electrons, creating a charge imbalance and drawing the electrons in towards the positive nucleus. (b) A chloride ion is larger because it has one more electron than a chlorine atom, in its outer shell. Also, the ion has 17 protons and 18 electrons, creating a charge imbalance, resulting in a lessening of the attraction of the electrons towards the nucleus. (c) A magnesium ion is larger than an aluminum ion. Both ions will have 10 electrons in their electron shells, but the aluminum ion will have a greater charge imbalance since it has 13 protons and the magnesium ion has 12 protons. The charge imbalance draws the electrons in more closely to the nucleus. (d) A sodium atom is larger than a silicon atom. Sodium and silicon are both in period 3. Going across a period, the radii of atoms decrease. (e) A bromide ion is larger than a potassium ion. The bromide ion has 35 protons and 36 electrons, creating a charge imbalance that results in a lessening of the attraction of the electrons towards the nucleus. The potassium ion has 19 protons and 18 electrons, creating a charge imbalance that results in the electrons being drawn more closely to the nucleus.

(a) Fe2+ has one electron more than Fe3+, so it will have a larger radius. (b) A potassium atom is larger than a potassium ion. They both have 19 protons. The potassium atom has 19 electrons. The potassium ion has 18 electrons, creating a charge imbalance that results in the electrons being drawn more closely to the nucleus. (c) A chloride ion is larger than a sodium ion. The chloride ion has 17 protons and 18 electrons, creating a charge imbalance that results in a lessening of the attraction of the electrons towards the nucleus. The sodium ion has 11 protons and 10 electrons, creating a charge imbalance that results in the electrons being drawn more closely to the nucleus.

– Chapter 11 –

(d) A strontium atom is larger than an iodine atom. Strontium and iodine are both in period 5. Going across a period, the radii of atoms decrease. (e) A rubidium ion is larger than a strontium ion. Both ions will have 36 electrons in their electron shells, but the strontium ion will have a greater charge imbalance since it has 38 protons and the rubidium ion has 37 protons. The charge imbalance draws the electrons in more closely to the nucleus. 5.

Cations form when atoms lose their valence electrons. Cations have fewer electrons than the neutral atom. In cations, the nucleus is attracting fewer electrons than in a neutral atom. These fewer electrons are held more closely. Additionally, the electron(s) lost by a cation are often in a higher energy level. Emptying this higher energy level also results in a smaller size.

Anions form when atoms gain extra electrons. Anions have more electrons than a neutral atom. In anions the nucleus is unable to hold the larger number of electrons as tightly, thus the size increases.

Ionization energy is the energy required to remove an electron from an atom. As you move across the periodic table from left to right, the ionization energy of the elements increase. As you move down the periodic table, the ionization energy of the elements decreases.

Electronegativity is the relative attraction that an atom has for a pair of shared electrons in a covalent bond. As you move across the periodic table from left to right, the electronegativity of the elements increases. As you move down the periodic table, the electronegativity of the elements decreases.

9. +

−

(a) H

(d)

(b) Rb

(e)

(c)

(f)

−

(a) H

(d)

(b) S

(e)

(c)

(f)

(a)

covalent

(b)

ionic

(d)

covalent

(a)

ionic

(b)

covalent (c) covalent (d)

covalent

10.

11. (c) ionic

12. 13. (a) Mg → Mg2+ + 2 e− (b) Br + l e− → Br− 14. (a) K → K+ + l e−

(b) S + 2 e− → S2−

– Chapter 11 –

15. (a) Se (6)

(b) P (5)

(d) Mg (2)

(e) He (2)

(f) As (5)

16. (a) Pb (4)

(b) Li (1)

(d) Cs (1)

(e) Ga (3)

(f) Ar (8)

17. Noble gas structures: (a) potassium atom, lose 1 e−

(c)

bromine atom, gain 1 e−

(b) aluminum ion, none

(d)

selenium atom, gain 2 e−

(a) sulfur atom, gain 2 e−

(c)

nitrogen atom, gain 3e−

(d)

iodide ion, none

18. (b) calcium atom, lose 2 e− 2

19. (a) 1s 2s 2p 3s 3p 4s 3d

(c) The valence electrons in the 4s sublevel are lost. (d) 1s2 2s2 2p6 3s2 3p6 3d3 20. (a) 1s2 2s2 2p6 3s2 3p6 4s2 3d5 (c) The valence electrons in the 4s sublevel are lost. (d) 1s2 2s2 2p6 3s2 3p6 3d5 21. (a) ionic, NaCl

(b)

molecular, CH4

(d) molecular, Br2

(e)

molecular, CO2

(c)

ionic, MgBr2

22. (a) Two oxygen atoms will form a nonpolar covalent compound. The formula is O2. (b) Hydrogen and bromine will form a polar covalent compound. The formula is HBr. (c) Oxygen and two hydrogen atoms will form a polar covalent compound. The formula is H2O. (d) Two iodine atoms will form a nonpolar covalent compound, the formula is I2. 23. (a) NaH, Na2O

(c)

A1H3, A12O3

(b) CaH2, CaO

(d)

SnH4, SnO2

(a) SbH3, Sb2O3

(c)

HCl, C12O7

(b) H2Se, SeO3

(d)

CH4, CO2

(a) covalent

(c)

ionic

(b) ionic

(d)

covalent

24.

25.

– Chapter 11 –

26. (a) covalent

(c)

covalent

(b) ionic

(d)

covalent

27. (a) covalent

(b)

covalent

(c)

covalent

28. (a) covalent

(b)

covalent

(c)

covalent

29. (a)

(b)

(c)

30. (a) 31.

(b)

(c)

(a)

(c)

(b) (d) 32. (a)

(c)

(d) (b) 33. (a) [Ba]2+

(d) (e)

(b) [Al]3+ (c)

34. (d) (a) (e) (b) (c)

– Chapter 11 –

35. (a) (b) 36. (a)

(b) 37. (a) CH3Cl, polar

(b) Cl2, non polar

(d) PBr3, polar

(a) H2, nonpolar (b) NI3, polar

38.

39. (a) 4 electron pairs, tetrahedral (b) 4 electron pairs, tetrahedral (c) 3 electron pairs, trigonal planar 40. (a) 3 electron pairs, trigonal planar (b) 4 electron pairs, tetrahedral (c) 4 electron pairs, tetrahedral 41. (a) tetrahedral (b) trigonal pyramidal

42. (a) tetrahedral (b) trigonal pyramidal 43.

Electron pair arrangement

Molecular structure

S (a)

tetrahedral

bent

C (b)

trigonal planar

N (c)

tetrahedral

trigonal pyramidal

C (d)

linear

Electron pair arrangement

Molecular structure

S (a)

trigonal planar

bent

C (b)

linear

N (c)

tetrahedral

O (d)

tetrahedral

bent

44.

– Chapter 11 –

45. (a) tetrahedral

(b) trigonal pyramidal

46. (a) tetrahedral

(b) bent

47. Oxygen 48. Potassium 49. Atom 1

Argon

Atom 2

Sodium

Atom 3

Cesium

Drawing 1

Sr2+

Drawing 2

Rb+

Drawing 3

All of these particles are isoelectronic, meaning they have the same number of electrons. The more protons in their nucleus, the smaller the particle.

Drawing 4

Br− ion

Drawing 5

Se2− ion

50.

51. (a) Cl− (b) Al (c) Na (d) Fe2+ 52. The outermost electron in potassium is farther away from the nucleus than the outermost electrons in calcium, so the first ionization energy of potassium is lower than that of calcium. However, once potassium loses one electron, it achieves a noble gas electron configuration and, therefore, taking a second electron away requires considerably more energy. For calcium, the second electron is still in the outermost shell and does not require as much energy to remove it. 53. The ionization energy is the energy required to remove an electron. A chlorine atom forms a chloride ion by gaining an electron to achieve a noble gas configuration. 54. The anion is Cl−; therefore, the cation is K+ and the noble gas is Ar. K+ has the smallest radius, while Cl− will have the largest. K loses an electron, and therefore, in K+, the remaining electrons are pulled in even closer. Cl was originally larger than Ar, and gaining an electron means that, since the nuclear charge is exceeded by the number of electrons, the radius will increase relative to a Cl atom. 55. hydrazine N2H4 14 e−

hydrozoic acid HN3 16 e−

– Chapter 11 –

56. (a)

NO 2− 18 e −

bent

(b) SO24− 32 e −

tetrahedral

trigonal pyramidal

(d) Cl2O 20 e−

bent

57.

(a) C2H6 14 e−

(b) C2H4 12 e− (c) C2H2 10 e− 58. (a) Be

(b) He

59. (a) Cl

(b) O

(d) F

(e) Fr

(f) Ne

60. The noble gases already have a full outer electron configuration and therefore there is no need for them to attract electrons. 61. Lithium has a +1 charge after the first electron is removed. It takes more energy to overcome that charge and to remove another electron than to remove a single electron from an uncharged He atom, 62. Yes. Each of these elements have an ns1 electron and they could lose that electron in the same way elements in Group 1A do. They would then form +1 ions and ionic compounds such as CuCl, AgCl, and AuCl. 63. SnBr2, GeBr2 64. The bond between sodium and chlorine is ionic. An electron has been transferred from a sodium atom to a chlorine atom. The substance is composed of ions not molecules. Use of the word molecule implies covalent bonding. 65. A covalent bond results from the sharing of a pair of electrons between two atoms, while an ionic bond involves the transfer of one or more electrons from one atom to another.

– Chapter 11 –

66. This structure shown is incorrect since the bond is ionic. It should be represented as:

67. The four most electronegative elements are F, O, N, Cl. 68. highest F, O, S, H, Mg, Cs lowest 69. It is possible for a molecule to be nonpolar even though it contains polar bonds. If the molecule is symmetrical, the polarities of the bonds will cancel (in a manner similar to a positive and negative number of the same size) resulting in a nonpolar molecule. An example is CO2 which is linear and nonpolar. 70. Both molecules contain polar bonds. CO2 is symmetrical about the C atom, so the polarities cancel. In CO, there is only one polar bond, therefore the molecule is polar. 71. (a) NO < CO < NaO

(b)

GeO < SiO < CO

(c)

BSe < BS < BO

72. (a) 109.5° (actual angle closer to 105°)

(b)

109.5° (actual angle closer to 107°)

(d)

109.5°

73. The structure shown in the question implies covalent bonds between Al and F, since the lines represent shared electrons. Solid AlF3 is an ionic compound and, therefore, probably exists as an Al3+ ion and three F− ions. Only valence electrons are shown in Lewis structures. 74. Carbon has four valence electrons; it needs four electrons to form a noble gas electron structure. By sharing four electrons, a carbon atom can form four covalent bonds. 75. The atom is Br (35e−), which should form a slightly polar covalent bond with sulfur. The Lewis structure of Br is: 76. (a) Both use the p orbitals for bonding. B uses one s and two p orbitals while N uses one s and three p orbitals for bonding. (b) BF3 is trigonal planar while NF3 is trigonal pyramidal. (c) BF3 has no lone pairs NF3 has one lone pair (d) BF3 has 3 very polar covalent bonds. NF3 has 3 polar covalent bonds 77. Fluorine’s electronegativity is greater than any other element. Ionic bonds form between atoms of widely different electronegativities. Therefore, Fr–F, Cs–F, Rb–F, or K–F would be ionic substances with the greatest electronegativity difference.

– Chapter 11 –

78. S

1.40 g = 0.0437 mol 32.07 g/mol

0.0437 = 1.00 0.0437

2.10 g = 0.131 mol 16.00 g/mol

0.131 = 3.00 0.0437

Empirical formula is SO3

79. We need to know the molecular formula before we can draw the Lewis structure. From the data, determine the empirical and then the molecular formula. C

14.5 g 1.21 = 1.21 mol = 1.00 12.01 g/mol 1.21 Cl 85.5 g 2.41 = 2.01 = 2.41 mol 35.45 g/mol 1.21 CCl2 is the empirical formula

empirical mass = 1(12.01 g ) + 2 ( 35.45 g ) = 82.91 g 166 g = 2.00 82.91 g Therefore, the molecular formula is (CCl2)2 or C2Cl4

80. (a) ionic

(b) both

(d) covalent

81. (a)

ionic;

sodium phosphide

(b)

both;

ammonium iodide

(c)

covalent;

sulfur dioxide

(d)

covalent;

hydrogen sulphide

(e)

both;

copper(II) nitrate

(f)

ionic;

magnesium oxide

– Chapter 11 –

82. (a)

(d)

(e)

(b)

(c)

83.

(f)

 1 mol   520 kJ  3   = 1.9 ×10 kJ 6.941 g mol   

( 25 g Li ) 

84. Removing the first electron from 1 mole of sodium atoms requires 496 kJ. To remove a second electron from 1 mole of sodium atoms requires 4,565 kJ. The conversions are:

(15 mol Na ) 

496 kJ  3  = 7.4 × 10 kJ mol  

(15 mol Na ) 

4565 kJ  4  = 6.8 × 10 kJ  mol 

( 7.4 ×10 kJ ) + ( 6.8 ×10 kJ ) = 7.5 ×10 kJ 3

85.

86.

– Chapter 11 –

87.

88. NCl3 is pyramidal. The presence of three pairs of electrons and a lone pair of electrons around the central atom (N) gives the molecule a tetrahedral structure and a pyramidal shape. BF3 has three pairs of electrons and no lone pairs of electrons around the central atom (B), so both the structure and the shape of the molecule are trigonal planar.

CHAPTER 12

THE GASEOUS STATE OF MATTER SOLUTIONS TO REVIEW QUESTIONS 1.

The pressure of a gas is the force that gas particles exert on the walls of a container. The pressure depends on the temperature, the number of molecules of the gas, and the volume of the container.

The air pressure inside the balloon is greater than the air pressure outside the balloon. The pressure inside must equal the sum of the outside air pressure plus the pressure exerted by the stretched rubber of the balloon.

The major components of dry air are nitrogen and oxygen.

1 torr = 1 mm Hg

The molecules of H2 at 100°C are moving faster. Temperature is a measure of average kinetic energy. At higher temperatures, the molecules will have more kinetic energy.

1 atm corresponds to 4 L.

The pressure times the volume at any point on the curve is equal to the same value. This is an inverse relationship as is Boyle’s law. (PV = k)

If T2 < T1, the volume of the cylinder would decrease (the piston would move downward).

N2(g) + O2(g) → 2 NO(g) 1 vol + 1 vol → 2 vol According to Avogadro’s law, equal volumes of nitrogen and oxygen at the same temperature and pressure contain the same number of molecules. In the reaction, nitrogen and oxygen molecules react in a 1:1 ratio. Since two volumes of nitrogen monoxide are produced, one molecule of nitrogen and one molecule of oxygen must produce two molecules of nitrogen monoxide. Therefore each nitrogen and oxygen molecule must be made up of two atoms (diatomic).

10. We refer gases to STP because some reference point is needed to relate volume to moles. A temperature and pressure must be specified to determine the moles of gas in a given volume, and 0°C and 760 torr are convenient reference points. 11. Gases are described by the following parameters: (a) pressure

(c)

temperature

(b) volume

(d)

number of moles

12. An ideal gas is one which follows the described gas laws at all P, V, and T and whose behavior is described exactly by the Kinetic Molecular Theory. 13. Boyle’s law: P1V1 = P2V2; ideal gas equation: PV = nRT If you have an equal number of moles of two gases at the same temperature, the right side of the ideal gas equation will be the same for both gases. You can then set PV for the first gas equal to PV for the second gas (Boyle’s law) because the right side of both equations will cancel. 1

– Chapter 12 –

14. Charles’ law: V1/T1 = V2/T2, ideal gas equation: PV = nRT Rearrange the ideal gas equation to: V/T = nR/P If you have an equal number of moles of two gases at the same pressure, the right side of the rearranged ideal gas equation will be the same for both. You can set V/T for the first gas equal to V/T for the second gas (Charles’ law) because the right side of both equations will cancel. 15. Basic assumptions of Kinetic Molecular Theory include: (a) Gases consist of tiny particles. (b) The distance between particles is great compared to the size of the particles. (c) Gas particles move in straight lines. They collide with one another and with the walls of the container with no loss of energy. (d) Gas particles have no attraction for each other. (e) The average kinetic energy of all gases is the same at any given temperature. It varies directly with temperature. 16. The order of increasing molecular velocities is the order of decreasing molar masses. increasing molecular velocity ⎯⎯⎯⎯⎯⎯⎯⎯ → Rn; F2 ,N 2 CH4 , He, H2 ⎯⎯⎯⎯⎯⎯⎯→ decreasing molar mass

At the same temperature the kinetic energies of the gases are the same and equal to ½ mv2. For the kinetic energies to be the same, the velocities must increase as the molar masses decrease. 17. Average kinetic energies of all these gases are the same, since the gases are all at the same temperature. 18. A gas is least likely to behave ideally at low temperatures. Under this condition, the velocities of the molecules decrease and attractive forces between the molecules begin to play a significant role. 19. A gas is least likely to behave ideally at high pressures. Under this condition, the molecules are forced close enough to each other so that their volume is no longer small compared to the volume of the container. Attractive forces may also occur here and sooner or later, the gas will liquefy. 20. Behavior of gases as described by the Kinetic Molecular Theory. (a) Boyle’s law. Boyle’s law states that the volume of a fixed mass of gas is inversely proportional to the pressure, at constant temperature. The Kinetic Molecular Theory assumes the volume occupied by gases is mostly empty space. Decreasing the volume of a gas by compressing it, increases the concentration of gas molecules, resulting in more collisions of the molecules and thus increased pressure upon the walls of the container. (b) Charles’ law. Charles’ law states that the volume of a fixed mass of gas is directly proportional to the absolute temperature, at constant pressure. According to Kinetic Molecular Theory, the kinetic energies of gas molecules are proportional to the absolute temperature. Increasing the temperature of a

– Chapter 12 –

gas causes the molecules to move faster, and in order for the pressure not to increase, the volume of the gas must increase. (c) Dalton’s law. Dalton’s law states that the pressure of a mixture of gases is the sum of the pressures exerted by the individual gases. According to the Kinetic Molecular Theory, there are no attractive forces between gas molecules; therefore, in a mixture of gases, each gas acts independently and the total pressure exerted will be the sum of the pressures exerted by the individual gases. 21. Conversion of oxygen to ozone is an endothermic reaction. Evidence for this statement is that energy (286 kJ/3 mol O2) is required to convert O2 to O3. 22. Oxygen atom = O

Oxygen molecule = O2

Ozone molecule = O3

An oxygen molecule contains 16 electrons. 23. The pressure inside the bottle is less than atmospheric pressure. We come to this conclusion because the water inside the bottle is higher than the water in the trough (outside the bottle). 24. The density of air is 1.29 g/L. Any gas listed below air in Table 12.4 has a density greater than air. For example: O2, H2S, HCl, F2, CO2. 25. Equal volumes of H2 and O2 at the same T and P: (a) have equal number of molecules (Avogadro’s law) (b) mass O2 = 16 times mass of H2 (c) moles O2 = moles H2 (d) average kinetic energies are the same (T same) (e) density O2 = 16 times the density of H2

 mass O2   mass H2  density O2 =   density H2 =    volume O2   volume H2  volume O2 = volume H2  mass O2   mass H2   mass O2     density O2 =   ( density H2 )  density O2   density H2   mass H2   32  density O2 =   ( density H2 ) = 16 ( density H2 )  2  26. Heating a mole of N2 gas at constant pressure has the following effects: (a) Density will decrease. Heating the gas at constant pressure will increase its volume. The mass does not change, so the increased volume results in a lower density. (b) Mass does not change. Heating a substance does not change its mass. (c) Average kinetic energy of the molecules increases. This is a basic assumption of the Kinetic Molecular Theory.

– Chapter 12 –

(d) Average velocity of the molecules will increase. Increasing the temperature increases the average kinetic energies of the molecules; hence, the average velocity of the molecules will increase also. (e) Number of N2 molecules remains unchanged. Heating does not alter the number of molecules present, except if extremely high temperatures were attained. Then, the N2 molecules might dissociate into N atoms resulting in fewer N2 molecules.

– Chapter 12 –

SOLUTIONS TO EXERCISES

Pressure conversions: Torr

inches Hg

kilopascals

768

30.2

102

752

29.6

100.

745

29.3

99.3

(a)

in. Hg → torr

( 30.2 in. Hg ) 

in. Hg → kPa

( 30.2 in. Hg ) 

 101.325 kPa   = 102 kPa  29.9 in. Hg 

(b) torr → in. Hg

( 752 torr ) 

torr → kPa

( 752 torr ) 

kPa → torr

( 99.3 kPa ) 

kPa → in. Hg

( 99.3 kPa ) 

(c)

 760 torr   = 768 torr  29.9 in. Hg 

29.9 in. Hg   = 29.6 in. Hg  760 torr  101.325 kPa   = 100. kPa  760 torr  760 torr   = 745 torr  101.325 kPa 

29.9 in.Hg   = 29.3 in. Hg  101.325 kPa 

Pressure conversions: mm Hg

lb/in.2

atmospheres

789

15.3

1.04

1700

2.2

1100

1.4  14.7 lb/in.2  2  = 15.3 lb/in. 760 mm Hg  

(a) mm Hg → lb/in.

( 789 mm Hg ) 

mm Hg → atm

( 789 mm Hg ) 



 1 atm  = 1.04 atm  760 mm Hg 

(b) lb/in.2 → mm Hg lb/in.2 → atm

mm Hg  (32 lb/in. )  760  = 1700 mm Hg 14.7 lb/in.  2

 (32 lb/in. )  14.71 atm  = 2.2 atm lb/in.  2

(1.4 atm ) 

760 mm Hg   = 1100 mm Hg  1 atm 

 14.7 lb/in.2  2 1.4 atm ( )  = 21 lb/in.  1 atm  5

– Chapter 12 –

(a) torr → kPa

( 953 torr ) 

(b) kPa → atm

( 2.98 kPa ) 

(c)

101.3 kPa   = 127 kPa  760 torr  1 atm   = 0.0294 atm  101.3 kPa 

atm → mm Hg

1 atm   = 0.489 atm  760 torr 

atm → cm Hg

76.0 cm Hg   = 214 cm Hg  1 atm 

( 649 torr ) 

101.3 kPa   = 86.5 kPa  760 torr 

(b) kpa → atm

(5.07 kPa ) 

1 atm   = 0.0500 atm  101.3 kPa 

atm → mm Hg

(d) torr → atm

( 2.81 atm ) 

(a) torr → kPa

(c)

760 mm Hg   = 2110 mm Hg  1 atm 

( 372 torr ) 

(d) torr → atm (e)

( 2.77 atm ) 

(3.64 atm ) 

760 mm Hg   = 2770 mm Hg 1 atm  

(803 torr ) 

1 atm   = 1.06 atm  760 torr 

(e)

atm → cm Hg

(1.08 atm ) 

76.0 cm Hg   = 82.1 cm Hg  1 atm 

(a)

lb/in.2 → atm

 (1920 lb/in. )  14.71 atm  = 131 atm lb/in.  2

(b) lb/in.2 → torr

760 torr  (1920 lb/in. )  14.7  = 9.93 × 10 torr lb/in. 

(c)

lb/in.2 → kPa

101.3 kPa  (1920 lb/in. )  14.7  = 1.32 × 10 kPa lb/in. 

(a)

lb/in.2 → atm

 (31 lb/in. )  14.71 atm  = 2.1 atm lb/in. 

(b) lb/in.2 → torr

760 torr  (31 lb/in. )  14.7  = 1600 torr lb/in. 

lb/in.2 → kPa

101.3 kPa  (31 lb/in. )  14.7  = 210 kPa lb/in. 

(c)

– Chapter 12 –

(a)

( 5570 torr ) 

1 atm   = 7.33 atm  760 torr 

(b)

(5570 torr ) 

(c)

(5570 torr ) 

14.7 psi   = 108 psi  760 torr  101.3 kPa   = 742 kPa  760 torr 

(d) YES! Don’t shake it! 8.

(a)

( 631 torr ) 

1 atm   = 0.830 atm  760 torr 

(b)

( 631 torr ) 

(c)

( 631 torr ) 

14.7 psi   = 12.2 psi  760 torr 

101.3 kPa   = 84.1 kPa  760 torr 

(d) Yes, the pressure is too low to get sufficient oxygen. 9.

P1V1 = P2 V2 or P2 =

(a)

P1V1 V2

( 825 torr )( 725 mL ) = 2110 torr ( 283 mL ) Change 2.87 L to mL

(b) 10. P1V1 = P2 V2 or P2 = (a)

(825 torr )( 725 mL ) = 208 torr ( 2.87 L )(1000 mL/1 L )

P1V1 V2

( 508 torr )( 486 mL ) = 1330 torr (185 mL )

(b) Change 6.17 L to mL

(508 torr )( 486 mL ) = 40.0 torr ( 6.17 L )(1000 mL/1L )

11. This is a Boyle’s law problem Original volume x ratio of pressures = new volume

(1.00 L ) 

2100 torr   = 3.8 L  550 torr 

12. Using Boyle’s law

P1V1 = P2 V2

P2 =

( 785 torr )(3.00 L ) = 9.42 ×103 torr PV 1 1 = V2 0.250 L

– Chapter 12 –

13.

14.

V1 V2 = T1 T2

or V2 =

V1T2 ; temperatures must be in Kelvin ( °C + 273 ) T1

(a)

(125 mL )( 268 K ) = 114 mL

(b)

(125 mL )( 308 K ) = 131 mL

(c)

(125 mL )(1095 K ) = 466 mL

294 K

V1 V2 = T1 T2

or V2 =

V1T2 ; temperatures must be in Kelvin ( °C + 273 ) T1

(a)

( 575 mL )( 298 K ) = 691 mL

(b)

( 575 mL )( 273 K ) = 633 mL

(c)

( 575 mL )( 318 K ) = 737 mL

248 K

15. Use the combined gas laws V2 =

or V2 =

P1V1T2 P2 T1

P1V1 P2 V2 = T1 T2

or V2 =

P1V1T2 P2 T1

( 678 torr )( 25.6 L )( 308 K ) = 30.8 L (595 torr )( 292 K )

17. Use the combined gas law V2 =

P1V1 P2 V2 = T1 T2

( 0.75 atm )(1025 mL )( 308 K ) = 544 mL (1.25 atm )( 348 K )

16. Use the combined gas laws V2 =

( 32°F = 0°C = 273 K )

P1V1 P2 V2 = T1 T2

or V2 =

P1V1T2 P2 T1

( 0.950 atm )(1400. L )( 275 K ) = 2.4 ×105 L ( 4.0 torr )(1 atm/760 torr )( 291 K )

18. Use the combined gas law

P1V1 P2 V2 = T1 T2

or T2 =

P2 V2 T1 P1V1

Change 765 torr to atmospheres.

( 765 torr )(1 atm/760 torr )(1.5 L )( 292 K ) = 120 K 120 K − 273 , K = −153°C ( ) (1.5 atm )( 2.5 L )

– Chapter 12 –

19. Ptotal = PO2 + PH2 O vapour = 772 torr

PH2 O vapour = 21.2 torr PO2 = 772 torr − 21.2 torr = 751 torr 20. Ptotal = PCH4 + PH2 O vapour = 749 mm Hg

PH2O = 30.0 torr = 30.0 mm Hg PCH4 = 749 mm Hg − 30.0 mm Hg = 719 mm Hg 21. Ptotal = PN2 + PH2 + PO2

= 200. torr + 600. torr + 300. torr = 1100. torr = 1.100×103 torr 22. Ptotal = PH2 + PN2 + PO2

= 325 torr + 475 torr + 650, torr = 1450, torr = 1.450×103 torr 23. Ptotal = PCH2 + PH2 O vapour ( Solublity of methane is being ignored.)

PH2O vapour = 23.8 torr PCH4 = 720. torr − 23.8 torr = 696 torr To calculate the volume of dry methane, note that the temperature is constant, so P1V1 = P2V2 can be used. V2 =

P1V1 ( 696 torr )( 2.50 L CH4 ) = = 2.29 L CH 4 P2 ( 760. torr )

24. Ptotal = PC3H8 + PH2 O vapour

C 3H8 is propane

PH2O vapour = 20.5 torr PC3H8 = 754 torr − 20.5 torr = 725 torr To calculate the volume of dry propane, note that the temperature is constant, so P1V1 = P2V2 can be used. V2 =

P1V1 ( 725 torr )(1.25 L C3H8 ) = = 1.19 L C3H8 P2 ( 760. torr )

25. PV = nRT  P  T  P1V1 P2 V2  760 torr   291 K  = → V2 = V1  1  2  = 0.143 L    = 0.166 L or 166 mL T1 T2  635 torr   300 K   P2  T1 

26. PV = nRT

 P  T  P1V1 P2 V2  755 torr  285 K  = → V2 = V1  1  2  = 32.7 L    = 9.07 L T1 T2  2577 torr  301 K   P2  T1  27. As you go to higher altitude, the atmospheric pressure decreases. With reduced pressure, the volume of the air in a bag of chips increases.

– Chapter 12 –

28. As the temperature decreases the air pressure in the tires also decreases. In order to bring the pressure back up you need to add more air to the tires. 29. (a)

L  ( 6.022 ×10 molecules CO )  6.022 ×1022.4molecules  = 22.4 L CO  23

(b)

( 2.5 mol CH 4 ) 

(c)

(12.5 g O2 ) 

22.4 L   = 56 L CH 4  mol 

 22.4 L   = 8.75 L O2  32.00 g 

22.4 L   1.80 ×10 molecules SO )  (  = 67.0 L SO 30. (a)  6.022 ×10 molecules  24

(b)

( 7.5 mol C 2H6 ) 

(c)

( 25.2 g Cl2 ) 

22.4 L   = 170 L C 2 H6  mol 

 22.4 L   = 7.96 L Cl2  70.90 g 

31.

( 725 mL NH3 ) 

1 L   1 mol   17.03 g     = 0.551 g NH3  1000 mL   22.4 L   mol 

32.

( 945 mL C 3H6 ) 

33.

(1025 molecules CO2 ) 

34.

(10.5 L CO2 ) 

1 L   1 mol   42.08 g     = 1.78 g C 3H6  1000 mL   22.4 L   mol  1 mol   22.4 L  −20   = 3.813 × 10 L CO 2 23  6.022 ×10 molecules   mol 

1 mol   6.022 ×1023 molecules  23  = 2.82 ×10 molecules CO2  mol  22.4 L   

35. Density of Cl2 gas = 3.17 g/L (from table 12.4)

 1L   = 3.15 L Cl2  3.17 g 

(10.0 g Cl2 ) 

36. Density of CH4 gas = 0.716 g/L (from table 12.4) (3.0 L CH4)(0.716 g/L) = 2.1 g CH4 37. PV = nRT V = nRT V = ( 75 mol NH3 )( 0.0821 L atm/mol K )( 295 K ) = 1.9 ×103 L NH 3 729 torr torr 760 atm

38. PV = nRT V = nRT V = ( P

105 mol CH 4 )( 0.0821 L atm/mol K )( 312 K ) = 1.8×103 L CH 4 1.5 atm

– Chapter 12 –

39. PV = nRT n =

(1.5 atm )( 2.50 L N 2 ) = 0.15 mol N PV = 2 L atm  RT  K 0.0821 303 ( )   mol K  

   505 torr  (10.45 L CH 4 )  torr   760  PV  atm  40. PV = nRT n = = = 0.25 mol CH 4 L atm  RT   0.0821  ( 338 K ) mol K  

41. PV = nRT T =

42. PV = nRT T =

PV nR

   732 torr  ( 645 L Xe )  torr   760  atm   = 300. K T= ( 25.2 mol Xe )( 0.0821 L atm/mol K )

PV nR

   675 torr  ( 725 L Ar )  torr   760  atm   T= = 209 K (37.5 mol Ar )( 0.0821 L atm/mol K )

43. (a) Density of Gases  4.003 g He   1 mol  d=   = 0.179 g/L He mol    22.4 L   20.01 g HF   1 mol  (b) d =    = 0.893 g/L HF mol    22.4 L 

(c)

 42.08 g C 3H6   1 mol  d=   22.4 L  = 1.89 g/L C 3H6 mol   

 120.9 g CCl 2 F2   1 mol  (d) d =    = 5.40 g/L CCl2 F2 mol    22.4 L 

44. (a)

 222 g Rn   1 mol  d=   = 9.91 g/L Rn  mol   22.4 L 

 46.01 g NO 2   1 mol  (b) d =   = 2.05 g/L NO 2  mol    22.4 L 

(c)

 80.07 g SO3   1 mol  d=  = 3.57 g/L SO3  mol    22.4 L 

 28.05 g C 2H 4   1 mol  (d) d =    = 1.25 g/L C 2H 4 mol    22.4 L 

– Chapter 12 –

45. (a) Assume 1.00 mol of NH3 and determine the volume using the ideal gas equation, PV = nRT. V= d=

nRT

(1.00 mol NH3 )( 0.0821 L atm/mol K )( 298 K )

P 17.03 g 20. L NH3

1.2 atm

= 20.L NH3 at 25°C and 1.2 atm

= 0.85 g/L NH3

(b) Assume 1.00 mol of Ar and determine the volume using the ideal gas equation, PV = nRT. nRT (1.00 mol Ar )( 0.0821 L atm/mol K )( 348 K ) = = 29.1 L Ar at 75°C and 745 torr 745 torr P torr 760 atm 39.95 g d= = 1.37 g/L Ar 29.1 L Ar V=

46. (a) Assume 1.00 mol C2H4 and determine the volume using the ideal gas equation, PV = nRT. nRT (1.00 mol C 2 H 4 )( 0.0821 L atm/mol K )( 305 K ) = = 33 L C 2H 4 at 32°C and 0.75 atm P 0.75 atm 28.05 g d= = 0.85 g/L C 2 H 4 33 L C 2 H 4 V=

(b) Assume 1.00 mol of He and determine the volume using the ideal gas equation, PV = nRT. nRT (1.00 mol He )( 0.0821 L atm/mol K )( 330. K ) = = 26.0 L He at 57°C and 791 torr 791 torr P torr 760 atm 4.003 g d= = 0.154 g/L He 26.0 L He V=

47. The balanced equation is CaCO3(s) → CaO(s) + CO2(g); 1 mol of a gas occupies 22.4 L at STP. (a)

 1 mol   1 mol CO2   22.4 L CO2      100.1 g   1 mol CaCO3   1 mol CO2 

( 6.24 g CaCO3 ) 

= 1.40 L CO 2 or 1.40 × 103 mL

(b)

 1 mol CO2   1n mol CaCO3    = 2.35 mol CaCO3  22.4 L CO2   1 mol CO2 

(52.6 L CO2 ) 

48. The balanced equation is Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g); 1 mol of a gas occupies 22.4 L at STP. (a)

 1 mol   1 mol H2   22.4 L   1000 mL  4     = 3.95 ×10 mL H2 1 mol 1 L 24.31 g 1 mol Mg     

( 42.9 g Mg ) 

– Chapter 12 –

(b)

1 L   1 mol H2   2 mol HCl   = 0.0737 mol HCl    1000 mL   22.4 L H2   1 mol H2 

(825 mL H2 ) 

49. The balanced equation is 4 NH3(g) + 5O2(g) → 4 NO(g) + 6 H2O(g) Remember that volume–volume relationships are the same as mole–mole relationships when dealing with gases at the same T and P.

 5 L O2   = 3.1 L O2  4 L NH3 

(a)

( 2.5 L NH3 ) 

(b)

( 25 L NH3 ) 

 6 L H2 O   1 mol   18.02 g     = 30. g H2 O  4 L NH3   22.4 L   1 mol 

( 25 L O2 ) 

 4 L NO   = 25 L NO  4 L NH3 

( 25 L NH3 ) 

Oxygen is the limiting reactant. 20. L NO is formed. 50. The balanced equation is C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) Remember that volume–volume relationships are the same as mole–mole relationships when dealing with gases at the same T and P.

 5 L O2   = 36 L O2  1 L C3H8 

(a)

( 7.2 L C3H8 ) 

(b)

(35 L C3H8 ) 

 3 L CO2   1 mol   44.01 g     = 210 g CO2  1 L C3H8   22.4 L   1 mol 

 4 L H2 O   = 60. L H2 O  1 L C3H8 

(15 L C3H8 ) 

 4 L H2 O   = 12 L H2 O  5 L O2 

(15 L O2 ) 

Oxygen is the limiting reactant. 12 L H2O is formed. 51. The balanced equation is 2 KClO3(s) → 2 KCl(s) + 3 O2(g)

 1000 g  1 mol   3 mol O2  22.4 L      = 237 L O2  1 kg  74.55 g   2 mol KCl   1 mol 

( 0.525 kg KCl ) 

– Chapter 12 –

52. The balanced equation is C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)

 1000 g  1 mol   6 mol CO2   22.4 L  3     = 1.12 ×10 L CO2  1 kg  180.2 g   1 mol C6H12 O6   1 mol 

(1.50 kg C6H12 O6 ) 

53. Like any other gas, water in the gaseous state occupies a much larger volume than in the liquid state. 54. During the winter the air in a car’s tires is colder, the molecules move slower and the pressure decreases. In order to keep the pressure at the manufacturer’s recommended psi, air needs to be added to the tire. The opposite is true during the summer. 55. Sample (b) has the greatest pressure because it has more molecules. Pressure is proportional to number of moles when temperature and volume remain constant. 56. Image (a) best represents the balloon because lowering the temperature of the gas will decrease its volume. The size of the balloon will also decrease, but the gas molecules will still be distributed throughout the volume of the balloon. 57. (a) the pressure will be cut in half (b) the pressure will double if the kelvin temperature is doubled (c) the pressure will be cut in half (d) the pressure will increase to 3.7 atm or 2800 torr

nRT V (1.5 mol )( 0.0821 L atm/mol K )( 303 K ) = 3.7 atm P= 10. L  760 torr  3 P = 3.7 atm   = 2.8 ×10 torr  1 atm  PV = nRt P =

58. 1)

Compressibility – b

Pressure – d

Expansion – b, c, d

Pressure of equal volumes of gases – f

59. a.

– Chapter 12 –

60. Law

Factors that are constant

Factors that are variable

Graph showing the relationship of variable factors

Boyle’s law

Temperature and number of moles

Pressure and volume

(c)

Charles’ law

Pressure and number of moles

Volume and temperature

(b)

Avogadro’s law

Pressure and temperature

Volume and number of moles

(b)

61. The can is a sealed unit and very likely still contains some of the aerosol. As the can is heated, pressure builds up in it eventually causing the can to explode and rupture with possible harm from flying debris. 62. First calculate the moles of H2 and then solve the gas equation for volume.  1 mol H2   = 7.5 mol H2  2.0 g 

(15 g ) 

nRT P L ⋅ atm  ( 7.5 mol )  0.0821  ( 298 K ) mol ⋅ K   V= = 1.5×10 2 L 1.2 atm

PV = nRT or V =

63. One mole of an ideal gas occupies 22.4 liters at standard conditions (0°C and 1 atm pressure). PV = nRT (1.00 atm)(V) = (1.00 mol)(0.0821 L atm/mol K)(273 K) V = 22.4 L 64. Solve for volume using PV = nRT (a)

(b) V =

( 0.2 mol Cl2 )( 0.0821 L atm/mol K )( 321 K ) = 5 L Cl 2 (80 cm/76 cm ) atm  1 mol   0.0821 L atm   ( 262 K )  mol K   17.03 g   = 8.2 L NH3 0.65 atm

( 4.2 g NH3 ) 

– Chapter 12 –

(c)

 1 mol   0.0821 L atm   ( 328 K )  mol K   80.07 g   = 6.5 L SO3 110 kPa kPa 101.3 atm

( 21 g SO3 ) 

4.2 g NH3 has the greatest volume 65. (a) 1 mol of a gas occupies 22.4 L at STP

 1 mol CH4  6022 ×1023 molecule CH4  22   = 2.69 ×10 molecule CH4 22.4 L CH 1 mol CH   4  4

(1 L CH4 ) 

(b) Temperature must be in K; must convert torr to atm and mol to molecules

100°C PV = 113°C; 113 + 273 = 386 K; PV = nRT or n = 180°F RT ( 952 torr )(1 atm/760 torr )( 3.29 L N 2 ) = 0.130 mol N 2 ( 0.0821 L atm/mol K )( 386 K )

( 235°F − 32°F )

 6.022 ×1023 molecule N 2  22  = 7.83 ×10 molecule N 2 1 mol N   2

( 0.130 mol N 2 ) 

0 + 273 = 273 K; PV = nRT or n =

PV RT

( 0.624 atm )(5.05 L Cl2 ) = 0.141 mol Cl 2 ( 0.0821 L atm/mol K )( 273 K )  6.022 ×1023 molecule Cl2  22 0.141 mol Cl (  = 8.49 ×10 molecule Cl2 2 ) 1 mol Cl2   The container with chlorine gas contains the largest number of molecules. 66. Assume 1 mol of each gas (a) SF6 = 146.1 g/mol SF6

 146.1 g   1 mol  d =   = 6.52 g/L SF6  mol SF6   22.4 L  (b) Assume 25°C and 1 atm pressure

 298 K  V ( at 25°C ) = ( 22.4 L C 2H6 )   = 24.5 L C 2 H6  273 K  C 2H6 = 30.07 g/mol  1 mol  30.07 g   d =  = 1.23 g/L C 2H6   mol   24.5 L C 2H6 

– Chapter 12 –

(1 mol He )( 0.0821 L atm/mol K )(193 K ) = 7.37 L He

2.15 atm  4.003 g   1 mol  d =   = 0.543 g/L He  mol   7.37 L He 

SF6 has the greatest density 67. (a) Sample 2 has the higher density because it has more molecules in the same volume. (b) Sample 2 has the higher density because the molar mass of the individual molecules is larger. 68. (a) Empirical formula. Assume 100. g starting material 80.0 g C 6.66 = 6.66 mol C = 1.00 12.01 g/mol 6.66 20.0 g H 19.8 = 19.8 mol H = 2.97 1.008 g/mol 6.66

Empirical formula = CH3 Empirical mass = 12.01 g + 3:024 g = 15.03 g (b) Molecular formula.

 2.01 g   22.4 L     = 30.g/mol ( molar mass )  1.5 L   mol 

30. g/mol = 2; Molecular formula is C 2 H6 15.03 g/mol

69. PV = nRT

 ( 790 torr )(1 atm )    ( 2.0 L ) = ( n )( 0.0821 L atm/mol K )( 298 K ) 760 torr (a)   n = 0.085 mol ( total moles ) (b) mol N 2 = toal moles − mol O2 − mol CO2 0.65 g O 2 0.58 g CO 2 − 32.00 g/mol 44.01 g/mol mol N 2 = 0.085 mol − 0.020 mol O 2 − 0.013 mol CO 2 = 0.052 mol N 2 = 0.085 mol −

( 0.052 mol N 2 ) 

28.02 g   = 1.5 g N 2  mol 

– Chapter 12 –

(c)

 0.020 mol O 2  2 PO 2 = ( 790 torr )   = 1.9 × 10 torr ( O 2 ) 0.085 mol  

 0.013 mol CO2  2 PCO 2 = ( 790 torr )   = 1.2 × 10 torr ( CO 2 )  0.085 mol   0.051 mol N 2  2 PN 2 = ( 790 torr )   = 4.7 × 10 torr ( N 2 )  0.085 mol  g   70. PV = nRT or PV =   RT  molar mass 

 1.4 g   1000 cm  3  = 1.4 ×10 g/L  3  L  cm    3





×10 g (1.3 ×10 atm ) (1.0 L ) =  1.4  ( 0.0821 L atm/mol K )( T ) 2.0 g/mol 9





(1.3 ×10 atm ) (1.0 L )( 2.0 g/mol ) = 2.3 ×10 K ( 0.0821 L atm/mol K ) (1.4 ×10 g ) 9

71. (a) Assume atmospheric pressure of 14.7 lb/in.2 to begin with. Total pressure in the ball = 14.7 lb/in.2 + 13 lb/in.2 = 28 lb/in.2 PV = nRT  ( 28 lb/in. )  14.71 atm  ( 2.24 L ) = ( n )( 0.0821 L atm/mol K )( 293 K ) lb/in.  2

n = 0.18 mol air (b) mass of air in the ball

molar mass of air is about 29 g/mol

 29 g  m = ( 0.18 mol )   = 5.2 g air  mol 

(c) Actually the pressure changes when the temperature changes. Since pressure is directly proportional to moles we can calculate the change in moles required to keep the pressure the same at 30°C as it was at 20°C. PV = nRT  ( 28 lb/in. )  14.71 atm  ( 2.24 L ) = ( n )( 0.0821 L atm/mol K )( 303 K ) lb/in.  2

n = 0.17 mol of air required to keep the pressure the same at 30°C. 0.01 mol air (0.18 − 0.17) must be allowed to escape from the ball.

