SOLUTIONS MANUAL FOR Fundamentals of Chemistry for Today General, Organic, and Biochemistry 1st Edit by StudyGuide

Fundamentals of Chemistry for Today General, Organic, and Biochemistry 1e Spencer Seager, Tiffiny Rye-McCurdy, Ryan Yoder (Solutions Manual All Chapters, 100% Original Verified, A+ Grade) Chapter 1

Chapter 1: Matter, Measurements, and Calculations CHAPTER OUTLINE 1.1 What is Matter?

1.7 Signiﬁcant Figures

1.2 Physical and Chemical Properties and Changes

1.8 Using Units in Calculations: An Introduction to

1.3 Classifying Matter

Dimensional Analysis

1.4 Measurement Units

1.9 Calculating Percentages

1.5 The Metric System

1.10 Density and Its Applications

1.6 Large and Small Numbers: An Introduction to Scientiﬁc Notation

LEARNING OBJECTIVES/ASSESSMENT When you have completed your study of this chapter, you should be able to: 1.

Explain what matter is. (Section 1.1; Exercise 1.2)

Explain the diﬀerence between the terms physical and chemical as they apply to the properties of matter and changes in matter. (Section 1.2; Exercises 1.8 and 1.10)

Classify matter as an element, compound, homogenous mixture, or heterogeneous mixture. (Section 1.3; Exercises 1.16, 1.20, and 1.22)

Describe four measurement units used in everyday activities. (Section 1.4; Exercise 1.26))

Convert measurements within the metric system into related units. (Section 1.5; Exercises 1.28 and 1.38))

Convert temperatures measured in Fahrenheit to Celsius and vice versa. (Section 1.5; Exercises 1.41 and 1.42)

Express numbers using scientiﬁc notation. (Section 1.6; Exercises 1.46 and 1.47)

Perform calculations with numbers expressed in scientiﬁc notation. (Section 1.6; Exercises 1.58 and 1.59)

Express measurements and calculations using the correct number of signiﬁcant ﬁgures. (Section 1.7; Exercises 1.62 and 1.64)

10. Use dimensional analysis to solve numerical problems. (Section 1.8; Exercise 1.80) 11. Perform calculations involving percentages. (Section 1.9; Exercise 1.90) 12. Perform calculations involving densities. (Section 1.10; Exercise 1.96)

LECTURE HINTS AND SUGGESTIONS 1.

When describing chemistry as the “central science,” explain how everything around us is somehow related to chemistry. Look around the classroom and point out things which are a result of the study of chemistry; such as the plastic materials which make up part of the furniture, the paint on the walls, the clothing that we have on, the paper that we write on, the ink that we write with, and even the biochemical reactions which take place in our bodies which keep us alive.

Stress that a pure substance contains only one kind of basic building block or one kind of constituent particle. Every constituent particle in a pure substance is the same. If there are two or more kinds of

Chapter 1 constituent particles present, it is a mixture. Sugar has sugar molecules; water has water molecules; and sugar water has both sugar molecules and water molecules. 3.

Emphasize that an important characteristic of a pure substance is a constant composition. Give some simple examples, such as water or salt, which when free of other substances, always have the same composition regardless of source. Simple common solutions such as salt water can be used as examples of mixtures. Also, stress that a mixture may have a varying composition. For example, salt water may contain a very small amount of salt or a lot of salt. Salt water is a mixture. If it is left out in an open dish, the water will evaporate (a physical process) leaving behind the salt.

Students sometimes miss the whole point behind significant figures. The most important point to convey is that all measured data have some uncertainty associated with them that is inherent in the measuring device. A simple demonstration is to have students measure the classroom width using a rope knotted at about one-meter intervals, a meter stick and a tape measure. Note: Since the knots in the rope are not numbered, students need to manually count them. Have three students perform the same counting. The results often differ significantly for a large classroom.

Explain that dimensional analysis is just a way to convert between units and it can really save time when solving complex numerical problems. Begin by showing students how equalities can be written as conversion factors (i.e., fractions) and then move to show how multiplying conversion factors together can eliminate unwanted units and solve for the answer in the unit of interest. Emphasize that learning this method may take some time, however, it can be used to solve quantitative problems presented in not only chemistry but all the natural sciences, and thus it is time well spent learning the method.

Providing a handout with commonly used conversion factors and equations is helpful when introducing unit conversion and dimensional analysis. The example handout titled Chapter 1: Unit Conversion on the next page could be used as a resource for students to reference as they problem solve.