( 0.01 mol air ) 

29 g   = 0.29 g or 0.3 g air must be allowed to escape.  mol 

– Chapter 12 –

72. Use the combined gas laws to calculate the bursting temperature (T2).

P1 = 65 cm

P1V1 P2 V2 = T1 T2 T2 =

 76 cm  P2 = 1.00 atm   = 76 cm  1 atm  V2 = 2.00 L

V1 = 1.75 L

T1 = 20°C ( 293 K ) T2 = T2

P2 V2 T1 ( 76 cm )( 2.00 L )( 293 K ) = = 392 K (119°C ) P1V1 ( 65 cm )(1.75 L )

73. To double the volume of a gas, at constant pressure, the temperature (K) must be doubled.

V1 V2 = T1 T2

V2 = 2 V1

V1 2 V1 = T1 T2

T2 =

2 V1T1 V1

T2 = 2 T1

T2 = 2 ( 300. K ) = 600. K = 327°C 74. V = volume at 22°C and 740 torr 2 V = volume after change in temperature (P constant) V = volume after change in pressure (T constant) Since temperature is constant, P1V1 = P2 V2

or P2 =

P1V1 V2

2 V 3 P2 = ( 740 torr )   = 1.5 × 10 torr ( pressure to change 2 V to V )  V 

75. Use the combined gas laws

( P1 )( V1 ) = ( P2 )( V2 ) or T = ( T1 )( P2 )( V2 ) 2 ( T1 ) ( T2 ) ( P1 )( V1 ) Since the volume stays constant, V1 = V2 and the equation reduces to T2 = T2 =

( T1 )( P2 ) ( P1 )

( 500. torr )( 295 K ) = 211 K = −62°C 700. torr

76. Use the combined gas laws

( P1 )( V1 ) = ( P2 )( V2 ) ( T1 ) ( T2 )

or T2 =

( T1 )( P2 )( V2 ) ( P1 )( V1 )

The volume of the tires remains constant (until they burst) so V1 = V2 and the equation reduces to T2 =

( T1 )( P2 ) ( P1 )

– Chapter 12 –

71.0°F = 21.7°C = 295 K

T2 =

( 44 psi )( 295 K ) = 433 K = 160°C = 320°F 30. psi

77. The molar mass of NO is 14.01 g/mol + 16.00 g/mol = 30.01 g/mol  30.01 g   1 mol  d =   = 1.34 g /L  mol   22.4 L 

78. 1 mol of a gas occupies 22.4 L at STP

 1 mol   22.4 SO2   = 3.20 L SO2   64.07 g  1 mol SO2 

( 9.14 g SO2 ) 

79. Use the combined gas law

P1V1 P2 V2 = T1 T2

V2 =

P1V1T2 P2 T1

First calculate the volume at STP. V2 =

( 400. torr )( 600. mL N 2 O )( 273 K ) = 275 mL N O = 0.275 L N O 2 2 ( 760. torr )(313 K )

At STP, a mole of any gas has a volume of 22.4 L

1 mol   6.022 ×1023 molecules  21  = 7.39 ×10 molecules N 2 O  1 mol  22.4 L   

( 0.275 L N 2 O ) 

Each molecule of N2O contains 3 atoms, so





3 atoms ( 7.39 ×10 molecules N O )  1 molecule  = 2.22 ×10 atoms NO 21





80. 950 atm – (346 atm O2 + 93 atm N2O) = 950 atm − 439 atm = 511 atm Note that these pressures are very high. If you were to land on this planet you would probably be crushed by the extreme pressure. 81. Use the combined gas laws

( P1 )( V1 ) = ( P2 )( V2 ) ( T1 ) ( T2 )

or P2 =

( P1 )( T2 )( V1 ) ( T1 )( V2 )

Since the volume stays constant (unless it does burst), V1 = V2 and the equation reduces to P2 =

( P1 )( T2 ) ( T1 )

T1 = 25°C + 273 = 298 K T2 = 212°F = 100°C = 373 K

(32 lb/in. ) (373 K ) = 40. lb/in. P = 2

298 K At 212°F the tire pressure is 40. lb/in.2 The tire will not burst. 20

– Chapter 12 –

82. A column of mercury at 1 atm pressure is 760 mm Hg high. The density of mercury is 13.6 times that of water, so a column of water at 1 atm pressure should be 13.6 times as high as that for mercury. (760 mm)(13.6) = 1.03 × 104 mm (33.8 ft) 83. Use the ideal gas equation

RT PV

PV = nRT n =

Change 2.20 × 103 lb/in.2 to atmosphere  ( 2.20 ×10 lb/in. )  14.71 atm  = 150. atm lb/in.  3

(150. atm )( 55 L O2 )

 0.0821 L ⋅ atm    ( 300. K ) mol ⋅ K  

= 3.3 × 10 2 mol O2

84. First calculate the moles of gas and then convert moles to molar mass.

( 0.560 L ) 

1 mol   = 0.0250 mol  22.4 L 

1.08 g = 43.2 g/mol ( molar mass ) 0.0250 mol

85. The conversion is: g/L → g/mol  1.78 g   22.4 L     = 39.9 g/mol ( molar )  L   mol 

86. PV = nRT V=

nRT ( 0.510 mol H2 )( 0.0821 L atm/mol K )( 320. K ) = = 8.4 L H2 P 1.6 atm

(b) V =

PV ( 600 torr )(1 atm/760 torr )(16.0 L CH 4 ) = = 0.513 mol CH 4 RT ( 0.0821 L atm/mol K )(300. K )

(a)

The molar mass for CH4 is 16.04 g/mol (16.04 g/mol)(0.513 mol) = 8.23 g CH4 (c)

PV = nRT, but n = Thus, PV =

gRT . To determine density, d = g/V. M

Solving PV = d=

g where M is the molar mass and g is the rams of the gas. M

gRT g g PM = . for produces V M V RT

g ( 4.00 atm )( 44.01 g/mol CO 2 ) = = 8.48 g/L CO 2 V ( 0.0821 L atm/mol K )( 253 K )

– Chapter 12 –

(d) Since d = M=

g PM = from part (c), solve for M (molar mass) V RT

dRT ( 2.58 g/L )( 0.0821 L atm/mol K )( 300. K ) = = 63.5 g/mol ( molar mass ) P 1.00 atm

87. C2H2(g) + 2HF(g) → C2H4F2(g) 1.0 mol C2H2 → 1.0 mol C2H4F2

( 5.0 mol HF ) 

1 mol C 2 H 4 F2   = 2.5 mol C 2 H 4 F2  2 mol HF 

C2H2 is the limiting reactant. 1.0 mol C2H4F2 forms, no moles C2H2 remain. According to the equation, 2.0 mol HF yields 1.0 mol C2H4F2. Therefore, 5.0 mol HF – 2.0 mol HF = 3.0 mol HF unreacted The flask contains 1.0 mol C2H4F2 and 3.0 mol HF when the reaction is complete. The flask contains 4.0 mol of gas. P=

88.

nRT ( 4.0 mol )( 0.0821 L atm/mol K )( 273 K ) = = 9.0 atm V 10.0 L

( 8.30 mol Al ) 

3 mol H 2   22.4 L    = 279 L H 2 at STP  2 mol Al   mol 

89. Assume 100. g of material to start with. Calculate the empirical formula. 7.14  1 mol   = 7.14 mol 7.14 = 1.00 mol  12.01 g 

(85.7 g ) 

(14.3 g ) 

 1 mol   = 14.2 mol  1.008 g 

14.2 = 1.99 mol 7.14

The empirical formula is CH2. To determine the molecular formula, the molar mass must be known.  2.50 g   22.4 L     = 56.0 g/mol ( molar mass )  L   mol 

The empirical formula mass is 14.0

56.0 =4 14.0

Therefore, the molecular formula is (CH2)4 = C4H8

2 CO(g) + O2 ( g ) → 2 CO2 ( g ) Determine the limiting reactant 90.

(10.0 mol CO ) 

2 mol CO2   = 10.0 mol CO2 ( from CO )  2 mol CO 

 2 mol CO2   = 16 mol CO2 ( from O2 )  1 mol O2 

(8.0 mol O2 ) 

CO: the limiting reactant,

O2: in excess, 3.0 mol O2 unreacted.

– Chapter 12 –

(a) 10.0 mol CO react with 5.0 mol O2 10.0 mol CO2 and 3.0 mol O2 are present, no CO will be present. (b) P =

nRT (13 mol ) ( 0.0821 L atm/mol K )( 273 K ) = = 29 atm V 10.L

Δ → 2 KCl ( s ) + 3 O2 ( g ) 91. 2 KClO3 ( s ) ⎯⎯

First calculate the moles of O2 produced. Then calculate the grams of KClO3 required to produce the O2. Then calculate the % KC1O3.

( 0.25 L O2 ) 

1 mol   = 0.011 mol CO 2  22.4 L 

 2 mol KClO3   122.6 g    = 0.90 g KClO3 in the sample  3 mol O 2   mol 

( 0.011 mol O2 ) 

 0.90 g    (100 ) = 75% KClO3 in the mixture  1.20 g 

92. Mass of CO2 in solution  PV  = ( molar mass )( moles ) = ( molar mass )    RT     1 atm ×1.40 L  44.01 g CO2   =  = 2.52 g CO2   0.08206 L atm mol   × 298 K  mol K    0.965 g  mass of soft drink = ( 345 mL )   = 333 g  mL    6 2.52 g ppm of CO2 =   (10 )  333 g + 2.52 g  = 7.51× 103 ppm 93. (a) The CO2 balloon will be heaviest, followed by the Ar balloon. The H2 balloon would be the lightest. Gases at the same temperature, pressure, and volume contain the same number of moles. All balloons will contain the same number of moles of gas molecules, so when moles are converted to mass, the order from heaviest to lightest is CO2, Ar, H2. (b) Molar mass: O2, 32.00; N2, 28.02; Ne, 20.18 Using equal masses of gas, we find that the balloon containing O2 will have the lowest number of moles of gas. Since pressure is directly proportional to moles, the balloon containing O2 will have the lowest pressure.

– Chapter 12 –

94. Assume 1.00 L of air. The mass of 1.00 L of air is 1.29 g.

P1V1 P2 V2 = T1 T2 V2 = d=

P1V1T2 ( 760 torr )(1.00 L )( 290 K ) = = 1.8 L P2 T1 ( 450 torr )( 273 K )

m 1.29 g = = 0.72 g/L V 1.8 L

95. Each gas behaves as though it were alone in a 4.0 L system. (a) After expansion: P1V1 = P2V2

For CO2 For H2

P2 =

P1V1 (150. torr )( 30. L ) = = 1.1×102 torr V2 4.0 L

P2 =

P1V1 ( 50. torr )(1.0 L ) = = 13 torr V2 4.0 L

(b) Ptotal = PH2 + PCO2 = 110 torr +13 torr = 120 torr ( 2 sig. figures ) 96. Use the combined gas laws

( P1 )( V1 ) = ( P2 )( V2 ) ( T1 ) ( T2 )

or P2 =

( P1 )( T2 )( V1 ) ( T1 )( V2 )

P1 = 40.0 atm P2 = unknown V1 = 50.0 L V2 = 50.0 L T1 = 25°C = 298 K T2 = 25°C + 125°C = 177°C = 450 K P2 =

( 40.0 atm )( 450 K ) ( 50.0 L ) = 60.4 atm ( 298 K ) ( 50.0 L )

(Note that the volume canceled out of the expression because it stays constant.) 97. You can identify the gas by determining its density. Mass of gas = 1.700 g – 0.500 g = 1.200 g volume of gas: Charles’ law problem. Correct volume to 273 K

V1 V2 = T1 T2

V2 =

V1T2 ( 0.4478 L )( 273 K ) = = 0.3785 L T1 323 K

m 1.200 g = = 3.170 g/L V 0.3785 L

gas is chlorine (see Table 12.4) 98. (a) If 15.0% escapes, 85.0% will remain. 0.850(5.27 × 105 mol He) = 4.48 × 105 mol He

– Chapter 12 –

 n  T   4.48×105 mol   304 K  (b) P2 = P1  2  2  = 976 torr    = 885 torr 5  5.27×10 mol   285 K   n1   T1  99. 1st step: Find the volume of CO2 that must be produced  55.0 L CO 2  volume CO 2 = ( volume of batter )( 55.0% ) = (1.32 L batter )    100 L batter  = 0.726 L CO 2

2nd step: Find the moles of CO2 needed to produce the necessary rise in the cupcakes PV RT 1 atm   P = ( 738 torr )   = 0.971 atm 760 torr   V = 0.726 L T = 325°F = 163°C = 436 K PV = nRT or n =

( 0.971 atm )( 0.726 L ) = 0.0197 mol CO2 ( 0.0821 L atm/mol K )( 436 K )

3rd step: Calculate the mass of sodium bicarbonate which will produce the CO2 needed 3 NaHCO3 + H3C 6 O7 → Na 3C 6 H5 O7 + 3 H 2 O + 3 CO 2  3 mol NaHCO3   84.02 g    = 1.66 g NaHCO3  3 mol CO 2   1 mol 

( 0.0197 mol CO 2 ) 

4th step: Calculate mass of sodium bicarbonate needed to add since only 63.7% will decompose to form carbon dioxide

 100 g NaHCO3 added   = 2.61 g NaHCO3 added  63.7 g NaHCO3 reacted 

(1.66 g NaHCO3 reacted ) 

CHAPTER 13

LIQUIDS SOLUTIONS TO REVIEW QUESTIONS 1.

At 0°C, all three substances, H2S, H2Se, and H2Te, are gases, because they all have boiling points below 0°C.

Liquids are made of particles that are close together, they are not compressible and they have definite volume. Solids also exhibit similar properties.

Liquids take the shape of the container they are in. Gases also exhibit this property.

The water in both containers would have the same vapor pressure, for it is a function of the temperature of the liquid.

Vapor pressure is the pressure exerted by a vapor when it is in equilibrium with its liquid. The liquid molecules on the surface and the gas molecules above the liquid are the important players in vapor pressure. One liter of water will evaporate at the same rate as 100 mL of water if they are at the same temperature and have the same surface area. The liquid and vapor in both cases will come to equilibrium in the same amount of time and they will have the same vapor pressure. Vapor pressure is a characteristic of the type of liquid, not the number of molecules of a liquid.

In Figure 13.5, it would be case (b) in which the atmosphere would reach saturation. The vapor pressure of water is the same in both (a) and (b), but since (a) is an open container the vapor escapes into the atmosphere and doesn’t reach saturation.

If ethyl ether and ethyl alcohol were both placed in a closed container, (a) both substances would be present in the vapor, for both are volatile liquids; (b) ethyl ether would have more molecules in the vapor because it has a higher vapor pressure at a given temperature.

Rubbing alcohol feels cold when applied to the skin, because the evaporation of the alcohol absorbs heat from the skin. The alcohol has a fairly high vapor pressure (low boiling point) and evaporates quite rapidly. This produces the cooling effect.

(a) Order of increasing rate of evaporation: mercury, acetic acid, water, toluene, benzene, carbon tetrachloride, methyl alcohol, bromine. (b) Highest boiling point is mercury. Lowest boiling point is bromine.

10. As temperature increases, molecular velocities increase. At higher molecular velocities, it becomes easier for molecules to break away from the attractive forces in the liquid. 11. The pressure of the atmosphere must be 1.00 atmosphere, otherwise the water would be boiling at some other temperature. 12. The thermometer would be at about 70°C. The liquid is boiling, which means its vapor pressure equals the confining pressure. From Table 13.7, we find that ethyl alcohol has a vapor pressure of 543 torr at 70°C. 13. At 30 torr, H2O would boil at approximately 29°C, ethyl alcohol at 14°C, and ethyl ether at some temperature below 0°C.

– Chapter 13 –

14. (a) At a pressure of 500 torr, water boils at 88°C. (b) The normal boiling point of ethyl alcohol is 78°C. (c) At a pressure of 0.50 atm (380 torr), ethyl ether boils at 16°C. 15. Water boils when its vapor pressure equals the prevailing atmospheric pressure over the water. In order for water to boil at 50°C, the pressure over the water would need to be reduced to a point equal to the vapor pressure of the water (92.5 torr). 16. In a pressure cooker, the temperature at which the water boils increases above its normal boiling point, because the water vapor (steam) formed by boiling cannot escape. This results in an increased pressure over water and, consequently, an increased boiling temperature. 17. Vapor pressure varies with temperature. The temperature at which the vapor pressure of a liquid equals the prevailing pressure is the boiling point of the liquid. 18. Ammonia would have a higher vapor pressure than SO2 at −40°C because it has a lower boiling point (NH3 is more volatile than SO2). 19. As the temperature of a liquid increases, the kinetic energy of the molecules as well as the vapour pressure of the liquid increases. When the vapor pressure of the liquid equals the external pressure, boiling begins with many of the molecules having enough energy to escape from the liquid. Bubbles of vapor are formed throughout the liquid and these bubbles rise to the surface, escaping as boiling continues. 20. 34.6°C, the boiling point of ethyl ether. (See Table 13.1) 21. The potential energy is greater in the liquid water than in the ice. The heat necessary to melt the ice increases the potential energy of the liquid, thus allowing the molecules greater freedom of motion. The potential energy of steam (gas) is greater than that of liquid water. 22. Based on Figure 13.8: (a) Line BC is horizontal because the temperature remains constant during the entire process of melting. The energy input is absorbed in changing from the solid to the liquid state. (b) During BC, both solid and liquid phases are present. (c) The line DE represents the change from liquid water to steam (vapor) at the boiling temperature of water. 23. Apply heat to an ice-water mixture, the heat energy is absorbed to melt the ice (heat of fusion), rather than warm the water, so the temperature remains constant until all the ice has melted. 24. Ice at 0°C contains less heat energy than water at 0°C. Heat must be added to convert ice to water, so the water will contain that much additional heat energy. 25. The boiling liquid remains at constant temperature because the added heat energy is being used to convert the liquid to a gas, i.e., to supply the heat of vaporization for the liquid at its boiling point.

– Chapter 13 –

26. Intermolecular forces are the attractive forces between molecules. These forces hold molecules together to form liquids and solids. Intramolecular forces are forces between atoms in a molecule. These forces hold the atoms in a molecule together. 27. Polar bonds are covalent bonds in which atoms do not share electrons equally. 28. The ability of a molecule to form instantaneous dipoles is most dependent on its number of electrons. 29. The heat of vaporization of water would be lower if water molecules were linear instead of bent. If linear, the molecules of water would be nonpolar. The relatively high heat of vaporization of water is a result of the molecule being highly polar and having strong dipole-dipole and hydrogen bonding attraction for other water molecules. 30. Ethyl alcohol exhibits hydrogen bonding; ethyl ether does not. This is indicated by the high heat of vaporization of ethyl alcohol, even though its molar mass is much less than the molar mass of ethyl ether. 31. Although a linear water molecule would be nonpolar due to its symmetry, the individual O–H bonds would still be polar meaning that the hydrogens would still be able to form hydrogen bonds. The number of possible hydrogen bonds would not change. The overall intermolecular attractive force would be lower though due to its nonpolar nature. 32. Water, at 80°C, will have fewer hydrogen bonds than water at 40°C. At the higher temperature, the molecules of water are moving faster than at the lower temperature. This results in less hydrogen bonding at the higher temperature. 33. H2NCH2CH2NH2 has two polar NH2 groups. It should, therefore, show more hydrogen bonding and a higher boiling point (117°C) versus 49°C for CH3CH2CH2NH2. 34. Water has a relatively high boiling point because there is a high attraction between molecules due to hydrogen bonding. 35. HF has a higher boiling point than HCl because of the strong hydrogen bonding in HF (F is the most electronegative element). Neither F2 nor Cl2 will have hydrogen bonding, so the compound, F2, with the lower molar mass, has the lower boiling point.

36. 37. Melting point, boiling point, heat of fusion, heat of vaporization, density, and crystal structure in the solid state are some of the physical properties of water that would be very different, if the molecules were linear and nonpolar instead of bent and highly polar. For example, the boiling point, melting point, heat of fusion and heat of vaporization would be lower because linear molecules have no dipole moment and the attraction among molecules would be much less. 38. Physical properties of water: (a) melting point, 0°C (b) boiling point, 100°C (at 1 atm pressure) (c) colorless 3

– Chapter 13 –

(d) odorless (e) tasteless (f)

heat of fusion, 335 J/g (80 cal/g)

(g) heat of vaporization, 2.26 kJ/g (540 cal/g) (h) density = 1.0 g/mL (at 4°C) (i)

specific heat = 4.184 J/g°C

39. For water, to have its maximum density, the temperature must be 4°C, and the pressure sufficient to keep it liquid, d = 1.0 g/mL. 40. Ice floats in water because it is less dense than water. The density of ice at 0°C is 0.915 g/mL. Liquid water, however, has a density of 1.00 g/mL. Ice will sink in ethyl alcohol, which has a density of 0.789 g/mL. 41. If the lake is in an area where the temperature is below freezing for part of the year, the expected temperature would be 4°C at the bottom of the lake. This is because the surface water would cool to 4°C (maximum density) and sink. 42. The formation of hydrogen and oxygen from water is an endothermic reaction, due to the following evidence: (a) Energy must continually be provided to the system for the reaction to proceed. The reaction will cease when the energy source is removed. (b) The reverse reaction, burning hydrogen in oxygen, releases energy as heat.

– Chapter 13 –

SOLUTIONS TO EXERCISES 1.

CH3Br < CH3Cl < CH3F

H2Se < H2S < H2O

CCl4, < CBr4 < CI4

CO2 < SO2 < CS2

In which of the following substances would you expect to find hydrogen bonding? (a) C3H7OH will hydrogen bond; one of the hydrogens is bonded to oxygen. (b) H2O2 will hydrogen bond; hydrogen is bonded to oxygen. (c) CHCl3 will not hydrogen bond; hydrogen is not bonded to fluorine, oxygen, or nitrogen. (d) PH3 will not hydrogen bond; hydrogen is not bonded to fluorine, oxygen, or nitrogen. (e) HF will hydrogen bond; hydrogen is bonded to fluorine.

(a) HI will not hydrogen bond; hydrogen is not bonded to fluorine, oxygen, or nitrogen. (b) NH3 will hydrogen bond; hydrogen is bonded to nitrogen. (c) CH2F2 will not hydrogen bond; hydrogen is not bonded to fluorine, oxygen, or nitrogen. (d) C2H5OH will hydrogen bond; one of the hydrogens is bonded to oxygen. (e) H2O will hydrogen bond; hydrogen is bonded to oxygen.

Surface tension is a measure of the ability of a liquid to minimize its surface area. As intermolecular forces increase, the molecules in a substance are more strongly attracted to each other. This stronger attraction keeps the molecules close together resulting in an increased surface tension. Thus, liquids with stronger intermolecular forces will have higher surface tensions.

The boiling point of a substance is the temperature where the vapor pressure of a liquid is equal to the external pressure above the liquid. In order to increase the vapor pressure, more molecules must be converted into the vapor phase. As intermolecular forces increase, the molecules in a substance are more strongly attracted to each other. In order for vaporization to occur, energy must be provided to overcome the intermolecular attractions and separate the molecules. As the average temperature increases, the average energy of the molecules also increases and more molecules will be able to convert to the vapor phase. Molecules with strong intermolecular forces will require more energy to vaporize than those with weak intermolecular forces and therefore will require higher temperatures in order for the vapor pressure to increase. Thus, liquids with stronger intermolecular forces require more energy to vaporize and will have higher boiling points.

– Chapter 13 –

9. (a)

C3H7OH

(b)

H2O2

(c)

10. (a) C2H5OH (b) NH3

(c) H2O 11. The adhesive forces between the cotton fabric of the T-shirt and the water are stronger than the cohesive forces between water molecules causing the water to absorb into the fabric. The cohesive forces between water molecules is stronger than the adhesive forces between the raincoat fabric and water causing the water to bead up on the raincoat. 12. Water forming beaded droplets is an example of cohesive forces. The water molecules have stronger attractive forces for other water molecules than they do for the surface. 13. Water has the stronger intermolecular forces which means that it will also have the higher boiling point. 14. Glycerol has stronger intermolecular forces because it has a higher surface tension. The glycerol should also have the higher boiling point. 15. a.

2595°C

2600°C

2589°C

2560°C

16. a.

240 torr

180 torr

680 torr

150 torr

– Chapter 13 –

17. Energy (Ea) to heat the water to steam from 15°C → 100.°C  4.184 J  4 Ea = ( m ) ( specific heat )( Δt ) = ( 275 g )   ( 85°C ) = 9.8 × 10 J ° g C   Energy ( E b ) to convert water at 100°C to steam: heat of vaporization = 2.26 ×103 J/g E b = ( m )( heat of vaporization ) = ( 275 g ) ( 2.26 ×103 J/g ) = 6.22 ×105 J Etotal = Ea + E b = ( 9.8 × 10 4 J ) + ( 6.22 ×105 J ) = 7.20 ×105 J

18. Energy (Ea) to cool the water from 35°C → 0°C

 4.184 J  4 Ea = ( m )( specific heat )( Δt ) = ( 325 g )   ( 35°C ) = 4.8 ×10 J  g°C  Energy (Eb) to convert water to ice: heat of fusion = 335 J/g Eb = (m)(heat of fusion) = (325 g)(335 J/g) = 1.09 × 105 J Etotal = Ea + Eb = (4.8 ×104 J) + (1.09 × 105 J) = 1.57 × 105 J 19. Energy released in cooling the water: 25°C to 0°C

 4.184 J  4 E = ( m )( specific heat )( Δt ) = ( 300. g )   ( 25°C ) = 3.1×10 J g ° C   Energy required to melt the ice E = (m)(heat of fusion) = (100. g)(335 J/g) = 3.35 × 104 J Less energy is released in cooling the water than is required to melt the ice. Ice will remain and the water will be at 0°C. 20. Energy to heat the water = energy to condense the steam

 4.184 J   (100.°C − 25°C ) = ( m )( 2259 J/g )  g°C 

(300. g ) 

m = 42 g (42 g of steam are required to heat 300. g of water to 100.°C.) Since only 35 g of steam are added to the system, the final temperature will be less than 100.°C. Not sufficient steam. 21. Energy lost by warm water = energy gained by the ice x = final temperature

 1000 mL  1.0 g  mass ( H2 O ) = (1.5 L H2 O )    = 1500 g L   mL   4.184 J    4.184 J  J (1500 g )   ( 75°C − x ) = ( 75 g )  335  + ( 75 g )   ( x − 0°C ) g  g°C    g°C 

( 4.707 ×10 J ) − ( 6276x J/°C ) = 2.51×10 J + 313.8x J/°C 5

4.46 ×105 J = 6.5898 × 103 x J/°C x = 68°C

– Chapter 13 –

22. E = (m)(heat of fusion) (500. g)(335 J/g) = 167,000 J needed to melt the ice 9560 J < 167,500 J Since 167,500 J are required to melt all the ice, and only 9560 J are available, the system will be at 0°C. It will be a mixture of ice and water. 23. Intermolecular forces and the forces that hold one molecule close to another. They are forces holding one molecule close to another molecule while covalent bonds hold one atom close to another within a molecule. 24. Cl2 is nonpolar. Nonpolar molecules such as chlorine gas have dispersion forces only. 25. Water forms droplets because of surface tension, or the desire for a droplet of water to minimize its ratio of surface area to volume. The molecules of water inside the drop are attracted to other water molecules all around them, but on the surface of the droplet the water molecules feel an inward attraction only. This inward attraction of surface water molecules for internal water molecules is what holds the droplets together and minimizes their surface area. 26. Steam molecules will cause a more severe burn. Steam molecules contain more energy at 100°C than water molecules at 100°C due to the energy absorbed during the vaporization stage (heat of vaporization). 27. The alcohol has a higher vapor pressure than water and thus evaporates faster than water. When the alcohol evaporates it absorbs energy from the water, cooling the water. Eventually the water will lose enough energy to change from a liquid to a solid (freeze). 28. When one leaves the swimming pool, water starts to evaporate from the skin of the body. Part of the energy needed for evaporation is absorbed from the skin, resulting in the cool feeling. 29. a. boiling point (condensation point)

b. gas in equilibrium with liquid

c. all-liquid state, temperature begins to fall

d. melting point (freezing point)

e. liquid in equilibrium with solid 30.

– Chapter 13 –

(a) From 0°C to 40.°C solid X warms until at 40.°C it begins to melt. The temperature remains at 40.0°C until all of X is melted. After that, liquid X will warm steadily to 65°C where it will boil and remain at 65°C until all of the liquid becomes vapor. Beyond 65°C, the vapor will warm steadily until 100°C. (b) Joules needed (0°C to 40°C)

= (60. g)(3.5 J/g°C)(40.°C) = 8400 J

Joules needed at 40°C

= (60. g)(80. J/g)

= 4800 J

Joules needed (40°C to 65°C)

= (60. g)(3.5 J/g°C)(25°C)

= 5300 J

Joules needed at 65°C

= (60. g)(190 J/g)

= 11,000 J

Joules needed (65°C to 100°C) = (60. g)(3.5 J/g°C)(35°C) Total Joules needed

= 7400 L 37,000 J

(each step rounded to two significant figures) 31. During phase changes (ice melting to liquid water or liquid water evaporating to steam), all the heat energy is used to cause the phase change. Once the phase change is complete the heat energy is once again used to increase the temperature of the substance. 32. As the temperature of a liquid increases, the molecules gain kinetic energy thereby increasing their escaping tendency (vapor pressure). 33. Since boiling occurs when vapor pressure equals atmospheric pressure, the graph in Figure 13.7 indicates that water will boil at about 78°C or 172°F at 330 torr pressure. 34. For the noble gases the boiling point increases as the molar mass increases. This suggests that boiling points are directly related to molar masses. This is consistent with an increase in dispersion forces with an increase in molar mass which would result in an increase in boiling point. 35. The elevation in Santa Fe is much higher than in Santa Barbara. This means that the air pressure is also lower. Since the boiling temperature of water is dependent on the atmospheric pressure, it will be lower in Santa Fe than in Santa Barbara. Boiling the eggs for 8 minutes at a lower pressure will result in undercooked eggs. 36. Water is a very polar molecule. Glass, an oxide of silicon, is composed of polar silicon-oxygen bonds. Because both water and glass are composed of polar molecules there are strong intermolecular forces between them which create a strong adhesive force. The water climbs up the walls of the graduated cylinder in order to maximize the water-glass interactions resulting in a downward curving meniscus. When water is poured into a plastic cylinder the intermolecular forces between water and the nonpolar hydrocarbons composing the plastic are very weak. The cohesive forces between water molecules are stronger than the adhesive forces with the plastic resulting in an almost flat meniscus which minimizes interactions between the water and the plastic.

– Chapter 13 –

37. T (°C)

drops/min

104

As the temperature increases the rate of flow increases. As the amount of energy in the honey increases, the individual molecules have enough energy to overcome the strong intermolecular forces holding them together allowing the liquid to flow more easily. 38. The wax on the floors is composed of long chain hydrocarbons which are very nonpolar. When polar water is spilled on the floor it will bead up to minimize contact with the nonpolar wax. When the nonpolar hexane is spilled on the floor it will spread out so that it has more contact with the nonpolar wax. 39. The compound with the lower boiling point should have the weaker intermolecular forces and thus the higher vapor pressure, so isobutyl propionate with the lower boiling point has weaker intermolecular forces and will have a higher vapor pressure. 40. Boiling point is the temperature where the liquid is converted into a vapor. The normal boiling point is the temperature where this transition occurs at atmospheric pressure. As intermolecular forces become stronger, the vapor pressure of a liquid decreases. This is because stronger intermolecular forces cause the molecules to be more strongly attracted to each other resulting in a decrease in the number of molecules which are able to escape from the liquid. In order to vaporize enough molecules to increase the vapor pressure to atmospheric pressure the temperature must be increased. Thus the boiling point will increase as intermolecular forces increase. 41. Dispersion forces are intermolecular forces between molecules or atoms due to the formation of temporary instantaneous dipoles. These dipoles act like little magnets attracting the particles together. C9H20 has greater dispersion forces because it has a higher molar mass and more electrons. The more electrons a molecule has the more polarizable it is meaning that it is more likely for the electrons to be unevenly distributed causing a temporary dipole to form.

– Chapter 13 –

42. (a) The boiling point of acetic acid is approximately 119°C and the melting point is approximately 18°C according to this diagram. (b) and (c)

43. (a) Melt ice: Ea = (m)(heat of fusion) = (225 g)(335 J/g) = 7.54 × 104 J (b) Warm the water: E b = ( m ) ( specific heat )( Δt )

 4.184 J  4 = ( 225 g )   (100.°C ) = 9.41×10 J g C °   (c) Vaporize the water: E c = ( m )( heat of vaporization ) = ( 225 g )( 2259 J/g ) = 5.08 × 105 J Etotal = Ea + E b + E c = 6.78 × 105 J 1 cal  ( 6.78 ×10 J )  4.184  = 1.62 × 10 cal J 5

44. The heat of vaporization of water is 2.26 kJ/g.

( 2.26 kJ/g ) 

18.02 g   40.7 kJ/mol  mol 

 0.096 cal  45. E = ( m )( specific heat )( Δt ) = ( 250. g )   (150. − 20.0°C )  g°C  = 3.1×103 cal ( 3.1 kcal ) 46. Energy liberated when steam at 100.0°C condenses to water at 100.0°C

18.02 g   2.26 kJ   1000 J  6    = 2.04 ×10 J  mol   g   kJ 

(50.0 mol steam ) 

– Chapter 13 –

Energy liberated in cooling water from 100.0°C to 30.0°C 18.02 g   4.184 J  5  (100.0°C − 30.0°C ) = 2.64 ×10 J  mol g ° C   

(50.0 mol H2 O ) 

Total energy liberated = 2.04 × 106 J + 2.64 ×105 J = 2.30 ×106 J

47. Energy to warm the ice from −10.0°C to 0°C

 2.01 J   (10.0°C ) = 2010 J  g°C 

(100. g ) 

Energy to melt the ice at 0°C (100. g)(335 J/g) = 33,500 J Energy to heat the water from 0°C to 20.0°C  4.184 J   ( 20.0°C ) = 8370 J  g°C 

(100. g ) 

Etotal = 2010 J + 33,500 J + 8370 J = 4.39 ×10 4 J = 43.9 kJ

48. 2 H2O(l) → 2 H2(g) + O2(g) The conversion is L O2 → mol O2 → mol H2O → g H2O

1 mol   2 mol H2 O   18.02 g     = 40.2 g H2 O  22.4 L   1 mol O2   mol 

( 25.0 L O2 ) 

49. The conversion is mol molecules molecules molecules molecules → → → → day day hr min s  1.00mol H2 O   6.022 ×10 23 molecules   1.00day   1 hr   1 min        day mol     24 hr   60 min   60 s  = 6.97 × 1018 molecules H2 O/s

50. Liquid water has a density of 1.00 g/mL. d=

m V

m 18.02 g = = 18.0 mL d 1.00 g/mL

( volume of 1 mole )

1.00 mole of water vapor at STP has a volume of 22.4 L (gas) 51. 2 H2(g) + O2(g) → 2 H2O(g) (a)

 1 mL O2   = 40.0 mL O2 react with 80.0 mL of H2  2 mL H2 

(80.0 mL H2 ) 

Since 60.0 mL of O2 are available, some oxygen remains unreacted. (b) 60.0 mL – 40.0 mL = 20.0 mL O2 unreacted.

– Chapter 13 –

52. Energy absorbed by the student when steam at 100.°C changes to water at 100.°C

 2.26 kJ   = 3.4 kJ  g 

(1.5 g steam ) 

(3.4 ×10 J ) 3

Energy absorbed when water cools from 100.°C to 20.0°C E = ( m ) ( specific heat )( Δt )  4.184 J  2 E = (1.5 g )   (100.°C − 20.0°C ) = 5.0 × 10 J  g°C  Etotal = 3.4 × 103 J + 5.0 × 102 J = 3.9 × 103 J 53. (a) won’t hydrogen bond; no hydrogen in the molecule (b) will hydrogen bond

(e) won’t hydrogen bond; hydrogen must be attached to N, O, or F 54. During the fusion (melting) of a substance the temperature remains constant so a temperature factor is not needed. 55. Energy needed to heat Cu to its melting point: E = (50.0 g)(0.385 J/g°C)(1083°C – 25.0°C) = 2.04 × 104 J Energy needed to melt the Cu:

E = ( 50.0 g )(134 J/g ) = 6.70 ×103 J Etotal = 2.04 ×10 4 J + 6.70 ×103 J = 2.71×10 4 J 56. Energy released by the soup to convert ice at 0°C to water at 0°C.

 335 J  4  = 2.5 ×10 J  g 

( 75 g ice ) 

Energy released by soup to increase temperature of water from 0°C to 87°C.

E = ( m )( specific heat )( Δt )  4.184 J  4 E = ( 75 g )   ( 87°C ) = 2.7 ×10 J g ° C   4 Etotal = 2.5 ×10 J + 2.7 × 10 4 J = 5.2 ×10 4 J 57. 2 H2O(l) → 2 H2(g) + O2(g)

– Chapter 13 –

58. At 500 torr, water will boil at approximately 88°C or 190°F At 300 torr, water will boil at approximately 76°C or 169°F At 100 torr, water will boil at approximately 51°C or 124°F 59. A lake freezes from the top down because the density of water as a solid is lower than that of the liquid, causing the ice to float on the top of the liquid. Since the liquid remains below the solid, marine and aquatic life continues. 60. Heat lost by warm water = heat gained by ice m = grams of ice to lower temperature of water to 0.0°C.

 4.184 J   ( 45°C − 0.0°C ) = ( m )( 335 J/g )  g°C 

(120.g ) 

68 g = m ( grams of ice melted ) 68 g of ice melted. Therefore, 150. g − 68 g = 82 g ice remain. 61. Zack went to Ely, Gaye went to the Dead Sea, and Lamont was in Honolulu. Zack’s boiling point was lowered, so he was in a region of lower atmospheric pressure (on a mountain). Lamont was basically at sea level, so his boiling point was about normal. Since Gaye’s boiling point was raised, she was at a place of higher atmospheric pressure and therefore was possibly in a location below sea level. The Dead Sea is below sea level. 62. The particles in solids and liquids are close together (held together by intermolecular attractions), and an increase in pressure is unable to move them significantly closer to each other. In a gas, the space between molecules is significant, and an increase in pressure is often accompanied by a decrease in volume.

63. a. Acetone

56°C

Isopropanol

83°C

Methanol

64°C

Ethanol

78°C

– Chapter 13 –

b. Acetone Isopropanol Methanol Ethanol c.

Isopropanol, methanol, and ethanol all have the ability to form hydrogen bonds which increase the boiling point.

Isopropanol, methanol, and ethanol all form hydrogen bonds so their molecular masses determine the relative boiling points. Isopropanol (M = 60 g/mol) with the highest molar mass has larger dispersion forces and the highest boiling point followed by ethanol (M = 46 g/mol) and methanol (M = 32 g/mol).

CHAPTER 14

SOLUTIONS SOLUTIONS TO REVIEW QUESTIONS 1.

A true solution is one in which the size of the particles of solute are between 0.1–1 nm. True solutions are homogeneous and the ratio of solute to solvent can be varied. They can be colored or colorless but are transparent. The solute remains distributed evenly in the solution, it will not settle out.

The two components of a solution are the solute and the solvent. The solute is dissolved into the solvent or is the least abundant component. The solvent is the dissolving agent or the most abundant component.

It is not always apparent which component in a solution is the solute. For example, in a solution composed of equal volumes of two liquids, the designation of solute and solvent would be simply a matter of preference on the part of the person making the designation.

The ions or molecules of a dissolved solute do not settle out because the individual particles are so small that the force of molecular collisions is large compared to the force of gravity.

Yes. It is possible to have one solid dissolved in another solid. Metal alloys are of this type. Atoms of one metal are dissolved among atoms of another metal.

Orange. The three reference solutions are KCl, KMnO4 and K2Cr2O7. They all contain K+ ions in solution. The different colors must result from the different anions dissolved in the solutions: MnO4− (purple) and Cr2O72− (orange). Therefore, it is predictable that the Cr2O72− ion present in an aqueous solution of Na2Cr2O7 will impart an orange color to the solution.

Air is considered to be a solution because it is a homogeneous mixture of several gaseous substances and does not have a fixed composition.

These diagrams are intended to illustrate the orientation of the water molecules about the ions, not the number of water molecules. 9.

From Table 14.2, approximately 26.8 g of KBr would dissolve in 50 g water at 0°C.

– Chapter 14 –

10. From Figure 14.4, solubilities in water at 25°C: (a)

NH4Cl

39 g/100 g H2O

(b)

CuSO4

22 g/100 g H2O

(c)

NaNO3

91 g/100 g H2O

11. Going down Group 1A, the solubilities of both the chlorides and bromides decrease. 12. From Fig. 14.4, the solubility, in grams of solute per 100 g of H2O: (a)

NH4Cl at 35°C, 43 g

(b)

NaNO3 at 15°C, 82 g

(b)

CuSO4 at 60°C, 39 g

(c)

SO2 gas at 30°C, 8 g at 1 atm

13. KCl, CuSO4, HCl, Li2SO4, KNO3 (all at 1 atm) 10°C

CuSO4, KNO3, KCl, Li2SO4, HCl

50°C

Li2SO4, CuSO4, KCl, HCl, KNO3

14. Li2SO4 15.

6.5 g ×100 = 30 mass percent. The solution will be unsaturated. 21.5 g

16.

40. g ×100 = 35 mass percent. The solution will be saturated. 115 g

17. A supersaturated solution of NaC2H3O2 may be prepared in the following sequence: (a) Determine the mass of NaC2H3O2 necessary to saturate a specific amount of water at room temperature. (b) Place a bit more NaC2H3O2 in the water than the amount needed to saturate the solution. (c) Heat and stir the solution until all the solid dissolves. (d) Cover the container and allow it to cool undisturbed. The cooled solution, which should contain no solid NaC2H3O2, is supersaturated. To test for supersaturation, add one small crystal of NaC2H3O2 to the solution. Immediate crystallization is an indication that the solution was supersaturated. 18. Hexane and benzene are both nonpolar molecules. There are no strong intermolecular forces between molecules of either substance or with each other, so they are miscible. Sodium chloride consists of ions strongly attracted to each other by electrical attractions. The hexane molecules, being nonpolar, have no strong forces to pull the ions apart, so sodium chloride is insoluble in hexane. 19. Coca Cola has two main characteristics, taste and fizz (carbonation). The carbonation is due to a dissolved gas, carbon dioxide. Since dissolved gases become less soluble as temperature increases, warm Coca Cola would be flat, with little to no carbonation. It is, therefore, unappealing to most people. 20. The solubility of gases in liquids is greatly affected by the pressure of a gas above the liquid. The greater the pressure, the more soluble the gas. There is very little effect of pressure regarding the dissolution of solids in liquids. 2

– Chapter 14 –

21. Four factors are particle size of a solute, stirring, temperature and solution concentration. 22. In a saturated solution, the net rate of dissolution is zero. There is no further increase in the amount of dissolved solute, even though undissolved solute is continuously dissolving, because dissolved solute is continuously coming out of solution, crystallizing at a rate equal to the rate of dissolving. 23. The champagne would spray out of the bottle all over the place. The rise in temperature and the increase in kinetic energy of the molecules by shaking both act to decrease the solubility of gas within the liquid. The pressure inside the bottle would be great. As the cork is popped, much of the gas would escape from the liquid very rapidly, causing the champagne to spray. 24. The rate of dissolving decreases. The rate of dissolving is at its maximum when the solute and solvent are first mixed. 25. A teaspoon of sugar would definitely dissolve more rapidly in 200 mL of hot coffee than in 200 mL of iced tea. The much greater thermal agitation of the hot coffee will help break the sugar molecules away from the undissolved solid and disperse them throughout the solution. Other solutes in coffee and tea would have no significant effect. The temperature difference is the critical factor. 26. For a given mass of solute, the smaller the particles, the faster the dissolution of the solute. This is due to the smaller particles having a greater surface area exposed to the dissolving action of the solvent. 27. When crystals of AgNO3 and NaCl are mixed, the contact between the individual ions is not intimate enough for the double displacement reaction to occur. When solutions of the two chemicals are mixed, the ions are free to move and come into intimate contact with each other, allowing the reaction to occur easily. The AgCl formed is insoluble. 28. The two solutions contain the same number of chloride ions. One liter of 1 M NaCl contains 1 mole of NaCl, therefore 1 mole of chloride ions. 0.5 liter of 1 M MgCl2 contains 0.5 mol of MgCl2 and 1 mole of chloride ions.

1 mol MgCl2   2 mol Cl −  −  = 1 mol Cl  L    1 mol MgCl2 

( 0.5 L ) 

29. The number of grams of NaCl in 750 mL of 5.0 molar solution is

( 0.75 L ) 

5.0 mol NaCl   58.44 g  2   = 2.2 × 10 g NaCl L 1 mol   

Dissolve the 220 g of NaCl in a minimum amount of water, then dilute the resulting solution to a final volume of 750 mL (0.75 L).

– Chapter 14 –

30. Ranking of the specified bases in descending order of the volume of each required to react with 1 liter of 1 M HCl. The volume of each required to yield 1 mole of OH− ion is shown. (a) 1 M NaOH

1 liter

(b) 0.6 M Ba(OH)2

0.83 liter

(c)

2 M KOH

0.50 liter

(d) 1.5 M Ca(OH)2

0.33 liter

31. The boiling point of a liquid or solution is the temperature at which the vapor pressure of the liquid equals the pressure of the atmosphere. Since a solution containing a nonvolatile solute has a lower vapour pressure than the pure solvent, the boiling point of the solution must be at a higher temperature than for the pure solvent. At the higher boiling temperature the vapor pressure of the solution equals the atmospheric pressure. 32. The freezing point is the temperature at which a liquid changes to a solid. The vapor pressure of a solution is lower than that of a pure solvent. Therefore, the vapor pressure curve of the solution intersects the vapor pressure curve of the pure solvent, at a temperature lower than the freezing point of the pure solvent (see Figure 14.8b). At this point of intersection, the vapor pressure of the solution equals the vapour pressure of the pure solvent. 33. Water and ice are different phases of the same substance in equilibrium at the freezing point of water, 0°C. The presence of the methanol lowers the vapor pressure and hence the freezing point of water. If the ratio of alcohol to water is high, the freezing point can be lowered by as much as 10°C or more. 34. Effectiveness in lowering the freezing point of 500. g water: (a) 100. g (2.17 mol) of ethyl alcohol is more effective than 100. g (0.292 mol) of sucrose. (b) 20.0 g (0.435 mol) of ethyl alcohol is more effective than 100. g (0.292 mol) of sucrose. (c) 20.0 g (0.625 mol) of methyl alcohol is more effective than 20.0 g (0.435 mol) of ethyl alcohol. 35. Both molarity and molality describe the concentration of a solution. However, molarity is the ratio of moles of solute per liter of solution, and molality is the ratio of moles of solute per kilogram of solvent. 36. 5 molal NaCl = 5 mol NaCl/kg H2O; 5 molar NaCl = 5 mol NaCl/L of solution. The volume of the 5 molal solution will be larger than 1 liter (1 L H2O + 5 mol NaCl). The volume of the 5 molar solution is exactly 1 L (5 mol NaCl + sufficient H2O to produce 1 L of solution). The molarity of a 5 molal solution is therefore, less than 5 molar. 37. A nonvolatile solute (such as salt) lowers the freezing point of water. Adding salt to icy roads in winter melts the ice because the salt lowers the freezing point of water. 38. Because the concentration of water is greater in the thistle tube, the water will flow through the membrane from the thistle tube to the urea solution in the beaker. The solution level in the thistle tube will fall.

– Chapter 14 –

39. A semipermeable membrane will allow water molecules to pass through in both directions. If it has pure water on one side and 10% sugar solutions on the other side of the membrane, there is a higher concentration of water molecules on the pure water side. Therefore, there are more water molecule impacts per second on the pure water side of the membrane. The net result is more water molecules pass from the pure water to the sugar solution. Osmotic pressure effect. 40. The urea solution will have the greater osmotic pressure because it has 1.67 mol solute/kg H2O, while the glucose solution has only 0.83 mol solute/kg H2O. 41. A lettuce leaf immersed in salad dressing containing salt and vinegar will become limp and wilted as a result of osmosis. As the water inside the leaf flows into the dressing where the solute concentration is higher the leaf becomes limp from fluid loss. In water, osmosis proceeds in the opposite direction flowing into the lettuce leaf maintaining a high fluid content and crisp leaf. 42. The concentration of solutes (such as salts) is higher in seawater than in body fluids. The survivors who drank seawater suffered further dehydration from the transfer of water by osmosis from body tissues to the intestinal tract.

– Chapter 14 –

SOLUTIONS TO EXERCISES 1.

2. 3.

Reasonably soluble:

(b) K2SO4

(d) Na3PO4

(d)

NaOH

Insoluble:

(a) AgCl

(c)

PbI2

(e)

SnCO3

Reasonably soluble:

(b) Cu(NO3)2

(d) NH4C2H3O2

(f)

AgNO3

Insoluble:

(a) Ba3(PO4)2

(c)

(e)

MgO

Fe(OH)3

(a)

Soluble in both because the molecule has polar and nonpolar regions making it soluble in both polar and nonpolar solvents.

(b)

Soluble in hexane only because this is a nonpolar molecule that is soluble only in nonpolar solvents such as hexane.

(c)

Soluble in water only because this is an ionic compound which is extremely polar and will be soluble in polar solvents such as water.

(a)

Soluble in hexane only because this is a nonpolar molecule that is soluble only in nonpolar solvents such as hexane.

(b)

Soluble in water only because this is an ionic compound which is extremely polar and will be soluble in polar solvents such as water.

(c)

Soluble in both because the molecule has polar and nonpolar regions making it soluble in both polar and nonpolar solvents.

(a)

not saturated

(b)

saturated

(c)

saturated

(d)

not saturated

(a)

saturated

(b)

not saturated

(c)

saturated

(d)

not saturated

Mass percent calculations (a)

35.0 g KCl + 100.0 g H2 O = 135.0 g solution  35.0 g    (100 ) = 25.9% KCl  135.0 g 

(b) 3.64 g Na3PO4 + 20.0 g H2 O = 23.6 g solution

 3.64 g    (100 ) =15.4% Na3PO4  23.6 g 

– Chapter 14 –

(c)

( 0.45 mol NH4C 2H3O2 ) 

77.09 g   = 35 g NH4C 2 H3O2  1 mol 

35 g NH4C2H3O2 + 230. g H2 O = 265. g solution  35 g    (100) =13% NH4C2H3O2  265 g  (d)

(1.70 mol NaOH ) 

40.00 g   = 68.0 g NaOH  1 mol 

( 27.0 mol H2O ) 

18.02 g   = 487 g H2 O  1 mol  68.0 g NaOH + 487 g H2 O = 555 g solution  68.0 g    (100 ) = 12.3% NaOH  555 g  8.

Mass percent calculations (a) 25.0 NaNO3 in 125.0 g H2O = 150.0 g solution

 25.0 g    (100 ) =16.7% NaNO3  150.0 g  (b) 1.25 g CaCl2 in 35.0 g H2O = 36.3 g solution

 1.25 g    (100 ) = 3.44% CaCl2  36.3 g  (c)

( 0.75 mol K 2CrO4 ) + 

194.2 g   = 150 g K 2CrO 4  1 mol 

150 g K 2CrO 4 + 225 g H2 O = 380 g solution  150 g    (100 ) = 39% K 2CrO4  380 g  (d)

(1.20 mol H2SO4 ) 

98.09 g   = 118 g H2SO 4  1 mol 

( 72.5 mol H2 O ) 

18.02 g  3  = 1.31×10 g H2 O  1 mol 

118 g H2SO4 + 1.31×103 g H2 O = 1.43×103 g solution  118 g    (100 ) = 8.25% H2SO4 3  1.43 ×10 g  9.

 100 g solution  5.23 g NaClO   = 24.3 g solution  21.5 g NaClO 

– Chapter 14 –

 100. g solution   1 mL solution  10. 55.0 g sucrose    = 137 ml solution  35.0 g sucrose   1.151 g solution  11. (a)

 7.5 g CaSO4   = 1.9 g CaSO4  100. g solution 

( 25 g solution ) 

(a) 25 g solution – 1.9 g solute = 23 g solvent 12. (a)

 12.0 g BaCl2   = 9.0 g BaCl2  100. g solution 

( 75 g solution ) 

(b) 75 g solution – 9.0 g solute = 66 g solvent 13. Mass/volume percent. (a)

 15.0 g C 2 H5 OH    (100 ) = 10.0% C 2 H5 OH  150.0 mL solution 

 25.2 g NaCl  (b)   (100 ) = 20.1% NaCl  125.5 mL solution 

14. Mass/volume percent. (a)

 175.2 g C12 H22 O11    (100 ) = 63.59% C12 H22 O11  275.5 mL solution 

 35.5 g of CH3OH  (b)   (100 ) = 47.3% CH3OH  75.0 mL solution 

15. Volume percent. (a)

 75.0 mL hexanol    (100 ) = 20.8% hexanol  360. mL solution 

 2.0 mL ethanol  (b)   (100 ) = 13% ethanol  15.0 mL solution 

(c)

 21.5 mL acetone    (100 ) = 7.82% acetone  275 mL solution 

16. Volume percent. (a)

 37.5 mL butanol    (100 ) = 13.6% butanol  275 mL solution 

 4.0 mL methanol  (b)   (100 ) = 16% methanol  25.0 mL solution 

(c)

 45.0 mL isoamyl alcohol    (100 ) = 6.00% isoamyl alcohol 750. mL solution  

– Chapter 14 –

mol   17. Molarity problems  M =  L  

(a)

 0.25 mol   1000 mL     = 3.3 M  75.0 mL   1 L 

 1.75 mol KBr  (b)   = 2.3 M KBr 0.75 L  

(c)

 35.0 g NaC2H3O2   1 mol   = 0.341 M NaC2H3O2   1.25 L    82.03 g 

 77 g CuSO4 ⋅5H2 O   1 mol  (d)   = 0.30 M CuSO4  1.0 L    249.7 g  mol   18. Molarity problems  M =  L  

(a)

 0.50 mol   1000 mL     = 4.0 M  125 mL   1 L 

 2.25 mol  (b)   = 1.50 M CaCl2  1.50 L 

(c)

 275 g   1 mol   1000 mL      =1.97 M C6H12 O6  775 mL   180.2 g   1 L 

(d)

 125 g MgSO4 ⋅7 H2O   1 mol   = 0.203 M MgSO4   2.50 L    246.5 g 

19. Molarity =

mol solute or mol solute = ( L solution ) ( Molarity ) L solution

(a)

( 2.6 L ) 

1.20 mol H2SO 4   = 3.1 mol H2SO 4 L  

(b)

( 25.0 mL ) 

(c)

( 475 mL ) 

1 L   0.0018 mol BaCl2  −5   = 4.5 × 10 mol BaCl2 1000 mL L   

1 L   0.35 mol K 3PO 4    = 0.166 mol K 3PO 4 L  1000 mL  

20. Molarity =

mol solute or mol solute = ( L solution ) ( Molarity ) L solution

(a)

( 0.75 L ) 

1.50 mol HNO3   = 1.1 mol HNO3 L  

(b)

(10.0 mL ) 

1 L   0.75 mol NaClO3  −3   = 7.5 × 10 mol NaClO3 1000 mL L   

– Chapter 14 –

(175 mL ) 

1 L   0.50 mol LiBr    = 0.088 mol LiBr L  1000 mL   

( 2.5 L ) 

0.75 mol K 2CrO 4   194.2 g    = 360 g K 2 CrO 4 L    1 mol  1 L   0.050 mol HC 2 H3 O 2   60.05 g    = 0.226 g HC 2 H3 O 2  L  1000 mL     1 mol 

(b)

( 75.2 mL ) 

(c)

( 250 mL ) 

22. (a)

1 L   16 mol HNO3   63.02 g    = 250 g HNO3  L  1000 mL     1 mol 

18 mol H2SO 4   98.09 g  3  = 2.1×10 g H2SO 4  L    1 mol 

(1.20 L ) 

 1 L   1.5 mol KMnO 4   158.0 g  (b) ( 27.5 mL )   = 6.52 g KMnO 4   L  1000 mL     1 mol 

 1 L   0.025 mol Fe2 ( SO4 )3   399.9 g  (c) (120 mL )     = 1.2 g Fe2 ( SO4 )3 L  1000 mL     1 mol  23. (a)

 0.1000 g NaCl   = 0.5250 g NaCl  100. g solution  525.0 g solution − 0.5250 g NaCl = 524.5 g water

(525.0 g solution ) 

(b)

(1.5 L solution ) 

18 L alcohol   = 0.27 L alcohol = 270 mL alcohol  100. L solution 

(c)

 3.8 mol H2SO 4   1000 mL   = 15 M H2SO 4   250. mL   1 L 

 2.6 mol KCl  (d)   = 0.65 M KCl 4.0 L  

24. (a)

 100 g solution  0.2755 g Kl   = 183.7 g solution  0.1500 g Kl 

 185 g Na 2S  0.992 g merlot   (b) 750. mL merlot   = 0.1376 g Na 2S   1 mL merlot   1,000,000 g merlot  (c)

 1 L solution   0.486 mol NaC 2 H3 O2   82.03 g NaC 2 H3 O2     1 L solution  1000 mL solution     1 mol NaC 2 H3 O2 

375 mL solution 

= 14.9 g NaC 2 H3 O 2

 2.000 g sucrose  (d) 1500 g solution   = 30.00 g sucrose  100 g solution  1500 g solution − 30.00 g sucrose = 1470 g water

– Chapter 14 –

25. (a)



  1000 mL  1L 2   = 2.0 ×10 mL  0.750 mol H3PO4   1 L 

( 0.15 mol H3PO4 ) 

  1000 mL   1 mol   1L    = 483 mL  97.99 g   0.750 mol H3PO4   1 L 

(b)

(35.5 g H3PO4 ) 

(c)

7.34 ×10 22 molecules H 3 PO 4 

  1000 mL  1 mol 1L      23  6.022 ×10 molecules   0.750 mol H3 PO 4   1 L 

= 163 mL

26. (a)



  1000 mL  1L 3   = 3.4 ×10 mL Cl 1 L 0.250 mol NH   4  

( 0.85 mol NH4Cl ) 

  1000 mL   1 mol   1L 3    = 1.88 × 10 mL Cl L 53.49 g 0.250 mol NH    4 

(b)

( 25.2 g NH4Cl ) 

(c)

 1 mol 1L   2.06 ×10 20 formula units NH 4Cl    23  6.022 ×10 formula units   0.250 mol NH 4Cl   1000 mL    = 1.37 mL  1L 

 g C3H8O3   311 g C3H8O3  27. (a)   (100 ) = 48.7% C3H8O3  (100 ) =   g solution   638 g solution   311 g C3H8O3   1 mol C3H8O3   1000 mL  (b)    = 6.14 M C3H8O3   550 mL   92.09 g C3H8O3   1 L   g HC 2H3O2  14.3 g HC 2H3O2  28. (a)   (100 ) = 2.65% HC 2H3O2  (100 ) =   g solution   540.6 g solution   14.3 g HC 2H3O2   1 mol HC 2 H3O3   1000 mL  (b)    = 0.441 M HC 2 H3O2  540.0 mL    60.05 g HC 2H3O2   1 L  29. Dilution problem V1M1 = V2M2 (a)

V1 = 125 mL V2 = (125 mL + 775 mL ) = 900. mL M1 = 5.0 M M2 = M2

(125 mL ) (5.0 M ) = ( 900. mL )( M 2 ) (125 mL ) (5.0 M ) = 0.694 M M = 2

(b)

900. mL

V1 = 250 mL M1 = 0.25 M

V2 = ( 250 mL + 750 mL ) = 1.00 × 103 mL M2 = M2

( 250 mL ) ( 0.25 M ) = (1.00 ×103 mL ) ( M 2 ) ( 250 mL ) ( 0.25 M ) = 0.063 M M = 2

1.00 ×103 mL

– Chapter 14 –

( 75 mL ) 

1 L   0.50 mol    = 0.038 mol HNO3  1000 mL   1 L 

( 75 mL ) 

1 L   1.5 mol    = 0.11 mol HNO3  1000 mL   1 L  Total mol = 0.15 mol Total volume = 75 mL + 75 mL = 150. mL = 0.150 L 0.150 mol = 1.00 M 0.150 L

30. Dilution problem V1M1 = V2M2 (a)

V1 = 175 mL V2 = (175 mL + 275 mL ) = 450. mL M2 = M2 M1 = 3.0 M

(175 mL ) ( 3.0 M ) = ( 450. mL )( M 2 ) (175 mL ) (3.0 M ) = 1.2 M M = 2

(b)

450. mL

V1 = 350 mL

V2 = ( 350 mL +150 mL ) = 5.0 ×102 mL

M1 = 0.10 M

M2 = M2

(350 mL ) ( 0.10 M ) = (5.0 ×102 mL ) ( M 2 ) (350 mL ) ( 0.10 M ) = 0.070 M M = 2

5.0 ×102 mL

(50.0 mL ) 

1 L   0.250 mol HCl    = 0.0125 mol HCl 1L  1000 mL   

( 25.0 mL ) 

1 L   0.500 mol HCl    = 0.0125 mol HCl 1L  1000 mL    Total mol = 0.0250 mol HCl Total volume = 50.0 mL + 25.0 mL = 75.0 mL = 0.0750 L 0.250 mol = 0.333 M 0.0750 L

31. V1M1 = V2M2 (a)

( V1 ) (15 M ) = ( 750 mL )(3.0 M ) ( 750 mL ) (3.0 M ) = 150 mL 15 M H PO V = 1

(b)

15 M

( V1 ) (16 M ) = ( 250 mL )( 0.50 M ) ( 250 mL ) ( 0.50 M ) = 7.8 mL 16 M HNO V = 1

16 M

– Chapter 14 –

32. V1M1 = V2M2 (a)

( V1 ) (18 M ) = ( 225 mL )( 2.0 M ) ( 225 mL ) ( 2.0 M ) = 25 mL 18 M H SO V = 1

(b)

18 M

( V1 ) (15 M ) = ( 75 mL )(1.0 M ) ( 75 mL ) (1.0 M ) = 5.0 mL 15 M NH V = 1

15 M

33. M1V1 = M2V2

M   6.05×10 −5 M  → V2 = V1  1  = (10.0 L )   = 0.526 L zingerone solution −3  1.15×10 M   M2  34. M1V1 = M2V2

M   7.15×10 −6 M  → V2 = V1  1  = ( 2.50 L )   = 0.00464 L anethole solution or 4.64 mL −3  3.85×10 M   M2  35. 2 KMnO4 ( aq ) + 16 HCl ( aq ) → 2 MnCl2 ( aq ) + 5 Cl2 ( g ) +8 H2O ( l ) + 2 KCl ( aq ) (a)

(15.0 mL HCl ) 

0.250 mol   8 mol H2 O  −3   = 1.88 ×10 mol H2 O  1000 mL   16 mol HCl 

(b)

(1.85 mol MnCl2 ) 

(c)

(125 mL KCl ) 

(d)

(15.60 mL KMnO4 ) 

 2 mol KMnO4   1 L KMnO4   = 12.3 L KMnO4  2 mol MnCl2   0.150 mol KMnO4 

0.525 mol   16 mol HCl   1000 mL     = 210. mL HCl  1000 mL   2 mol KCl   2.50 mol 

0.250 mol   16 mol HCl   = 0.0312 mol HCl   1000 mL   2 mol KMnO 4 

 0.0312 mol HCl  M =  = 1.41 M HCl  0.02220 L 

(e)

mL HCl → mol HCl → mol Cl2 → L Cl 2 ( gas at STP ) 2.5 mol   5 mol Cl 2   22.4 L     = 2.2 L Cl 2  1000 mL   16 mol HCl   mol 

(125 mL HCl )  (f)

Limiting reactant problem. Convert volume of both reactants to liters of Cl2 gas.

(15.0 mL HCl ) 

0.750 mol   5 mol Cl2   22.4 L    = 0.0788 L Cl2   1000 mL   16 mol HCl   mol  0.550 mol   5 mol Cl2   22.4 L     = 0.373 L Cl2  1000 mL   2 mol KMnO 4   mol 

(12.0 mL KMnO4 ) 

HCl is limiting reactant. 0.0788 L of Cl2 is produced.

– Chapter 14 –

36. K2CO3 ( aq ) + 2 HC2H3O2 ( aq ) → 2 KC2H3O2 ( aq ) + H2O ( l ) + CO2 ( g )

0.150 mol   1 mol H2 O  −3  = 1.88 ×10 mol H2 O  1000 mL 2 mol HC H O   2 3 2 

(a)

( 25.0 mL HC 2H3O2 ) 

(b)

(17.5 mol KC2H3O2 ) 

(c)

( 75.2 mL K 2CO3 ) 

 1 mol K 2CO3   1 L K 2CO3   = 41.7 L K 2CO3  2 mol KC 2 H3O2   0.210 mol K 2CO3 

0.750 mol   2 mol HC 2H3O2  1000 mL HC 2H3O2      1000 mL   1 mol K 2CO3   1.25 mol HC 2H3O2 

= 90.2 mL HC 2H3O2 (d)

0.250 mol   2 mol HC 2 H3O 2  −3  = 9.25 ×10 mol HC 2H3O 2  1000 mL CO 1 mol K   2 3 

(18.50 mL K 2CO3 ) 

 9.25×10 −3 mol HC 2H3O 2  M =  = 0.911 M HC 2 H3O2 0.01015 L  

(e)

mL HC 2H3O2 → mol HC 2H3O2 → mol CO2 → L CO2 ( gas at STP )   22.4 L  1.5 mol   1 mol CO2    = 1.8 L CO2  1000 mL   2 mol HC 2H3O2   mol 

(105 mL of HC2H3O2 )  (f)

Limiting reactant problem. Convert volume of both reactants to liters of CO2 gas.

0.350 mol   1 mol CO2   22.4 L    = 0.196 L CO2   1000 mL   1 mol K 2CO3   mol 

( 25.0 mL K 2CO3 ) 

0.250 mol   1 mol CO2   22.4 L     = 0.0700 L CO2  1000 mL   1 mol HC 2H3O2   mol 

( 25.0 mL HC 2H3O2 ) 

HC2H3O2 is the limiting reactant. 0.0700 L of CO2 is produced. 37. Molality = m = (a)

mol solute kg solvent

 2.0 mol HCl   1000 g     = 11 m HCl  175 g H2 O   kg 

 14.5 g C12 H22 O11   1000 g   1 mol  (b)     = 0.0770 m C12H22 O11  550.0 g H2 O   kg   342.3 g  (c)

 25.2 mL CH3OH   0.791 g   1000 g   1 mol      = 0.985 m CH3OH   595 g CH3CH2 OH   mL   kg   32.04 g 

– Chapter 14 –

38. Molality = m = (a)

mol solute kg solvent

 125 mol CaCl2   1000 g     = 1.67 m CaCl2  750.0 g H2 O   kg 

 2.5 g C 6H12 O6   1000 g   1 mol  (b)     = 0.026 m C 6H12 O6  525 g H2 O   kg  180.2 g  (c)

 17.5 mL ( CH3 )2 CHOH   0.785 g   1 mL  1000 g  1 mol        35.5 mL H2 O    mL   1.00 g   kg  60.09 g  = 6.44 m ( CH3 )2 CHOH

 2.68 g C10 H8     1000 g  1 mol 39. (a)     = 0.544 m  38.4 g C 6H6   128.2 g C10H8   kg  (b)

Kf ( for benzene ) =

5.1°C Freezing point of benzene = 5.5°C m

 5.1°C  Δt f = ( 0.544 m )   = 2.8°C  m 

Freezing point of solution = 5.5°C – 2.8°C = 2.7°C (b)

K b ( for benzene ) =

2.53°C Boiling point of benzene = 80.1°C m

 2.53°C  Δt b = ( 0.544 m )   = 1.38°C  m 

Boiling point of solution = 80.1°C + 1.38°C = 81.5°C 40. (a)

(b)

 100.0 g C 2H6 O2   1 mol   1000 g      = 10.74 m C 2H6 O2  150.0 g H2 O  62.07 g  kg   1.86°C  Δt f = mK f = (10.74 m )   = 20.0°C ( Decrease in freezing point )  m 

Freezing point = 0.00°C − 20.0°C = −20.0°C

(c)

 0.512°C  Δt b = mK b = (10.74 m )   = 5.50°C ( Increase in boiling point ) m  

Boiling point = 100.00°C + 5.50°C = 105.50°C

– Chapter 14 –

41. Freezing point of acetic acid is 16.6°C K f acetic acid =

3.90°C m

Δt f = 16.6°C −13.2°C = 3.4°C Δt f = mK f

3.4°C = 0.87 m 3.90°C/m

Convert 8.00 g unknown/60.0 g HC2H3O2 to g/mol (molar mass) Conversion:

g unknown g unknown g → → g HC 2H3O2 kg HC 2 H3O2 mol

 8.00 g unknown  1000 g   1 kg HC 2 H3O2      = 153 g/mol  60.0 g HC 2H3O2   kg   0.87 mol unknown 

42. Δt f = 2.50°C Kf ( for H2 O ) =

1.86°C m

Δt f = mK f m=

2.50°C = 1.34 m 1.86°C/m

Convert 4.80 g unknown/22.0 g H2O to g/mol (molar mass)

 4.80 g unknown   1000 g   1 kg H2 O      = 163 g/mol  22.0 g H2 O   kg   1.34 mol unknown  43. Jasmine tea and gasoline are true solutions. 44. Concord grape juice and stainless steel are true solutions. 45. Chromium metal is not a true solution because it is only a single substance. Muddy water is not a true solution because the dirt particles will settle out. 46. Red paint and oil and vinegar salad dressing are not true solutions because they will separate out if they are not shaken or stirred. 47. (a) The teaspoon of sugar will dissolve faster because it has a smaller particle size. (b) The copper(II) sulfate will dissolve in the pure water faster because solutes dissolve in a pure solvent faster than in a solvent with some solute already dissolved. (c) The sweetener will dissolve faster in a cup of hot tea because the rate of dissolving increases with increasing temperature. (d) The silver nitrate will dissolve faster in the sloshing water because agitation increases the rate of dissolution. 48. (a) The amino acid will dissolve faster in the solvent at 75°C because the rate of dissolving increases with increasing temperature. (b) The powdered sodium acetate will dissolve faster because it has a smaller particle size.

– Chapter 14 –

(c) The carton of table salt will dissolve faster because it has both a smaller particle size and the temperature is higher. (d) The acetaminophen will dissolve faster in the infant pain medication because it has less acetaminophen already dissolved in it. mol   49. Molarity Problem  M =  L  

 396.1 g KI   1 mol   1000 mL      = 3.182 M KI  750.0 mL   166.0 g   1 L  mol   50. Molarity Problem  M =  L  

 74.15 g HgCl2   1 mol   1000 mL      = 1.092 M HgCl2  250.0 mL   271.5 g   1 L  51. Salt (NaCl) is an ionic compound. When it is dissolved in water the sodium and chloride ions separate or dissociate. The polar water molecules are attracted to the sodium and chloride ions and are hydrated (surrounded by water molecules). The sodium and chloride ions are separated from one another and distributed throughout the water in this way. Na+ and Cl− are hydrated in aqueous solution: See Question 8 in the Review Questions. 52. Sugar molecules are not ionic; therefore they do not dissociate when dissolved in water. However, sugar molecules are polar, so water molecules are attracted to them and the sugar becomes hydrated. The water molecules help separate the sugar molecules from each other and distribute them throughout the solution. 53. Sugar and salt behave differently when dissolved in water because salt is an ionic compound and sugar is a molecular compound. 54. Solids and liquids are generally more soluble at high temperatures. Gases are always more soluble at low temperatures.

 3.84 g NaCl   mass NaCl  55. % NaCl =   (100 ) = 7.63% NaCl  (100 ) =   total mass   50.320 g solution  56. An isotonic sodium chloride solution has the same osmotic pressure as human blood plasma. When blood cells are placed in an isotonic solution the osmotic pressure inside the cells is equal to the osmotic pressure outside the cells so there is no change in the appearance of the blood cells. 57. The KMnO4 crystals give the solution its purple color. The purple streaks are formed because the solute has not been evenly distributed throughout that solvent yet. The MnO4− has a purple color in solution. 58. The line for KNO3 slopes upward, because the solubility increases as the temperature increases. KNO3 has the steepest slope of all the compounds given in the diagram. It exhibits the greatest increase in the number of grams of solute that is able to dissolve in 100 g of water than any other compound in the diagram as the temperature increases.

– Chapter 14 –

 9.0 g NaCl   1 mol  59.   = 0.15 M NaCl  1L    58.44 g  60. (a)

1000 mL   1.06 g   15.0 g sugar  2  = 1.6 × 10 g sugar   L    mL   100.g syrup 

(1.0 L syrup ) 

 1.6 ×102 g C12H22 O11   1 mol  (b)    = 0.47 M C12H22 O11 L   342.3 g  m=

(c)

mol sugar 15% sugar by mass = 15.0 g C12 H22 O11 + 85.0 g H2 O kg H2 O

 15.0 g C12H22 O11   1000 g  1 mol      = 0.516 m C12H22 O11  85.0 g H2 O   1 kg  342.3 g 

5.1°C Δt f = 0.614°C m  3.84 g C 4 H2 N   1000 g  15.4 g C 4 H2 N   = kg C 6H6  250.0 g C 6H6   kg 

61. Kf =

Δt f = mK f m=

0.614°C 0.12 mol C 4 H2 N = 0.12 m = 5.1°C/m kg C 6H6

 15.4 g C 4H2 N    1 kg C 6 H6 2    = 128 g/mol = 1.3×10 g/mol  kg C 6 H6   0.12 mol C 4 H2 N  Empirical mass ( C 4 H2 N ) = 64.07 g 130 g = 2.0 ( number of empirical formulas per molecular formula ) 64.07 g

Therefore, the molecular formula is twice the empirical formula, or C8H4N2. 62. First calculate the moles of sodium chloride in 700. mL of water from the Great Salt Lake  3.42 mole NaCl  700. mL Lake water   = 2.39 mol NaCl  1000 mL Lake water 

Next calculate the volume of fresh water needed to dilute this to a salt concentration of 0.0171 M  1 L drinkable water  2.39 mol NaCl   = 140. L drinkable water  0.0171 mol NaCl 

700. mL of water must be diluted to 140. L with fresh water. 63.

1 qt   946.1 mL   14 mL ethyl alcohol     qt  32 oz     100 mL witch hazel  = 66 mL ethyl alcohol

(16 fl. oz witch hazel ) 

– Chapter 14 –

64. Verification of Kb for water

Δt b = mK b

Δt b = 101.62°C − 100°C = 1.62°C K b =

Δt b m

First calculate the molality of the solution.

16.10 g C 2H6 O2 3.16 mol C 2H6 O2 = kg H2 O ( 62.07 g/mol )( 0.0820 kg H2 O )

Kb = 65. (a)

Δt b 0.513°C kg H2 O 1.62°C = = m 3.16 mol/kg H2 O mol

(500.0 mL solution ) 

0.90 g NaCl   = 4.5 g NaCl  100. mL solution 

 4.5 g NaCl  (b)   (100 ) = 9.0% x = volume of 9.0% solution  x mL  4.5 g NaCl x= (100 ) = 50. mL ( 4.5 g NaCl in solution ) 9.0% 500. mL − 50. mL = 450. mL H 2 O must evaporate

66.

67.



 1 mol C8H10 N 4 O2     1000 mg C8H10 N 4 O2   194.2 g C8H10 N 4 O2   1000 mL energy drink    = 62.9 mL energy drink  0.02080 mol C8H10 N 4 O2 

( 254 mg C8H10 N4 O2 ) 

1 g C8H10 N 4 O2

( 0.414 mL soln ) 

0.5298 mol methyl jasmonate   1000 mL solution  

 6.022 ×10 23 moleculesmethyl jasmonate    1 mol methyl jasmonate   20 = 1.32 ×10 molecules methyl jasmonate

68. Mass percent of glucose solution is 47.6%; molality is 5.05 m. Blood glucose molality is 0.50 m. The molar mass of glucose is 180.2 g/mol. A solution of 1.00 g glucose in 1.10 mL of H 2 O is equal to a mass percent concentration of 1.00 g glucose ×100 = 47.6% (Remember, water has a density of 1.00 g/mL.) (1.10 g H2 O ) + (1.00 glucose ) 1.00 g glucose 1000. g water × = 909 g glucose dissolve in 1.000 kg water 1.10 g water Converting this to moles

The molality is

909 g 180.2 g /mol

= 5.04 mol glucose dissolve in 1.000 kg water

5.04 mol glucose = 5.04 m 1.000 kg water

– Chapter 14 –

Human blood has a concentration of 0.090% or 0.90 g per 1000 g of blood. Using the assumption that 1000 g of blood is equivalent to 1000 g of solvent, the molality can be calculated as follows: 0.90 g 180.2 g /mol molality =

= 0.0050 mol glucose

0.0050 mol = 0.0050 m 1 kg solvent

69. From Figure 14.4, the solubility of KNO3 in H2O at 20°C is 32 g per 100. g H2O.  100. g H2 O   = 156 g H 2 O to produce a saturated solution.  32.0 g KNO3 

(50.0 g KNO3 ) 

175 g H2 O − 156 g H2 O = 19 g H2 O must be evaporated.   1 L  100 mL solution 70. 7.35 mL oil of wintergreen    = 2.94 L solution  0.25 mL oil of wintergreen   1000 mL 

71. (a)

(b)

1000 mL solution   1.21 g   35.0 g HNO3   = 424 g HNO3   L solution   mL   100. g solution 

(1.00 L solution) 

 1000 mL solution   1.00 L    = 1.18 L solution  424 g HNO3   1000 mL 

(500. g HNO3 ) 

72. Molarity = M =

mol solute L solution

 85 g H3PO4   1.7 g solution   1000 mL   1 mol H3PO4     = 15 M H3PO4   L   97.99 g   100 g solution   mL solution   73. First calculate the molarity of the solution

 80.0 g H2SO4  1000 mL   1 mol   = 1.63 M H2SO4    L   98.09 g   500. mL   M1V1 = M 2 V2

(1.63 M )( 500. mL ) = ( 0.10 M )( V2 ) (1.63 M )( 500. mL ) = 8.2 ×103 mL = 8.2 L V = 2

0.10 M

 4 qt   1 L   4.28 mol C 3H8O3   92.09 g   1 lb  74. 30.0 gal   = 98.6 lb C 3H8O3     1L   1 mol   453.6 g   1 gal   1.057 qt  

75. Mg + 2 HCl → MgCl 2 + H 2 ( g ) mL HCl → mol HCl → mol H 2

(a)

( 200.0 mL HCl ) 

3.00 mol   1 mol H 2    = 0.300 mol H 2  1000 mL   2 mol HCl 

– Chapter 14 –

(b)

PV = nRT  1 atm  P = ( 720 torr )   = 0.95 atm  760 torr  T = 27°C = 300. K n = 0.300 mol V=

nRT ( 0.300 mol ) ( 0.0821 L atm/mol K )( 300. K ) = = 7.8 L H2 P 0.95 atm

76. Mg ( OH)2 + 2 HCl → MgCl2 + 2 H2O

Al ( OH)3 + 3 HCl → AlCl3 + 3 H2O

Calculate the moles of HCl neutralized by each base. 



1 mol  2 mol HCl  = 0.0411 mol HCl (1.20 g Mg ( OH ) )  58.33   g 1 mol Mg ( OH )  2











1 mol  3 mol HCl  = 0.0385 mol HCl (1.00 g Al ( OH ) )  78.00  g  1 mol Al ( OH )  3







1.20 g Mg(OH)2 reacts with more HCl than 1.00 g Al(OH)3. Therefore, Mg(OH)2 is more effective in neutralizing stomach acid. 77. (a) With equal masses of CH3OH and C2H5OH, the substance with the lower molar mass will represent more moles of solute in solution. Therefore, the CH3OH will be more effective than C2H5OH as an antifreeze. (b) Equal molal solutions will lower the freezing point of the solution by the same amount. 78. Calculate molarity and molality. Assume 1000 mL of solution to calculate the amounts of H2SO4 and H2O in the solution.

(1000 mL solution ) 

1.29 g  3  = 1.29 ×10 g solution  mL 

38 g H SO  (1.29 ×10 g solution )  100  = 4.9 ×10 g H SO g solution 3

  3 2 1.29 ×10 g solution − 4.9 × 10 g H2SO4 = 8.0 × 102 g H2 O in the solution  490 g H2SO4  1000 g  1 mol  m=    = 6.2 m H2SO 4 2  8.0 ×10 g H2 O   kg  98.09 g   4.9 × 102 g H2SO 4  1 mol  M =   = 5.0 M H2SO4 L    98.09 g  79. Freezing point depression is 5.4°C (a)

Δt f = mKf Δt 5.4°C m= f = = 2.9 m K f 1.86°C kg solvent/mol solute

– Chapter 14 –

(b)

0.512°C kg solvent 0.512°C = mol solvent m  0.512°C  Δt b = mK b = ( 2.9 m )   = 1.5°C m  

K b ( for H2 O ) =

Boiling point = 100°C + 1.5°C = 101.5°C 80. Freezing point depression = 0.372°C Kf =

1.86°C m

Δt f = mK f 0.372°C = 0.200 m 1.86°C / m  1 mol  ( 6.20 g C 2H6 O2 )   = 0.100 mol C 2H6 O2  62.07 g  m=



 1000 g H2 O  1 kg H2 O   = 500. g H2 O  0.200 mol C 2 H6 O2   kg H2 O 

( 0.100 mol C 2H6 O2 ) 

81. (a) Freezing point depression = 20.0°C  1000 mL   1.00 g  4 12.0 L H2 O    = 1.20 ×10 g H2 O L    mL  Δt f = mK f 20.0°C = 10.8 m 1.86°C/m  8 mol C 2H6 O2   62.07 g  3 (1.20 ×104 g H2O )  10.1000   = 8.04 ×10 g C 2H6 O 2 g H O mol   2   m=

(b)

mL  (8.04×10 g C H O )  1.00  = 7.24 ×10 mL C H O 1.11 g

(c)

°F = 1.8 ( °C ) +32 =1.8 ( −20.0 ) +32 = −4.0°F





82. HNO3 + NaHCO3 → NaNO3 + H2 O + CO2 First calculate the grams of NaHCO3 in the sample.

mLHNO3 → L HNO3 → mol HNO3 → mol NaHCO3 → g NaHCO3 1 L  0.055 mol   1 mol NaHCO3   84.01 g      L  1000 mL     1 mol HNO3   mol 

(150 mL HNO3 ) 

= 0.69 g NaHCO3 in the sample  0.69 g NaHCO3    (100 ) = 47% NaHCO3  1.48 g sample 

– Chapter 14 –

 866 mg C 9H8O   1 g C 9H8O   1 mol C9H8O   1000 mL      83. (a)  600 mL   1000 mg C 9H8O   132.2 g C 9H8O   1 L  = 0.0109 M C 9H8O

 g C9H8O   0.866 g C9H8O  (b)   (100 ) =   (100 ) = 0.140% C9H8O  g solution   618 g solution  (c)

 g C 9 H8 O   0.866 g C 9H8O    (100 ) =   (100 ) = 0.144 m/v% C 9H8O  mL solution   600 mL solution 

84. M1V1 = M2 V2

V   6.04 mL  → M 2 = M1  1  = 2.75 M   = 0.0111 M Ca ( C 2 H3O2 )2  1500 mL   V2  85. (a) Dilution problem: M1V1 = M2 V2

(1.5 M ) ( 8.4 L ) = (17.8 M )( V2 ) (1.5 M ) ( 8.4 L ) = 0.71 L H SO V = 2

17.8 M

If 0.71 L of 17.8 M H2SO4 are required, then 7.7 L of water must also be added to bring the total volume to 8.4 L.  17.8 mol H2SO 4  (b)   (1.00 mL ) = 0.0178 mol H2SO 4 1000. mL  

(c)

 1.5 mol H2SO 4    (1.00 mL ) = 0.0015 mol H 2SO 4 in each mL  1000. mL 

86. moles HNO3 total = moles HNO3 from 3.00 M + moles HNO3 from 12.0 M

MTVT = M3.00 M V3.00 M + M12.0 M V12.0 M Assume preparation of 1000. mL of 6 M solution Let y = volume of 3.00 M solution; volume of 12.0 M = 1000. mL – y

( 6.00 M )(1000. mL ) = ( 3.00 M )( y ) + (12.0 M )(1000. mL − y ) 6000. mL = 3.00 y mL + 12,000 mL − 12.0 y 6000. mL 6000. mL = 9.00 y y = = 667 mL 3 M 9.00 1000. mL − 667 mL = 333 mL 12 M

Mix together 667 mL 3.00 M HNO3 and 333 mL of 12.0 M HNO3 to get 1000. mL of 6.00 M HNO3. 87. HBr + NaOH → NaBr + H2 O First calculate the molarity of the diluted HBr solution.

– Chapter 14 –

The reaction is 1 mol HBr to 1 mol NaOH, so

M A VA = M B VB

( M A )(100.0 mL ) = ( 0.37 M )(88.4 mL ) ( 0.37 M )( 88.4 mL ) = 0.33 M HBr diluted solution MA = ( ) 100.00 mL

Now calculate the molarity of the HBr before dilution. M1V1 = M 2 V2

( M1 )( 20.0 mL ) = ( 0.33 M )( 240. mL ) ( 0.33 M )( 240. mL ) = 4.0 M HBr original solution M1 = ( ) 20.0 mL

88. Ba ( NO3 )2 + 2 KOH → Ba ( OH)2 + 2 KNO3 This is a limiting reactant problem. First calculate the moles of each reactant and determine the limiting reactant.  moles  M ×L =   ( L ) = moles  L   0.642 mol    ( 0.0805 L ) = 0.0517 mol Ba ( NO3 )2 L    0.743 mol    ( 0.0445 L ) = 0.0331 mol KOH L  

According to the equation, twice as many moles of KOH as Ba(NO3)2 are needed, so KOH is the limiting reactant. The mass of Ba(OH)2 formed will be 2.84 g.

 1 mol Ba ( OH )2   171.3 g Ba ( OH )2  0.0331 mol KOH   = 2.84 g Ba ( OH )2    2 mol KOH   1 mol Ba ( OH )2   0.25 mol  89. (a)   ( 0.0458 L ) = 0.011 mol Li2 CO3 L    0.25 mol   73.89 g  (b)   = 14 g Li2 CO3  ( 0.75 L )  L    mol 

(c)

 1 mol   1000. mL  2   = 3.2 ×10 mL solution  73.89 g   0.25 mol 

( 6.0 g Li2CO3 ) 

(d) Assume 1000. mL solution

 1.22 g    (1000. mL ) = 1220 g solution  mL   0.25 mol   73.89 g Li2CO3     = 18 g Li2CO3 per L solution L mol     g solute   18 g  %=  (100 ) =   (100 ) = 1.5% ( mass percent )  g solution   1220 g  24

– Chapter 14 –

mol   90. Molarity Problem  M =  L  

 0.625 g EuCl3 ⋅ 6 H2 O   1 mol   1000 mL  −3   = 3.40 ×10 M EuCl3 ⋅ 6 H2O   500.0 mL 366.5 g 1 L     91. Picture A represents potassium chloride, picture B represents sucrose, and picture C represents sodium phosphate. 1.86°C   92. Freezing Point Depression  freezing point of water is 0°C, K f water =  m  

Δt f = 0°C − ( −12.7°C ) = 12.7°C Δt f = mK f mions =

12.7°C = 6.83 m ions 1.86°C/m

To determine molality of sodium chloride remember there are 2 ions formed when sodium chloride dissociates in water.