Chapter 1

CHAPTER 1: UNIT CONVERSION TABLE 1.2 Common Preﬁxes of the Metric System

Temperature Scales = °C

= °F

5 ( °F − 32 ) 9

9 ( °C ) + 32° 5

°C =

K − 273

K =

°C + 273

TABLE 1.5 Commonly Used Conversion Factors

A cm3 is commonly abbreviated “cc.” bA foot-pound is the energy it takes to push with one pound-force

(lbf) for a distance of one foot. cA BTU (British thermal unit) is the amount of heat required to increase the temperature of 1 pound of water by 1 °F.

Chapter 1

SOLUTIONS TO ALL END-OF-CHAPTER EXERCISES What follows are more complete explanations/full solutions to the EOC exercises whose answers are published in shorter form at the end of the textbook

SECTION 1.1 WHAT IS MATTER? 1.1

If a heavy steel ball is suspended by a thin wire and hit from the side with a hammer on the moon, the heavy steel ball will hardly move, just like on earth. This experiment depends only on the mass of the ball and the hammer, not their weights.

1.2

All matter occupies space and has mass. Mass is a measurement of the amount of matter in an object. The mass of an object is constant regardless of where the mass is measured. Weight is a measurement of the gravitational force acting on an object. The weight of an object will change with gravity; therefore, the weight of an object will be different at different altitudes and on different planets.

1.3

To prove to a doubter that air is matter, precisely weigh a deflated balloon, then inflate it and weigh it again. The mass of the inflated balloon will be greater than the mass of the deflated balloon because the air in the inflated balloon has mass. The volume of the air is also clearly evident in the increased size of the balloon.

1.4

The distance you can throw a bowling ball will change more than the distance you can roll a bowling ball on a ﬂat, smooth surface. When throwing a ball, gravity pulls the ball towards the ground and air resistance slows its decent. The gravitational force on the moon is approximately 1/6th the gravitational force that is present on the earth; therefore, when throwing a ball on the moon, you should be able to throw it further than you can on earth. The moon does not have air resistance. When rolling a ball, friction helps to slow down the ball. If the ﬂat, smooth surface is the same on the earth and the moon, the amount of friction should remain constant.

1.5

If you were transported from a deep mine to the top of a tall mountain, your mass would not be changed by the move because mass is independent of gravity. b. If you were transported from a deep mine to the top of a tall mountain, your weight would decrease because weight depends on gravity and gravity decreases with distance from the earth’s center. A mountaintop is further from the earth’s center than a deep mine; therefore, your weight will be less on the mountaintop.

1.6

The attractive force of gravity for objects near the earth’s surface increases as you get closer to the center of the earth (Exercise 1.5). If the earth bulges at the equator, the people at the equator are further from the center of the earth than people at the North Pole. If two people with the same mass were weighed at the equator and at the North Pole, the person at the equator would weigh less than the person at the North Pole because the gravitational force at the North Pole is stronger than the gravitational force at the equator.

Chapter 1

SECTION 1.2 PHYSICAL AND CHEMICAL PROPERTIES AND CHANGES 1.7

a. The plum’s color, smell, and taste have changed. This was a change in composition; therefore, it is a chemical change. b. The water vapor can be condensed into liquid water and its properties will not have changed by the boiling. The composition of the water has not changed by boiling; therefore, it is a physical change. c. The glass pieces still have the same chemical composition as the original glass window. This was a change that did not involve composition; therefore, it is a physical change. d. The food is broken down into components that can be used by the body. This is a change that involves composition; therefore, it is a chemical change.

1.8

a. The two pieces of the stick still have the same chemical composition as the original stick. This was a change that did not involve composition; therefore, it is a physical change. b. As the candle burns, it produces carbon dioxide, water, soot, and other products. This is a change that involves composition; therefore, it is a chemical change. c. The pieces of rock salt have the same chemical composition as the original larger piece of rock salt. This was a change that did not involve composition; therefore, it is a physical change. d. Many tree leaves are green in the spring and summer because of the green chlorophyll that is used in photosynthesis to produce energy for the tree. During these seasons, the tree stores the extra energy so that in autumn when the days grow shorter, the chlorophyll is no longer needed. As the leaves in the cell stop producing chlorophyll, the other colors present in the leaves become more visible. This change involves composition; therefore, it is a chemical change.