 6.83 mol ions   1 mol NaCl  mNaCl =    = 3.42 m NaCl  kg water   2 mol ions  93. Salt will cause the loss of water from a snail due to osmosis. When salt is sprinkled on a snail it dissolves in the watery coating of the snail. Because the concentration of salt outside the snail is now greater than the concentration of salt on the inside, water will flow out of the snail to dilute the more concentrated solution on its surface. 94. A slice of eggplant coated with a layer of salt will release water as a result of osmosis. When salt is sprinkled on the surface of a slice of eggplant, the water inside the slice will flow out onto the surface to dilute the salt. As the water flows out of the slice of eggplant it will also draw out some of the compounds which are responsible for the bitter flavor of the eggplant. 0.512°C   95. Boiling point Elevation  boiling point of water is 100°C, K b water =  m  

Δtb = 127°C − 100°C = 27°C Δtb = mK b m=

27°C = 53 m suger 0.512°C/m

Taffy is made from a very concentrated sugar solution! 96. The balanced equation is

2 HCl ( aq ) + Na 2SO3 ( aq ) → 2 NaCl ( aq ) + H2 O ( l ) + SO2 ( g )

(125 mL HCl ) 

2.50 mol   1 mol SO2    = 0.156 mol SO2  1000 mL  2 mol HCl  1.75 mol   1 mol SO2   = 0.131 mol SO2   1000 mL   1 mol Na 2SO3 

( 75 mL Na 2 SO3 ) 

– Chapter 14 –

Na2SO3 is the limiting reactant; 0.131 mol of SO2 gas will be produced. The gas is at nonstandard conditions, so use PV = nRT to find the liters of SO2.

nRT P

( 0.131 mol )( 0.0821 L atm/mol K )( 295 K ) = 3.11 L SO 2 ( 775 torr )(1 atm/760 torr )

97. mass of solute = mass of container and solute − mass of container − mass of water mass of water = ( 5.49 moles ) (18.02 g/mol ) = 100.0 g mass of solute =563 g − 375 g − 100.0 g = 88 g solubility in water = g solute/100 g H 2O at 20°C Using this data, solubility is 88 g solute/100.0 g water = NaNO3 (see Table 14.3). 98. 10% KOH m/v solution contains 10 g KOH in 100 mL solution. 10% KOH by mass solution contains 10 g KOH + 90 g H2O. The 10% by mass solution is the more concentrated solution and therefore would require less volume to neutralize the HCl. 99.

0.355 mol = 0.470 M 0.755 L

100. The lower pathway represents the evaporation of water; only a phase change occurs; no new substances are formed. The upper path represents the decomposition of water. The middle path is the ionization of water. 101. For most reactions to occur, molecules or ions need to collide. In the solid phase, the particles are immobile and therefore do not collide. In solution or in the gas phase, particles are more mobile and can collide to facilitate a chemical reaction. 102. Δtb = 81.48°C − 80.1°C = 1.38°C °C kg solvent K b = 2.53 mol solute Δtb = mK b °C kg solvent   1.38°C = m  2.53  mol solute   mol solute =m 0.545 kg solvent

Now we convert molality to molarity.

 5.36 g solute  1000 g benzene   1 kg benzene      = 128. g /mol  76.8 g benzene  1 kg benzene   0.545 mol solute 

CHAPTER 15

ACIDS, BASES, AND SALTS SOLUTIONS TO REVIEW QUESTIONS 1.

The Arrhenius definition is restricted to aqueous solutions, while the BrønstedLowry definition is not.

By the Arrhenius theory, an acid is a substance that produces hydrogen ions in aqueous solution. A base is a substance that produces hydroxide ions in aqueous solution. By the Brønsted-Lowry theory, an acid is a proton donor, while a base accepts protons. Since a proton is a hydrogen ion, then the two theories are very similar for acids, but not bases. A chloride ion can accept a proton (producing HCl), so it is a Brønsted-Lowry base, but would not be a base by the Arrhenius theory, since it does not produce hydroxide ions. By the Lewis theory, an acid is an electron pair acceptor, and a base is an electron pair donor. Many individual substances would be similarly classified as bases by Brønsted-Lowry or Lewis theories, since a substance with an electron pair to donate, can accept a proton. But, the Lewis definition is almost exclusively applied to reactions where the acid and base combine into a single molecule. The BrønstedLowry definition is usually applied to reactions that involve a transfer of a proton from the acid to the base. The Arrhenius definition is most often applied to individual substances, not to reactions. According to the Arrhenius theory, neutralization involves the reaction between a hydrogen ion and a hydroxide ion to form water. Neutralization, according to the Brønsted-Lowry theory, involves the transfer of a proton to a negative ion. The formation of a covalent bond constitutes a Lewis neutralization.

Neutralization reactions: Arrhenius:

Brønsted-Lowry: Lewis:

( H + OH → H O ) HCl + KCN → HCN + KCl ( H + CN → HCN )

HCl + NaOH → NaCl + H2 O

−

AlCl3 + NaCl → AlCl 4 − + Na +

4. These ions are considered to be bases according to the Brønsted-Lowry theory, because they can accept a proton at any of their unshared pairs of electrons. They are considered to be bases according to the Lewis acid-base theory, because they can donate an electron pair. 5.

(a) base

(b)

base

(c)

acid

– Chapter 15 –

Metals that lie above hydrogen in the activity series will form hydrogen gas when they react with an acid.

NaNO3, sodium nitrate; Ca(NO3)2, calcium nitrate; Al(NO3)3, aluminium nitrate. These are three of the many possible salts which can be formed from nitric acid.

LiCl, lithium chloride; Li2SO4, lithium sulfate; Li3PO4, lithium phosphate. These are three of the many possible salts which can be formed from lithium hydroxide.

An electrolyte must be present in the solution for the bulb to glow.

10. Electrolytes include acids, bases, and salts. (Electrolytes are any compound that conducts electricity in solution.) 11. (a) E

(b) NE

(d) E

(e) NE

(f) E

(g) E

12. First, the orientation of the polar water molecules about the Na+ and Cl− is different. The positive end (hydrogen) of the water molecule is directed towards Cl−, while the negative end (oxygen) of the water molecule is directed towards the Na+. Second, more water molecules will fit around Cl−, since it is larger than the Na+ ion. 13. The electrolytic compounds are acids, bases, and salts. 14. Hydrogen chloride dissolved in water conducts an electric current. HCl reacts with polar water molecules to produce H3O+ and Cl− ions, which conduct an electric current. Hexane is a nonpolar solvent, so it cannot pull the HCl molecules apart. Since there are no ions in the hexane solution, it does not conduct an electric current. HCl does not ionize in hexane. 15. In their crystalline structure, salts exist as positive and negative ions in definite geometric arrangement to each other, held together by the attraction of the opposite charges. When dissolved in water, the salt dissociates as the ions are pulled away from each other by the polar water molecules. 16. Testing the electrical conductivity of the solutions shows that CH3OH is a nonelectrolyte, while NaOH is an electrolyte. This indicates that the OH group in CH3OH must be covalently bonded to the CH3 group. 17. Molten NaCl conducts electricity because the ions are free to move. In the solid state, however, the ions are immobile and do not conduct electricity. 18. Dissociation is the separation of already existing ions in an ionic compound. Ionization is the formation of ions from molecules. The dissolving of NaCl is a dissociation, since the ions already exist in the crystalline compound. The dissolving of HCl in water is an ionization process, because ions are formed from HCl molecules and H2O. 19. Strong electrolytes are those which are essentially 100% ionized or dissociated in water. Weak electrolytes are those which are only slightly ionized in water. 20. (a) w

(b) w

(d) w

(e) s

(f)

(g) s

(h) s

(i) w

(j) s

21. Ions are hydrated in solution because there is an electrical attraction between the charged ions and the polar water molecules.

– Chapter 15 –

22. The main distinction between water solutions of strong and weak electrolytes is the degree of ionization of the electrolyte. A solution of an electrolyte contains many more ions than does a solution of a nonelectrolyte. Strong electrolytes are essentially 100% ionized. Weak electrolytes are only slightly ionized in water. 23. The pH for a solution with a hydrogen ion concentration of 0.003 M will be between 2 and 3. 24. Tomato juice is more acidic than blood, since its pH is lower. 25. (a) In a neutral solution, the concentration of H+ and OH− are equal. (b) In an acid solution, the concentration of H+ is greater than the concentration of OH−. (c) In a basic solution, the concentration of OH− is greater than the concentration of H+. 26. Pure water is neutral because when it ionizes it produces equal molar concentrations of acid [H+] and base [OH−] ions. 27. A neutral solution is one in which the concentration of acid is equal to the concentration of base [H+] = [OH−]. An acidic solution is one in which the concentration of acid is greater than the concentration of base [H+] > [OH−]. A basic solution is one in which the concentration of base is greater than the concentration of acid [H+] < [OH−]. 28. A titration is used to determine the concentration of a specific substance (often an acid or a base) in a sample. A titration determines the volume of a reagent of known concentration that is required to completely react with a volume of a sample of unknown concentration. An indicator is used to help visualize the endpoint of a titration. The endpoint is the point at which enough of the reagent of known concentration has been added to the sample of unknown concentration to completely react with the unknown solution. An indicator color change is visible when the endpoint has been reached. 29. The net ionic equation for an acid-base reaction in aqueous solutions is: H + + OH − → H 2 O

– Chapter 15 –

SOLUTIONS TO EXERCISES 1. Conjugate acid-base pairs: (a)

NH3 − NH +4 ; H 2 O − OH −

(b) HC 2 H3O 2 − C 2 H3O 2− ; H 2 O − H3O + (c)

− H2 PO 4− − HPO 2− 4 ; OH − H 2 O

(d) HCl − Cl − ;H 2 O − H3 O + 2.

Conjugate acid-base pairs: (a)

H2S − HS− ; NH3 − NH 4 +

(b)

+ HSO −4 − SO 2− 4 ; NH3 − NH 4

(c)

HBr − Br − ; CH3 O − − CH3 OH

(d) HNO3 − NO3 − ; H2 O − H3O + 3.

(a)

Zn ( s ) + 2 HCl ( aq ) → ZnCl 2 ( aq ) + H 2 ( g

(b) 2 Al ( OH )3 ( s ) + 3 H 2SO 4 ( aq ) → Al 2 ( SO 4 )3 ( aq ) + 6 H 2 O ( l ) (c)

Ca ( HCO 3 )2 ( s ) + 2 HBr ( aq ) → CaBr2 ( aq ) + 2 H 2 O ( l ) + 2 CO 2 ( g )

(d) 3 KOH ( aq ) + H 3 PO 4 ( aq ) → K 3 PO 4 ( aq ) + 3 H 2 O ( l ) 4.

Complete and balance these equations: (a) Mg ( s ) + 2 HClO 4 ( aq ) → Mg ( ClO 4 )2 ( aq ) + H 2 ( g ) (b) 2 Al ( s ) + 3 H 2SO 4 ( aq ) → Al 2 ( SO 4 )3 ( aq ) + 3 H 2 ( g ) (c)

2 NaOH ( aq ) + H 2 CO 3 ( aq ) → NaCO3 ( aq ) + 2 H 2 O ( l )

(d) Ba ( OH )2 ( s ) + 2 HClO 4 ( aq ) → Ba ( ClO 4 )2 ( aq ) + 2 H 2 O ( l ) 5.

(a)

Zn + ( 2 H + + 2 Cl − ) → ( Zn2+ + 2 Cl − ) + H2 Zn + 2 H + → Zn2+ + H 2

3+ (b) 2 Al ( OH )3 + ( 6 H + + 3 SO 2− + 3 SO 2− 4 ) + 6 H2 O 4 ) → ( 2 Al

2 Al ( OH )3 + 6 H + 2 Al3+ + 6 H2 O or Al ( OH )3 + 3 H+ → Al3+ + 3 H2 O

– Chapter 15 –

Ca ( HCO3 )2 + ( 2 H+ + 2 Br − ) → ( Ca 2+ + 2 Br − ) + 2 H2 O+2 CO 2

(c)

Ca ( HCO 3 )2 + 2 H + → Ca 2+ + H 2 O + CO 2

(3 K + 3 OH ) + H PO → (3 K + PO ) + 3 H O +

(d)

−

3− 4

−

3 OH + H3PO 4 → PO + 3 H 2 O

3− 4

Mg + ( 2 H + + 2 ClO 4− ) → ( Mg 2+ + 2 ClO 4− ) + H2

(a)

Mg + 2 H + → Mg 2+ + H 2 3+ (b) 2 Al + ( 6 H+ → 3 SO2− + 3 SO2− 4 ) → ( 2 Al 4 ) + 3 H2

2 Al + 6 H + → 2 Al3+ + 3 H 2

( 2 Na + 2 OH ) + H CO → ( 2 Na + CO ) + 2 H O +

(c)

−

2− 3

2 OH + H 2 CO3 → CO + 2 H 2 O 2− 3

(d) Ba ( OH )2 + ( 2 H + + 2 ClO 4− ) → ( Ba 2+ + 2 ClO −4 ) + 2 H2 O Ba ( OH )2 + 2 H + → Ba 2+ + 2 H 2 O

HC3 H5O2 + NH3 → C3 H5O2 − + NH 4 −       acid

base

conjugate acid

HClO4 + O H − → ClO4 − + H 2O       base congjugate base

acid

congjugate base

(a)

conjugate acid

HNO3 + NaOH → H2 O + NaNO3

(b) 2 HC 2 H3 O 2 + Ba ( OH )2 → 2 H 2 O + Ba ( C 2 H 3 O 2 )2 (c)

HClO 4 + NH4 OH → H2 O + NH4ClO 4

10. (a)

2 HBr + Mg ( OH )2 → 2 H 2 O + MgBr2

(b) H3PO4 + 3 KOH → 3 H2 O + K3PO 4 (c)

H 2SO 4 + 2 NH 4 OH → 2 H 2 O + ( NH 4 )2 SO 4

11. (a) H2SO4 (b) SO42− 12. (a) H2CO3 (b) CO32− 13. The following compounds are electrolytes: (a)

SO3, acid in water

(e)

CuBr2, salt

(b)

K2CO3, salt

(f)

HI, acid in water

– Chapter 15 –

14. The following compounds are electrolytes: (b)

P2O5, acid in water

(d)

LiOH, base

(c)

NaClO, salt

(f)

KMnO4, salt

15. Molarity of ions. (a)

 1 mol Cu 2+  2+  = 1.25 M Cu 1 mol CuBr  2 

(1.25 M CuBr2 ) 

 2 mol Br −  − 1.25 M CuBr (  = 2.50 M Br 2 )  1 mol CuBr2  (b)



1 mol Na +  +  = 0.75 M Na  1 mol NaHCO3 

( 0.75 M NaHCO3 ) 

 1 mol HCO3−  −  = 0.75 M HCO3 1 mol NaHCO 3  

( 0.75 M NaHCO3 )  (c)



3 mol K +  +  = 10.5 M K 1 mol K AsO 3 4  

(3.50 M K3 AsO4 ) 

 1 mol AsO3−  3− 4  = 3.50 M AsO 4 l K AsO 1 mo 3 4  

(3.50 M K3 AsO4 )  (d)





+ 4

NH  = 1.3 M NH ( 0.65 M ( NH ) SO )  1 mol2 mol ( NH ) SO  4 2



4 2



2− 4



4 2



+ 4



SO  = 0.65 M SO ( 0.65 M ( NH ) SO )  1 mol1 mol ( NH ) SO  4

4 2



16. Molarity of ions. (a)

 1 mol Fe3+  3+  = 2.25 M Fe  1 mol FeCl3 

( 2.25 M FeCl3 ) 

 3 mol Cl −  −  = 6.75 M Cl 1 mol FeCl 3  

( 2.25 M FeCl3 )  (b)

 1 mol M g 2+  2+  = 1.20 M M g 1 mol MgSO  4 

(1.20 M MgSO4 ) 

 1 mol SO2−  2− 4 1.20 M MgSO (  = 1.20 M SO4 4 )  1 mol MgSO4  (c)

  1 mol Na + + ( 0.75 M NaH2 PO4 )   = 0.75 M Na  1 mol NaH2 PO 4   1 mol H2 PO 4−  − ( 0.75 M NaH2 PO4 )   = 0.75 M H2 PO4  1 mol NaH2 PO 4 

2− 4

– Chapter 15 –

(d)





3 2

+ −pH H  = 10

+ −5 H   = 1×10

+ −7 (b) H   = 2 × 10

(c)

+ −8 H   = 1×10

+ −10 (d) H   = 2 ×10

18. pH = −log H +  (a)

+ −pH H  = 10

+ −7 H   = 1× 10

(b) H+  = 5 ×10 −5 (c)

+ −6 H   = 2 × 10

+ −11 (d) H   = 5 × 10

19. (a)

3 2    2 mol ClO3−  −  = 0.70 M ClO3 ( 0.35 M Ca ( ClO3 )2 )  1 mol  Ca ( ClO3 )2  

17. pH = −log H+  (a)

1 mol Ca  = 0.35 M Ca ( 0.35 M Ca ( ClO ) )  1 mol Ca ( ClO ) 

(85.5 mL ) 

0.50 mol HCl   = 0.043 mol HCl  1000 mL 

( 75.0 mL ) 

1.25 mol HCl   = 0.0938 mol HCl  1000 mL 

Total mol HCl = 0.043 mol + 0.0938 mol = 0.137 mol HCl Total volume = 0.0855 L + 0.0750 L = 0.1605 L 0.137 mol HCl = 0.854 M HCl 0.1605 L  1 mol H+  + ( 0.854 M HCl )   = 0.854 M H  1 mol HCl   1 mol Cl −  −  = 0.854 M Cl  1 mol HCl 

( 0.854 M HCl ) 

– Chapter 15 –

(b)

(125 mL ) 

0.75 mol CaCl2   = 0.094 mol CaCl2 1000 mL  

0.25 mol CaCl2   = 0.031 mol CaCl2 1000 mL   Total mol CaCl2 = 0.094 mol + 0.031 mol = 0.125 mol CaCl2

(125 mL ) 

Total volume = 0.125 L + 0.125 L = 0.250 L 0.125 mol CaCl2 = 0.500 M CaCl2 0.250 L  1 mol Ca 2+  2+ 0.500 M CaCl (  = 0.500 M Ca 2 )  1 mol CaCl2   2 mol Cl −  −  = 1.00 M Cl  1 mol CaCl2 

( 0.500 M CaCl2 )  (c)

NaOH + HCl → NaCl + H2 O

( 65.0 mL ) 

0.333 mol NaOH   = 0.0216 mol NaOH 1000 mL  

( 22.5 mL ) 

0.250 mol HCl   = 0.00563 mol HCl  1000 mL 

0.00563 mol HCI reacts with 0.00563 mol NaOH. 0.0160 mol NaOH remains unreacted and 0.00563 mol NaCl is produced. The final volume is 0.0875 L and contains 0.0160 mol NaOH and 0.00563 mol NaCl. Moles of ions are: (0.0160 mol Na+ + 0.00563 mol Na+) = 0.0216 mol Na+, 0.0160 mol OH−, and 0.00563 mol Cl−. Concentrations of ions are: 0.0216 mol Na + = 0.247 M Na + 0.0875 L 0.00563 mol Cl − = 0.0643 M Cl − 0.0875 L

0.0160 mol OH − = 0.183 M OH− 0.0875 L

(d) H2SO 4 + 2 NaOH → Na 2SO 4 + 2 H2 O

(12.5 mL ) 

0.500 mol H2SO 4   = 0.00625 mol H2SO 4 1000 mL  

( 23.5 mL ) 

0.175 mol NaOH   = 0.00411 mol NaOH 1000 mL  

( 0.00411 mol NaOH ) 

1 mol H2SO 4   = 0.00206 mol H2SO 4 reacted  2 mol NaOH 

0.00206 mol H2SO4 reacts with 0.00411 mol NaOH. 0.00419 mol H2SO4 remains unreacted and 0.00206 mol Na2SO4 is produced. The final volume is 0.0360 L and contains 0.00206 mol Na2SO4 and 0.00419 mol H2SO4. Moles of ions are 0.00412 mol Na+, 0.00838 mol H+, and (0.00206 + 0.00419) = 0.00625 mol SO 2− 4 . Concentration of ions are:

– Chapter 15 –

0.0412 mol Na + = 0.114 M Na + 0.0360 L 0.00625 mol SO24− = 0.174M SO 24− 0.0360 L

20. (a)

0.00838 mol H+ = 0.233 M H + 0.0360 L

( 45.5 mL ) 

0.10 mol NaCl   = 0.0046 mol NaCl  1000 mL 

( 60.5 mL ) 

0.35 mol NaCl   = 0.021 mol NaCl  1000 mL 

Total mol NaCl = 0.0046 mol + 0.021 mol = 0.026 mol NaCl Total volume = 0.0455 L + 0.0605 L = 0.1060 L

0.026 mol NaCl = 0.25 M NaCl 0.1060 L  1 mol Na +  + ( 0.25 M NaCl )   = 0.25 M Na 1 mol NaCl    1 mol Cl −  − 0.25 M NaCl ( )  = 0.25 M Cl  1 mol NaCl  (b)

( 95.5 mL ) 

1.25 mol HCl   = 0.119 mol HCl  1000 mL 

(125.5 mL ) 

2.50 mol HCl   = 0.314 mol HCl  1000 mL 

Total mol HCl = 0.119 mol + 0.314 mol = 0.433 mol HCl Total volume = 0.0955 L + 0.1255 L = 0.2210 L 0.433 mol HCl = 1.96 M HCl 0.2210 L  1 mol H+  + M 1.96 HCl ( )  = 1.96 M H  1 mol HCl   1 mol Cl −  − M 1.96 HCl ( )  = 1.96 M Cl  1 mol HCl  (c)

 0.10 mol Ba ( NO3 )2   = 0.0016M Ba ( NO3 )2 1000 mL   0.20 mol AgNO3  (10.5 mL )   = 0.0021 M AgNO3 1000 mL  

(15.5 mL ) 

Number of moles of each substance: 0.0016 mol Ba2+, 0.0021 mol Ag+, and (0.0032 mol + 0.0021 mol) = 0.0053 mol NO3−

– Chapter 15 –

Total volume = 0.0155 L + 0.0105 L = 0.0260 L 0.0016 mol Ba 2+ = 0.062M Ba 2+ 0.0260 L 0.0021 mol Ag + = 0.081 M Ag + 0.0260 L 0.0053 mol NO3− = 0.20M NO3− 0.0260 L (d)

( 25.5 mL ) 

0.25 mol NaCl   = 0.0064 mol NaCl  1000 mL 

 0.15 mol Ca ( C 2H3O2 )2   = 0.0023 mol Ca ( C 2H3O2 )2 1000 mL  

(15.5 mL ) 

Number of moles of each substance: 0.0064 mol Na+, 0.0064 mol Cl−, 0.0023 mol Ca2+, 0.0046 mol C 2 H3O 2− . Total volume = 0.0255 L + 0.0155 L = 0.0410 L 0.0064 mol Na + = 0.16 M Na + 0.0410 L 0.0064 mol Cl − = 0.16 M Cl− 0.0410 L 0.0023 mol Ca 2+ = 0.056 M Ca 2+ 0.0410 L 0.0046 mol C 2H3O2− = 0.11 M C 2H3O2− 0.0410 L + − 21. HNO3 ( aq ) + H2 O ( l ) → H3O ( aq ) + NO3 ( aq )

Or H2 O HNO3 ( aq ) ⎯⎯⎯ → H + ( aq ) + NO3 − ( aq )

Because nitric acid ionizes completely it would be both a strong electrolyte and a strong acid. + − 22. HCN ( aq ) + H2 O ( l )  H3 O ( aq ) + CN ( aq )

Or + − 2   HCN ( aq )    H ( aq ) + CN ( aq ) H O

HCN is only partially ionized and so it would be a poor electrolyte and a weak acid. 23. The reaction of HCl and NaOH occurs on a 1:1 mole ratio.

HCl + NaOH → NaCl + H2 O At the endpoint in these titration reactions, equal moles of HCl and NaOH will have reacted. Moles = (molarity) (volume). At the endpoint, mol HCl = mol NaOH.

– Chapter 15 –

Therefore, at the endpoint, M A VA = M B VB

MA =

M B VB VA

(a)

(37.70 mL )( 0.728 M ) = 1.10 M HCl

(b)

( 25.77 mL )( 0.306 M ) = 0.415 M HCl

(c)

(18.00 mL )( 0.555 M ) = 0.241 M HCl

24.87 mL

19.00 mL

41.52 mL

24. The reaction of HCl and NaOH occurs on a 1:1 mole ratio.

HCl + NaOH → NaCl + H2 O At the endpoint in these titration reactions, equal moles of HCl and NaOH will have reacted. Moles = (molarity)(volume). At the endpoint, mol HCl = mol NaOH. Therefore, at the endpoint, M A VA = M B VB

MA =

M B VB VA

(a)

(37.19 mL )( 0.126 M ) = 0.147 M NaOH

(b)

( 48.04 mL )( 0.482 M ) = 0.964 M NaOH

(c)

(13.13 mL )(1.425M ) = 0.4750 M NaOH

25. (a)

31.91 mL

24.02 mL

39.39 mL

2+ 2 PO3− ( aq ) → Ca3 ( PO3−4 ) ( s ) 4 ( aq ) + 3 Ca 2

(b) 2 Al ( s ) + 6 H ( aq ) → 3 H 2 ( g ) + 2 Al +

( aq )

CO32 − ( aq ) + 2 H + ( aq ) → H 2 O ( aq ) + CO 2 ( g ) Mg ( s ) + Cu 2 + ( aq ) → Cu ( s ) + Mg 2 + ( aq )

(b) H + ( aq ) + OH − ( aq ) → H 2 O ( l ) (c)

SO 32 − ( aq ) + 2 H + ( aq ) → H 2 O ( l ) + SO 2 ( g )

– Chapter 15 –

27. (a) 1 molar H2SO4 is more acidic. The concentration of H+ in 1 M H2SO4 is greater than 1 M since there are two ionizable hydrogens per mole of H2SO4. In HCl the concentration of H+ will be 1 M, since there is only one ionizable hydrogen per mole HCl. (b) 1 molar HCl is more acidic. HCl is a strong electrolyte, producing more H+ than HC2H3O2 which is a weak electrolyte. 28. (a) 2 molar HCl is more acidic. 2 M HCl will yield 2 M H+ concentration. 1 M HCl will yield 1 M H+ concentration. (b) 1 molar H2SO4 is more acidic. Both are strong acids. The concentration of H+ in 1 M H2SO4 is greater than in 1 M HNO3 because H2SO4 has two ionizable hydrogens per mole whereas HNO3 has only one ionizable hydrogen per mole. 29. 2 HClO 4 ( aq ) + Ca ( OH )2 ( s ) → Ca ( ClO 4 )2 ( aq ) + 2 H 2 O ( l ) g Ca ( OH )2 → mol Ca ( OH )2 → mol HClO 4 → mL HClO 4 



mol  2 mol HClO  1000 mL   ( 61.85 g Ca ( OH ) )  74.10    = 3.18 ×10 mL HClO  g 1 mol Ca ( OH )  0.525 mol  2







30. 3 HCl ( aq ) + Al ( OH )3 ( s ) → AlCl3 ( aq ) + 3 H 2 O ( l )

mL HCl → mol HCl → mol Al ( OH )3 → g Al ( OH )3 0.125 mol   1 mol Al ( OH )3   78.00 g     = 0.894 g Al ( OH )3  1000 mL   3 mol HCl   mol 

( 275 mL HCl ) 

31. NaOH + HCl → NaCl + H2 O First calculate the grams of NaOH in the sample. L HCl → mol HCl → mol NaOH → g NaOH

( 0.01825 L HCl ) 

0.4311 mol  1 mol NaOH  40.00 g     L  1 mol HCl  mol   = 0.3147 g NaOH in the sample

 0.3147 g NaOH    (100 ) = 78.6% NaOH  0.400 g sample  32. NaOH + HCl → NaCl + H2 O L HCl → mol HCl → mol NaOH → g NaOH

( 0.04990 L HCl ) 

0.466 mol  1 mol NaOH  40.00 g     = 0.930 g NaOH in the sample L   1 mol HCl  mol  1.00 g sample − 0.930 g NaOH = 0.070 g NaCl in the sample  0.070 g NaCl    (100 ) = 7.0% NaCl in the sample  1.00 g sample 

– Chapter 15 –

33. Zn + 2 HCl → ZnCl2 + H2 This is a limiting reactant problem. First find the moles of Zn and HCl from the given data and then identify the limiting reactant. g Zn → mol Zn

 1 mol   = 0.0765 mol Zn  65.39 g 

(5.00 g Zn ) 

0.350 mol   = 0.0350 mol HCl L  

( 0.100 L HCl ) 

Therefore Zn is in excess and HCl is the limiting reactant.

( 0.0350 mol HCl ) 

1 mol H2   = 0.0175 mol H2 produced in the reaction  2 mol HCl 

 1 atm  T = 27°C = 300. K P = ( 700. torr )   = 0.921 atm  760 torr  PV = nRT V=

nRT ( 0.0175 mol H2 )( 0.0821 L atm/mol K )( 300. K ) = = 0.468 L H2 P 0.921 atm

34. Zn + 2 HCl → ZnCl2 + H2 This is a limiting reactant problem. First find moles of Zn and HCl from the given data and then identify the limiting reactant. g Zn → mol Zn

 1 mol   = 0.0765 mol Zn  65.39 g 

(5.00 g Zn ) 

( 0.200 L HCl ) 

0.350 mol   = 0.0700 mol HCl L  

Zn is in excess and HCl is the limiting reactant.

( 0.0700 mol HCl ) 

1 mol H2   = 0.0350 mol H2  2 mol HCl 

 1 atm  T = 27°C = 300. K P = ( 700. torr )   = 0.921 atm  760 torr  PV = nRT V=

nRT ( 0.0350 mol H2 O )( 0.0821 L atm/mol K )( 300. K ) = = 0.936 L H2 P 0.921 atm

35. pH = − log H+  (a) [H+] = 0.057 M;

pH = −log(0.057) = 1.24

(b) [H+] = 1.42 M;

pH = −log(1.42) = −0.152

pH = −log(2.0 × 10−8) = 7.70

– Chapter 15 –

36. pH = −log H+  (a) [H+] = 0.0020 M; +

−8

(b) [H ] = 7.0 × 10 M; (c)

[H+] = 3.0 M;

pH = −log(0.0020) = 2.70 pH = −log(7.0 × 10−8) = 7.15 pH = −log(3.0) = −0.48

37. (a) Orange juice = 3.7 × 10−4 M H+ pH = −log (3.7 × 10−4) = 3.43 (b) Vinegar = 2.8 × 10−3 M H+ pH = −log(2.8 × 10−3) = 2.55 (c) Shampoo = 2.4 × 10−6 M H+ pH = −log(2.4 × 10−6) = 5.62 (d) Dishwashing detergent = 3.6 × 10−8 M H+ pH = −log (3.6 × 10−8) = 7.44 38. (a) Black coffee = 5.0 × 10−5 M H+ pH = −log(5.0 × 10−5) = 4.30 (b) Limewater = 3.4 × 10−11 M H+ pH = −log(3.4 × 10−11) = 10.47 (c) Fruit punch = 2.1 × 10−4 M H+ pH = −log(2.1 × 10−4) = 3.68 (d) Cranberry apple drink = 1.3 × 10−3 M H+ pH = −log(1.3 × 10−3) = 2.89 39. (a)

NH3 is a weak base

+ − 2   NH3 ( aq )   NH4 ( aq ) + OH ( aq )

(b)

HCl is a strong acid

H2 O HCl ( aq ) ⎯⎯ ⎯ → H + ( aq ) + Cl − ( aq )

(c)

KOH is a strong base

H2 O KOH ⎯⎯ ⎯ → K + ( aq ) + OH − ( aq )

(d)

HC 2H3O2 is a weak acid

H2 O + −   HC 2 H3O2 ( aq )    H ( aq ) + C 2H3O2 ( aq )

acid

H O

40. (a)

H2C2O4 is a weak acid

+ − 2   H2C 2 O 4 ( aq )   H ( aq ) + HC 2 O 4 ( aq )

(b)

Ba(OH)2 is a strong base

H2 O Ba ( OH )2 ⎯⎯ ⎯ → Ba 2 + ( aq ) + 2 OH − ( aq )

(c)

HClO4 is a strong acid

H2 O HClO 4 ( aq ) ⎯⎯ ⎯ → H + ( aq ) + ClO 4− ( aq )

(d)

HBr is a strong acid

HBr ( aq ) → H + ( aq ) + Br − ( aq )

H O

– Chapter 15 –

41. (a) CH3 NH2 + H+ → CH3 NH3 + base

conjugate acid

Note that a neutral base forms a positively charged conjugate acid. (b) HS− + H + → base

2 conjugate acid

Note that a negatively charged base forms a neutral acid upon adding a positively charged hydrogen ion. (c)

CO32− + H+ → HCO3−

42. (a) HBrO3 − H + → BrO3 − acid

conjugate base

(b) NH 4 + − H + → acid

(c)

3 conjugate base

H2 PO 4 − − H + → HPO 4 2− acid

conjugate base

43. HC 2H3 O2 + NH3 → NH4 + + C 2 H3O 2 − acid

base

conjugate acid

conjugate base

Note that in any acid base reaction, the original acid and base react to form a new acid and base. 44. S2 − ( aq ) + H 2 O ( l ) → HS − ( aq ) + OH − ( aq ) The sulfide ion is able to act as a Bronsted-Lowry base by accepting a proton from a water molecule. Note that Bronsted-Lowry bases will often cause the formation of hydroxide ions in aqueous solution. 45. Mg ( s ) + 2 HCl ( aq ) → MgCl 2 ( aq ) + H 2 ( g ) H2 O 46. Na 2SO 4 ( aq ) ⎯⎯ ⎯ → 2 Na + ( aq ) + SO 4 2 − ( aq )

47. (a) H3PO 4 + H 2 O → H3O + + H 2 PO 4 − (b) H 2 PO 4 − + H 2 O → H3O + + HPO 4 2 − (c)

HPO 4 2 − + H2 O → H3O + + PO 43−

In each case H3O+ and H2O are a conjugate acid-base pair. For reaction (1) H3PO4 is the acid and H2PO4− is the base. For reaction (2) H2PO4− is the acid and HPO42− is the base while in (3) HPO42− is the acid and PO43− is the base. 48. (a)

Basic

(d) acidic

(b)

Acidic

(e)

acidic

(c)

neutral

(f)

basic

– Chapter 15 –

49. (a)

CaCl 2 ( s ) → Ca 2 + ( aq ) + 2 Cl − ( aq )

For each CaCl2 ionic compound, 1 calcium ion and 2 chloride ions result.

(b) KF ( s ) → K + ( aq ) + F − ( aq ) For each KF ionic compound, 1 potassium ion and 1 fluoride ion result.

(c)

AlBr3 ( s ) → Al3+ ( aq ) + 3 Br − ( aq )

For each AlBr3 ionic compound, 1 aluminum ion and 3 bromide ions result.

50. (a) Dissociation (D). Ions are present in the crystalline salt, and the water molecules act only as the separation agent. Molten salt as well as a solution of salt conducts electricity. (b) Ionization (I). Water molecules are necessary to form ions in reaction with HC2H3O2. Pure HC2H3O2 is a very weak electrolyte. (c) Ionization (I). When ammonia gas dissolves in water, it undergoes reaction with water molecules to form a solution we call ammonium hydroxide (NH4OH). (d) Dissociation (D). (e) Ionization (I). The bond in HCl (g) is predominantly covalent. 51. AlBr3 → Al3+ + 3 Br −

 0.648 mol Br −  1 mol Al3+   0.216 mol Al3+  3+ =  = 0.216 M Al   −  L 3 mol Br L     

– Chapter 15 –

52. H2SO4 + 2 NaOH → Na 2SO4 + 2 H2 O  mol H2SO 4  mL NaOH → mol NaOH → mol H2SO 4 ;   = M H2SO 4 L   0.313 mol   1 mol H2SO 4  (35.22 mL NaOH )    = 0.00551 mol H2SO 4  1000 mL   2 mol NaOH   0.00551 mol H2SO 4    = 0.218 M H2SO 4 0.02522 L  

53. A hydronium ion is a hydrated hydrogen ion. H+

( hydrogen ion )

+ H2O →

H3 O +

( hydronium ion )

54. (a) 100°C pH = −log (1×10 −6 ) = 6.0

25°C pH = − log (1× 10−7 ) = 7.0

pH of H2 O is greater at 25°C

(b) 1×10−6 > 1×10−7 , so H+ concentration is higher at 100°C. (c) The water is neutral at both temperatures because the H2O ionizes into equal concentrations of H+ and OH− at any temperature. 55. As the pH changes by 1 unit, the concentration of H+ in solution changes by a factor of 10. For example, the pH of 0.10 M HCl is 1.00, while the pH of 0.0100 M HCl is 2.00. 56. H2SO4 + 2 KOH → K 2SO4 + 2 H2 O g KOH → mol KOH → mol H2SO 4 → M H2SO 4   1 mol KOH   1 mol H2SO 4   1000 mL     56.11 g KOH   2 mol KOH   0.4233 mol H2SO 4 

( 6.38 g KOH ) 

= 134 mL of 0.4233 M H2SO 4

57. KOH + HNO3 → KNO3 + H2 O L HNO3 → mol HNO3 → mol KOH → g KOH 0.240 mol   1 mol KOH   56.11 g     = 0.673 g KOH L    1 mol HNO3   mol 

( 0.05000 L HNO3 ) 

58. pH of 1.0 L solution containing 0.1 mL of 1.0 M HCl

( 0.1 mL ) 

1.0 L  1 mol HCl  −4   = 1×10 mol HCl added L  1000 mL  

1×10 −4 mol HCl = 1×10 −4 M HCl 1.0 L −4 1×10 M HCl produces 1× 10 −4M H+ pH = − log (1×10 −4 ) = 4.0

– Chapter 15 –

59. Dilution problem: V1M1 = V2M 2 M1 =

V2M 2 V1

M1 =

(10.0 mL )(12 M ) = 0.462 M HCl ( 260.0 mL )

60. NaOH + HCl → NaCl + H2 O

 1 mol   = 0.075 mol NaOH  40.00 g 

(3.0. g NaOH ) 

(500. mL HCl ) 

1 L   0.10 mol    = 0.050 mol HCl L  1000 mL   

This solution is basic. The NaOH will neutralize the HCl with an excess of 0.025 mol of NaOH remaining unreacted. 61. Ba ( OH )2 ( aq ) + 2 HCl ( aq ) → BaCl 2 ( aq ) + 2 H 2 O ( l )

( 0.38 L Ba ( OH ) )  0.35Lmol  = 0.13 mol Ba ( OH ) 2

0.13 mol Ba ( OH )2 → 0.26 mol OH−

( 0.500 L HCl ) 

0.65 mol   = 0.33 mol HCl L  

0.33 mol HCl → 0.33 mol H+ 0.33 mol H+ will neutralize 0.26 mol OH− and leave 0.07 mol H+ (0.33 – 0.26) remaining in solution.

Total volume = 500. mL + 380 mL = 880 mL ( 0.88 L ) 0.07 mol H+ = 0.08M H+ 0.88 L pH = − log H+  = − log ( 8 ×10 −2 ) = 1.1 H+  in solution =

The solution is acidic. 62.

( 0.05000 L HCl ) 

0.2000 mol  +  = 0.01000 mol HCl = 0.01000 mol H in 50.00 mL HCl L  

(a) no base added: pH = log(0.2000) = 0.700 (b) 10.00 mL base added: 0.2000 mol  ( 0.01000 L )   = 0.002000 mol NaOH L   − = 0.002000 mol OH

( 0.01000 mol H ) − ( 0.002000 mol OH ) = 0.00800 mol H in 60.00 mL solution +

+ H   =

0.00800 mol 0.0600 L

−

 0.00800  pH = − log   = 0.880  0.06000 

– Chapter 15 –

( 0.02500 L ) 

0.2000 mol  −  = 0.005000 mol NaOH = mol OH L  

( 0.01000 mol H ) − ( 0.005000 mol OH ) = 0.00500 mol H in 75.00 solution +

0.00500 mol + H   = 0.07500 L

−

 0.00500  pH = − log   = 1.2  0.07500 

(d) 49.00 mL base added:

( 0.04900 L ) 

0.2000 mol  −  = 0.009800 mol NaOH = mol OH L  

( 0.01000 mol H ) − ( 0.009800 mol OH ) = 0.00020 mol H in 99.00 mL solution +

0.00020 mol + H   = 0.09900 L

−

 0.00020  pH = − log   = 2.69  0.09900 

(e) 49.90 mL base added:

( 0.04990 L ) 

0.2000 mol  −  = 0.009980 mol NaOH = mol OH L  

( 0.01000 mol H )( 0.009800 mol OH ) = 2 ×10 mol H in 99.00 mL solution +

2 ×10 −5 mol + H  =   0.09990 L (f)

−

−5

 2 ×10 −5  pH = − log   = 3.7  0.09990 

49.99 mL base added:

( 0.04999 L ) 

0.2000 mol  −  = 0.009998 mol NaOH = mol OH L  

( 0.01000 mol H ) − ( 0.009998 mol OH ) = 2 ×10 mol H in 99.99 mL solution +

2 × 10 −6 H+  = 0.09999 L

−

−6

 2 ×10 −6  pH = − log  = 4.7 −2   9.999 ×10 

(g) 50.00 mL of 0.2000 M NaOH neutralizes 50.00 mL of 0.2000 M HCl. No excess acid or base is in the solution. Therefore, the solution is neutral with a pH = 7.0.

– Chapter 15 –

63. (a)

2 NaOH ( aq ) + H 2SO 4 ( aq ) → Na 2SO 4 ( aq ) + 2 H 2 O ( l )

(b) mol H2SO4 → mol NaOH → mL NaOH

 2 mol NaOH   1000 mL  2   = 1.0 ×10 mL NaOH 1 mol H SO 0.10 mol    2 4 

( 0.0050 mol H2SO4 )  (c)

 1 mol Na 2SO4   142.1 g    = 0.71 g Na 2SO4  1 mol H2SO4   mol 

( 0.0050 mol H2SO4 ) 

64. HNO3 + KOH → KNO3 + H2 O M A VA = M B VB

( M A )( 25 mL ) = ( 0.60 M )(50.0 mL ) M A = 1.2 M ( diluted solution ) Dilution problem M1V1 = M 2 V2

( M A ) (10.0 mL ) = (1.2 M )(100.00 mL ) M A = 12M HNO3 ( original solution ) 65. Yes, adding water changes the concentration of the acid, which changes the concentration of the [H+], and changes the pH. The pH will rise. No, the solution theoretically will never reach a pH of 7, but it will approach pH 7 as water is added. 66. H2SO4 + 2 NaOH → Na 2SO4 + 2 H2 O

( 0.425 L H2SO 4 ) 

0.94 mol H 2SO 4   = 0.40 mol H 2SO 4 L  

0.40 mol H2SO 4 → 0.80 mol H +

( 0.750 L NaOH ) 

0.83 mol NaOH   = 0.62 mol NaOH L  

0.62 mol NaOH → 0.62 mol OH −

0.80 mol H+ will neutralize 0.62mol OH− and leave 0.18mol (0.80 − 0.62) of H+ remaining in solution; so the solution will be acidic.

Total volume = 425 mL + 750 mL = 1175 mL (1.175 L ) 0.18 mol H+ = 0.15M H+ 1.175 L + pH = − log H  = − log ( 0.15 ) = 0.82 H+  in solution =

67. (a) 1st determine kind of substance Copper(II) sulfate is a soluble salt so it will dissociate completely in water 2nd write the dissociation/ionization equation H2 O CuSO4 ( aq ) ⎯⎯⎯ → Cu2+ ( aq ) + SO4 2− ( aq ) .

– Chapter 15 –

3rd Analyze the equation to determine the number of ions formed. Each CuSO4 will produce 1 Cu2+ ion and 1 SO42− ion, so [Cu2+] = [SO42−] = 1 M (b) 1st determine kind of substance Nitric acid is a strong acid so it will ionize completely in water 2nd write the dissociation/ionization equation H2 O HNO3 ( aq ) ⎯⎯ ⎯ →H+ ( aq ) + NO3− ( aq )

3rd Analyze the equation to determine the number of ions formed. Each HNO3 will produce 1 H+ ion and 1 NO3− ion, so [H+] = [NO3−] = 1 M (c) 1st determine kind of substance Sulfuric acid is a strong acid so it will ionize completely in water 2nd write the dissociation/ionization equation H2 O H2SO4 ( aq ) ⎯⎯→2 ⎯ H+ ( aq ) + SO4 2− ( aq )

3rd Analyze the equation to determine the number of ions formed. Each H2SO4 will produce 2 H+ ions and 1 SO42− ion, so [H+] = 2 M and [SO42−] =1M (d) 1st determine kind of substance Calcium sulfide is an insoluble salt so it will not dissociate in water. 2nd write the dissociation/ionization equation H2 O CaS ( s ) ⎯⎯→ ⎯ no reaction

3rd Analyze the equation to determine the number of ions formed. CaS is insoluble so very few of the particles will dissociate and the concentration of all ions will be close to 0 M. (e) 1st determine kind of substance Acetic acid is a weak acid so it will ionize slightly in water 2nd write the dissociation/ionization equation H2 O + −   HC 2 H3O 2 ( aq )   H ( aq ) + C 2 H3 O 2 ( aq )

3rd Analyze the equation to determine the number of ions formed. Each HC2H3O2 will produce some H+ ions and C2H3O2− ions, so [H+] and [C2H3O2−] will be between 0 and 1 M. 68. (a) Formula equation Total ionic equation Net ionic equation

. HNO3 ( aq ) + LiOH ( aq ) + H2 O ( l ) + LiNO3 ( aq ) . H+ ( aq ) + NO3 − ( aq ) + Li + ( aq ) + OH− ( aq ) → H2 O ( l ) + Li + ( aq ) + NO3 − ( aq )

H+ ( aq ) + OH− ( aq ) → H2O ( l )

– Chapter 15 –

(b) Formula equation

2 HBr ( aq ) + Ba ( OH)2 ( aq ) → 2 H2O ( l ) + BaBr2 ( aq ) Total ionic equation. Net ionic equation (c) Formula equation Total ionic equation.

2 H+ ( aq ) + 2 Br − ( aq ) + Ba 2+ ( aq ) + 2 OH− ( aq ) → 2 H2 O ( l ) + Ba 2+ ( aq ) + 2 Br − ( aq )

H+ ( aq ) + OH− ( aq ) → H2O ( l ) HF ( aq ) + NaOH ( aq ) → H2 O ( l ) + NaF ( aq ) HF ( aq ) + Na + ( aq ) + OH − ( aq ) → H2 O ( l ) + Na + ( aq ) + F − ( aq )

(Note that HF is a weak acid and so it does not ionize to any appreciable extent and is written in its unionized form.) Net ionic equation

HF ( aq ) + OH− ( aq ) → H2 O ( l ) + F− ( aq ) 69. [H+] needs to be 0.00158M to give the final solution of sodium rhidizonate a pH of 2.800. V1M1 =V2 M 2

(500.0 mL )( 0.00158 M ) =V2 ( 0.60 M ) (500.0 mL )( 0.00158 M ) = 1.3 mL 0.60 M HCl V2 = ( 0.60 M ) To make the solution put 1.3mL of 0.60 M HCl in a graduated cylinder and fill the cylinder to the 500mL mark with sodium rhidizonate solution. 70. V1M1 = V2M 2

(5.00 L )( 3.25 M ) = V2 (12.1 M ) (5.00 L )(3.25 M ) = 1.34 L of 12.1 M HCl V2 = (12.1 M ) 71. . CaCO3 ( s ) + 2 HCl ( aq ) → CaCl2 ( aq ) + H 2O ( l ) + CO2 ( g ) . 72. First determine the molarity of the two HCl solutions. Take the antilog of the pH value to obtain the [H+]. pH = 0.300; H+ = 2.00 M = 2.00M HCl

pH = 0.150; H+ = 1.41M = 1.41M HCl Now treat the calculation as a dilution problem. VM V1M1 = V2M 2 V2 = 1 1 M2

( 200 mL HCl )( 2.00M ) = 284 mL solution

1.41M 284 mL − 200 mL = 84 mL H2O to be added 22

– Chapter 15 –

73. mol acid = mol base

( lactic acid has one acidic H )

1.0 g acid  0.65 mol NaOH  = ( 0.017 L )   = 0.011 mol NaOH molar mass L   mol acid = mol base mol HC3H5 O3 = 0.011 mol 1.0 g = 91 g/mol = molar mass 0.011 mol molar mass ( 91 g/mol ) = mass of empirical formula ( 90.08 g/mol )

Therefore, the molecular formula HC3H5O3 is the same as the empirical formula. 74. (a)

pH = − log H +  = − log [ 0.10 ] = 1.00

(b) mol HCl = 0.050 L ×

0.10 mol = 0.0050 mol HCl L

Flask A

Zn ( s ) + 2 HCl ( aq ) → ZnCl2 ( aq ) + H2 ( g ) HCl is the limiting reactant, so no HCl will remain in the product. pH = 7.0

Flask B No reaction occurs in flask B, so the pH does not change. pH = 1.00 75. H2SO4 is limiting, theoretical yield is 1.2 g BaSO4, percent yield is 83%. Net ionic equation is Ba 2+ + SO4 2− → BaSO4 ( s )  mol  Number of moles Ba 2+ =  0.43  0.018 L = 0.0077 mol L  

(

)

 mol  Number of moles SO 4 2− =  0.15  0.035 L = 0.0053 mol L   g Molar mass BaSO 4 = 233 mol  g  Amount BaSO 4 =  233  0.0053 mol = 1.2 g mol  

(

Percent yield =

)

1.0 g Actual yield ×100 = ×100 = 83% Theoretical yield 1.2 g

CHAPTER 16

CHEMICAL EQUILIBRIUM SOLUTIONS TO REVIEW QUESTIONS 1.

At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction.

The rate of a reaction increases when the concentration of one of the reactants increases. The increase in concentration causes the number of collisions between the reactants to increase. The rate of a reaction, being proportional to the frequency of such collisions, as a result, will increase.

An increase in temperature causes the rate of reaction to increase, because it increases the velocity of the molecules. Faster moving molecules increase the number and effectiveness of the collisions between molecules resulting in an increase in the rate of the reaction.

At 25°C both tubes would appear the same.

The reaction is endothermic because the increased temperature increases the concentration of product (NO2) present at equilibrium.

If pure HI is placed in a vessel at 700 K, some of it will decompose. Since the  2 HI , HI molecules will react to produce H2 reaction is reversible H2 + I 2 

(

)

and I2. 7.

Acids stronger than acetic acid are: benzoic, cyanic, formic, hydrofluoric, and nitrous acids (all equilibrium constants are greater than the equilibrium constant for acetic acid). Acids weaker than acetic acid are: carbolic, hydrocyanic, and hypochlorous acids (all have equilibrium constants smaller than the equilibrium constant for acetic acid). All have one ionizable hydrogen atom.

In an endothermic process heat is absorbed or used by the system so it should be placed on the reactant side of a chemical equation. In an exothermic process heat is given off by the system so it belongs on the product side of a chemical equation.

A catalyst speeds up the rate of a reaction by lowering the activation energy. A catalyst is not used up in the reaction.

10. It is very important to specify the temperature because equilibrium constants change with changing temperatures. For example, Kw for H2O at 25°C = 14.00 and at 50°C = 13.26. 11. A very large equilibrium constant means the equilibrium lies far to the right. 12. A very small equilibrium constant means the equilibrium lies far to the left. 13. Free protons (H+) do not exist in water because they are hydrated forming H3O+. 14. The sum of the pH and the pOH is 14. A solution whose pH is −1 would have a pOH of 15. 15. At different temperatures, the degree of ionization of water varies, being higher at higher temperatures. Consequently, the pH of water can be different at different temperatures.

– Chapter 16 –

16. In pure water, H+ and OH− are produced in equal quantities by the ionization of the  H+ + OH− . Since pH = −log[H+], and pOH = −log [OH−], water molecules, H2 O  they will always be identical for pure water. At 25°C, they each have the value of 7, but at higher temperatures, the degree of ionization is greater, so the pH and pOH would both be less than 7, but still equal.

 H3O + + C 2H3O2− 17. HC 2H3O2 + H2 O  As water is added (diluting the solution from 1.0 M to 0.10 M), the equilibrium shifts to the right, giving more ions and thus yielding a higher percent ionization. 18. The statement does not contradict Le Chatelier’s Principle. The previous question deals with the case of dilution. If pure acetic acid is added to a dilute solution, the reaction will shift to the right, producing more ions in accordance with Le Chatelier’s Principle. But, the concentration of the unionized acetic acid will increase faster than the concentration of the ions, thus yielding a smaller percent ionization. 19. In water the silver acetate dissociates until the equilibrium concentration of ions is reached. In nitric acid solution, the acetate ions will react with hydrogen ions to form acetic acid molecules. The HNO3 removes acetate ions from the silver acetate equilibrium allowing more silver acetate to dissolve. If HCl is used, a precipitate of silver chloride would be formed, since silver chloride is less soluble than silver acetate. Thus, more silver acetate would dissolve in HCl than in pure water.  Ag + ( aq ) + C 2 H3 O 2 − ( aq ) AgC 2 H3O 2 ( s ) 

20. When the salt, sodium acetate, is dissolved in water, the solution becomes basic. The dissolving reaction is H2 O NaC 2H3O2 ( s ) ⎯⎯ ⎯ → Na + ( aq ) + C 2H3O2 − ( aq )

The acetate ion reacts with water. The reaction does not go to completion, but some OH− ions are produced and at equilibrium the solution is basic.  OH − ( aq ) + HC 2 H3O 2 ( aq ) C 2 H3O 2 − ( aq ) + H 2 O ( l ) 

21. The order of solubility will correspond to the order of the values of the solubility product constants of the salts being compared. This occurs because each salt in the comparison produces the same number of ions (two in this case) for each formula unit of salt that dissolves. This type of comparison would not necessarily be valid if the salts being compared gave different numbers of ions per formula unit of salt dissolving. The order is: AgC2H3O2, PbSO4, BaSO4, AgCl, BaCrO4, AgBr, AgI, PbS. 22. (a)

K spMn ( OH )2 = 2.0 ×10 −13 ; K sp Ag 2CrO4 = 1.9 ×10 −2 Each salt gives 3 ions per formula unit of salt dissolving. Therefore, the salt with the largest Ksp (in this case Ag2CrO4) is more soluble.

(b) K spBaCrO 4 = 8.5 ×10 −11; K sp Ag 2CrO 4 = 1.9 ×10 −12. Ag 2CrO4 has a greater molar solubility than BaCrO4, even though its Ksp is smaller, because the Ag2CrO4 produces more ions per formula unit of salt dissolving than BaCrO4.

– Chapter 16 –

 Ba 2+ + CrO 4 2− BaCrO 4 ( s ) 

2+ 2−  CrO  K sp = Ba 4   

Let y = molar solubility of BaCrO4 K sp = [ y ][ y ] = 8.5 ×10 −11 y = 8.5 × 10 −11 = 9.2 ×10 −6 mol BaCrO 4 /L  2 Ag + + CrO 4 2− Ag 2 CrO 4 ( s ) 

K sp =  Ag +  CrO 4 2− 

Let y = molar solubility of Ag2CrO4 K sp = [ 2y ] [ y ] = 1.9 ×10 −12 2

1.9 ×10 −12 = 7.8 ×10 −5 mol Ag 2CrO 4 /L 4

Ag2CrO4 has the greater solubility. 23. In a saturated sodium chloride solution, the equilibrium is  Na + ( aq ) + Cl − ( aq ) NaCl ( s ) 

Bubbling in HCl gas increases the concentration of Cl−, creating a stress, which will cause the equilibrium to shift to the right, precipitating solid NaCl. 24. All four Keq expressions are equilibrium constants. They describe the ratio between the concentrations of products and reactants for different types of reactions when at equilibrium. Ka is the equilibrium constant expression for the ionization of a weak acid, Kb is the equilibrium constant expression for the ionization of a weak base, Kw is the equilibrium constant expression for the ionization of water and Ksp is the equilibrium constant expression for a slightly soluble salt.

 H+ + C 2H3O2− 25. HC 2H3O2  Initial concentrations HC2H3O2 0.10 M + 1.8 × 10−5 M H C2H3O2− 0.10 M pH 4.74

Added ---- 0.010 mol ----

Concentration after equilibrium shifts 0.11 M 1.9 × 10−5 M 0.09 M 4.72

The initial concentration of H+ in the buffer solution is very low (1.8 × 10−5 M) because of the large excess of acetate ions. 0.010 mol of HCl is added to one liter of the buffer solution. This will supply 0.010 M H+. The added H+ creates a stress on the right side of the equation. The equilibrium shifts to the left, using up almost all the added H+, reducing the acetate ion by approximately 0.010 M, and increasing the acetic acid by approximately 0.010 M. The concentration of H+ will not increase significantly and the pH is maintained relatively constant.

– Chapter 16 –

26. A buffer solution contains a weak acid or base plus a salt of that weak acid or base, such as dilute acetic acid and sodium acetate.  H + ( aq ) + C 2 H3O 2 − ( aq ) HC 2 H3O 2 ( aq )  NaC 2 H3O 2 ( aq ) ⎯⎯ → Na + ( aq ) + C 2H3O 2 − ( aq )

When a small amount of a strong acid (H+) is added to this buffer solution, the H+ reacts with the acetate ions to form un-ionized acetic acid, thus neutralizing the added acid. When a strong base, OH−, is added it reacts with un-ionized acetic acid to neutralize the added base. As a result, in both cases, the approximate pH of the solution is maintained. 27. A + B  C + D When A and B are initially mixed, the rate of the forward reaction to produce C and D is at its maximum. As the reaction proceeds, the rate of production of C and D decreases because the concentrations of A and B decrease. As C and D are produced, some of the collisions between C and D will result in the reverse reaction, forming A and B. Finally, an equilibrium is achieved in which the forward rate exactly equals the reverse rate.

– Chapter 16 –

SOLUTIONS TO EXERCISES 1.

Reversible systems. (a)

 K + ( aq ) + MnO−4 ( aq ) KMnO4 ( s ) 

 CO2 ( g ) (b) CO2 ( s )  2.

Reversible systems. (a)

 I2 ( s )   I2 ( g )

+ −  (b) NaNO3 ( s )   Na ( aq ) + NO3 ( aq )

Equilibrium system.

 SiF4 ( g ) + 2 H2 O ( g ) + 103.8 kJ   SiO2 ( s ) + 4 HF ( g ) (a) The reaction is endothermic with heat being absorbed. (b) The addition of HF will shift the reaction to the left until equilibrium is re-established. The concentrations of SiF4, H2O, and HF will be increased. The concentration of SiO2 will be decreased. (c) The addition of heat will shift the reaction to the right. 4.

Equilibrium system.

 4 HCl ( g ) + O2 ( g )   2 H2O ( g ) + 2 Cl2 ( g ) +114.4 kJ (a) The reaction is exothermic with heat being released. (b) The addition of O2 will shift the reaction to the right until equilibrium is re-established. The concentrations of O2, H2O, and Cl2 will be increased. The concentration of HCl will be decreased. (c) The addition of heat will cause the reaction to shift to the left. 5.

 N2 ( g ) + 3 H2 ( g )   2 NH3 ( g ) + 92.5 kJ Direction of reaction, left or right, to re-establish equilibrium right left right

Changes in number of moles N2 H2 NH3 I D I I D D D D I

Change or stress imposed on the system at equilibrium (a) Add N2 (b) Remove H2 (c) Decrease volume of reaction vessel left I (d) Increase temperature I = Increase; D = Decrease; N = No change; ? = Insufficient information to determine

– Chapter 16 –

 N2 ( g ) + 3 H2 ( g )   2 NH3 ( g ) + 92.5 kJ Change or stress imposed on Direction of reaction, left or Changes in number of moles the system at right, to re-establish NH3 N2 H2 equilibrium equilibrium (a) Add NH3 left I I I left I I D (b) Increase volume of reaction vessel no change N N N (c) Add a catalyst ? ? I I (d) Add H2 and NH3 I = Increase; D = Decrease; N = No change; ? = Insufficient information to determine

(1) The conditions that caused the stress to the system are italicized. Amount Amount Equilibrium shift Amount O2 SO3 (Left, Right, or none) SO3 SO2 is a. shifts right decreases increases added b. shifts right (fewer decreases decreases increases molecules) c. no effect no effect no effect no effect d.

shifts right

decreases

increases

Other XXXXXX

pressure is increased catalyst is added vol of container is decreased

a. b. c. d. 9.

Equilibrium shift (Left, Right, or none) shifts right shifts right shifts left shifts left (towards having more molecules)

Amount O2 O2 is added decreases increases increases

Amount O3 increases ozone is removed decreases decreases

Equilibrium shifts

 CO2 ( g ) + 2 H2 O ( g ) + 802.3 kJ CH4 ( g ) + 2 O2 ( g )  (a) left (b) none (c) right (d) none

Other XXXXXX XXXXXX reaction is cooled pressure is decreased

– Chapter 16 –

10. Equilibrium shifts

 C 2 N 2 ( g ) + 2 O2 ( g ) 2 CO2 ( g ) + N 2 ( g ) +1095.9 kJ  (a) none (b) left (c) right (d) right 11. (a)

2+ 2−  SO K eq = Pb   4 

PbSO4 is a solid; therefore, it is left out of the Keq expression.

 Ag ( NH3 )+  2 (b) K eq =  + 2  Ag  [ NH3 ] K eq =

[ O3 ] 3 [ O2 ]

(d) K eq =

[CO ] [CO 2 ]

(c)

Carbon is in the solid state; therefore, it does not change concentration. It can be left out of the equilibrium expression.

[CO2 ][H2 O] K eq = 2 [CH4 ][O2 ]

12. (a)

H +  N O 2−  (b) K eq =    [HNO 2 ]

(c)

 Al ( OH )2+   K eq =  3+ −  Al  OH 

(d)

[SO3 ] K eq = 2 [SO2 ] [O2 ]

(a)

K sp = Ag +  Cl − 

(c)

Ksp = Zn2+  OH− 

(b)

 K sp =  Pb 2+  CrO 2− 4 

(d)

Ksp = Ca 2 +  PO34 − 

13.

– Chapter 16 –

14. 3

(a)

K sp = Mg 2+  CO 32 − 

(c)

Ksp = Tl3+  OH− 

(b)

 K sp = Ca 2 +  C 2 O 2− 4 

(d)

Ksp = Pb2 +  AsO34 − 

15. If the H+ ion concentration is decreased: (a) pH is increased (b) pOH is decreased (c) [OH−] is increased (d) Kw remains the same. Kw is a constant at a given temperature. 16. If the H+ ion concentration is increased: (a) pH is decreased (pH of 1 is more acidic than that of 4) (b) pOH is increased (c) [OH−] is decreased (d) Kw remains unchanged. Kw is a constant at a given temperature. −

17. When excess acid (H+) gets into the blood stream it reacts with HCO3 to form un ionized H2CO3, thus neutralizing the acid and maintaining the approximate pH of the blood. 18. When excess base gets into the bloodstream it reacts with H+ to form water. Then H2CO3 ionizes to replace H+, thus maintaining the approximate pH of the blood. 19. (a)

 H+ ( aq ) + HCO3− ( aq ) H2CO3 ( aq )  Let x = molarity of H+ + − H   = HCO3  = x [H2CO3 ] = 1.25M − x = 1.25 ( since x is small ) + H  HC O3−  x2   = = 4.4 × 10 −7 Ka = 1.25 [H2CO3 ]

x 2 = (1.25 ) ( 4.4 ×10 −7 ) x=

(1.25 ) ( 4.4 ×10 −7 ) = 7.4 ×10 −4 M = H+ 

(b) pH = − log H +  = − log ( 7.4 ×10 −4 M ) = 3.13 (c) Percent ionization + −4 H  100 =  7.4 ×10  100 = 0.059% ( )  ) ( [H2CO3 ]  1.25 

– Chapter 16 –

 H + ( aq ) + C 3H5 O 2− ( aq ) 20. (a) HC 3H5 O 2 ( aq ) 

Let x = molarity of H+ + − H  = C  3H5 O2  = x [H2CO3 ] = 0.025M − x = 0.025

( since x is small )

H+  C3H5 O2−  x2 = = 8.4 ×10 −4 Ka =    [HC3H5O2 ] 0.025 x 2 = ( 0.025 ) ( 8.4 ×10 −4 ) x=

( 0.025 ) (8.4 ×10 −4 ) = 4.6 ×10 −3 M = H+ 

(b) pH = − log H+  = − log ( 4.6 ×10 −3 M ) = 2.34 (c) Percent ionization +  4.6 ×10 −3  H  (100 ) =   (100 ) = 18% [HC 3H5 O 2 ]  0.025 

 H + + A − 21. HA  + − −4 H  = A   = ( 0.025M )( 0.0045 ) = 1.1×10 M [HA ] = 0.025M − 0.00011M = 0.025M

H +   A −  (1.1× 10 −4 )  K a = K eq = = = 4.8 ×10 −7 HA 0.025 [ ] 2

 H + + A − 22. HA  + − −3 H   = A   = ( 0.500M )( 0.0068 ) = 3.4 ×10 M [HA ] = 0.500M − 0.0034M = 0.497M

H+   A −  ( 3.4 ×10 −3 ) = = 2.3 ×10 −5 Ka = 0.497 [HA ] 2

 H+ ( aq ) + C 6H5o − ( aq ) 23. HC 6H5 O ( aq )  + H  C 6 H5 O −   = 1.3 ×10 −10 Ka = [HC 6H5 O ]

Let x = molarity of H+ [HC6H5O] = initial concentration − x = initial concentration Since Ka is small, the degree of ionization is small. Therefore, the approximation, initial concentration – x = initial concentration, is valid.

– Chapter 16 –

(a)

H+  = C 6H5 O −  = x [HC 6H5 O] = 1.0M ( x )( x ) = 1.3 ×10−10 1.0 2 x = 1.3×10 −10

(1.0 ) (1.3 ×10−10 ) = 1.1×10−5 M

 1.1×10 −5 M    (100 ) = 0.0011% ionized  1.0M  pH = − log (1.1×10 −5 ) = 4.96

(b)

[HC6H5O] = 0.10M ( x )( x ) = 1.3 ×10 −10

0.10 x 2 = ( 0.10 ) (1.3 ×10 −10 ) x=

( 0.10 ) (1.3 ×10 −10 ) = 3.6 ×10 −6 M

 3.6 ×10 −6 M    (100 ) = 0.00036% ionized  1.0M  pH = − log ( 3.6 ×10 −6 ) = 5.44

(c)

[HC6H5O] = 0.010M ( x )( x ) = 1.3 ×10 −10

0.010 x 2 = ( 0.010 ) (1.3 ×10 −10 ) x=

( 0.010 ) (1.3 ×10 −10 ) = 1.1 ×10 −6 M

 1.1×10 −6 M    (100 ) = 0.011% ionized  0.010M  pH = − log (1.1×10 −6 ) = 5.96

 H + ( aq ) + C 7 H5 O 2− ( aq ) 24. HC 7 H5 O 2 ( aq ) 

(a)

+ − H  = C  7 H5 O 2  = x ( x )( x ) = 6.3×10−5 1.0

[HC 7H5O2 ] = 1.0M

x 2 = (1.0 ) ( 6.3 ×10 −5 ) x=

(1.0 ) ( 6.3 ×10−5 ) = 7.9 ×10−3 M

 7.9 ×10 −3 M    (100 ) = 0.79% ionized  1.0M  pH = − log ( 7.9 ×10 −3 ) = 2.10

– Chapter 16 –

(b)

[HC7H5O2 ] = 0.10M ( x )( x ) = 6.3 ×10 −5

0.10 x 2 = ( 0.10 ) ( 6.3 ×10 −5 ) x=

( 0.10 ) ( 6.3 ×10 −5 ) = 2.5 ×10 −3 M

 2.5 ×10 −3 M    (100 ) = 2.5% ionized  0.10M  pH = − log ( 2.5 × 10 −3 ) = 2.60

(c)

[HC7H5O2 ] = 0.010M ( x )( x ) = 6.3 ×10−5

0.010 x 2 = ( 0.010 ) ( 6.3 ×10 −5 ) x=

( 0.010 ) ( 6.3 ×10−5 ) = 7.9 ×10−4 M

 7.9 ×10 −4 M    (100 ) = 7.9% ionized  0.010M  pH = − log ( 7.9 ×10 −4 ) = 3.10

H +   A −  Ka =     [HA ] First, find the [H+]. This is calculated from the pH expression, pH = −log[H+] = 3.7.

25. HA  H + + A −

+ −4 H   = 2 × 10 + − −4 H   = A   = 2 × 10

[HA ] = 0.37

H+   A −  ( 2 × 10 −4 )( 2 ×10 −4 )  = = 1× 10 −7 Ka = 0.37 [HA ] +

26. HA  H + A

−

+ H   A −   Ka = [HA ]

pH = 2.89

− log H +  = 2.89 H +  = 1.3 ×10 −3 H +  = 1.3 ×10 −3 =  A − 

[HA ] = 0.23

+ H   A −  (1.3 ×10 −3 )(1.3 ×10 −3 )  Ka = = = 7.3 ×10 −6 0.23 [HA ]

– Chapter 16 –

27. 1.0M NaOH yields OH −  = 1.0M (100% ionized ) pOH = − log 1.0 = 0.00 pH = 14 − pOH = 14.00 Kw 1.0 ×10 −14 + H  = = = 1× 10 −14   OH −  1.0   + 28. 3.0M HNO3 yields H   = 3.0M (100% ionized ) pH = − log 3.0 = −0.48

pOH = 14 − pH = 14 − ( −0.48 ) = 14.48 Kw 1×10 −14 OH−  = = = 3.3 ×10 −15 + 3.0 H  29. pH + pOH = 14.0 pOH = 14.0 − pH (a) 0.250 M HBr yields [H+] = 0.250 M.(100% ionized) pH = −log [H+] = −log (0.250) = 0.602 pOH = 14.0 – 0.602 = 13.4 (b) 0.333 M KOH yields [OH−] = 0.333 M (100% ionized) pOH = −log [OH+] = −log(0.333) = 0.478 pH = 14.0 – 0.478 = 13.5 (c)

HC 2H3O2  H+ + C 2H3O2− 0.895 M

30. pH + pOH = 14.0 pOH = 14.0 – pH (a) 0.0010 M NaOH yields [OH−] = 0.0010 M (100% ionized) pOH = −log [OH−] = −log(0.0010) = 3.00 pH = 14.00 – 3.00 = 11.00 (b) 0.125 M HCl yields [H+] = 0.125 M (100% ionized) pH = −log[H+] = −log(0.125) = 0.903 pOH = 14.00 – 0.903 = 13.10 (c)

HC 6H5 O  H+ + C 6H5 O − 0.0250 M

H  C 2H3C  = 1.8 × 10 −5 Ka =    [HC 2H3O2 ] +

− 2

( x )( x ) = 1.8×10−5 0.895

x 2 = ( 0.895 ) (1.8 ×10 −5 ) x = 1.6 ×10 −5 + x = 4.0 ×10 −3 = H  

pH= − log ( 4.0 ×10 −3 ) = 2.40 pOH = 14.0 − 2.40 = 11.6

– Chapter 16 –

H+  C6H5O −  Ka =    = 1.3×10 −10 HC H O [ 6 5 ]

( x )( x ) = 1.3×10−10

0.0250

x 2 = ( 0.0250 ) (1.3×10 −10 ) x = 3.3×10 −12 x = 1.8 ×10 −6 = H+ 

pH= − log (1.8 ×10 −6 ) = 5.74 pOH = 14.00 − 5.74 = 8.26 Kw − −  OH  = 31. Calculate the OH   H + 

H +  = 1.0 ×10 −2

1.0 ×10 −14  OH −  = = 1.0 ×10 −12 M −2 1.0 ×10

(b) H +  = 3.2 ×10 −7

1.0 ×10 −14 − −8 OH  =   3.2 ×10 −7 = 3.1× 10 M

(a)

(c) KOH is a strong base; 1.25 M = [OH−] = 1.25 M (d) First find [H+] of 0.75 M HC2H3O2 + − H  = C 2H3O2  = x x2 = 1.8 ×10 −5 0.75 x 2 = ( 0.75 ) (1.8 ×10 −5 )

Ka = 1.8 × 10−5

[HC 2H3O2 ] = 0.75M

( 0.75 ) (1.8 ×10−5 ) = 3.7 ×10 −3 M

+ −3 H  = 3.7 ×10

32. Calculate the [OH−].

1.0 ×10 −14 − −12 OH  =   3.7 ×10 −3 = 2.7 ×10 M Kw − OH  =  + H  

+ −9 H  = 4.0 ×10

1.0 ×10 −14  OH −  = = 2.5 ×10 −6 −9 4.0 ×10

(b) H +  = 1.2 ×10 −5

1.0 ×10 −14  OH −  = = 8.3 ×10 −10 −5 1.2 ×10

(a)

– Chapter 16 –

x2 = 4.8×10 −10 1.25 x 2 = (1.25 ) ( 4.0 ×10 −10 ) x=

(1.25 ) ( 4.0 ×10−10 ) = 2.2 ×10 −5 M 1.0 ×10 −14 − −10 OH  =   2.2 ×10 −5 = 4.5×10 M

+ −5 H  = 2.2 ×10

(d) NaOH is a strong base; 0.333 M = [OH−] = 0.333 M Kw + H  = OH −   

33. Calculate the [H+]

(a)

− OH  = 1.0 ×10 −8 

(b)  OH−  = 2.0 ×10 −4

1.0 ×10 −14 H+  = = 5.0 ×10 −11 −4 2.0 ×10 Kw H +  = OH − 

34. Calculate the [H+]

(a)

1.0 ×10 −14 + −6 H  =   1.0 ×108 = 1.0 ×10

− −2  OH  = 4.5 ×10

(b) OH−  = 5.2 ×10 −9

1.0 ×10 −14 H +  = = 2.2 ×10 −13 −2 4.5 ×10 1.0 ×10 −14 + −6 H  =   5.2 ×10 −9 = 1.9 ×10

35. The molar solubilities of the salts and their ions are indicated below the formulas in the equilibrium equations. (a) BaSO 4 ( s )  Ba 2+ + SO 2− 4 −5 3.9 × 10

3.9 ×10 −5

K sp = Ba 2 +  SO4 2 −  ( 3.9×10 −5 ) = 1.5 ×10 −9 2

(b) Ag 2CrO 4 ( s )  2 Ag + + CrO 24

(

2 7.8 ×10 −5

K sp = Ag +  CrO4 2− 

) 7.8 ×10−5

(15.6 ×10 ) ( 7.8 ×10 ) = 1.9×10 −5 2

(c) First change g/L ⎯⎯→ mol/L  0.67 g CaSO 4   1 mol  −3  = 4.9 ×10 M CaSO 4   L    136.1 g   Ca 2+ + SO 2− CaSO 4 ( s )  4 −3 4.9 ×10

4.9 ×10−3

K sp = Ca 2+  SO 4 2−  = ( 4.9 ×10 −3 ) = 2.4 ×10 −5 2

−5

−12

– Chapter 16 –

(d) First change g/L ⎯⎯→ mol/L  0.0019 g AgCl   1 mol  −5  = 1.3 ×10 M AgCl   L    143.4 g   Ag + + Cl − AgCl ( s )  −5 1.3 ×10−5

1.3×10

K sp =  Ag +  Cl −  = (1.3 ×10 −5 ) = 1.7 ×10 −10 2

36. The molar solubilities of the salts and their ions are indicated below the formulas in the equilibrium equations. (a)

 Zn2+ + ZnS ( s )  −12 3.5×10

S2− −12

3.5×10

2+  S2−  = ( 3.5 ×10−12 ) = 1.2×10−23 K sp = Zn  2

 Pb 2+ + 2IO3− (b) Pb ( IO3 )2 ( s )  −5 4.0 ×10

(

2 4.0 ×10 −5

)

K sp = Pb 2+  IO3−  = ( 4.0 × 10 −5 )( 8.0 ×10 −5 ) = 2.6 × 10 −13 2

(c) First change g/L ⎯⎯→ mol/L  6.73 ×10 −3 g Ag3 PO 4   1 mol  −5    = 1.61× 10 M Ag3PO4 L 418.7 g    +  3 Ag + PO3− Ag3PO 4 ( s )  4 31.61×10−5  1.61×10−5 



K sp =  Ag  PO 4  = ( 4.83×10 −5 ) (1.61×10 −5 ) = 1.81× 10 −18 +

3−

(d) First change g/L ⎯⎯→ mol/L

 2.33×10 −4 g Zn ( OH )2   1 mol  −6     = 2.34 ×10 M Zn ( OH )2 L    99.43 g   Zn2+ + 2 OH− Zn ( OH ) ( s )  2.34 ×10−6

2( 2.34 ×10−6 )

K sp = Zn2+  OH−  = ( 2.34 ×10 −6 )( 4.68 ×10 −6 ) = 5.13 ×10 −17 2

37. The molar solubilities of the salts and their ions will be represented in terms of x below their formulas in the equilibrium equations.

 Ca 2+ + 2 F− (a) CaF2  2x x K sp = Ca 2+  F−  = ( x )( 2x ) = 4x3 = 3.9 ×10 −11 2

x=3

3.9 ×10 −11 = 2.1×10 −4 M 4

– Chapter 16 –

 Fe3+ + 3 OH− (b) Fe ( OH )3  x

K sp = Fe3+  OH  = ( x )( 3x ) = 27x 4 = 6.1×10 −38 − 3

x=4

6.1×10 −38 = 2.2 ×10 −10 M 27

38. The molar solubilities of the salts and their ions will be represented in terms of x below their formulas in the equilibrium equations. (a)

 Pb2 + + SO 4 2 − PbSO 4  x

2 −8 K sp = Pb   SO  = ( x )( x ) = x = 1.3 ×10 2+

2− 4

x = 1.3 ×10 −8 = 1.1× 10 −4 M  Ba 2+ + CrO 4 2− (b) BaCrO4  x

K sp = Ba  CrO  = ( x )( x ) = x 2 = 8.5 ×10 −11 2+

2− 4

x = 8.5 ×10 −11 = 9.2 ×10 −6 M 39. K sp = 1.03×10 −9 K sp = Sn2+  CO32− 

First calculate the molar mass of SnCO3, using a list of atomic masses, molar mass of SnCO3 = (118.7 g ) + (12.01 g ) + ( 3×16.00 g ) = 178.7

g . mol

Next, determine the molarity of the saturated solution of SnCO3, then substitute the values in the solubility product equation. Molarity =

g/L mol = g/mol L

Substituting in the above equation M=

5.73 ×10 −3 g/L 178.7 g/mol

= 3.21× 10 −5 M

2− −5  K sp = Sn2+  CO32−  , Sn 2+  = CO 3  = 3.21 ×10 M 

= 3.21×10 −5  3.21×10 −5  = 10.3 ×10 −10 K sp = 1.03 ×10 −9

– Chapter 16 –

40. K sp = 4.4 ×10 −11π molar mass of ZnS = 97.44

g mol

6.4 ×10 −4 g/L g/L = = 6.6 ×10 −6 M g/mol 97.44 g/mol

2+ 2− −6  CO32−  ,  Zn 2+  = S K sp = Sn    = 6.6 ×10 M −6 −6 = 6.6  ×10  6.6 ×10 

= 44 ×10 −12 K sp = 4.4 ×10 −11

41. The molar concentrations of ions, after mixing, are calculated and these concentrations are substituted into the equilibrium expression. The value obtained is compared to the Ksp of the salt. If the value is greater than the Ksp, precipitation occurs. If the value is less than the Ksp, no precipitation occurs. 100. mL 0.010 M Na 2SO 4 ⎯⎯ →100. mL 0.010 M SO 24 − 100. mL 0.001 M Pb ( NO3 )2 ⎯⎯ →100. mL 0.001 M Pb 2 +

Volume after mixing = 200. mL Concentrations after mixing : SO4 2− = 0.0050 M

Pb2+ = 0.0005 M

 Pb 2+  SO 4 2−  = ( 5.0 ×10 −3 )( 5×10 −4 ) = 3×10 −6

Ksp = 1.3 × 10−8 which is less than 3 × 10−6, therefore, precipitation occurs. 42. The molar concentrations of ions, after mixing, are calculated and these concentrations are substituted into the equilibrium expression. The value obtained is compared to the Ksp of the salt. If the value is greater than the Ksp, precipitation occurs. If the value is less than the Ksp, no precipitation occurs.

50.0 mL 1.0 ×10 −4 M AgNO3 ⎯⎯ → 50.0 mL 1.0 × 10−4M Ag + 100. mL 1.0 ×10−4M NaCl ⎯⎯ →100. mL 1.0×10 −4M Cl− Volume after mixing = 150. mL Concentration after mixing: 50.0 mL  (1.0 ×10 M Ag )  150.  = 3.3 ×10 M Ag mL  −4

−5

mL  (1.0 ×10 M Cl )  100.0  = 6.7 ×10 M Cl 150. mL  −4

−

−5

−

+ Ag  Cl −  = ( 3.3 ×10 −5 )( 6.7 ×10 −5 ) = 2.2 ×10 −9 

2.2 × 10−9 is greater than the Ksp of 1.7 × 10−10 therefore, precipitation occurs.

– Chapter 16 –

43. The concentration of Br− = 0.10 M in 1.0 L of 0.10 M NaBr. Substitute this Br− concentration in the Ksp expression and solve for the [Ag+] in equilibrium with 0.10 M Br−.

K sp =  Ag +  Br −  = 5.2 ×10 −13 +  Ag  =

5.2 ×10 −13 5.2 ×10 −13 = = 5.2 ×10 −12 M 0.10 Br − 

 5.2 ×10 −12 mol Ag +  1 mol AgBr  1.0 L ) = 5.2 ×10 −12 mol AgBr will dissolve   + ( L   1 mol Ag  − −  0.10 mol MgBr2   2 mol Br   0.20 mol Br  − 44.  =     = 0.20 M Br in solution  L L    1 mol MgBr2    Substitute the Br− concentration in the Ksp expression and solve for [Ag+] in equilibrium with 0.20 M Br−.

+ Ag   =

5.2 ×10 −13 5.2 ×10 −13 = = 2.6 ×10 −12 M − 0.20 Br 

 2.6 ×10 −12 mol Ag +  1 mol AgBr  1.0 L ) = 2.6 ×10 −12 mol AgBr will dissolve   + ( L 1 mol Ag   

 H+ + C 2H3O2− 45. HC 2H3O2  H+  C 2H3O2−   = 1.8 ×10 −5 Ka = [HC 2H3O2 ]  [HC 2 H3 O2 ]   [HC 2H3O2 ] = 0.20M H+  = K a   C 2H3O2−        0.20  −5 H+  = (1.8 × 10 −5 )   = 3.6 × 10 M 0.10  

C 2H3O2 −  = 0.10M

pH = − log ( 3.6 × 10 −5 ) = 4.44

 H+ + C 2H3O2− 46. HC 2H3O2  + H  C 2 H3O2−   = 1.8×10 −5 Ka = [HC 2H3O2 ]

 [HC 2H3O2 ]  + H    [HC 2H3O2 ] = 0.20M = K a    C 2H3O2−       + −5  0.20  −5 H   = (1.8 ×10 )  0.20  = 1.8×10 M  

pH = − log (1.8 × 10 −5 ) = 4.74

− C  2H3O2  = 0.20M

– Chapter 16 –

47. Initially, the solution of NaCl is neutral. [H+] = 1 × 10−7 pH = − log (1×10 −7 ) = 7.0

Final H+ = 2.0 ×10 −2 M

pH = − log ( 2.0 ×10 −2 ) = 1.70

Change in pH = 7.0 – 1.70 = 5.3 units in the unbuffered solution. 48. Initially, H+  = 1.8 ×10 −5 pH = − log (1.8 ×10 −5 ) = 4.74

H +  =

K a [HC 2H3O2 ] C 2H3O2 − 

Final H +  = 1.9 ×10 −5

pH = − log (1.9 ×10 −5 ) = 4.72

Change in pH = 4.74 – 4.72 = 0.02 units in the buffered solution. 49. The concentration of solid salt is not included in the Ksp equilibrium constant because the concentration of solid does not change. It is constant and part of the Ksp. 50. The energy diagram represents an exothermic reaction because the energy of the products is lower than the energy of the reactants. This means that energy was given off during the reaction. 51.

 2 HI 52. H2 + I 2  The reaction is a 1 to 1 mole ratio of hydrogen to iodine. The data given indicates that hydrogen is the limiting reactant.  2 mol HI  ( 2.10 mol H2 )   = 4.20 mol HI  1 mol H 2   2 HI 53. H2 + I 2  (a) 2.00 mol H2 and 2.00 mol I2 will produce 4.00 mol HI assuming 100% yield. However, at 79% yield you get 4.00 mol HI × 0.79 = 3.16 mol HI.

– Chapter 16 –

(b) The addition of 0.27 mol I2 makes the iodine present in excess and the 2.00 mol H2 the limiting reactant. The yield increases to 85%.  2 mol HI   ( 0.85 ) = 3.4 mol HI  1 mol H 2 

( 2.00 mol H2 ) 

There will be 15% unreacted H2 and I2 plus the extra I2 added. (0.15)(2.0 mol H2) = 0.30 mol H2 present; also 0.30 mol I2. In addition to the 0.30 mol of unreacted I2, will be the 0.27 mol I2 added. 0.27 mol + 0.30 mol = 0.57 mol I2 present.

[HI ] K eq = [H2 ][I 2 ] 2

(c)

The formation of 3.16 mol HI required the reaction of 1.58 mol I2 and 1.58 mol H2. At equilibrium, the concentrations are: 3.16 mol HI; 2.00 −1.58 = 0.42 mol H2 = 0.42 mol I 2

(3.16 ) = 57 K eq = ( 0.42 )( 0.42 ) 2

In the calculation of the equilibrium constant, the actual number of moles of reactants and products present at equilibrium can be used in the calculation in place of molar concentrations. This occurs because the reaction is gaseous and the liters of HI produced equals the sum of the liters of H2 and I2 reacting. In the equilibrium expression, the volumes will cancel.

 2 HI 54. H2 + I 2   1 mol   = 0.500 mol HI present  127.9 g 

( 64.0 g HI ) 

( 0.500 mol HI ) 

1 mol I 2   = 0.250 mol I 2 reacted  2 mol HI  1 mol H2   = 0.250 mol H2 reacted  2 mol HI 

( 0.500 mol HI ) 

 1 mol   = 2.98 mol H2 initially present  2.016 g 

( 6.00 g H2 ) 

 1 mol   = 0.788 mol I 2 initially present  253.8 g 

( 200. g I2 ) 

At equilibrium, moles present are: 0.500 mol HI;

2.98 – 0.250 = 2.73 mol H2

0.788 – 0.250 = 0.538 mol I2

– Chapter 16 –

 PCl5 ( g ) 55. PCl3 ( g ) + Cl2 ( g ) 

K eq =

[PCl5 ] [PCl3 ][Cl2 ]

The concentrations are: 0.22 mol = 0.011M 20. L 0.10 mol = 0.0050M PCl3 = 20. L 1.50 mol = 0.075M Cl2 = 20. L 0.011 = 29 K eq = ( 0.0050 )( 0.075 ) PCl5 =

56. 100°C − 30°C =70°C temperature increase. This increase is equal to seven 10°C increments. The reaction rate will be increased by 27 = 128 times. 57. NH 4+   NH3 + H + 0.30 M

K eq = 5.6 × 10 −10

+ y = H   = [ NH3 ]

[ NH3 ] H+  [ y][ y] = = 5.6 ×10 −10 + NH  [0.30] 4   y 2 = ( 5.6 ×10 −10 ) ( 0.30 ) = 1.7 ×10 −10 y = H+  = 1.3 ×10 −5

pH = − log (1.3 ×10 −5 ) = 4.89 ( an acidic solution ) 58. pH of an acetic acid-acetate buffer

 H+ + C 2 H3O2 − HC 2H3O2  + H  C 2H3O 2 −  H+  [ 0.20 ]  Ka = = = 1.8× 10 −5 0.30 [HC 2H3O2 ]

(1.8×10 ) ( 0.30 ) = 2.7 ×10 H = −5

−5

0.20 pH = −log ( 2.7 ×10 −5 ) = 4.57

59. Concentration of Ba2+ in solution  BaCrO 4 ( s ) + NaCl ( aq ) BaCl2 ( aq ) + Na 2CrO 4 ( aq )   Ba 2+ + CrO 4 2− K sp = 8.5 ×10 −11 BaCrO 4 

– Chapter 16 –

Determine the moles of Ba2+ and CrO42− in solution

 0.10 mol Ba 2 +  2+   ( 0.050 L ) = 0.0050 mol Ba L    0.15 mol CrO4 2 −  2−   ( 0.050 L ) = 0.0075 mol CrO4 L   Excess CrO42− in solution = 0.0025 mol (2.5 × 10−3) after 0.005 mol BaCrO4 precipitate. Concentration of CrO42− in solution (total volume = 100 mL) 2.5 ×10 −3 mol CrO 4 2− = 2.5 ×10 −2 M CrO 4 2− 0.100 L

Now using the Ksp, calculate the Ba2+ remaining in solution.