1.9

a. Physical: a state of matter b. Chemical: binding indicates a change in composition c. Chemical: corrosion indicates a change in composition d. Chemical: neutralizes indicates a change in composition e. Physical: color is easily observed

1.10

a. The phase of matter at room temperature is a physical property because the composition does not change while making this observation. b. The reaction between two substances is a chemical property because the composition of the products diﬀers from the reactants. The products for the reaction between sodium metal and water are sodium hydroxide and hydrogen gas. (Note: Predicting the products for this type of chemical reaction is covered in Section 9.6) c. Freezing point is a physical property because the composition does not change while making this observation. d. The inability of a material to form new products by rusting is a chemical property because rust would have a diﬀerent chemical composition than gold. Attempting to change the chemical composition of a material is a test of chemical property regardless of whether the attempt is

Chapter 1 successful. e. The color of a substance is a physical property because the composition does not change while making this observation. 1.11

a. b.

Dying your hair is a chemical property because a chemical reaction alters the pigments in your hair which result in a color change. When sugar is spun to make cotton candy, it is a physical property because the sugar is only heated to change the state and then spun rapidly in the machine to convert the liquid sugar to thin solid strands. Using hydrogen peroxide to disinfect a wound forms bubbles in a chemical reaction. The hydrogen peroxide comes into contact with an enzyme called catalase, which decomposes into water and oxygen, which forms the bubbles. The use of magnesium chloride as a deicer is a physical property because the compound dissolves in water, which forms a solution that lowers the freezing point of water, which helps prevent ice from forming.

SECTION 1.3 CLASSIFYING MATTER 1.12

Carbon dioxide is heteroatomic. If oxygen and carbon atoms react to form one product, then carbon dioxide must contain these two types of atoms.

1.13

Hydrogen peroxide is heteroatomic. If water (which contains hydrogen and oxygen atoms) and oxygen gas can be produced from hydrogen peroxide, then hydrogen peroxide must contain both hydrogen and oxygen atoms.

1.14

Water is heteroatomic. If breaking water apart into its components produces both hydrogen gas and oxygen gas, then water must contain two types of atoms.

1.15

Heteroatomic: If the products contain hydrogen (in H2) and carbon (in CO2), the hydrogen and carbon must have come from glucose, making it heteroatomic.

1.16

a. b. c.

1.17

a. b. c.

1.18

Substance A is a compound because it is composed of molecules that contain more than one type of atom. Substance D is an element because it is composed of molecules that contain only one type of atom. Substance E is a compound because it is a pure substance that can break down into at least two different materials. Substances G and J cannot be classified because no tests were performed on them. Substance L is a compound. It is formed by combining two elements. Substances M and Q cannot be classified. Without further testing it is impossible to tell if the substances are elements or compounds. Substance X cannot be classified. The absence of a change is not conclusive evidence that a substance is an element or a compound. Substance R might appear to be an element based on the tests performed. It has not decomposed into any simpler substances based on these tests; however, this is not an exhaustive list of tests that could be performed on Substance R. Substance R cannot be

Chapter 1

b. c.

classified as an element or a compound based on the information given. Substance T is a compound. It is composed of at least two different elements because it produced two different substances on heating. The solid left in part b cannot be classified as an element or a compound. No tests have been performed on it.

1.19

Early scientists incorrectly classified calcium oxide (lime) as an element for a number of years. It is possible this mistake in classification was made because calcium oxide was the product of decomposing limestone (calcium carbonate) and it was difficult to further decompose the lime into the elements of calcium and oxygen.

1.20

a. heterogeneous b. homogeneous c. homogeneous d. heterogeneous e. homogeneous f. homogeneous g. heterogeneous

1.21

Muddy ﬂood water

It is heterogeneous because it does not have the same composition throughout (concentration of mud/debris depends on water depth).

Gelatin dessert

Normal urine

Smog-ﬁlled air

An apple

Mouthwash

Petroleum jelly

It is homogeneous because it has the same composition throughout. It is homogeneous because it has the same composition throughout. It is heterogeneous because it does not have the same composition throughout (concentration of smog, oxygen, other gases depend on the altitude.) It is heterogeneous because it does not have the same composition throughout (skin, meat, seeds.) It is homogeneous because it has the same composition throughout. It is homogeneous because it has the same composition throughout.

a. b.