Ba 2 +  CrO4 2 −  = Ba 2 +  2.5 ×10 −2  = 8.5×10 −11 8.5×10 −11 Ba 2 +  = = 3.4 ×10 −9 mol/L −2 2.5 ×10 60. Hypochlorous acid

 H + + OCl − HOCl 

Equilibrium concentrations: H +  = OCl −  = 5.9 ×10 −5 M a 2 + b 2

[HOCl] = 0.1 − 5.9 ×10−5 = 0.10M ( neglecting 5.9 ×10 −5 ) H +  OCl −  ( 5.9 ×10 −5 )( 5.9 ×10 −5 ) = = 3.5 ×10 −8 Ka = 0.10 [HOCl]

Propanoic acid

 H+ + C3H5 O2 − HC3H5 O2 

Equilibrium concentrations: + − −3 H  = C  3H5 O 2  = 1.4 ×10 M

[HC3H5O2 ] = 0.15 −1.4 ×10 −3 = 0.15M ( neglecting 1.4 ×10−3 ) H +  C 3H5 O 2 −  (1.4 ×10 −3 )(1.4 ×10 −3 )  = = 1.3 ×10 −5 Ka = 0.15 [HC3H5O2 ]

Hydrocyanic acid

 H + + CN − HCN 

Equilibrium concentrations: + − −6 H  = CN  = 8.9 ×10 M

[HCN ] = 0.20 − 8.9 ×10−6 = 0.20M ( neglecting 8.9×10−6 ) H+  CN −  ( 8.9×10 −6 ) = = 4.0 ×10 −10 Ka =    0.20 [HCN ] 2

– Chapter 16 –

61. Let y = M CaF2 dissolved

 Ca 2+ + 2 F− CaF2 ( s )  y

y = molar solubility (a)

K sp = Ca 2+  F−  = ( y )( 2y ) = 4y3 = 3.9 ×10 −11 2

y=3

3.9 × 10 −11 = 2.1×10 −4 M ( CaF2 dissolved ) 4

 2.1× 10 −4 mol CaF2   1 mol Ca 2+  −4 2+  = 2.1× 10 M Ca   L   1 mol CaF2   2.1× 10 −4 mol CaF2   2 mol F−  −4 −  = 4.2 ×10 M F   L   1 mol CaF2   2.1×10 −4 mol CaF2   78.08 g  −3 (b)   ( 0.500 L )   = 8.2 ×10 g CaF2 L mol    

62. The molar concentrations of ions, after mixing, are calculated and these concentrations are substituted into the equilibrium expression. The value obtained is compared to the Ksp of the salt. If the value is greater than the Ksp, precipitation occurs. If the value is less than the Ksp, no precipitation occurs. (a) 100. mL 0.010M Na 2SO 4 ⎯⎯ →100. mL 0.010M SO 4 2 − 100. mL 0.001M Pb ( NO3 )2 ⎯⎯ →100. mL 0.001M Pb 2 +

Volume after mixing = 200. mL Concentrations after mixing:

SO4 2− = 0.0050M Pb2+ = 0.0005M

 Pb 2+  SO 4 2−  = ( 5.0 ×10 −3 )( 5×10 −4 ) = 3×10 −6

Ksp = 1.3 × 10−8 which is less than 3 × 10−6, therefore, precipitation occurs. (b) 50.0 mL 1.0 ×10 −4M AgNO3 ⎯⎯ → 50.0 mL 1.0 ×10−4M Ag +

100. mL 1.0 ×10−4M NaCl ⎯⎯ →100. mL 1.0 ×10 −4M Cl− Volume after mixing = 150. mL Concentrations after mixing: 50.0 mL  (1.0 ×10 )  150.  = 3.3 ×10 M Ag mL  −4

−5

100. mL  (1.0 ×10 )  150.  = 6.7 ×10 M Cl mL  −4

−5

−

+ Ag  Cl −  = ( 3.3 ×10 −5 )( 6.7 ×10 −5 ) = 2.2 ×10 −9 

Ksp = 1.7 × 10−10 which is less than 2.2 × 10−9, therefore, precipitation occurs.

– Chapter 16 –

(c) Convert g Ca(NO3)2 to g Ca2+   1.0 g Ca ( NO3 )2   1 mol   1 mol Ca 2+ 2+  = 0.041M Ca     0.150 L    164.1 g   1 mol Ca ( NO3 )2  250 mL 0.01M NaOH → 250 mL 0.01M OH −

Final volume = 4.0 × 102 mL Concentration after mixing:

150 mL  ( 0.041M Ca )  4.0×10  = 0.015M Ca mL  2+

mL  ( 0.01M OH )  4.0250  = 0.0063M OH ×10 mL  −

−

Ca 2+  OH−  = ( 0.015 )( 0.0063) = 6.0 ×10 −7 2

Ksp = 1.3 × 10−6 which is greater than 6.7 × 10−7, therefore, no precipitation occurs. 63. With a known Ba2+ concentration, the SO42− concentration can be calculated using the Ksp value.  Ba 2+ + SO 4 2− BaSO 4 ( s ) 

K sp = Ba 2+  SO 4 2−  = 1.5 ×10 −9

(a)

Ba 2+ = 0.050M

K sp 1.5 ×10 −9 2− SO  = = = 3.0 ×10 −8 M SO4 2− in solution  4  Ba 2+  0.050  

(b) M SO4 2− = M BaSO4 in solution  3.0 ×10 −8 mol BaSO 4   233.4 g    ( 0.100 L )   L  mol    = 7.0 ×10 −7 g BaSO 4 remain in solution

64. If [Pb2+][Cl−]2 exceeds the Ksp, precipitation will occur. 2

2+  Cl −  = 2.0 ×10 −5 K sp = Pb 

→ 0.050M Pb 2+ 0.050M Pb ( NO3 )2 ⎯⎯ 0.010M NaCl ⎯⎯ → 0.010M Cl −

( 0.050 )( 0.010 ) = 5.0 ×10 −6 2

[Pb2+][Cl−]2 is smaller than the Ksp value. Therefore, no precipitate of PbCl2 will form. 65

HC2H3O2 and NaC2H3O2 , Na2HPO4 and H2 PO4 − , H2CO3 and NaHCO3

– Chapter 16 –

 2 SO3 ( g ) 66. 2 SO 2 ( g ) + O 2 ( g )  2 SO3 ] 11.0 ) ( [ K eq = = = 1.1×104 2 2 −3 [SO2 ] [O2 ] ( 4.20 ) ( 0.60 ×10 ) 2

67.

 1 mol  −4  = 2.7 ×10 mol BaF2 175.3 g  

( 0.048 g BaF2 ) 

 2.7 ×10 −4 mol  −2   = 1.8×10 M BaF2 dissolved  0.015 L   Ba 2+ + 2 F− BaF2 ( s )  −2 1.8 ×10

( molar concentration )

21.8 ×10−2  



K sp = Ba 2+  F −  = (1.8 ×10 −2 )( 3.6 ×10 −2 ) = 2.3 ×10 −5 2

 2 NH3 68. N 2 + 3 H2 

[ NH3 ] = 4.0 K eq = 3 [ N 2 ][H2 ] 2

Let y = [ NH3 ]

4.0 =

y 2 = 64

( 2.0 )( 2.0 ) y = 8.0M = [ NH3 ] 3

y = 64

69. Total volume of mixture = 40.0 mL (0.0400 L) K sp = Sr 2+  SO 4 2−  = 7.6 ×10 −7

(1.0 ×10 M ) ( 0.0250 L ) = 6.3×10 M Sr  = −3





2− SO  4  =

−4

0.0400 L ( 2.0 ×10−3 M ) ( 0.0150 L )

= 7.5 ×10 −4 M

0.0400 L −4 −4 −7  = ( 6.3 ×10 )( 7.5 ×10 ) = 4.7 ×10

Sr   SO 4 4.7 × 10−7 < 7.6 × 10−7 no precipitation should occur. 2+

2−

→mol Hg 2 I2 70. First change g Hg2 I2 ⎯⎯  3.04 ×10 −7 g Hg 2 I 2  1 mol  −10    = 4.64 ×10 M Hg 2 I 2 L 655.0 g    2+  Hg 2 Hg 2 I 2  + 2 I− 4.64 ×10−10 M

( molar solubility )

2 4.64 ×10−10 M  



K sp = Hg 2 2+  I −  = ( 4.64 ×10 −10 )( 9.28 ×10 −10 ) = 4.00 ×10 −28 2

 2 O3 ( g ) 71. 3 O 2 ( g ) + heat 

Three ways to increase ozone (a) increase heat (b) increase amount of O2

– Chapter 16 –

HC2H3O2, acetic acid, is strongest HCIO, hypochlorous acid, is next HCN, hydrocyanic, is weakest CaF2, calcium fluoride, is most soluble (Even though CaF2 has the lowest solubility product, if the molar solubility is calculated, it is higher than the others.)

Salts

PbSO4, lead(II) sulfate, is next BaSO4, barium sulfate, is least soluble  H 2 O ( g ) 73. H 2 O ( l ) 

Conditions on the second day (a) the temperature could have been cooler (b) the humidity in the air could have been higher (c) the air pressure could have been greater  CO 2 ( g ) + H 2 ( g ) 74. CO ( g ) + H 2 O ( g ) 

[CO2 ][H2 ] = 1 [CO][H2O ]

With equal concentrations of products and reactants, the Keq value will equal 1. 75. (a)

[O ] K eq = 3 3 [O2 ]

(b)

H 2 O ( l )  K eq =  H  2 O ( g ) 

(c)

K eq = CO 2 ( g ) 

(d)

+ H   K eq = 3 3+ 2 Bi  [H 2S]

76. 2A

 

1.0 M

Initial conditions

1.0 − 2(0.30)

1.0 − 0.30

0.30

Equilibrium concentrations

0.4 M

0.7 M

0.30 M

K eq =

[C ] = 0.30 = 3 2 2 [ A ] [B] ( 0.4 ) ( 0.7 )

– Chapter 16 –

77. Since the second reaction is the reverse of the first, the Keq value of the second reaction will be the reciprocal of the Keq value of the first reaction. [I ][Cl2 ] = 2.2 ×10 −3 first reaction K eq = 2 ( ) 2 [ICl]

[ICl] K eq = [I2 ][Cl2 ] 2

K eq =

1 = 450 2.2 ×10 −3

 H + ( aq ) + NO 2 − ( aq ) 78. HNO 2 ( aq ) 

OH− reacts with H+ and equilibrium shifts to the right. (a) After an initial increase, [OH−] will be neutralized and equilibrium shifts to the right. (b) [H+] will be reduced (reacts with OH−). Equilibrium shifts to the right. (c) [NO2−] increases as equilibrium shifts to the right. (d) [HNO2] decreases and equilibrium shifts to the right. 79. (a)

2 A3X  2 A 2 X + A 2

( A 2 X ) ( A 2 ) = ( 3) ( 6 ) = 3.375 K eq = 2 2 ( A3X ) (4) 2

(b) The equilibrium lies to the right. Keq > 1 80. (a)

X 2 + 2G  X 2G 2 K eq =

( X 2G 2 ) = (1) = 8.33×10−2 2 2 ( X 2 )( G 2 ) ( 3)( 2 )

(b) Exothermic. An increase in the amount of reactants means that the equilibrium shifted to the left. (c) An increase in pressure will cause an equilibrium to shift by reducing the number of moles of gas in the equilibrium. If the equilibrium shifts to the right, there must be fewer moles of gas in the product than in the reactants. 81. CaSO 4 ( s )  Ca 2 + ( aq ) + SO 4 2 − ( aq ) K sp = Ca 2 +  SO 4 2 −  = 2.0 ×10 −4 Let x = moles CaSO4 that dissolve per L = Ca 2 +  = SO 4 2 − 

( x )( x ) = 2.0 ×10−4

x = 2.0 ×10 −4

x = 0.014M CaSO 4 M → moles → grams  0.014 mol CaSO 4   136.2 g    ( 0.600 L )   = 1.1 g CaSO 4 L    mol 

– Chapter 16 –

82.

PbF2 ( s )  Pb2+ + 2 F− change g PbF2 → mol PbF2  0.098 g PbF2   1 mol  −3 −3  = 1.0 ×10 mol/L = 1.0 ×10 M PbF2    0.400 L   245.2 g  K sp = ( Pb2+ )( F− )

Pb2  = 1.0 ×10 −3 ; F−  = 2 (1.0 ×10 −3 ) = 2.0 ×10 −3

K sp = (1.0 ×10 −3 )( 2.0 ×10 −3 ) = 4.0 ×10 −9 2

83. Equilibrium shifts HCOOH ( g )  CO ( g ) + H2 O ( g ) (a) Reaction will shift to the right (b) Reaction will shift to the left (c) Reaction will shift to the right (d) Reaction will shift to the left 84. C 7 H 6 O 3 + C 4 H 6 O 3  C 9 H 8 O 4 + C 2 H 4 O 2 Molar cost of salicylic acid    138.1 g C 7 H6 O3  $58.90 $16.27   =  500 g C 7H 6 O3   1 mol C 7H 6 O3  mol C 7 H6 O3

Molar cost of acetic anhydride    102.1 g C 4 H6 O3  $53.50 $5.46   =  1000 g C 4H6 O3   1 mol C 4 H6 O3  mol C 4 H6 O3

It is more economical to use the acetic anhydride to drive the reaction because it has a lower cost per mole. 85. (a)

HCO3 − ( aq ) + H+ ( aq )  H2CO3 ( aq )

(b) HCO3 − ( aq ) + OH− ( aq )  H2 O ( l ) + CO32− ( aq ) 86. Treat this is an equilibrium where W = whole nuts, S = shell halves, and K = kernels W  2S + K 144 0 0 amount before cracking 144 − x 2x x x = number of kernels after cracking

144 – x + 2x + x = 194 total pieces 144 + 2x = 194; 2x = 50 x = 25 kernels; 50 shell halves;

K eq

119 whole nuts left

( 2x ) ( x ) = (50 ) ( 25 ) = 5.3×102 = 2

144 − x

119

– Chapter 16 –

87. SO2 ( g ) + NO2 ( g )  SO3 ( g ) + NO ( g )

0.50M 0.50M 0 0 Initial conditions 0.50 − x 0.50 − x x x Equilibrium concentrations K eq =

[SO3 ][ NO] = x 2 = 81 [SO2 ][ NO2 ] ( 0.50 − x )2

Take the square root of both sides x = 9.0 x = 0.45M 0.50 − x [SO3 ] = [ NO] = 0.45M

[SO2 ] = [ NO2 ] = 0.05M 88. (a)

2 NaOH ( aq ) + H2S ( aq ) → Na 2S ( aq ) → 2 H2 O ( l )

(b) H 2S  2H + + S2 − ( aqueous solution ) Na 2S → 2 Na + + S2 − ( aqueous solution )

The addition of S2− to a solution of H2S will shift the equilibrium to the left, reducing the [H+] and thereby increasing the pH (more basic). 89. (a) Yes. The Keq of AgCN indicates that it is slightly soluble in water, so a precipitate will form. Net ionic equation: Ag + ( aq ) + CN − ( aq )  AgCN ( s ) (b) NaCN is a salt of a weak acid and a strong base and will hydrolyze in water. CN − ( aq ) + H2 O ( l )  HCN ( aq ) + OH− ( aq ) The solution will be basic due to increased OH− concentration.

CHAPTER 17

OXIDATION-REDUCTION SOLUTIONS TO REVIEW QUESTIONS 1.

(a) Iodine is oxidized. Its oxidation number increases from 0 to +5. (b) Chlorine is reduced. Its oxidation number decreases from 0 to −1.

The oxidation number for an atom in an ionic compound is the same as the charge of the ion that resulted when that atom lost or gained electrons to form an ionic bond. In a covalently bonded compound electrons are shared between the two atoms making up the bond. Those shared electrons are assigned to the atom in the bond with a higher electronegativity giving it a negative oxidation number.

Oxidation and reduction are complementary processes because one does not occur without the other. The loss of e− in oxidation is accompanied by a gain of e− in reduction.

Redox reactions are usually very difficult to balance by inspection so other more efficient methods are utilized to balance them.

Mn2+ → Mn+6 4 electrons are needed on the right-hand side of the equation.

O0 → O2− 2 electrons are needed on the left-hand side of the equation.

Break the reaction into half reactions; one for the oxidation reaction and one for the reduction reaction.

2 IO3− +12 H+ +10 e − → I 2 + 6 H2 O To balance this equation in base you would add 12 OH− ions to both sides of the reaction. On the left side, these OH−s would combine with the H+ to form water. Cancel 6 H2O from each side of the equation. 2 IO3− + 6 H2 O +10 e − → I 2 +12 OH−

Oxidation of a metal occurs when the metal loses electrons. The easier it is for a metal to lose electrons, the more active the metal is.

10. The higher metal on the activity series list is more reactive. (a) Ca

(b) Fe

11. If the free element is higher on the activity series than the ion with which it is paired, the reaction occurs. The higher metal on the activity series is more reactive. (a)

2 Al ( s ) + 3 ZnCl2 ( aq ) → 3 Zn ( s ) + 2 AlCl3 ( aq )

(b) Sn ( s ) + 2HCl ( aq ) → H2 ( g ) + SnCl2 ( aq ) (c)

Ag ( s ) + H2SO4 ( aq ) → no reaction H2 is more reactive than Ag

(d) Fe ( s ) + 2 AgNO3 ( aq ) → 2 Ag( s ) + Fe ( NO3 )2 ( aq ) (e)

2 Cr( s ) + 3 Ni2 + ( aq ) → 3 Ni ( s ) + 2 Cr 3+ (aq)

– Chapter 17 –

(f)

Mg ( s ) + Ca 2 + ( aq ) → no reaction

Ca is more reactive than Mg

(g) Cu( s ) + H+ ( aq ) → no reaction

H2 is more reactive than Cu

(h) Ag ( s ) + Al3+ ( aq ) → no reaction

Al is more reactive than Ag

12. If a copper wire is placed into a solution of lead(II) nitrate, no reaction will occur. Lead is more active than copper and undergoes oxidation more readily than copper. Lead is already oxidized, therefore will stay oxidized in the presence of copper. 13. (a)

2 Al + Fe2 O3 → Al2 O3 + 2 Fe + Heat

(b) Al is above Fe in the activity series, which indicates Al is more active than Fe. (c) No. Iron is less active than aluminum and will not displace aluminum from its compounds. (d) Yes. Aluminum is above chromium in the activity series and will displace Cr3+ from its compounds. 14. (a)

2 Al ( s ) + 6 HCl ( aq ) → 2 AlCl3 ( aq ) + 3 H2 ( g )

2 Al ( s ) + 3 H2SO4 ( aq ) → Al2 ( SO4 )3 ( aq ) + 3 H2 ( g )

(b) 2 Cr ( s ) + 6 HCl ( aq ) → 2 CrCl3 ( aq) + 3H2 ( g )

2 Cr ( s ) + 3H2SO4 ( aq ) → Cr2 (SO4 )3 ( aq ) + 3H2 ( g )

(c)

Au ( s ) + HCl( aq ) → no reaction

Au ( s ) + H2SO 4 ( aq ) → no reaction (d) Fe ( s ) + 2 HCl ( aq ) → FeCl2 ( aq) + H2 ( g )

Fe ( s ) + H2SO 4 ( aq ) → FeSO4 ( aq ) + H2 ( g )

(e)

Cu( s ) + HCl ( aq ) → no reaction

Cu ( s ) + H2SO4 ( aq ) → no reaction (f)

Mg ( s ) + 2 HCl( aq ) → MgCl2 ( aq) + H2 ( g )

Mg ( s ) + H2SO4 ( aq ) → MgSO4 ( aq) + H2 ( g )

(g) Hg(l ) + HCl ( aq ) → no reaction

Hg(l ) + H2SO4 ( aq ) → no reaction

(h) Zn ( s ) + 2 HCl ( aq ) → ZnCl2 ( aq ) + H2 ( g )

Zn ( s ) + H2SO4 ( aq ) → ZnSO4 ( aq ) + H2 ( g )

15. In an electrolytic cell the anode is the positively charged electrode and attracts negatively charged ions (anions). The cathode is the negatively charged electrode and attracts positively charged ions (cations). In a voltaic cell the anode is the negatively charged electrode where oxidation occurs. The cathode is the positively charged electrode where reduction occurs.

– Chapter 17 –

16. (a) Oxidation occurs at the anode. The reaction is

2 Cl − ( aq ) → Cl2 ( g ) + 2 e − (b) Reduction occurs at the cathode. The reaction is

Ni2+ (aq) + 2 e− → Ni( s ) (c) The net chemical reaction is electrical Ni 2+ (aq) + 2 Cl − ( aq ) ⎯⎯⎯ ⎯ → Ni( s ) + Cl 2 ( g ) energy

17. In Figure 17.4, electrical energy is causing chemical reactions to occur. In Figure 17.5, chemical reactions are used to produce electrical energy. 18. (a) It would not be possible to monitor the voltage produced, but the reactions in the cell would still occur. (b) If the salt bridge were removed, the reaction would stop. Ions must be mobile to maintain an electrical neutrality of ions in solution. The two solutions would be isolated with no complete electrical circuit. 19. Ca 2+ + 2 e − → Ca −

2 Br → Br2 + 2 e

cathode reaction, reduction −

anode reaction, oxidation

20. (a) voltaic (b) electrolytic (c) voltaic 21. During electroplating of metals, the metal is plated by reducing the positive ions of the metal in the solution. The plating will occur at the cathode, the source of the electrons. With an alternating current, the polarity of the electrode would be constantly changing, so at one instant the metal would be plating and the next instant the metal would be dissolving. 22. Since lead dioxide and lead(II) sulfate are insoluble, it is unnecessary to have salt bridges in the cells of a lead storage battery. 23. The electrolyte in a lead storage battery is dilute sulfuric acid. In the discharge cycle, SO42−, is removed from solution as it reacts with PbO2 and H+ to form PbSO4(s) and H2O. Therefore, the electrolyte solution contains less H2SO4 and becomes less dense. 24. If Hg2+ ions are reduced to metallic mercury, this would occur at the cathode, because reduction takes place at the cathode. 25. In both electrolytic and voltaic cells, oxidation and reduction reactions occur. In an electrolytic cell an electric current is forced through the cell causing a chemical change to occur. In voltaic cells, spontaneous chemical changes occur, generating an electric current. 26. In some voltaic cells, the reactants at the electrodes are in solution. For the cell to function, these reactants must be kept separated. A salt bridge permits movement of ions in the cell. This keeps the solution neutral with respect to the charged particles (ions) in the solution.

– Chapter 17 –

SOLUTIONS TO EXERCISES

Oxidation numbers of each element in the compound (a)

CuCO3 Cu = +2 C = +4 O = −2

(b) CH4 (c)

C = −4 H = +1

IF I = +1 F = −1

C = 0 H = +1 Cl = −1

(d) CH2Cl2

(e)

SO2 S = +4 O = −2

(f)

Rb2CrO4

Oxidation numbers of each element in the compound. (a) CHF3

C = +2 H = +1 F = −1

(b) P2 O5

P = +5 O = −2

(c)

SF6 S = +6 F = −1

(d) SnSO4

Rb = +1 Cr = +6 O = −2

Sn = +2 S = +6 O = −2

(e)

CH3OH C = −2 H = +1 O = −2

(f)

H3PO4 H = +1 P = +5 O = −2

(a) +7; sum of oxidation numbers = charge on species; Cl + 4(−2) = −1, so Cl = −1 + 8 = +7 (b) +5; 2 P + 5(−2) = 0, so 2 P = +10, so P = +5 (c) +3; N + 2(−2) = −1, so N= −1 + 4 = +3 (d) 0; 6 C + 12(+1) + 6(−2) = 0, so C = 0 (e) +5; Br + 3(−2) = −1, so Br = +5 (f)

+3; 2 (C) + 4(−2) = −2, so C = +3

(g) +4; 2(+1) + S + 3(−2) = 0, so S = +4 4.

(a) +6 (each O is −2 so the oxidation number of the three O’s is −6. CrO3 is neutral so the oxidation numbers sum to zero. Cr must have an oxidation number of +6) or Cr + 3(−2) = 0, so Cr = +6 (b) +4; C + 4(−1) = 0, so C = +4 (c) +1; 2 (N) + (−2) = 0, so 2 N = +2, N = +1 (d) +7; (+1) + Mn + 4(−2) = 0, so Mn = +7 (f)

+2; Zn + 2(−2) = −2, so Zn = +2

(e) +5; As + 4(−2) = −3, so As = +5 (g) −4; C + 4(+1) = 0, so C = −4

– Chapter 17 –

(a)

Na2CrO4 Cr + 6

(b) K3 [Fe(CN)6 ] Fe + 3 (c)

CoCl2 Co + 2; Cl −1

(d) NiCl2 ⋅ 6 H2 O Ni + 2 6.

(a)

CuSO4 ⋅5 H2 O Cu + 2; S + 6

(b) KMnO4 (c)

MnO2 Mn + 4

(d) K 2Cr2 O7 7.

Mn + 7

Cr + 6

Balanced half-reaction Element Changing

Type of Reaction

(a)

Na → Na+ + 1 e−

oxidation

(b)

C2O42− → 2 CO2 + 2 e−

oxidation

(c)

2 I−→I2+2 e−

oxidation

(d)

Cr2O72− + 14 H+ + 6 e− → 2 Cr3+ + 7 H2O

reduction

Element Changing

Type of Reaction

8. Balanced half-reaction

(a)

Cu2+ + 1 e− → Cu+

reduction

(b)

F2 + 2 e−→ 2 F−

reduction

(c)

2 IO4− + 16 H+ + 14 e− → I2 + 8 H2O

reduction

(d)

Mn → Mn2+ + 2 e−

oxidation

(a) Hydrogen gas is oxidized and is the reducing agent. Chromium is reduced and is the oxidizing agent. Cr2 O3( s ) + 3 H2 ( g ) ⎯Δ⎯ → 2 Cr( s ) + 3 H2 O( l )

First, assign oxidation numbers. O = −2 and H = +1. Hydrogen gas = 0. Chromium metal = 0. Cr2 O3 + 3 H2 → 2 Cr+3 H2 O +3

−2

+1 −2

What is Cr in Cr2O3? If O = −2, then each Cr will have to be +3 in order to balance the total amount of negative oxidation value of 3 oxygens, Which element has been oxidized (lost electrons)? Hydrogen has changed from 0 to +1, so it has been oxidized. Cr has changed from +3 to 0, so it has been reduced.

– Chapter 17 –

(b) Iodine is oxidized and is the reducing agent. Nitrogen is reduced and is the oxidizing agent.

2 HNO2( aq ) + 2 HI( aq ) → I2( s ) + 2 NO( g ) + 2 H2 O(l ) +1 +3 −2

+1 −1

+2 −2

+1 − 2

Iodine changes from −1 to 0, which is a loss of one electron (oxidation). Nitrogen has gained one electron in changing from +3 to +2. 10. (a) Hydrogen gas oxidized and is the reducing agent. Oxygen gas is reduced and is the oxidizing agent. 2 H2 ( g ) + O 2( g ) ⎯Δ⎯ → 2 H 2 O( g ) +1 −

Each hydrogen atom has lost an electron, and each oxygen atom has gained two electrons. (b) Bromine is both oxidized and reduced.

5 NaBr + NaBrO3 + 3H2SO 4 → 3Br2 + 3 Na 2SO4 + 3H2 O +1 −1

+1 + 6 − 2

+1 +5 − 2

+6 −2

+1 − 2

The only atoms changing oxidation number are Br. Five Br atoms change from −1 to 0.This is oxidation, a loss of electrons. NaBr is the reducing agent. The other Br atom changes from +5 to 0, which is a gain of electrons, or reduction. NaBrO3 is the oxidizing agent. 11. (a) correctly balanced (b) correctly balanced (c) incorrectly balanced

Mg( s ) + 2 HCl ( aq ) → Mg 2+ ( aq ) + 2 Cl− ( aq ) + H2 ( g ) (d) incorrectly balanced

3 CH3OH ( aq ) + Cr2 O7 2− ( aq ) + 8 H+ ( aq ) → 2 Cr3+ ( aq ) + 3 CH2O ( aq ) + 7 H2O ( l ) 12. (a) incorrectly balanced

3 MnO2 ( s ) + 4 Al ( s ) → 3 Mn ( s ) + 2 Al2 O3 ( s ) (b) correctly balanced (c) correctly balanced (d) incorrectly balanced

8H2 O ( l ) + 2 MnO4 − (aq) + 7S2− ( aq ) → 2 MnS ( s ) +16 OH− ( aq ) + 5 S ( s ) 13. Balancing oxidation-reduction equations using the change-in-oxidation number method: (a) Cu + O2 → CuO ox Cu 0 → Cu 2+ + 2 e − −

O2 + 4 e → 2 O 0

red

Multiply by 2

2−

2 Cu + O2 → 2 Cu + 2 O 0

2−

Add the equations; the 4 e − cancel

– Chapter 17 –

Transfer the coefficients to the original equation appropriately.

2 Cu + O2 → 2CuO (b) KClO3 → KCl + O2 ox 2 O 2− → O 2 0 + 4 e − red

Cl5+ + 6 e − → Cl 2 Cl + 6 O 5+

2−

Multiply by 3

−

→ 2 Cl + 3 O 2

Multiply by 2, add, the 12 e − cancel

Transfer the coefficients to the original equation appropriately.

2 KClO3 → 2 KCl + 3 O2 (c)

Ca + H2 O → Ca(OH)2 + H2 ox Ca → Ca 2+ + 2 e − red

2 H+ + 2 e − → H2 +

Add equations together; the 2 e − cancel

Ca + 2 H → Ca + H2 2+

Balance the equation by inspection.

Ca + 2 H2O → Ca(OH)2 + H2 (d) PbS + H2 O2 → PbSO4 + H2O ox red

S2− → S6+ + 8 e − 2 O − + 2 e − → 2 O 2− −

S +8 O →S +8 O 2−

2−

Multiply by 4, add, the 8 e− cancel

Transfer the coefficients to the original equation and complete the balancing by inspection.

PbS + 4 H2O2 → PbSO4 + 4 H2O (e)

CH4 + NO2 → N2 + CO2 + H2 O ox red

C 4− → C 4+ + 8e − 2 N 4+ + 8e − → N 2 0 C +2N 4−

→ C + N2 4+

Add equations together; the 8 e − cancel

Transfer the coefficients to the original equation and complete the balancing by inspection.

CH4 + 2 NO2 → N2 + CO2 + 2 H2O

– Chapter 17 –

14. Balancing oxidation-reduction equations using the change-in-oxidation number method: (a) Cu + AgNO3 → Ag + Cu ( NO3 )2 ox red

Cu 0 → Cu 2+ + 2 e − Ag+ +1 e − → Ag 0 +

Cu + 2 Ag → Cu + 2 Ag 0

Multiply by 2, add, the 2 e − cancel

Transfer the coefficients to the original equation.

Cu + 2 AgNO3 → 2 Ag + Cu ( NO3 )2 (b) MnO2 + HCl → MnCl2 + Cl2 + H2 O

ox red

2 Cl − → Cl 2 0 + 2 e −

Add equations together; the 2 e − cancel

Mn 4+ + 2 e − → Mn2+ 2 Cl − + Mn4+ → Cl2 0 + Mn2+

Transfer the coefficients to the original equation and complete the balancing by inspection.

MnO2 + 4 HCl → MnCl2 + Cl2 + 2 H2 O (c)

HCl + O2 → Cl2 + H2 O red ox

O 2 0 + 4 e − → 2 O 2− 2 Cl − → Cl2 0 + 2 e − −

4 Cl + 2 O → 2 Cl 2 + 2 O 0

2−

Multiply by 2, add, the 4 e − cancel

Transfer the coefficients to the original equation and complete the balancing by inspection.

4 HCl + O2 → 2 Cl2 + 2 H2 O (d) Ag + H2S + O2 → Ag 2S + H2 O red O2 0 + 4 e − → 2 O2− ox

Ag → Ag + +1e − +

4 Ag + O2 → 4 Ag + 2 O 0

2−

Multiply by 4, add, the 4 e − cancel

Transfer the coefficients to the original equation and complete the balancing by inspection.

4 Ag + 2 H2S + O2 → 2 Ag 2S + 2 H2 O (e)

KMnO4 + CaC2 O4 + H2SO4 → K2SO4 + MnSO4 + CaSO4 + CO2 + H2 O red Mn7+ + 5 e − → Mn 2+ 2 C → 2 C +2 e 3+

10 C + 2 Mn 3+

Multiply by 2;

−

→ 10 C + 2 Mn 4+

Multiply by 5, add, the 10 e − cancel

– Chapter 17 –

Transfer the coefficients to the original equation and complete the balancing by inspection. 2 K MnO 4 +5 CaC 2 O 4 +8 H 2SO 4 → K 2SO 4 + 2 MnSO 4 + 5 CaSO 4 +10 CO 2 + 8 H 2 O

15. Balancing ionic redox equations (a) Zn + NO3−→ Zn2+ + NH4+ (acidic solution) Step 1

Write half-reaction equations. Balance except H and O Zn → Zn2+ NO3−→NH4+

Step 2

Balance H and O using H2O and H+ Zn → Zn2+ 10 H++NO3−→NH4++3 H2O

Step 3

Balance electrically with electrons Zn → Zn2++2 e− 10 H++NO3−+ 8 e− → NH4++3 H2O

Step 4

Equalize the loss and gain of electrons 4(Zn → Zn2++2 e−) 10 H++NO3−+8 e− → NH4++3 H2O

Step 5

Add the half-reactions; electrons cancel 10 H++4 Zn+NO3− → 4 Zn2++NH4++3 H2O

(b) NO3−+ S → NO2 + SO42− (acidic solution) Step 1

Write half-reaction equations. Balance except H and O. S → SO42− NO3− → NO2

Step 2

Balance H and O using H2O and H+ 4 H2O + S → SO42−+8 H+ 2 H+ + NO3− → NO2 + H2O

Step 3

Balance electrically with electrons 4 H2O + S → SO42−+8 H+ + 6 e− 2 H+ + NO3−+ e− → NO2 + H2O

– Chapter 17 –

Steps 4 and 5 Equalize the loss and gain of electrons; add the half reactions 4 H2O + S → SO42− + 8 H+ + 6 e− 6(2 H+ + NO3− + e− → NO2 + H2O) 4 H+ + S + 6 NO3− → 6 NO2 + SO42− + 2 H2O 4 H2O, 8 H+ and 6 e−canceled from each side (c)

PH3 + I2 → H3PO2 + I− (acidic solution) Step 1

Write half-reaction equations. Balance except H and O PH3 → H3PO2 I2 → 2 I −

Step 2

Balance H and O using H2 O and H+ 2 H2 O + PH3 → H3PO2 + 4 H+ I2 → 2 I −

Step 3

Balance electrically with electrons 2 H2 O + PH3 → H3PO2 + 4 H+ + 4 e − I2 + 2 e− → 2 I−

Steps 4 and 5 Equalize the loss and gain of electrons; add the half-reactions 2 H2 O + PH3 → H3PO2 + 4 H+ + 4 e − 2(I 2 + 2 e − → 2 I − ) PH3 + 2 H2 O + 2 I 2 → H3PO2 + 4 I − + 4 H+ − 2+ (d) Cu + NO3 → Cu + NO (acidic solution)

Step 1

Write half-reaction equations. Balance except H and O Cu → Cu 2+ NO3 − → NO

Step 2

Balance H and O using H2 O and H+ Cu → Cu 2+ 4 H+ + NO3 − → NO + 2 H2 O

Step 3

Balance electrically with electrons Cu → Cu 2+ + 2 e − 4 H+ + NO3 − + 3e − → NO + 2 H2 O

Steps 4 and 5

Equalize the loss and gain of electrons; add the half-reactions 3(Cu → Cu 2+ + 2 e − ) 2(4 H+ + NO3 − + 3e − → NO + 2 H2 O) 3 Cu + 8 H+ + 2 NO3 − → 3 Cu 2+ + 2 NO+4 H2 O

– Chapter 17 –

(e)

ClO3− + Cl− → Cl2 (acidic solution) Step 1

Write half-reaction equations. Balance except H and O Cl − → Cl0 ClO3 − → Cl0

Step 2

Balance H and O using H2 O and H+ Cl − → Cl0 6 H+ + ClO3− → Cl0 + 3 H2 O

Step 3

Balance electrically with electrons Cl − → Cl0 + e − 6 H+ + ClO3− + 5 e − → Cl0 + 3 H2 O

Steps 4 and 5

Equalize the loss and gain of electrons; add the half-reactions 5(Cl − → Cl0 + e − ) 6 H+ + ClO3− + 5 e − → Cl0 + 3 H2 O 6 H+ + ClO3− + 5 Cl − → 3 Cl2 + 3 H2 O

− − − 16. (a) ClO3 + I → I2 + Cl (acidic solution)

Step 1

Write half-reaction equations. Balance except H and O 2 I− → I2 ClO3 − → Cl −

Step 2

Balance H and O using H2 O and H+ 2 I− → I2 6 H+ + ClO3 − → Cl − + 3 H2 O

Step 3

Balance electrically with electrons 2 I− → I2 + 2 e− _

6 H+ + ClO3 − + 6 e − → Cl + 3 H2 O Step 4 and 5

Equalize the loss and gain of electrons; add the half-reactions 3(2 I − → I 2 + 2 e − ) _

6 H+ + ClO3 − + 6 e − → Cl + 3 H2 O 6 H+ + ClO3 − + 6 I − → 3 I 2 + Cl − + 3 H2 O

– Chapter 17 – 2− 2+ 3+ 3+ (b) Cr2O7 + Fe → Cr + Fe (acidic solution)

Step 1

Write half-reaction equations. Balance except H and O Fe2+ → Fe3+ Cr2 O7 2− → 2 Cr3+

Step 2

Balance H and O using H2 O and H+ Fe2+ → Fe3+ 14 H+ + Cr2 O7 2− → 2 Cr 3+ + 7 H2 O

Step 3

Balance electrically with electrons Fe2+ → Fe3+ + e − 14 H+ + Cr2 O7 2− + 6 e − → 2 Cr 3+ + 7 H2 O

Steps 4 and 5

Equalize the loss and gain of electrons; add the half-reactions 6(Fe2+ → Fe3+ + e − ) 14 H+ + Cr2 O7 2− + 6 e − → 2 Cr 3+ + 7 H2 O 14 H+ + Cr2 O7 2− + 6 Fe2+ → 2 Cr3+ + 6 Fe3+ + 7 H2 O

(c)

MnO4 − +SO2 → Mn2+ + SO42− (acidic solution) Step 1

Write half-reaction equations. Balance except H and O SO2 → SO4 2− MnO4 − → Mn2+

Step 2

Balance H and O using H2 O and H+ 2 H2 O+SO2 → SO4 2− + 4 H+ 8 H+ +MnO4 − → Mn2+ + 4 H2 O

Step 3

Balance electrically with electrons 2 H2 O+SO2 → SO4 2− + 4 H+ + 2 e − 8 H+ +MnO4 − + 5 e − → Mn2+ + 4 H2 O

Steps 4 and 5 Equalize the loss and gain of electrons; add the half-reactions 5(2 H2 O+SO2 → SO 4 2− + 4 H+ + 2 e − ) 2(8 H+ +MnO4 − + 5 e − → Mn2+ + 4 H2 O) 2 H2 O + 2 MnO4 − + 5 SO2 → 4 H+ + 2 Mn2+ + 5 SO4 2− 8 H2 O, 16 H+, and 10 e − canceled from each side

– Chapter 17 – − 2+ (d) H3AsO3 + MnO4 → H3AsO4 +Mn (acidic solution)

Step 1

Write half-reaction equations. Balance except H and O H3 AsO3 → H3 AsO 4 MnO 4 − → Mn2+

Step 2

Balance H and O using H2 O and H+ H2 O+H3 AsO3 → 2 H+ + H3 AsO 4 8 H+ +MnO 4 − → Mn2+ + 4 H2 O

Step 3

Balance electrically with electrons H2 O+H3 AsO3 → 2 H+ + H3 AsO 4 + 2 e − 8 H+ +MnO 4 − + 5 e − → Mn2+ + 4 H2 O

Steps 4 and 5

Equalize the loss and gain of electrons; add the half-reactions 5(H2 O+H3 AsO3 → 2 H+ + H3 AsO4 + 2 e − ) 2(8 H+ +MnO 4 − + 5 e − → Mn2+ + 4 H2 O) 6 H+ + 5 H3 AsO3 + 2 MnO 4 − → 5 H3 AsO 4 + 2 Mn2+ + 3 H2 O 5 H2 O, 10 H+ , and 10 e − canceled from each side

(e)

Cr2O72− +H3AsO3 → Cr3+ +H3AsO4 (acidic solution) Step 1

Write half-reaction equations. Balance except H and O H3 AsO3 → H3 AsO 4 Cr2 O7 2− → 2 Cr3+

Step 2

Balance H and O using H2 O and H+ H2 O+H3 AsO3 → 2 H+ + H3 AsO 4 14 H+ +Cr2 O7 2− → 2 Cr 3+ + 7 H2 O

Step 3

Balance electrically with electrons H2 O+H3 AsO3 → 2 H+ + H3 AsO 4 + 2 e − 14 H+ +Cr2 O7 2− + 6 e − → 2 Cr 3+ + 7 H2 O

Steps 4 and 5

Equalize the loss and gain of electrons; add the half-reactions 3(H2 O+H3 AsO3 → 2 H+ + H3 AsO 4 + 2 e − ) 14 H+ +Cr2 O7 2− + 6 e − → 2 Cr 3+ + 7 H2 O 8 H+ + Cr2 O7 2− + 3 H3 AsO3 → 2 Cr 3+ + 3 H3 AsO 4 + 4 H2 O 3 H2 O, 6 H+ , and 6 e − canceled from each side

– Chapter 17 – − − − 17. (a) Cl2 + IO3 → Cl + IO4 (basic solution)

Step 1

Write half-reaction equations. Balance except H and O IO3− → IO4 − Cl2 → 2 Cl −

Step 2

Balance H and O using H2 O and H+ H2 O + IO3− → IO4 − + 2 H+ Cl2 → 2 Cl −

Step 3

Add OH− ions to both sides (same number as H + ions) 2 OH− + H2 O+IO3 − → IO4 − + 2 H+ + 2 OH− Cl2 → 2 Cl −

Step 4

Combine H+ and OH− to form H2 O; cancel H2 O where possible 2 OH− + H2 O + IO3 − → IO 4 − + 2 H2 O Cl2 → 2 Cl − 2 OH− + IO3 − → IO 4 − + H2 O (1 H2 O canceled) Cl2 → 2 Cl

Step 5

Balance electrically with electrons 2 OH− + IO3 − → IO 4 − + H2 O + 2 e − Cl2 + 2 e − → 2 Cl

Step 6 Step 7

Electron loss and gain is balanced Add half-reactions 2 OH− + IO3 − + Cl2 → IO 4 − + 2 Cl − + H2 O

− − − (b) MnO4 + ClO2 → MnO2 + ClO4 (basic solution)

Step 1

Write half-reaction equations. Balance except H and O ClO2 − → ClO 4 − MnO 4 − → MnO2

Step 2

Balance H and O using H2 O and H+ 2 H2 O + ClO2 − → ClO 4 − + 4 H+ MnO 4 − + 4 H + → MnO2 + 2 H2 O

Step 3

Add OH− ions to both sides (same number as H+ ions) 4 OH − + 2 H2 O + ClO 2 − → ClO 4 − + 4 H+ + 4 OH− 4 OH − + MnO 4 − + 4 H+ → MnO2 + 2 H2 O + 4 OH−

– Chapter 17 –

Step 4

Combine H + and OH − to form H2 O; cancel H2 O where possible 4 OH− + 2 H2 O + ClO 2 − → ClO 4 − + 4 H2 O 4 H2 O + MnO 4 − → MnO2 + 2 H2 O + 4 OH− 4 OH− + ClO2 − → ClO4 − + 2 H2 O (2 H2 O canceled) 2 H2 O + MnO 4 − → MnO2 + 4 OH − (2 H2 O canceled)

Step 5

Balance electrically with electrons 4 OH− + ClO 2 − → ClO 4 − + 2 H2 O + 4 e − 2 H2 O + MnO 4 − + 3 e − → MnO 2 + 4 OH−

Steps 6 and 7

Equalize gain and loss of electrons; add half-reactions 3(4 OH − + ClO2 − → ClO 4 − + 2 H2 O + 4 e − )

4 ( 2 H2 O + MnO 4 − + 3 e − → MnO2 + 4 OH− )

2 H2 O + 4 MnO 4 − + 3 ClO2 − → 4 MnO2 + 3 ClO 4 − + 4 OH − 6 H2 O, 12 OH− , and 12 e − canceled from each side

(c)

Se → SeO32− + Se2− (basic solution) Step 1

Write half reaction equations. Balance except H and O.

Se → SeO32− Se → Se2− Step 2

Balance H and O using H2O and H+

3 H2 O + Se → SeO32− + 6 H+ Se → Se2− Step 3

Add OH− ions to both sides (same number as H+ ions)

6 OH− + 3 H2O + Se → SeO32− + 6 H+ + 6 OH− Se → Se2− Step 4

Combine H+ and OH− to form H2O; cancel H2O where possible 6 OH − + 3 H2 O + Se → SeO32− + 6 H2 O Se → Se 2− 6 OH− + Se → SeO32− + 3 H2 O ( 3 H2 O canceled )

Step 5

Balance electrically with electrons

6 OH− + Se → SeO32− + 3 H2 O + 4 e− Se + 2 e− → Se2−

– Chapter 17 –

Steps 6 and 7

Equalize gain and loss of electrons; add half-reactions

6 OH− + Se → SeO32− + 3 H2 O + 4 e − 2 ( Se + 2 e − → Se2− )

6 OH− + 3 Se → SeO32− + 2 Se2− + 3 H2 O (d) Fe3O4 + MnO4 − + Fe2 O3 ⎯⎯ →MnO2 (basic solution) Step 1

Write half reaction equations. Balance except H and O 2 Fe3 O 4 → 3 Fe 2 O3 MnO 4 − → MnO 2

Step 2

Balance H and O using H2O and H+ H 2 O + 2 Fe3 O 4 → 3 Fe 2 O3 + 2 H + 4 H + + MnO 4 − → MnO 2 + 2 H 2 O

Step 3

Add OH− ions to both sides (same number as H+ ions) 2 OH − + H2 O + 2 Fe3 O 4 → 3 Fe 2 O3 + 2 H + + 2 OH − 4 OH − + 4 H + + MnO 4 − → MnO 2 + 2 H 2 O + 4 OH −

Step 4

Combine H+ and OH− to form H2O; cancel H2O where possible 2 OH− + H2 O + 2 Fe3O4 → 3 Fe2 O3 + 2 H2 O 4 H2 O + MnO 4 − → MnO2 + 2 H2 O + 4 OH − 2 OH− + 2 Fe3O 4 → 3 Fe 2 O3 + H2 O (1 H2 O canceled ) 2 H2 O + MnO 4 − → MnO2 + 4 OH− ( 2 H2 O canceled )

Step 5

Balance electrically with electrons 2 OH − + 2 Fe3 O 4 → 3 Fe 2 O3 + H 2 O + 2 e − 2 H 2 O + MnO 4 − + 3 e − → MnO 2 + 4 OH −

Steps 6 and 7

Equalize gain and loss of electrons; add half-reactions

3 ( 2 OH− + 2 Fe3O4 → 3 Fe2 O3 + H2 O + 2 e − ) 2 ( 2 H2 O + MnO 4 − + 3 e − → MnO2 + 4 OH− )

H2 O + 6 Fe3O4 + 2 MnO4 − → 9 Fe2 O3 + 2 MnO2 + 2 OH− 3 H2O, 6 OH−, and 6 e− canceled from each side (e)

BrO − + Cr(OH)4 − → Br − + CrO 4 2− (basic solution) Step 1

Write half reaction equations. Balance except H and O

Cr(OH)4 − → CrO4 2 − BrO− → Br −

– Chapter 17 –

Step 2

Balance H and O using H2O and H+ Cr(OH)4 − → CrO 4 2− + 4 H + 2 H + + BrO − → Br − + H 2 O

Step 3

Add OH− ions to both sides (same number as H+ ions) 4 OH − + Cr(OH)4 − → CrO 4 2− + 4 H + + 4 OH − 2 OH − + 2 H + + BrO − → Br − + H 2 O + 2 OH −

Step 4

Combine H+ and OH− to form H2O; cancel H2O where possible

4 OH− + Cr(OH)4 − → CrO4 2− + 4 H2 O 2 H2 O + BrO− → Br − + H2 O + 2 OH− H2 O + BrO− → Br − + 2 OH− (1 H2 O canceled ) Step 5

Balance electrically with electrons 4 OH − + Cr(OH)4 − → CrO 4 2− + 4 H 2 O + 3 e − H2 O + BrO − + 2 e − → Br − + 2 OH −

Steps 6 and 7

Equalize gain and loss of electrons; add half-reactions

2 ( 4 OH− + Cr(OH)4 − → CrO 4 2− + 4 H2 O + 3 e − )

3 ( H2 O + BrO − + 2 e − → Br − + 2 OH− )

2 OH− + 3 BrO − + 2 Cr(OH)4 − → 3 Br − + 2 CrO 4 2− + 5 H2 O 3 H2O, 6 OH− and 6 e− canceled from each side 18. (a) MnO4 − + SO32− → MnO2 + SO4 2− (basic solution) Step 1

Write half reaction equations. Balance except H and O SO32− → SO 4 2− MnO 4 − → MnO 2

Step 2

Balance H and O using H2O and H+ H 2 O + SO32− → SO 4 2− + 2 H + MnO 4 − + 4 H + → MnO 2 + 2 H 2 O

Step 3

Add OH− ions to both sides (same number as H+ ions) 2 OH − + H 2 O + SO3 2− → SO 4 2− + 2 H + + 2 OH − 4 OH − + MnO 4 − + 4 H + → MnO 2 + 2 H 2 O + 4 OH −

– Chapter 17 –

Step 4

Combine H+ and OH− to form H2O; cancel H2O where possible 2 OH− + H2 O + SO32− → SO 4 2− + 2 H2 O MnO4 − + 4 H2 O → MnO2 + 2 H2 O + 4 OH− 2 OH− + SO32− → SO4 2− + H2 O (1H2 O canceled ) MnO 4 − + 2 H2 O → MnO2 + 4 OH− ( 2 H2 O canceled )

Step 5

Balance electrically with electrons 2 OH − + SO32− → SO 4 2− + H2 O + 2 e − 3 e − + MnO 4 − + 2 H 2 O → MnO 2 + 4 OH −

Steps 6 and 7

Equalize gain and loss of electrons; add half-reactions

3 ( 2 OH− + SO32− → SO4 2− + H2 O + 2 e − )

2 ( MnO4 − + 2 H2 O + 3e − → MnO2 + 4 OH− )

H2 O + 2 MnO4 − + 3 SO32− → 2 MnO2 + 3 SO4 2− + 2 OH− 3 H2O, 4 OH−, and 6 e− canceled from each side (b) ClO 2 + SbO 2 − → ClO 2 − + Sb ( OH )6 Step 1

−

(basic solution)

Write half reaction equations. Balance except H and O.

SbO2 − → Sb ( OH )6

−

ClO2 → ClO2 − Step 2

Balance H and O using H2O and H+ 4 H2 O + SbO2 − → Sb ( OH )6 + 2 H + −

ClO2 → ClO 2 − Step 3

Add OH− ions to both sides (same number as H+ ions) 2 OH− + 4 H2 O + SbO2 − → Sb ( OH )6 + 2 H+ + 2 OH− −

ClO2 → ClO2 − Step 4

Combine H+ and OH− to form H2O; cancel H2O where possible

2 OH− + 4 H2 O + SbO2 − → Sb ( OH )6 + 2 H2 O −

ClO2 → ClO2 − 2 OH− + 2 H2 O + SbO2 − → Sb ( OH )6 ( 2 H2 O canceled ) −

– Chapter 17 –

Step 5

Balance electrically with electrons 2 OH− + 2 H2 O + SbO2 − → Sb ( OH )6 + 2 e − −

ClO2 + e − → ClO2 − Steps 6 and 7

Equalize gain and loss of electrons; add half-reactions 2 H 2 O + 2 OH − + SbO 2 − → Sb ( OH )6 + 2 e − −

2 ( ClO 2 + e − → ClO 2 − )

2 H 2 O + 2 ClO 2 + 2 OH − + SbO 2 − → 2 ClO 2 − + Sb ( OH )6

(c)

Al + NO 3 − → NH 3 + Al ( OH )4

Step 1

−

(basic solution)

Write half-reaction equations. Balance except H and O Al → Al ( OH )4

−

NO3 − → NH3 Step 2

Balance H and O using H2O and H+ 4 H2 O + Al → Al ( OH )4 + 4 H+ −

9 H+ + NO3 − → NH3 + 3 H2 O Step 3

Add OH− ions to both sides (same number as H+ ions) 4 OH − + 4 H2 O + Al → Al ( OH )4 + 4 H+ + 4 OH − −

9 OH− + 9 H+ + NO3− → NH3 + 3 H2 O + 9 OH − Step 4

Combine H+ and OH− to form H2O; cancel H2O where possible 4 OH − + 4 H2 O + Al → Al ( OH )4 + 4 H2 O −

9 H2 O + NO3 − → NH3 + 3 H2 O + 9 OH − 4 OH − + Al → Al ( OH )4

( 4 H2 O canceled ) 6 H2 O + NO3 − → NH3 + 9 OH − ( 3 H2 O canceled ) Step 5

−

Balance electrically with electrons 4 OH − + Al → Al ( OH )4 + 3 e − −

6 H2 O + NO3 − + 8 e − → NH3 + 9 OH−

−

– Chapter 17 –

Steps 6 and 7 Equalize gain and loss of electrons; add half-reactions

(

8 4 OH − + Al → Al ( OH )4 + 3 e − −

)

3 ( 6 H 2 O + NO3 − + 8 e − → NH3 + 9 OH − ) 8 Al + 3 NO3 − +18 H 2 O + 5 OH − → 3 NH3 + 8 Al ( OH )4

−

27 OH− and 24 e− canceled from each side (d) P4 → HPO32− + PH3 (basic solution)

Step 1

Write half-reaction equations. Balance except H and O P4 → 4 HPO32− P4 → 4 PH3

Step 2

Balance H and O using H2O and H+ 12 H 2 O + P4 → 4 HPO3 2− + 20 H + 12 H + + P4 → 4 PH3

Step 3

Add OH− ions to both sides (same number as H+ ions) 20 OH − +12 H 2 O + P4 → 4 HPO3 2− + 20 H + + 20 OH − 12 OH − +12 H + + P4 → 4 PH3 +12 OH −

Step 4

Combine H+ and OH− to form H2O; cancel H2O where possible 20 OH− +12 H2 O + P4 → 4 HPO32− + 20 H2 O 12 H2 O + P4 → 4 PH3 +12 OH− 20 OH− + P4 → 4 HPO32− + 8 H2 O (12 H2 O canceled )

Step 5

Balance electrically with electrons 20 OH − + P4 → 4 HPO3 2− + 8 H 2 O +12 e − 12 H 2 O + P4 +12 e − → 4 PH3 +12 OH −

Steps 6 and 7

Loss and gain of electrons are equal; add half-reactions 8 OH− + 4 H2 O + 2 P4 → 4 HPO32− + 4 PH3 Divide equation by 2 4 OH− + 2 H2 O + P4 → 2 HPO32− + 2 PH3

– Chapter 17 –

(e)

Al + OH − → Al ( OH )4 + H 2 (basic solution) −

Step 1

Write half-reaction equations. Balance except H and O Al → Al ( OH )4

−

OH− → H2 Step 2

Balance H and O using H2O and H+ 4 H2 O + Al → Al ( OH )4 + 4 H+ −

3 H+ + OH− → H2 + H2 O Step 3

Add OH− ions to both sides (same number as H+ ions) 4 OH− + 4 H2 O + Al → Al ( OH )4 + 4 H+ + 4 OH− −

3 OH− + 3 H+ + OH− → H2 + H2 O + 3 OH− Step 4

Combine H+ and OH− to form H2O; cancel H2O where possible

4 OH− + 4 H2 O + Al → Al ( OH )4 + 4 H2 O −

3 H2 O + OH− → H2 + H2 O + 3 OH− 4 OH− + Al → Al ( OH )4

( 4 H2 O canceled ) 2 H2 O + OH− → H2 + 3 OH− (1 H2 O canceled ) Step 5

−

Balance electrically with electrons 4 OH− + Al → Al ( OH )4 + 3 e − −

2 H2 O + OH− + 2 e − → H2 + 3 OH− Steps 6 and 7 Equalize gain and loss of electrons; add half-reactions

(

2 4 OH − + Al → Al ( OH )4 + 3 e − −

)

3 ( 2 H 2 O + OH − + 2 e − → H 2 + 3 OH − ) 2 Al + 6 H 2 O + 2 OH − → 2 Al ( OH )4 + 3 H 2 −

9 OH− and 6 e− canceled each side 19. (a)

IO3 − ⎯⎯ → I − + I 2 (acidic solution)

Step 1

Write half-reaction equations. Balance except H and O 2 IO3 − → I 2 2 I− → I2

Step 2

Balance H and O using H2O and H+ 12 H + + 2 IO3 − → I 2 + 6 H 2 O 2 I− → I2

– Chapter 17 –

Step 3

Balance electrically with electrons 12 H + + 2 IO3 − +10 e − → I 2 + 6 H 2 O 2 I− → I2 + 2 e−

Steps 4 and 5 Equalize the loss and gain of electrons; add the halfreactions. 12 H+ + 2 IO3 − +10 e − → I 2 + 6 H2 O 5 ( 2 I− → I2 + 2 e− )

12 H+ + 2 IO3 − +10 I − → 6 I 2 + 6 H2 O Each side has 12 H, 6 O, 12 I, and no charge (b) Mn 2+ + S2 O 8 2− → MnO 4 − + SO 4 2− (acid solution) Step 1

Write half-reaction equations. Balance except H and O Mn 2+ → MnO 4 − S2 O8 2− → 2 SO 4 2−

Balance H and O using H2O and H+

Step 2

4 H 2 O + Mn 2+ → MnO 4 − + 8 H + S2 O8 2− → 2 SO 4 2−

Step 3

Balance electrically with electrons 4 H 2 O + Mn 2+ → MnO 4 − + 8 H + + 5 e − 2 e − + S2 O82− → 2 SO 4 2−

Steps 4 and 5

Equalize the loss and gain of electrons; add the halfreactions 2 ( 4 H2 O + Mn2+ → MnO 4 − + 8 H+ + 5 e − )

5 ( 2 e − → S2 O82− → 2 SO 4 2− )

2 Mn 2+ + 5 S2 O82− + 8 H2 O → 2 MnO 4 − +10 SO 4 2− +16 H+

Each side has 2 Mn, 10 S, 16 H, and 48 O and a −6 charge. (c)

Co ( NO 2 )6 + MnO 4 − → Co 2+ + Mn 2+ + NO3 − (acidic solution) 3−

Step 1

Write half-reaction equations. Balance except H and O Co ( NO2 )6 → Co 2+ + 6 NO3− 3−

MnO 4 − → Mn2+

– Chapter 17 –

Step 2

Balance H and O using H2O and H+ 6 H2 O + Co ( NO2 )6 → Co 2+ + 6 NO3 − +12 H + 3−

8 H+ + MnO 4 − → Mn2+ + 4 H2O Step 3

Balance electrically with e− 6 H2 O + Co ( NO2 )6 → Co 2+ + 6 NO3 − +12 H+ +11 e − 3−

5 e − + 8 H+ + MnO 4 − → Mn2+ + 4 H2O Step 4

Equalize the loss and gain of electrons.

(

5 6 H2 O + Co ( NO 2 )6 → Co 2+ + 6 NO3 − +12 H+ +11e − 3−

11( 5 e − + 8H+ + MnO 4 − → Mn2+ + 4 H2O ) Step 5

)

Add the half-reactions

5 Co ( NO2 )6 +11MnO4 − + 28 H+ → 5 Co 2+ + 30 NO3 − +11 Mn2+ 3−

+14 H2O Each side has 5 Co, 30 N, 11 Mn, 28 H, 104 O and a +2 charge. 20. (a)

Mo 2 O3 + MnO 4 − → MoO 3 + Mn 2+ (acid solution)

Step 1

Write half-reaction equations. Balance except H and O Mo 2 O3 → 2 MoO3 MnO 4 − → Mn 2+

Step 2

Balance H and O using H2O and H+ 3 H 2 O + Mo 2 O3 → 2 MoO3 + 6 H + 8 H + + MnO 4 − → Mn 2+ + 4 H 2 O

Step 3

Balance electrically with electrons 3 H 2 O + Mo 2 O3 → 2 MoO3 + 6 H + + 6 e − 5 e − + 8 H + + MnO 4 − → Mn 2+ + 4 H 2 O

Steps 4 and 5

Equalize the loss and gain of electrons; add the halfreactions. 5 ( 3 H2 O + Mo2 O3 → 2 MoO3 + 6 H+ + 6 e − ) 6 ( 5 e − + 8 H + + MnO 4 − → Mn2+ + 4 H2 O )

5 Mo2 O3 + 6 MnO 4 − +18 H+ → 10 MoO3 + 6 Mn2+ + 9 H2 O

Each side has 18 H, 39 O, 10 Mo, 6 Mn, and a + 12 charge.

– Chapter 17 –

(b) BrO − + Cr(OH)4 − → Br − + CrO 4 2− (basic solution) Step 1

Write half-reaction equation. Balance except H and O BrO − → Br − Cr(OH)4 − → CrO 4 2−

Step 2

Balance H and O using H2O and H+ 2 H + + BrO − → Br − + H 2 O Cr(OH)4 − → CrO 4 2− + 4 H +

Step 3

Add OH− ions to both sides (same number as H+ ) 2 OH − + 2 H + + BrO − → Br − + H 2 O + 2 OH − 4 OH − + Cr(OH)4 − → CrO 4 2− + 4 H + + 4 OH −

Step 4

Combine H+ and OH− to form H2O; cancel H2O where possible 2 H2 O + BrO − → Br − + H2 O + 2 OH − 4 OH − + Cr(OH)4 − → CrO 4 2− + 4 H2 O H2 O + BrO − → Br − + 2 OH − (1H2 O canceled )

Step 5

Balance electrically with electrons

2 e − + H2 O + BrO− → Br − + 2 OH− 4 OH− + Cr(OH)4 − → CrO 4 2− + 4 H2 O + 3 e − Step 6 and 7

Equalize loss and gain of electrons; add the half-reactions

3(2 e − + H2 O + BrO − → Br − + 2 OH − ) 2 (4 OH − + Cr(OH)4 − → CrO 4 2− + 4 H2 O + 3 e − ) 3 BrO − + 2 Cr(OH)4 − + 2 OH − → 3 Br − + 2 CrO 4 2− + 5 H2 O Each side has 3 Br, 13 O, 2 Cr, 10 H, and a −7 charge. (c)

S2 O 3 2 − + MnO 4 − → SO 4 2− + MnO 2 (basic solution)

Step 1

Write half-reaction equation. Balance except H and O S2 O3 2− → 2 SO 4 2− MnO 4 − → MnO 2

Step 2

Balance H and O using H2O and H+ 5 H 2 O + S2 O32− → 2 SO 4 2− +10 H + 4 H + + MnO 4 − → MnO 2 + 2 H 2 O

– Chapter 17 –

Step 3

Add OH− ions to both sides (same number as H+ ) 10 OH − + 5 H 2 O + S2 O3 2− → 2 SO 4 2− +10 H + +10 OH − 4 OH − + 4 H + + MnO 4 − → MnO 2 + 2 H 2 O + 4 OH −

Step 4

Combine H+ and OH− to form H2O; cancel H2O where possible 10 OH − + 5 H2 O + S2 O3 2− → 2 SO 4 2− +10 H2 O 4 H2 O + MnO 4 − → MnO2 + 2 H2 O + 4 OH− 10 OH− + S2 O3 2− → 2 SO 4 2− + 5 H2 O (5 H2 O canceled) 2 H2 O + MnO 4 − → MnO2 + 4 OH − (2 H2 O canceled)

Step 5

Balance electrically with electrons 10 OH − + S2 O32− → 2 SO 4 2− + 5 H 2 O + 8 e − 3 e − + 2 H 2 O + MnO 4 − → MnO 2 + 4 OH −

Steps 6 and 7

Equalize loss and gain of electrons; add half-reactions. 3(10 OH − + S2 O32− → 2 SO 4 2− + 5 H2 O + 8 e − ) 8(3 e − + 2 H2 O + MnO 4 − → MnO2 + 4 OH − ) 3 S2 O3 2− + 8 MnO 4 − + H2 O → 6 SO 4 2− + 8 MnO2 + 2 OH − Each side has 6 S, 8 Mn, 2 H, 42 O and a −14 charge.

21. Iron is oxidized from the elemental state to 2+. This involves a loss of two electrons and takes place at the anode. Nickel is reduced from the 3+ state to 2+ state at the cathode. 22. In hydrogen peroxide (H2O2), oxygen is in a 1 − oxidation state whereas in water oxygen is its usual 2−.This is reduction and occurs at the cathode. I− loses electrons to reach the elemental state or zero oxidation state, I2. This is oxidation or the anode reaction. heat 23. Molten NiBr2 will exist as the following ions NiBr2 ⎯⎯⎯ → Ni 2+ + 2 Br − .

Oxidation, which is a loss of electrons, occurs at the anode. Br− is the ion capable of losing electrons. The Ni2+ ion needs 2 e− (reduction) to become a Ni atom. Therefore, anode reaction 2 Br − → Br2 + 2 e − The Ni2+ ion needs 2 e− (reduction) to become a Ni atom. Cathode reaction Ni2 + + 2 e− → Ni0 Overall reaction Ni 2 + + 2 Br − → Ni 0 + Br2 24. LiCl will exist as Li+ and Cl− ions in the molten state. Oxidation at the anode will involve the Cl− ion, but 2 Cl− ions are needed to balance the equation since Cl2 is produced. Anode 2 Cl − → Cl 2 + 2 e −

– Chapter 17 –

Reduction at the cathode will use the 2 e− available to reduce 2 Li+ ions to lithium atoms. Cathode 2 e− + 2 Li + → 2 Li0 Overall 2 Li + + 2 Cl − → 2 Li0 + Cl 2 25. (b), (d), and (e) will not react because the elemental substances are all less good at losing electrons than the ionized element in each pair. 26. Potassium is the most easily oxidized element and Ag is the least easily oxidized element. 27. Hydrogen and iron. 28. 2 C6H8O6 + O2 → 2 C6H6 O6 + 2 H2 O 29.

Nitrogen and sulfur are reduced. Carbon is oxidized. 30.

31. (a)

Pb + SO 4 2− → PbSO 4 + 2 e − PbO 2 + SO 4 2− + 4 H + + 2 e − → PbSO 4 + 2 H 2 O

(b) The first reaction is oxidation (Pb0 is oxidized to Pb2+). The second reaction is reduction (Pb4+ is reduced to Pb2+). (c) The first reaction (oxidation) occurs at the anode of the battery. 32. (a) The oxidizing agent is KMnO4. (b) The reducing agent is HCl. 5 moles of electrons 5 e − + Mn7+ → Mn 2+

(c)

 5 mol e −   6.022 ×10 23 e −  electrons 24    = 3.011× 10 − mol e mol KMnO 4   mol KMnO 4  

– Chapter 17 –

33. Zinc is a more reactive metal than copper so when corrosion occurs the zinc preferentially reacts. Zinc is above hydrogen in the Activity series of metals; copper is below hydrogen. 34. 3 Cu + 8 HNO3 → 3 Cu ( NO3 )2 + 2 NO + 4 H2 O  1 mol Cu   2 mol NO   22.4 L     = 17.7 L NO  63.55 g   3 mol Cu   mol 

( 75.5 g Cu ) 

 1 mol HNO3  2 mol NO   22.4 L     = 4.89 L NO  63.02 g  8 mol HNO3   mol 

(55.0 g HNO3 ) 

4.89 L of NO gas at STP will be produced; HNO3 is limiting. Cu is oxidized; N is reduced. 35. K2S2 O8 + H2C2 O4 → K2SO4 + H2SO4 + 2 CO2  1 mol K 2S2 O8  2 mol CO2   22.4 L     = 4.23 L CO2  270.3 g  1 mol HNO3   mol 

( 25.5 g K 2S2 O8 ) 

 1 mol H2 C 2 O 4   2 mol CO2   22.4 L     = 17.7 L CO2 90.04 g    1 mol H2C 2 O 4   mol 

( 35.5 g H2C 2 O4 ) 

4.23 L CO2 will be produced; K2S2O8 is limiting. C in H2C2O4 is oxidized and S in K2S2O8 is reduced. 36. C6H8O6 ( aq) + KI3 ( aq) → C6H6 O6 (aq ) + KI( aq ) + 2 HI( aq )  0.03741 mol KI3   1mol C 6H8O6  176.1 g  0.03261 L    = 0.2148 g in 100 g pepper  1L    1 mol KI3  1 mol   0.2148 g C 6H8O6  1000 mg  2.148 mg C 6H8O6  =  g pepper  100 g pepper   1 g 

37. CH3CH 2 OH + Cr2 O 7 2− → CH 3COOH + Cr 3+ 3 ( CH3CH2 OH + H2 O → CH3COOH + 4 H+ + 4 e − ) 2 ( Cr2 O7 2− +14 H+ + 6 e − → 2 Cr 3+ + 7 H2 O )

3 CH3CH2 OH + 3 H2 O + 2 Cr2 O7 2− + 28 H+ → 3 CH3CO2 H +12 H+ + 4 Cr 3+ +14 H2 O Which simplifies to 3 CH3CH2 OH + 2 Cr2 O7 2− +16 H+ → 3 CH3COOH + 4 Cr 3+ +11 H2 O

– Chapter 17 –

38. 5 H2 O2 + 2 KMnO4 + 3 H2SO4 → 5 O2 + 2 MnSO4 + K2SO4 + 8 H2 O mL H2 O2 → g H2 O 2 → mol H2 O 2 → mol KMnO 4 → g KMnO 4   1 mol  9.0 g H2 O 2 1.031 g       mL   100. g H 2 O 2 solution   34.02 g 

(100. mL H2 O2 solution ) 

 2 mol KMnO 4   158.0 g     = 17 g KMnO 4  5 mol H2 O 2   mol 

39. 3 Zn + 2 Fe3+ → 3 Zn2+ + 2 Fe 1.2 mol FeCl3   3 mol Zn   65.39 g    = 2.94 g Zn  L    2 mol FeCl3   mol 

( 0.0250 L FeCl3 ) 

40. Cr2 O 7 2− + 6 Fe 2+ +14 H + → 2 Cr 3+ + 6 Fe3+ + 7 H 2 O mL FeSO 4 → mol FeSO 4 → mol Cr2 O7 2 − → mL Cr2 O7 2 −  0.200 mol  1 mol Cr2 O7 2 −   1000 mL  60.0 mL FeSO (    4 )  1000 mL  6 mol FeSO 4  0.200 mol  = 10.0 mL of 0.200 M K 2 Cr2 O7 41. 2 Al + 2 OH − + 6 H 2 O → 2 Al(OH)4 − + 3 H 2 g Al → mol Al → mol H2  1 mol Al   3 mol H2    = 5.560 mol H2  26.98 g   2 mol Al 

(100.0 g Al ) 

42. (a) Cu+ → Cu2+ is an oxidation, but when electrons are gained reduction should occur.

Cu + + e − → Cu 0

Cu + → Cu 2+ + e −

(b) When Pb2+ is reduced, it requires two individual electrons. Pb2++2 e− → Pb0. An electron has only a single negative charge (e−). 43. The electrons lost by the species undergoing oxidation must be gained (or attracted) by another species which then undergoes reduction. 44.

A( s ) + B2+ (aq) → NR

B2+ cannot take e− from A

A( s ) + C + ( aq) → NR

C+ cannot take e− from A

D(s ) + 2 C + ( aq) → 2 C( s ) + D2+ ( aq )

C+ takes e− from D

B( s ) + D2+ (aq ) → D( s ) + B2+ (aq )

D2+ takes 2 e− from B

Therefore, B2+ is least able to attract e−, then D2+, then C+, then A+

– Chapter 17 –

45. Sn 4 + + 2 e − → Sn 2+

Sn4+ can only be an oxidizing agent.

Sn 4 + + 4 e − → Sn 0 Sn 0 → Sn 2 + + 2 e −

Sn can only be a reducing agent.

Sn 0 → Sn 4+ + 4 e −

Sn can be both oxidizing and reducing.

Sn 2+ + 2 e − → Sn 0 (oxidizing agent) Sn 2+ → Sn 4 + + 2 e − (reducing agent)

46. Mn(OH)2

MnF3

MnO2

K2MnO4

KMnO4

KMnO4 is the best oxidizing agent of the group, since its greater oxidation number (+7) makes it very attractive to electrons.

47. Equations (a) and (b) represent oxidation (a) Mg → Mg2+ + 2 e− (b) SO2 → SO3 ; (S4 + → S6 + + 2 e − ) 48. (a) MnO 2 + 2 Br − + 4 H + → Mn 2+ + Br2 + 2 H 2 O (b) mL Mn 2+ → mol Mn 2+ → mol MnO 2 → g MnO 2 

















0.05 mol 1 mol MnO  86.94 g  (100.0 mL Mn )  1000    = 0.4 g MnO mL 1 mol Mn  mol  2+

(c)

2 2+





0.05 mol 1 mol Br (100.0 mL Mn )  1000   = 0.005 mol Br mL 1 mol Mn 2+

2 2+



nRT P  0.005 mol   0.0821 L atm  V=   ( 323 K ) = 0.09 L Br2 vapor mol K   1.4 atm  

PV = nRT

49. (a) F2 + 2 Cl − → 2 F − + Cl 2 (b) Br2 + Cl − → NR (c)

I 2 + Cl − → NR

(d) Br2 + 2 I − → 2 Br − + I 2

– Chapter 17 –

50. Mn( s ) + 2 HCl( aq) → Mn 2+ ( aq) + H 2 ( g ) + 2 Cl − ( aq ) 51. 4 Zn + NO3 − +10 H + → 4 Zn 2+ + NH 4 + + 3 H 2 O

See Exercise 15(a).

52. 2

C oxidized

S oxidized

N oxidized

S oxidized

O22− oxidized

O2 reduced

N reduced

Cu reduced

O22− reduced

O2, O.A.

HNO3, O.A.

CuO, O.A.

H2O2, O.A.

C3H8, R.A.

H2S, R.A.

NH3, R.A.

Na2SO3, R.A.

H2O2, R.A.

S2− → S0

N3− → N20

S4+ → S6+

O22− → O20

N5+ → N2+

Cu2+ → Cu0

O22− → O2−

e f

2 2 +

C 3 → C 4+ O0 → O2−

Equation 1

O.A. = Oxidizing agent R.A. = Reducing agent 53. The Al3+ ions migrate toward the cathode, and the O2− ions migrate toward the anode. To balance the two half-reactions, balance the electron gain and loss. Al3+ + 3 e − → Al O 2− + C → CO + 2 e −

If you multiply the Al3+ half-reaction by 2 and the O2− half-reaction by 3, the electrons will be balanced. 2 Al3 + + 6 e − → 2 Al 3 O 2 − + 3 C → 3 CO + 6 e − Added together: 2 Al3+ + 3 O 2 − + 3 C → 2 Al + 3 CO

54. Pb + 2 Ag + → 2 Ag + Pb2+ (a) Pb is the anode (b) Ag is the cathode (c) Oxidation occurs at Pb (anode) (d) Reduction occurs at Ag (cathode) (e) Electrons flow from the lead electrode through the wire to the silver electrode.

– Chapter 17 –

(f)

Positive ions flow through the salt bridge towards the negatively charged cathode of silver; negative ions flow toward the positively charged anode of lead.

55.

56. The pH of the solution is 4.5. (Acidic solution) 5 Fe 2+ + MnO 4− + 8 H + → 5 Fe3+ + Mn 2+ + 4 H 2 O

57. 8 KI + 5 H2SO4 → 4 I2 + H2S + 4 K2SO 4 + 4 H2O start with grams of I 2 and work toward g of KI g I 2 → mol I 2 → mol KI → g KI  1mol   8 mol KI   166.0 g     = 3.65 g KI in sample  253.8 g   4 mol I 2   mol 

( 2.79 g I2 ) 

 3.65 g KI    (100 ) = 91.3% KI  4.00 g sample 

– Chapter 17 –

58. 3 Ag + 4 HNO3 → 3 AgNO3 + NO + 2 H2 O

mol Ag → mol NO  1 mol NO   = 0.167 mo l NO  3 mol Ag 

( 0.500 mol Ag ) 

nRT P  1 atm  P = (744 torr)   = 0.979 atm  760. torr  T = 301 K PV = nRT

V= 59. (a)

( 0.167 mol NO )( 0.0821 L atm /mol K )( 301 K ) = 4.22 L NO (0.979 atm)

MnO 4 − + 5 VO 2+ + H 2 O → 5 VO 2 + + 2 H + + Mn 2+

(b) P4 + 3 H2 O + 3 OH − → PH3 + 3 H2 PO 2 − (c)

MnO 2 + SO32− + H2 O → Mn(OH)2 + SO 4 2−

(d) 5 H2 O 2 + 2 MnO 4 − + 6 H + → 2 Mn2+ + 5 O2 + 8 H2 O (e)

2 Mn 2+ + 5 HBiO3 + 9H + → 2 MnO 4 − + 5 Bi3+ + 7 H2 O

(f)

Zn + 2 OH − + 2 H2 O → Zn(OH)2− 4 + H2

CHAPTER 18

NUCLEAR CHEMISTRY SOLUTIONS TO REVIEW QUESTIONS 1.

Contributions to the early history of radioactivity include: (a) Henri Becquerel: He discovered radioactivity. (b) Marie and Pierre Curie: They discovered the elements polonium and radium. (c) Wilhelm Roentgen: He discovered X rays and developed the technique of producing them. While this was not a radioactive phenomenon, it triggered Becquerel’s discovery of radioactivity. (d) Ernest Rutherford: He discovered alpha and beta particles, established the link between radioactivity and transmutation, and produced the first successful man-made transmutation. (e) Otto Hahn and Fritz Strassmann: They were first to produce nuclear fission.

Chemical reactions are caused by atoms or ions coming together, so are greatly influenced by temperature and concentration, which affect the number of collisions. Radioactivity is a spontaneous reaction of an individual nucleus, and is independent of such influences.

The term isotope is used with reference to atoms of the same element that contains different masses. For example, 126 C and 146 C . The term nuclide is used in nuclear chemistry to infer any isotope of any atom.  1 half-life  4. ( 5×109 years )   = 70 half-lives 7  7.6 ×10 year  Even if plutonium-244 had been present in large quantities five billion years ago, no measureable amount would survive after 70 half-lives.

In living species, the ratio of carbon-14 to carbon-12 is constant due to the constant C-14/C-12 ratio in the atmosphere and food sources. When a species dies, life processes stop. The C-14/C-12 ratio decreases with time because C-14 is radioactive and decays according to its half-life, while the amount of C-12 in the species remains constant. Thus, the age of an archaeological artifact containing carbon can be calculated by comparing the C-14/C-12 ratio in the artifact with the C-14/C-12 ratio in the living species.

The half-life of carbon-14 is 5730 years.  1 half-life  2 (4 ×106 years)   = 7 ×10 half-lives  5730 years  700 half-lives would pass in 4 million years. Not enough C-14 would remain to allow detection with any degree of reliability. C-14 dating would not prove useful in this case.

(a) Gamma radiation requires the most shielding. (b) Alpha radiation requires the least shielding.

– Chapter 18 –

Alpha particles are deflected less than beta particles while passing through a magnetic field, because they are much heavier (more than 7000 times heavier) than beta particles.

9. Alpha Beta Gamma

charge mass +2 4 amu −1 1 1837 amu 0 0

nature of particles He nucleus electron

penetrating power low moderate

electromagnetic radiation

high

10. Decay of bismuth-211 211 83

Bi → 42 He + 20781Tl

Tl → −10 e + 207 82 Pb

207 81

11. Pairs of nuclides that would be found in the fission reaction of U-235. Any two nuclides, whose atomic numbers add up to 92 and mass numbers (in the range of 70-160) add up to 230-234. Examples include: 90 38

Sr and 141 54 Xe

139 56

Ba and 94 36 Kr

101 42

Mo and 131 50 Sn

12. Natural radioactivity is the spontaneous disintegration of those radioactive nuclides found in nature. Artificial radioactivity is the spontaneous disintegration of radioactive isotopes produced synthetically, otherwise it is the same as natural radioactivity. 13. A radioactive disintegration series starts with a particular radionuclide and progresses stepwise by alpha and beta emissions to other radionuclides, ending at a stable nuclide. For example: 238 92

14 steps U ⎯⎯ ⎯→ 206 82 Pb (stable)

14. Transmutation is the conversion of one element into another by natural or artificial means. The nucleus of an atom is bombarded by various particles (alpha, beta, protons, etc.). The fast moving particles are captured by the nucleus, forming an unstable nucleus, which decays to another kind of atom. For example: 9 4

15. 16.

Be + 24 He → 126 C + 01 n α −β −β α −α −α Th ⎯−⎯ → 228 → 228 → 22890Th ⎯−⎯ → 224 → 220 → 88 Ra ⎯⎯ 89 Ac ⎯⎯ 88 Ra ⎯⎯ 86 Rn ⎯⎯

232 90

216 84

237 93

−α −β −β −α Po ⎯⎯ → 212 → 212 → 212 → 208 82 Pb ⎯⎯ 83 Bi ⎯⎯ 84 Po ⎯⎯ 82 Pb

Np loses seven alpha particles and four beta particles.

Determination of the final product: 209 83 Bi nuclear charge = 93 – 7(2) + 4(1) = 83 mass = 237 – 7(4) = 209 17. Radioactivity could be used to locate a leak in an underground pipe by using a water soluble tracer element. Dissolve the tracer in water and pass the water through the pipe. Test the ground along the path of the pipe with a Geiger counter until radioactivity from the leak is detected. Then dig.

– Chapter 18 –

18. A scintillation counter is a radiation detector which contains molecules that emit light when they are struck by ionizing radiation. The number of light flashes are recorded as a numerical output of the radiation level. 19. The curie describes the amount of radioactivity produced by an element. One curie is equal to 3.7 × 1010 disintegrations/sec 20. REM stands for roentgen equivalent to man and measures the effective exposure to ionizing radiation. 21. Two Germans, Otto Hahn and Fritz Strassmann, were the first scientists to report nuclear fission. The fission resulted from bombarding uranium nuclei with neutrons. 22. Natural uranium is 99+% U-238. Commercial nuclear reactors use U-235 enriched uranium as a fuel. Slow neutrons will cause the fission of U-235, but not U-238. Fast neutrons are capable of a nuclear reaction with U-238 to produce fissionable Pu-239. A breeder reactor converts nonfissionable U-238 to fissionable Pu-239, and in the process, manufactures more fuel than it consumes. 23. The fission reaction in a nuclear reactor and in an atomic bomb are essentially the same. The difference is that the fissioning is “wild” or uncontrolled in the bomb. In a nuclear reactor, the fissioning rate is controlled by means of moderators, such as graphite, to slow the neutrons and control rods of cadmium or boron to absorb some of the neutrons. 24. A certain amount of fissionable material (a critical mass) must be present before a self-sustaining chain reaction can occur. Without a critical mass, too many neutrons from fissions will escape, and the reaction cannot reach a chain reaction status, unless at least one neutron is captured for every fission that occurs. 25. The disadvantages of nuclear power include the danger of contamination from radioactive material and the radioactive waste products that accumulate, some having half-lives of thousands of years. 26. The hazards associated with an atomic bomb explosion include shock waves, heat and radiation from alpha particles, beta particles, gamma rays, and ultraviolet rays. Gamma rays and X-rays cause burns, sterilization, and gene mutation. If the bomb explodes close to the ground, radioactive material is carried by dust particles and is spread over wide areas. 27. Heavy elements undergo fission and lighter elements undergo fusion. 28. In a nuclear power plant, a controlled nuclear fission reaction provides heat energy that is used to produce steam. The steam turns a turbine that generates electricity. 29. The mass defect is the difference between the mass of an atom and the sum of the masses of the number of protons, neutrons, and electrons in that atom. The energy equivalent of this mass defect is known as the nuclear binding energy. 30. When radioactive rays pass through normal matter, they cause that matter to become ionized (usually by knocking out electrons). Therefore, the radioactive rays are classified as ionizing radiation. 31. Some biological hazards associated with radioactivity are: (a) High levels of radiation can cause nausea, vomiting, diarrhea, and death. The radiation produces ionization in the cells, particularly in the nucleus of the cells. 3

– Chapter 18 –

(b) Long-term exposure to low levels of radiation can weaken the body and cause malignant tumors. (c) Radiation can damage DNA molecules in the body causing mutations, which by reproduction, can be passed on to succeeding generations. 32. Strontium-90 has two characteristics that create concern. Its half-life is 28 years, so it remains active for a long period of time (disintegrating by emitting β radiation). The other characteristic is that Sr-90 is chemically similar to calcium, so when it is present in milk Sr-90 is deposited in bone tissue along with calcium. Red blood cells are produced in the bone marrow. If the marrow is subjected to beta radiation from strontium-90, the red blood cells will be destroyed, increasing the incidence of leukemia and bone cancer. 33. A radioactive “tracer” is a radioactive material, whose presence is traced by a Geiger counter or some other detecting device. Tracers are often injected into the human body, animals, and plants to determine chemical pathways, rates of circulation, etc. For example, use of a tracer could determine the length of time for material to travel from the root system to the leaves in a tree.

– Chapter 18 –

SOLUTIONS TO EXERCISES

1. Pb

Protons 82

Neutrons 125

Nucleons 207

Protons 52

Neutrons 76

Nucleons 128

(a)

207 82

(b)

70 31

2. (a)

128 52

(b)

32 16

When a nucleus loses an alpha particle, its atomic number decreases by two, and its mass number decreases by four.

When a nucleus loses a beta particle, its atomic number increases by one, and its mass number remains unchanged.

Equations for alpha decay:

(a)

210 83

Bi → 24 He + 206 81 Ra

(b)

238 92

U → 24 He + 23490Th

Equations for alpha decay: Th → 42 He + 234 88 Ra

(a)

238 90

(b)

239 94

Pu → 42 He + 235 92 U

Equations for beta decay: N → −10 e + 138 O

(a)

13 7

(b)

234 90

Th → −01 e + 234 91 Pa

Equations for beta decay: 28 Al → −10 e + 14 Si

(a)

28 13

(b)

239 93

Np → −01 e + 239 94 Pu

(a) alpha-emission

(b) beta-emission then gamma-emission

10. (a) gamma-emission (b) alpha-emission then beta-emission (c) alpha-emission then gamma-emission 11.

26 13

Al → +01 e + 26 12 Mg

12.

32 15

P → −01 e + 32 16 S

13. (a)

66 29

66 Zn + −10 e Cu → 30

(b)

0 −1

e + 47 Be → 73 Li 5

– Chapter 18 –

(c)

27 13

1 Al + 42 He → 30 14 Si + 1 H

(d)

85 37

4 Rb + 01 n → 82 35 Br + 2 He

14. (a)

27 13

1 Al + 24 He → 30 15 P + 0 n

(b)

27 14

Si → +01 e + 27 13 Al

(c)

12 6

C + 21 H → 137 N + 01 n

(d)

82 35

82 Kr + −01 e Br → 36

15. (a) Nuclear fusion – two light atoms combine to form a heavier nucleus. (b) Nuclear fission – heavy, unstable nucleus splits into two smaller fragments under bombardment with neutrons. (c) Nuclear fusion 16. (a) Nuclear fusion

(b) Nuclear fission

(C)

Nuclear fusion

17. Loss of mass: 233 − 225 = 8, equivalent to 2 alpha particles. Loss in atomic number: 91 − 89 = 2, equivalent to 1 alpha particle. With a loss of 2 alpha particles the loss of atomic number should be 4. Therefore, there must also be a loss of 2 beta particles to increase the atomic number by 2. So one possible series is: 233 91

−β −α α −β Pa ⎯⎯ → 233 → 22990Th ⎯−⎯ → 225 → 225 88 Ra ⎯⎯ 89 Ac 92 U ⎯⎯

18. Loss of mass: 228 − 212 = 16, which is equivalent to 4 alpha particles. Loss in atomic number 90 − 82 = 8, which is equivalent to 4 alpha particles. This looks like a total loss of 4 alpha particles. Therefore the series is: −α −α −α −α Th ⎯⎯ → 224 → 220 → 216 → 212 88 Ra ⎯⎯ 86 Rn ⎯⎯ 84 Po ⎯⎯ 82 Pb

228 90

19. (a)

235 92

139 1 U + 01 n → 94 38 Sr + 54 Xe + 3 0 n + energy

Mass loss = mass of reactants − mass of products Mass of reactants = 235.0439 amu +1.0087 amu = 236.0526 amu Mass of products = 93.9154 amu +138.9179 amu + 3(1.0087 amu) = 235,8594 amu Mass lost = 236.0526 amu − 235.8594 amu = 0.1932 amu 13 1.000g   9.0 ×10 J  −11  = 2.9 ×10 J/atom U-235  23 6.022 ×10 amu 1.00g   

( 0.1932 amu ) 

 2.9 × 10 −11 J  6.022 × 10 23 atoms  13 (b)    = 1.7 ×10 J/mol atom mol    (c)

 0.1932 amu    (100 ) = 0.08185% mass loss  236.0526 amu 

– Chapter 18 –

20. (a)

1 1

H + 21 H → 23 He + energy

Mass loss = mass of reactants − mass of products Mass of reactants = 1.00794 g/mol + 2.01410 g/mol = 3.02204 g/mol Mass of products = 3.01603 g/mol Mass lost = 3.02204 g/mol − 3.01603 g/mol = 0.00601 g/mol  0.00601g   9.0 ×10 J  11  = 5.4 ×10 J/mol   mol g    13

 0.00601 g  (b)   (100 ) = 0.199% mass loss  3.02204 g  21. (a) Chromium-51: 24 protons; 27 neutrons; 24 electrons (b) Holmium-166: 67 protons; 99 neutrons; 67 electrons (c) Palladium-103: 46 protons; 57 neutrons; 46 electrons (d) Strontium-89: 38 protons; 51 neutrons; 38 electrons 22.

Th → 208 82 Pb

232 90

−α −β β −α −α Th ⎯⎯ → 228 → 22889Th ⎯−⎯ → 228 → 224 → 220 90 Ac ⎯⎯ 88 Ra ⎯⎯ 88 Ra ⎯⎯ 86 Rn

232 90

−α −β −α −β −α ⎯⎯ → 216 → 216 → 212 → 212 → 208 84 Po ⎯⎯ 85 At ⎯⎯ 83 Bi ⎯⎯ 84 Po ⎯⎯ 82 Pb

23. The nuclide lost 2 alpha particles and 1 beta particle in any order. One series could −α −β −α 219 215 → 219 → 215 be 223 87 Fr ⎯⎯→ 85 At ⎯⎯ 86 Rn ⎯⎯ 84 Po. 84 Po is the resulting nuclide. 24.

1 × 25.0 = 12.5 g left after one half of the sample disintegrates. 2

 12 mo  10  = 1.50 ×10 months. 1 yr  

(1 half-life ) (1.25 ×109 years )  25.

249 98

Cf + 157 N → +4 01 n + 260 105 Db

26.

226 88

Ra contains 138 neutrons and 88 electrons

mass of neutron = 1.0087 amu mass of electron = 0.00055 amu

(138)(1.0087 amu ) 100 = 61.59% neutrons by mass ( )

226 amu (88) ( 0.00055 amu ) (100) = 0.021% electrons by mass 226 amu 27.

 226.0 g Ra   $90,000    = $685  296.9 g RaCl2  1 g Ra 

( 0.0100 g RaCl2 ) 

28. 100% to 25% requires 2 half-lives. The half-life of C-14 is 5730 years. The specimen will be the age of two half-lives: (2)(5730 years) = 11,460 years old.

– Chapter 18 –

29. Approximately 2. The half-life for 146 C is 5730 years. 7 3

30. (a)

Li is made up of 3 protons, 4 neutrons, and 3 electrons.

Calculated Mass 3 protons

3(1.0073 g)

3.0219 g

4 neutrons

4(1.0087 g)

4.0348 g

3 electrons

3(0.00055 g)

0.0017 g

calculated mass

7.0584 g

Mass defect = calcluated mass − actual mass Mass defect = 7.0584 g − 7.0160 g = 0.0424 g/mol

(b) Binding energy

 0.0424 g   9.0 ×10 J  12  = 3.8×10 J/mol   g  mol    13

31.

235 92

U → 207 82 Pb

Mass loss: 235 – 207 = 28 Net proton loss (atomic number): 92 p – 82 p = 10 p The mass loss is equivalent to 7 alpha particles (28/4). A loss of 7 alpha particles gives a loss of 14 protons. A decrease in the atomic number to 78 (14 protons) is due to the loss of 7 alpha particles (92 −14 = 78). Therefore, a loss of 4 beta particles is required to increase the atomic number from 78 to 82. The total loss = 7 alpha particles and 4 beta particles. 32. (a) Geiger counter: Radiation passes through a thin glass window into a chamber filled with argon gas and containing two electrodes. Some of the argon ionizes, sending a momentary electrical impulse between the electrodes to the detector. This signal is amplified electronically and read out on a counter or as a series of clicks. (b) Scintillation counter: Radiation strikes a scintillator, which is composed of molecules that emit light in the presence of ionizing radiation. A light sensitive detector counts the flashes and converts them into a digital readout. (c) Film badge: Radiation penetrates a film holder. The silver grains in the film darken when exposed to radiation. The film is developed at regular intervals. 33.

(3 days )( 24 hours/day ) = 72 hours 72 hr + 6 hr = 78 hr 6

78 hr 1 = 6 half-lives   (10 mg) = 0.16 mg remaining hr 2 13 t1 2

– Chapter 18 –

34. First change micrograms to nanograms. 15.0 μg (1000 ng/μg ) = 15, 000 ng 15,000 ng → 7,500 ng → 3,750 ng → 1,875 ng → 937 ng → 468 ng → 234 ng → 117 ng → 59 ng → 29 ng → 15 ng → 7 ng → 4 ng → <2 ng → < 1 ng

Each arrow represents 1 half-life. There are 14 half-lives for a total of 112 days, (14 half-lives)(8 days/half-life) = 112 days, for the iodine − 131 to decay to less than a nanogram. 35. If you were to start with a 100 μg sample, more than 99% would decay after 7 halflives. This would take 5 ½ hours for the bismuth-213, 28 days for the rhenium-186, and 35 years for the cobalt-60. The best choice is the rhenium-186 because it will be detectable long enough for the termites to move the sample, but not so long that it becomes a hazard. 36.

241 95

Am → 24 α + 237 93 Np

37. 96 g → 48 g → 24 g →12 g → 6 g → 3 g → 1.5 g Each change involves a half-life. Six (6) half-lives are required to go from 96 g to 1.5 g. The half-life is about four days. 38. Fission is the splitting of a heavy nuclide into two or more intermediate-sized fragments with the conversion of some mass into energy. Fission occurs in nuclear reactors, or atomic bombs. 1 143 90 1 Example: 235 92 U + 0 n → 54 Xe + 38 Sr + 3 0 n

Fusion is the process of combining two relatively small nuclei to form a single larger nucleus. Fusion occurs on the sun, or in a hydrogen bomb. Example: 31 H + 21 H → 24 He + 01 n + energy 39.

The graph produces a curve for radioactive decay which never actually crosses the x-axis (where mass = 0), it simply approaches that point. 40. (a)

235 92

1 90 U + 01 n → 143 54 Xe + 3 0 n + 38 Sr

(b)

235 92

U + 01 n → 10239Y + 3 01 n + 131 53 I

(c)

14 7

N + 01 n → 11 H + 146 C

– Chapter 18 –

41. (a)

(b)

H2 O(l ) → H2 O(g ) Energy 2 : Weakest bond changes requires the least energy. 2H2 ( g ) + O2 (g ) → 2 H2 O(g ) Energy1: medium-sized value involved in interatomic bonds. 2 1

(c)

H + 21 H → 31 H + 11 H

Energy3: Nuclear process; greatest amount of energy involved. 42.

236 92

1 143 U → 90 38 Sr + 3 0 n + 54 Xe

43. Mg → −01 e + 29 13 Al

(a)

beta emission:

29 12

(b)

alpha emission:

150 60

(c)

positron emission:

72 33

Nd → 24 He + 146 58 Ce

As → +01 e + 72 32 Ge

44. 12 μg Days P-32 remaining % remaining

0 1500 μg 100

14.3 750 μg 50

28.6 380 μg 25

42.9 190 μg 12

57.2 94 μg 6.3

71.5 47 μg 3.1

45. 1 curie = 3.7 ×1010 disintegrations/sec 1 becquerel = 1 disintegration/sec Therefore there are 3.7 ×1010 becquerels/1 curie  3.7 × 1010 becquerel  10   (1.24 curies ) = 4.6 ×10 becquerels 1 curie  

46. 1.00 g Co-60 (a) one half-life:

1.00g = 0.500 g left 2

(b) two half-lives:

0.500 g = 0.250 g left 2

1 left 16

(d) ten half-lives: 210 = 1024; 47. (a)

11 5

B → 24 He + 73 Li

(b)

88 38

Sr → −01 e + 8839Y

(c)

107 47

(d)

41 19

(e)

116 51

1 left 1024

1.00 g = 0.0625 g left 16

1.00 g = 9.77 × 10 −4 g left 1024

Ag + 01 n → 108 47 Ag

K → 11 H + 40 18 Ar Sb + −01 e → 116 50 Sn 10

85.8 23 μg 1.6

100 12 μg 0.78

– Chapter 18 –

48. C-14 content is 1/16 of that in living plants. This means that four half-lives have passed. 14 C half-life is 5730 years.  5730 years  4   ( 4 half-lives ) = 22,920 years (2.29 ×10 years) half-life  

49. Yes. All living things contain small amounts of carbon-14. Some people may have also been exposed to radioactive elements in other forms and have higher levels of radioactivity in their bodies. 50. Ionizing radiation can change the genetic material (DNA). These changes can then be passed on to future generations. 51. Long-term exposure to low-level ionizing radiation can cause tumors, cancer, and damage to blood-producing cells. 52. Scientists can feed a plant with radiophosphorus labeled phosphates and by recording the rate of increase of radiation emitted from the plant can gauge the rate of uptake of these phosphates by the plant. 53. (a) The process is transmutation. (b) Transmutation produces a different element by bombarding an element with a small particle such as hydrogen, deuterium, or a neutron. Radioactive decay is the formation of a different element by the loss of an alpha or beta particle from the nucleus of a radioactive element. (c) No. Nuclear fission involves the bombardment of a heavy element with a neutron, causing the element to split (fission) into two or more lighter elements and produces two or more neutrons. A large amount of energy also is formed in a nuclear fission. 54. There are a total of 11 α decays and 6 β decays; γ-radiation will have no effect.11 alpha decays results in a loss of 44 in mass number and 22 in atomic number. Six beta decays results in no change in atomic mass and a gain of 6 in atomic number. a. Atomic number will go down by 16. b. Atomic mass will be reduced by 44. 55.

t 1 , hours

12.5

25.0

37.5

50.0

62.5

75.0

87.5

100.

15.4

7.7

3.85

1.93

0.965

0.483

0.242

0.121 0.0605

Amount, mg

Fraction of K-42 remaining:

0.0605 mg = 0.00393(or 0.393%) 15.4 mg

No. After an additional eight half-lives there would be less than one microgram (0.000001 g) remaining.

( 200 hrs ) 

1 half-life   = 16 half-lives  12.5 hrs  16

1 Amount remaining =   2

 103 μg   = 0.235 μ g after 16 half-lives  mg 

(15.4 mg ) 

CHAPTER 19

INTRODUCTION TO ORGANIC CHEMISTRY SOLUTIONS TO REVIEW QUESTIONS 1.

The vital-force theory states that all organic compounds must come from living organisms. This theory was disproved by Friedrich Wöhler in 1828.

All organic compounds contain carbon.

Fuels, fabrics, cosmetics, soaps and detergents, medicines, and plastics are among the many consumer products which are composed of organic compounds.

Carbon atoms have the characteristic of bonding extensively with one another. They form organic compounds containing carbon chains of varying lengths and structure. Consequently, a great many compounds of carbon exist.

The most common geometric pattern of covalent bonds about carbon atoms is the tetrahedral arrangement of bonds. A simple example is the methane molecule, CH4, with hydrogen atoms at the corners of the tetrahedron and the carbon atom at the center.

In addition to single bonds, carbon atoms can also form double and triple bonds. For examples, see Question 7(b)

Lewis structures for:

Names and formulas of the first 10 normal alkanes

– Chapter 19 –

Alkyl groups

– Chapter 19 –

10. The single most important reaction of alkanes is combustion. 11. Adding a methyl group to carbon-1 of pentane makes the longest carbon chain of the compound have six carbon atoms so the compound is no longer a pentane. The correct name is hexane. 12. Compound (a) is an alkene and belongs to a homologous series of compounds represented by the formula CnH2n. Compounds (b), (c), and (d) are all alkanes, CnH2n+2 13. The word ethylene represents a compound containing a double bond. All the other words represent structures having no double bonds. 14. Structure of vinyl acetylene, C4H4, CH2=CH–C≡CH 15. Formula of C6H8 The formula would be C6H14 if the compound were a saturated hydrocarbon. This formula is 6 H atoms short of being saturated. Removing two H atoms from a saturated hydrocarbon forms a carbon-carbon double bond. Removing four H atoms can form a carbon-carbon triple bond. Therefore, C6H8 can contain three carbon-carbon double bonds or one double bond and one triple bond. 16. Benzene 17. The prefix ortho- means the two substituents are adjacent to each other, the prefix meta- means the two substituents are separated by one carbon atom, and the prefix para- means the two substituents are on opposite sides of the benzene molecule. 18. Alcohols (Phenols and carboxylic acids also have an –OH group, but we generally associate –OH groups with alcohols.) 19. Aldehydes and ketones (Carboxylic acids and esters also have a carbonyl group, but we generally associate carbonyl groups with aldehydes and ketones.) 20. Carboxylic acids

– Chapter 19 –

21. Ethylene glycol is superior to methyl alcohol as an antifreeze because of its low volatility. Methyl alcohol is much more volatile than water. If the radiator leaks under pressure (normally steam), it would primarily leak methanol vapor, thus losing the antifreeze. Ethylene glycol has a lower volatility and a higher boiling point than water, so it does not present this problem. 22. (a) Methanol, taken internally, is poisonous and capable of causing blindness and death. Breathing methanol vapors is also very dangerous. (b) Physiologically, ethanol acts as a food, as a drug, and as a poison. It is a food, in a limited sense. The body is able to metabolize small amounts of it to carbon dioxide and water, resulting in the production of energy. As a drug, ethanol is often mistakenly considered to be a stimulant, but it is, in fact, a depressant. In moderate quantities, ethanol causes drowsiness and depresses brain functions. In larger quantities, ethanol causes nausea, vomiting, impaired perception, and a lack of coordination. Consumption of very large quantities may cause unconsciousness, and even death. 23. R − O − R′ 24. The –yl suffix of an alkyl group is replaced with an –oxy suffix (i.e. methyl becomes methoxy). 25. Alcohols have one alkyl group bonded to an oxygen (R—O—H) and ethers have two alkyl groups bonded to an oxygen ( R − O − R′ ). 26. Aldehydes and ketones contain a carbonyl group. 27. The carbonyl group of an aldehyde is at the beginning of a carbon chain and the carbonyl group of a ketone is in the middle of a carbon chain. 28. The suffix –al indicates a compound is an aldehyde. 29. The suffix –one indicates a compound is a ketone. 30.

(–COOH and –CO2H are alternate ways to express this structure.)

31. The suffix –ic acid indicates a compound is a carboxylic acid. 32. The common name for methanoic acid is formic acid. This name comes from formica, the Latin word for ant. Formic acid is the compound that causes ant bites to sting. 33. Esters have an R group in place of the H atom of the –COOH group (–COOR). 34. The suffix –oate or –ate indicates a compound is an ester. 35. Esters often have very pleasant flavors and odors. The odors of banana, raspberries, and oranges are due to the presence of esters. 36. Monomers 37. The ethene molecule is the parent molecule of many of the monomers that are used to synthesize polymers.

– Chapter 19 –

SOLUTIONS TO EXERCISES 1.

(a) organic

(b) inorganic (c) organic

(a) inorganic (b) organic

Organic compounds are compounds in which carbon combines covalently with hydrogen, oxygen, nitrogen, sulfur and/or other elements. Each compound identified as organic above contains some combination of carbon with hydrogen, oxygen, and/or sulfur.

(d) organic

(e) inorganic (e) organic

Compounds in (a), (c), and (d) contain carbon bonded to carbon, hydrogen, oxygen, and sulfur atoms. 4.

Names of alkyl groups (a) C5H11–

pentyl

(b) C7H15–

heptyl

Names of alkyl groups (a) C8H17–

octyl

(b) C10H21–

decyl

Hexanes

Heptanes

– Chapter 19 –

IUPAC names (a) CH3CH2Cl

Chloroethane

(b) CH3CHClCH3

2-chloropropane

1-chloro-2-methylpropane

10. IUPAC names (a) CH3CH2CH2Cl

1-chloropropane

(b) (CH3)3CCl

2-chloro-2-methylpropane

2-chlorobutane

11. (a) 2-methylbutane (c) 2-chlorobutane 12. (a) 3-iodo-2-methylpentane (c) 2-fluoro-2-methylpropane

(b) 1-chloro-3,4,4-trimethylhexane (d) 2-bromo-1-chloropropane (b) 1-chloro-2-methylpropane (d) 3,3-dimethylpentane

13. Structural formulas (a) 2,4-dimethylpentane

(b) 2,2-dimethylpentane

(c)

3-isopropyloctane

14. (a) 4-ethyl-2-methylhexane

(b) 4-t-butylheptane

– Chapter 19 –

(c)

4-ethyl-7-isopropyl-2,4,8-trimethyldecane

15. (a)

3-methylbutane is an incorrect name because the carbon atoms were numbered from the wrong end of the chain. The correct name is 2-methylbutane. (b)

2-ethylbutane is incorrect. The longest carbon chain in the molecule was not used in determining the root name. The correct name is 3-methylpentane 16. (a)

2-dimethylbutane is incorrect. Each methyl group is attached to carbon 2 on the chain and requires a number to identify its location on the chain. The correct name is 2,2-dimethylbutane. (b)

2-ethyl-4-methylpentane is incorrect. The longest continuous chain in the molecule was not used to name the compound. The correct name is 2,4 dimethylhexane. 17. (a) (1 isomer)

CH3Br

(b) (1 isomer)

CH2Cl2

CH3CH2Cl

(d) (2 isomers)

CH3CH2CH2Br

– Chapter 19 –

18. (a)

(b) (4 isomers) CH3CH2CHCl2

CH3CCl2CH3

CH3CHClCH2Cl

CH2ClCH2CH2Cl

CH3CHClCH2Br

CH3CHBrCH2Cl

CH2ClCH2CH2Br

CH3CClBrCH3 (d)

19. (a)

CH3CH2CH=CH2 + Cl2 → CH3CH2CHClCH2Cl

(b) CH2 =CH2 + HBr → CH3CH2Br 20. (a) CH3CH=CH2 + Br2 → CH3CHBrCH2Br (b) CH3CH=CH2 + HBr → CH3CHBrCH3 21. (a)

CH2=CH2

alkene

(b) CH≡CH

alkyne

alkyl halide

(d) CH3CH2OH

alcohol

22. (a) CH3OCH3

ether

(b) CH3CHO

aldehyde

carboxylic acid

(d) HCOOCH3

ester

– Chapter 19 –

23. (a) ethene (b) ethyne 24. (a) methoxymethane (b) ethanal 25. (a) chloromethane

(b) vinyl chloride

CH2=CHCl

CHCl3

(d) 1,1-dibromoethene

CH2=CBr2

(a) hexachloroethane

CCl3CCl3

(b) iodoethyne

CH≡CI

26.

(b)

(c)

28. (a) 1,2-diphenylethene

(b) 3-pentene-1-yne

– Chapter 19 –

29. (a) 2-methylpropene (b) 1-pentyne (c) 3,4-dimethyl-2-pentene 30. (a) 5-methyl-l-hexene (b) 2-methyl-2-pentene (c) 2,3-dimethyl-l-butene 31. Complete the reactions and name the products. (a)

CH 2 =CHCH 3 + Br2 → CH 2 BrCHBrCH 3

(b) CH 2 =CH 2 + HBr → CH 3CH 2 Br (c)

( bromoethane )

1 atm CH3CH=CHCH3 + H 2 ⎯⎯ ⎯⎯ →CH3CH 2CH 2CH3 Pt, 25 ° C +

32. (a) CH 2 =CH2 + H2 O ⎯H⎯→CH3CH2 OH

( butane )

( ethanol )

(1,1, 2, 2-tetrabromoethane )

(b) CH ≡ CH + 2 Br2 → CHBr2 CHBr2 (c)

(1, 2-dibromopropane )

1 atm CH 2 =CH 2 + H2 ⎯⎯ ⎯⎯ → CH3CH3 Pt, 25 ° C

33. Names of aromatic compounds

34.

( ethane )

– Chapter 19 –

35.

36.

37. (a)

(b)

(c)

– Chapter 19 –

38. (a) (c) (b) 39. Trichlorobenzene, C6H3Cl3

40. Dichlorobromobenzene, C6H3Cl2Br (6 isomers)

41. (a) secondary (b) secondary (c) polyhydroxy (d) primary 42. (a) primary (b) polyhydroxy (c) primary (d) tertiary 43. Structural formulas (a) 2-pentanol

(b) isopropyl alcohol

– Chapter 19 –

44. (a) 2,2-dimethyl-1-heptanol (b) 1,3-propanediol

HOCH2CH2CH2OH

45. Names of aldehydes (a) H2C=O

methanal (formaldehyde)

(b)

Butanal

(c)

3-methylbutanal

46.

benzaldehyde

butanedial

3-hydroxybutanal 47. Names of ketones

propanone, acetone, dimethyl ketone

2-butanone, methyl ethyl ketone

1-phenyl-1-propanone, phenyl ethyl ketone 48.

3,3-dimethyl-2-butanone (methyl t-butyl ketone)

– Chapter 19 –

2,5-hexanedione

4-hydroxy-4-methyl-2-pentanone 49. (a) CH3CHBrCOOH

2-bromopropanoic acid

(b) CH2=CHCH2COOH

3-butenoic acid

butanoic acid

(d) salicylic acid o-hydroxybenzoic acid 50. (a) 2-methylbutanoic acid (b) phenylethanoic acid (phenylacetic acid) (c) CH3CH2CH2CH2COOH

pentanoic acid

(d) benzoic acid 51. Esters

(a) Ethyl formate

(b) Methyl ethanoate

(a) n-nonyl acetate

– Chapter 19 –

(b) Ethyl benzoate

ethyl ethanoate

phenyl ethanoate

isopropyl butanoate 54. isopropyl methanoate

methyl benzoate

ethyl-3-methylbutanoate 55. (a) CH3COOH + NaOH → CH3COO − Na + + H2 O

(b) 56. (a) CH3COOH + KOH → CH3COO− K + + H2 O

(b) 57. polyethylene

polyvinyl chloride

– Chapter 19 –

58.

polyacrylonitrile Teflon 59. (a) incorrect, should be 2-methylbutane

(d) incorrect, should be 3-pentanol

(b) incorrect, should be meta-dibromobenzene (e) correct (c) correct 60. (a) correct (b) incorrect, should be 2-butanone

(d)

incorrect, should be methyl propanoate

(e)

incorrect, should be 3-methylpentane

(c) incorrect, should be 4-methyl-2-pentene 61. (a) hydrocarbon (b) aromatic (c) alkyne (d) alcohol (e) carboxylic acid 62. (a) alkyl halide (b) ketone (c) alkene (d) ester (e) hydrocarbon 63. There are several possible 4 carbon saturated hydrocarbons. One example is shown below:

64. There are several possible 4 carbon unsaturated hydrocarbons. One example is shown below:

65. Bond angles are 109.5°. 66. Bond angles are 120°. 67. H − C ≡ C − H Bond angles are 180°.

– Chapter 19 –

68. Alkanes are nonpolar molecules. Therefore, there is little attraction between them because there are no partial positive and partial negative charges to attract one another. As a result, it takes very little energy to cause lower molar mass alkanes to boil because there are no intermolecular attractive forces to be overcome. 69. (a) C6H14, 5 isomers

(b) C5H10, 5 isomers

70. (a) Isomers of pentyne, C5H8 (3 isomers)

– Chapter 19 –

(b) Isomers of hexyne, C6H10 (7 isomers)

71. Isomeric alcohols, formula C5H11OH (8 isomers)

– Chapter 19 –

72. The molar mass of myricyl alcohol, an open chain saturated alcohol, contains 30 carbon atoms. The first three alcohols in the homologous series, CH3OH, C2H5OH, and C3H7OH, lead us to the formula C30H61OH, (CnH2n+1OH). molar mass = (30)(12.01 g/mol) + (62)(1.008 g/mol) + (1)(16.00 g/mol) = 438.8 g/mol 73. (a) methyl ethyl ether

CH3CH2OCH3

(b) dimethyl ether

CH3OCH3

CH3CH2OCH3

74. Prostaglandin A2 has two alkenes, an alcohol, a ketone, and a carboxylic acid (or carboxylate) group. 75. Water interacts mainly by hydrogen bonding, and molecules that can participate significantly in hydrogen bonding generally dissolve in water. The alcohol, CH3CH2CH2OH, is the only compound listed that can significantly participate in hydrogen bonding as both a donor and an acceptor and should be significantly soluble in water. 76. Structural isomers methyl propyl ether and 2-butanol; butanoic acid and methyl propanoate (b) methyl ethyl ether or methoxy ethane 77. (a) diethyl ether or ethoxy ethane (c) ethyl isopropyl ether or ethoxy isopropane 78.

Butanal and butanone are isomers (C4H8O). Both compounds have the same molecular formula. 79. (a) propanoic acid (b) ethanoic acid or acetic acid (c) 2-methylbutanoic acid (d) 3 chlorobenzoic acid 80. Isomers of hexanoic acid, CH3(CH2)4COOH (8 isomers)

– Chapter 19 –

81. Preparation of esters

– Chapter 19 –

82. (a)

(b) Molar mass of ethylene monomer = 28.05 g/mol molar mass of polymer = number of monomer units molar mass of monomer 35,000 g/mol = 1.2 ×103 ethylene units 28.05 g/mol

83.

84. (a) (b) (c)

85. (a) pentyl ethanoate (b) ethyl butanoate (c) isobutyl methanoate

– Chapter 19 –

86. Assume 100. g of material and calculate the empirical formula. C

24.3 g = 2.02 mol 12.01 g/mol

2.02 =1 mol 2.02

4.1 g = 4.1 mol 1.008 g/mol

4.1 = 2 mol 2.02

71.7 g = 2.02 mol 35.45 g/mol

2.02 =1 mol 2.02

Empirical formula = CH2Cl (mass = 49.48 g/mol) PV = nRT n =

PV RT

  740 mm Hg ×1 atm   ( 0.1403 L ) −3 n=  = 4.5×10 mol   760 mm Hg 0.0821 L atm/mol K 373 K ( )( )    0.442 g molar mass = = 98 g/mol 4.5×10 −3 mol 98 g/mol dividing = 2.0 49.48 g/mol therefore the molecular formula is 2× empirical formula = C2H4Cl2 Possible isomers are

87. Calculate the moles of each element.

24 g C = 2 mol C The empirical formula is CH2 O 12.01 g/mol 4gH = 4 mol H 1.008 g/mol 32 g O = 2 mol O 16.00 g/mol The only structure given that could match this formula is (e) CH3COOH. 88. (a) CH3CH2OH (b) CH3I (c) CH3CH2CH2CHClCH3 (d) CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3 (e)

– Chapter 19 – +

89. (a) HCOOH + CH3OH ⎯H⎯→ HCOOCH3 + H2 O methyl formate

H (b) CH3CH2CH2COOH + CH3CH2CH2CH2 OH ⎯⎯→ CH3CH2CH2COOCH2CH2CH3 + H2 O

butyl butanoate

CH3 ( CH2 )4 COOH + CH3 ( CH2 )4 CH2 OH ⎯H⎯→ +

(c)

CH3 ( CH2 )4 COOCH2 ( CH2 )4 CH3 + H2 O hexyl hexanoate

90.

91.

(a) ester (b) ether (c) alkyl halide (d) alcohol

– Chapter 19 –

92.

(a) ketone (b) aldehyde (c) carboxylic acid (d) ether

CHAPTER 20

INTRODUCTION TO BIOCHEMISTRY SOLUTIONS TO REVIEW QUESTIONS AND EXERCISES 1.

The three principal classes of animal food are carbohydrates, lipids and proteins.

The four major classes of biomolecules are carbohydrates, lipids, proteins, and nucleic acids.

In Table 20.1, the sweetest disaccharide is sucrose. The sweetest monosaccharide is fructose.

Monosaccharides, disaccharides, and polysaccharides. The simplest type of carbohydrate is the monosaccharide.

An aldose is a monosaccharide containing an aldehyde group on one carbon atom and a hydroxyl group on each of the other carbon atoms. An aldotetrose is an aldose containing four carbon atoms. A ketose is a monosaccharide containing a ketone group on one carbon atom and a hydroxyl group on each of the other carbon atoms. A ketohexose is a ketone containing six carbon atoms. Examples: aldose, glucose; aldotetrose, erythrose; ketose, fructose; ketohexose, fructose.

(a) mono (b) poly (c) mono (d) mono (e) di

Open chain formulas

– Chapter 20 –

Cyclic structural formulas

Properties and sources Ribose: Ribose is a white, crystalline, water soluble pentose sugar, present in adenosine triphosphate (ATP), one of the chemical energy carriers in the body. Ribose and one of its derivatives, deoxyribose, are also important components of the nucleic acids, DNA and RNA, the genetic information carriers in the body. Glucose: Glucose is an aldohexose and is found in the free state in plant and animal tissues. Glucose is commonly known as dextrose. It is a component of the disaccharides sucrose, maltose, and lactose, and it is also the monomer of the polysaccharides starch, glycogen, and cellulose. Glucose is the key sugar of the body and is circulated by the bloodstream to provide energy to all parts of the body. Fructose: Fructose, also called levulose, is a ketohexose and occurs in fruit juices as well as in honey.Fructose is also a constituent of sucrose. Fructose is the sweetest of all sugars, being about twice as sweet as glucose. This accounts for the sweetness of honey. The enzyme invertase, present in bees, splits sucrose into glucose and fructose. Fructose is metabolized directly, but it is also readily converted to glucose in the liver. Galactose: Galactose is an aldohexose and occurs, along with glucose, in lactose and in many oligo- and polysaccharides, such as pectin, gums, and mucilages. Galactose is synthesized in the mammary glands to make the lactose of milk.

10. Lactic acid has a hydroxyl group and an acid group. It is not a carbohydrate, because it has neither an aldehyde nor a ketone group, and will not yield one upon hydrolysis. 11. (a) maltose is 2 glucose units (b) lactose is galactose plus glucose (c) sucrose is glucose plus fructose (d) starch is glucose units

– Chapter 20 –

(e) Cellulose; a polysaccharide made from many units of glucose. (f)

Glycogen; a polysaccharide made from many units of glucose.

12. Cyclic structural formulas for sucrose and maltose.

13. The formation of a disaccharide, C12H22O11, from a monosaccharide, C6H12O6, involves combining two monosaccharide units with a molecule of water splitting out between them. crase ⎯→ Glucose + Fructose 14. (a) Sucrose + H2 O ⎯su⎯

– Chapter 20 – se (b) Maltose + H2 O ⎯ma ⎯lta⎯ → Glucose + Glucose

15. a.

fructose

glucose

ribose

fructose

galactose

ribose

glucose

fructose

glucose

16. Starch and cellulose have their basic composition in common. They contain many glucose units joined in long chains forming polysaccharide molecules with a higher molar mass. The difference in properties, which are highly significant, are due to the different manner in which the glucose units are attached to each other. The primary use of starch is for food. Humans cannot utilize cellulose as food due to a lack of the necessary enzymes to hydrolyze cellulose to usable glucose.

17. Carbohydrates are stored in the human body as glycogen, a polysaccharide of glucose. 18. Carbohydrate metabolism ase amylase maltase Starch ⎯amyl ⎯⎯ → Dextrins ⎯⎯⎯ ⎯ → Maltose ⎯⎯⎯ ⎯ → Glucose mouth intestines intestines

Glucose is absorbed through the intestinal walls into the bloodstream. From there the glucose may be stored in the liver as glycogen or utilized as energy by oxidation to carbon dioxide and water. Sugars, such as sucrose and lactose, are converted to monosaccharides by specific enzymes in the intestines.

– Chapter 20 –

19. Natural sources of sucrose, maltose, lactose, and starch are: Sucrose: Sucrose is found in the free state throughout the plant kingdom. Sugar cane contains 15% to 20% sucrose, while sugar beets contain 10% to 17% sucrose. Maple syrup and sorghum are also good sources of sucrose. Maltose: Maltose is found in sprouting grain, but occurs much less commonly in nature than either sucrose or lactose. Lactose: Lactose, also known as milk sugar, is found free in nature mainly in the milk of mammals. Human milk contains about 6.7% lactose and cow milk about 4.5% of this sugar. Starch: Starch is a polymer of glucose. It is found mainly in the seeds, roots, and tubers of plants. Corn, wheat, potatoes, rice, and cassava are the chief sources of starch. 20. Invert sugar is sweeter than sucrose, from which it comes, because it contains free fructose which is far sweeter than sucrose. The relative sweetnesses are: fructose, 100; sucrose, 58; glucose, 43. 21. (a) Fructose contains a C=O group in its carbon chain.

(b) Glucose contains a

group at the beginning of its carbon chain.

22. The simplest empirical formula for a carbohydrate is CH2O. 23. 2 O = −4

22 H =+22 C12H22O11 11 O = −22

CO2

1 C = +4

12 C = 0 C12H22O11 + 12 O2 → 12 CO2 + 11 H2O The change in oxidation state of carbon is from 0 to +4.

– Chapter 20 –

24. Molar mass of C6H10O5 = 162.1 g/mol Cellulose: Starch:

6.0 ×105 g/mol = 3.7 ×103 monomer units 162.1g/mol

4.0 ×103 g/mol = 25 monomer units 162.1g/mol

Cellulose: ( 3.7 × 103 units )( 5.0 × 10 −10 m/unit ) = 1.9 × 10 −6 m long Starch: ( 25 units ) ( 5.0 × 10 −10 m/unit ) = 1.3 × 10 −8 m long 25. Saturated fats are composed of single bonds between carbon atoms. Unsaturated fats are composed of both single and double bonds between carbon atoms. Saturated fats are solids and unsaturated fats are liquids at room temperature. 26. Fatty acids in vegetable oils are more unsaturated than fatty acids in animal fats. This is because vegetable oils contain higher percentages of oleic and linoleic (unsaturated) acids than animal fats. 27. Substances are classified as lipids on the basis of their solubility in nonpolar solvents such as diethyl ether, benzene, and chloroform, their insolubility in water, and their greasy feeling. 28. Structural formulas

29. Fats are solid and vegetable oils are liquid at room temperature. Fats contain higher amounts of saturated fatty acids and vegetable oils contain higher amounts of unsaturated fatty acids. 30. A triacylglycerol (triglyceride) is a triester of glycerol. Most animal fats are triacylgycerols. Tristearin, in the next exercise, is a triacylglycerol. 31. Tristearin

– Chapter 20 –

32. Triacylglycerol

There are two other formulas using the same three acids. In one, the linoleic acid would be in the middle and in the other, oleic acid would be in the middle. The top and bottom positions are equivalent. 33. (a) Tripalmitin is a fat in which all the fatty acid units are palmitic acid.

Sodium palmitate is a soap. (b)

The soaps top to bottom are: sodium linoleate, sodium stearate, and sodium oleate. 34. Vegetable oils can be solidified by hydrogenation, which adds hydrogen to the double bonds, to saturate the double bonds and form fats. Solid fats are preferable to oils for the manufacture of soaps and for certain food products. Hydrogenation extends the shelf-life of oils, because it is oxidation at the points of unsaturation that leads to rancidity of fats and oils.

– Chapter 20 –

35. Fats are an important food source for humans. They normally account for 25 to 50 percent of caloric intake. Fats are the major constituent of adipose tissue, which is distributed throughout the body. In addition to being a source of reserve energy, fat deposits function to insulate the body against loss of heat and protect vital organs against mechanical injury. 36. The three essential fatty acids are linoleic, linolenic, and arachidonic acids. Diets lacking these fatty acids lead to impaired growth and reproduction and skin disorders, such as eczema and dermatitis. 37. The structural formula of cholesterol is

38. The ring structure common to all steroids is

39.

40. The N-terminal residue is the amino-group (—NH2) of one end of a linear peptide. The C-terminal residue is the carboxyl-group (—COOH) of the other end of a linear peptide. 41. Of the common amino acids listed in Table 20.3, two, aspartic acid and glutamic acid, have more than one carboxyl group. The following amino acids have more than one amino group: arginine, histidine, lysine, and tryptophan. 42. There are three disulfide linkages in each molecule of beef insulin. 43. Some common foods with high (over 10%) protein content are gelatin, fish, beans, nuts, cheese, eggs, poultry and meat of all kinds. 44. Amino acids contain a carboxylic acid group (—COOH) and an amine group (—NH2).

– Chapter 20 –

45. The amino acids in proteins are called α-amino acids because the amine group is always attached in the α position, that is, the first carbon next to the carboxylic acid group. 46. (a) phenylalanine, tryptophan, tyrosine (b) cysteine, methionine (c) serine, threonine, tyrosine 47.

48.

49. All possible tripeptides of glycine (gly), phenylalanine (phe), and leucine (leu). gly - phe - leu

leu - phe - gly

gly - leu - phe

phe - gly - leu

leu - gly - phe

phe - leu - gly

– Chapter 20 –

50. Aspartame: two possible structures

The circled H is replaced by CH3 in aspartame. 51. Essential amino acids are those which are needed by the human body but cannot be synthesized by the body. Therefore, it is essential that they be included in the diet. They are: arginine, histidine, isoleucine, leucine, lysine, methionine, phenylalanine, threonine, tryptophan, and valine. 52. The proteins consumed by a human are converted by digestive enzymes into smaller peptides and amino acids. These smaller units are utilized in many ways: (a) to replace and repair body tissues (b) to synthesize new proteins (c) to synthesize other nitrogen-containing substances, such as enzymes, certain hormones, and heme molecules (d) to synthesize nucleic acids (e) to synthesize other necessary foods, such as carbohydrates and fats. Proteins are catabolized (degraded) to carbon dioxide, water, and urea. Urea, containing nitrogen, is eliminated from the body in the urine. 53. Tissue proteins are continuously being broken down and resynthesized. Protein is continually needed in a balanced diet because the body does not store free amino acids. They are needed to: (a) replace and repair body tissue (b) synthesize new proteins (c) synthesize other nitrogen-containing substances, such as enzymes, some hormones, and bone (d) synthesize nucleic acids (e) synthesize other necessary foods, such as carbohydrates and fats 54. Polypeptides are numbered starting with the N-terminal amino acid. N-terminal tyr

C-terminal val

– Chapter 20 –

55. 5

tyr - gly -

his -

phe - val N-terminal amino acid.

Polypeptides are numbered starting with the

56. The bond that connects one amino acid to another in a protein is called a peptide bond. 57. An amino acid contains an amino group (—NH2) on the carbon chain in addition to the carboxylic acid group. The amino group cannot be bonded to a group to be an amino acid. For example

58. Enzymes are proteins that act as catalysts by generally lowering the activation energy of specific biochemical reactions. With the assistance of enzymes, these chemical reactions proceed at high speed at normal body temperature. 59. Enzymes are usually specific for one particular reaction because the substrate (substance acted upon by the enzyme) usually fits exactly into a small part of the enzyme (known as the “active site”) to form an intermediate enzyme-substrate complex. Most substrates do not fit any other enzyme. 60. In the lock and key hypothesis the active site of an enzyme exactly fits the complementary-shaped part of a substrate to form an enzyme-substrate reaction complex on the way to forming the products. In the flexible site hypothesis (induced-fit) the enzyme changes its shape to fit the shape of the substrate to form the enzyme-substrate reaction complex. 61. Enzyme specificity means that an enzyme binds to only one substrate. Enzyme specificity is due to the particular shape of a small segment of the enzyme known as the “active site.” This site fits a complementary shape of the substrate with which the enzyme is reacting. 62. Enzymes act as catalysts for biochemical reactions. Their function is to increase the rate of biochemical reactions by lowering the activation energy of these biochemical reactions. 63. Chromosomes are composed of proteins and nucleic acids. 64. In DNA, the nitrogen bases are off to the side while the deoxyribose and phosphoric acid are part of the backbone chain. 65. In the double stranded helix structure of DNA (Figure 20.7), the hydrogen bonding of the nitrogen bases is as follows: guanine and cytosine are hydrogen bonded to each other as are adenine and thymine.

– Chapter 20 –

66. The component parts that make up DNA

67. (a) The three units that make up a DNA nucleotide are a phosphate, deoxyribose, and one of the four nitrogen-containing bases (A, T, G, C). (b) In DNA, the four types of nucleotides are phosphate - deoxyribose - thymine phosphate - deoxyribose - cytosine phosphate - deoxyribose - adenine phosphate - deoxyribose - guanine (c) Structure and name of one of the nucleotides phosphate - deoxyribose – cytosine

68. The stucture of DNA, as proposed by Watson and Crick, is in the form of a double helix with both strands coiled around the same axis. Along each strand, phosphate and deoxyribose units alternate. Each deoxyribose unit has one of the four bases attached, which in turn, is hydrogen-bonded to a complementary base on the other strand. Thus, the two strands are linked at each deoxyribose unit by two bases. 69. The two helices of the double helix are joined together by hydrogen bonds between bases. The structure of the bases is such that one base will hydrogen bond to only one other specific base. That is, adenine is always hydrogen-bonded to thymine and cytosine is always bonded to guanine. Therefore, the hydrogen bonding requires a specific structure on the adjoining helix. 12

– Chapter 20 –

70. Complementary bases are the pairs that “fit” to each other by hydrogen bonds between the two helices of DNA. For DNA, the complementary pairs are thymine with adenine and cytosine with guanine, or T-A and C-G. 71. If a segment of a DNA strand has a base sequence C-G-A-T-T-G-C-A, the other strand of the double helix will have the base sequence G-C-T-A-A-C-G-T. 72. Replication of DNA begins with the unwinding of the double helix by breaking the hydrogen bonds between the bases to form two separate strands. Each strand then combines with the proper free nucleotides to produce two identical replicas of the original double helix. This replication of DNA occurs just before the cell divides, giving each daughter cell the full genetic code of the parent cell. 73. DNA contains the genetic code of life. For any individual, the sequence of bases and the length of the nucleotide chains in the DNA molecules contain the coded messages that determine all the characteristics of the individual, including the reproduction of that species. Because of the mechanics of human reproduction, the offspring is a combination of the chromosomes of each parent, thus will not be a carbon copy of either parent. 74. The structural differences between DNA and RNA are: (a) RNA exists in the form of a single-stranded helix, whereas DNA is a double helix. (b) RNA contains the pentose sugar ribose, whereas DNA contains deoxyribose. (c) RNA contains the base uracil, whereas DNA contains 2-deoxyribose thymine. 75. In ordinary cell divisions, known as mitosis, each DNA molecule forms a duplicate by uncoiling to single strands. Each strand then assembles the complementary portion from available free nucleotides to form duplicates of the original DNA molecule. In most higher forms of life, reproduction takes place by the union of the sperm with the egg. Cell splitting to form the sperm cell and the egg cell occurs by a different and more complicated process called meiosis. In meiosis, the sperm cell carries only one half of the chromosomes from its original cell, and the egg cell also carries one half of its original chromosomes. Between them, they form a new cell that once again contains the correct number of chromosomes and all the hereditary characteristics of the species. 76. The location of protein synthesis is a ribosome. 77. Proteins are polymers of amino acids. 78. The red dotted lines in Figure 20.7 represent hydrogen bonds between complementary base pairs. 79. Galactosemia is the inability of infants to metabolize galactose to glucose. It is usually caused by a deficiency of an enzyme, resulting in a build-up of galactose in the blood and urine. Galactosemia causes vomiting, diarrhea, enlargement of the liver, and often mental retardation. Newborn infants are routinely checked for galactosemia.

– Chapter 20 –

80. (a) An ester can be formed from a carboxylic acid and an alcohol. Alcohol group: Ser, Tyr, or Thr. Carboxylic acid group: Asp or Glu. (b)

(c) Cholesterol is a lipid. Lipids are water-insoluble molecules. The alcohol group in cholesterol is only a small component of a large molecule. 81. (a) The zinc ion is called an activator. (b) RNA contains ribose sugar; DNA contains deoxyribose sugar. RNA contains the base uracil; DNA contains the base thymine. RNA is generally single stranded; DNA is generally double stranded.