Blood Liquid eye drops

An aspirin tablet

A urine sample with kidney stones Intravenous saline

1.22

This is a pure substance because it contains one substance. This is a solution because it contains many substances (water, salt, glycerin, etc.) This is a solution because it contains many substances (aspirin, starch, preservatives, etc.) This is a solution because it contains urine and kidney stones. This is a solution because it contains water and salt.

Chapter 1

1.23

f. g.

An antibiotic ointment Curdled milk

This is a solution because it contains many substances. This is a pure substance because the curdles are still milk.

a. b. c.

Muddy ﬂood water Gelatin dessert Normal urine

e. f. g.

An apple Mouthwash Petroleum jelly

This is a solution because it contains many substances. This is a solution because it contains many substances. This is a solution because it contains many substances (urea, water, dissolved salts, etc.) This is a pure substance because it contains one substance. This is a solution because it contains many substances. This is a solution because it contains many substances.

SECTION 1.4 MEASUREMENT UNITS 1.24

Modern society is complex and interdependent. Accomplishing projects like building a bridge, constructing a house, or machining an engine may require diﬀerent people to participate. Some people design the project, others supply the necessary materials, and yet another group does the construction. In order for the project to be successful, all of these people need a common language of measurement. Measurement is also important for giving directions, keeping track of the time people work, and keeping indoor environments at a comfortable temperature and pressure.

1.25

In the distant past, 1 in. was deﬁned as the length resulting from laying a speciﬁc number of grain kernels, such as corn, in a row. The size of 1 in. would vary in this system because the size of the individual kernels as well as the tightness of the packing would vary. The unit obtained by this system would not be consistent in all situations.

1.26

The amount of weight that a horse could carry or drag might have been measured in stones. It could also be used to measure people or other items in the 50-500 pound range. It is likely that a large stone was picked as the standard weight for the “stone” unit. Stones may have also been used as counterweights on an old-fashioned set of balances.

SECTION 1.5 THE METRIC SYSTEM 1.27

a. Nonmetric b. Metric c. Nonmetric d. Metric e. Metric f. Nonmetric

1.28

The metric units are (a) degrees Celsius, (b) liters, (d) milligrams, and (f) seconds. The English units are (c) feet and (e) quarts.

1.29

Grams, milligrams Meters Degrees Celsius

Chapter 1 1.30

1.31

1.32

1.33

1.34

1.35

1.36

Meters are a metric unit that could replace the English unit feet in the measurement of the ceiling height. Liters are a metric unit that could replace the English unit quarts in the measurement of the volume of a cooking pot.

Chapter 1

1.37

1.38

1.39

1.40

1.41

1.42

1.43

Chapter 1

SECTION 1.6 LARGE AND SMALL NUMBERS: AN INTRODUCTION TO SCIENTIFIC NOTATION 1.44

1.45

1.46

1.47

a. b. c.

02.7 x 10-3 4.1 x 102 71.9 x 10-6

Improper form because no leading zero is necessary. (2.7 x 10-3) Correct. Improper form because only one digit should be to the left of the decimal point. (7.19 x 10-5) Improper form because a nonexponential term should be written before the exponential term. (1 x 103) Improper form because one nonzero digit should be to the left of the decimal point. (4.05 x 10-4)

103

0.0405 x 10-2

0.119

Improper form because one nonzero digit should be to the left of the decimal point and an exponential term should be to the right of the nonexponential term. (1.19 x 10-1)

a. b. c.

4.2 x 103 6.84 202 x 10-3

0.026 x 10-2

10-2

74.5 x 105

Correct. Improper form because the “x 10” factor is missing. (6.8 x 104) Improper form because only one digit should be to the left of the decimal point. (2.02 x 10-3) Improper form because only one non-zero digit should be to the left of the decimal point. (2.6 x 10-2) Improper form because a nonexponential term should be written before the exponential term. (1 x 10-2) Improper form because only one digit should be to the left of the decimal point (7.45 x 105)

a. 14 thousand b. 365 c. 0.00204 d. 461.8 e. 0.00100 f. 9.11 hundred

= 14,000 =

a. Three hundred b. 4003 c. 0.682 d. 91.86 e. 6000 f. 400

= 300 =

= 9.11 x 100 =

1.4 x 104 3.65 x 102 2.04 x 10-3 4.618 x 102 1.00 x 10-3 9.11 x 102 3 x 102 4.003 x 103 6.82 x 10-1 9.186 x 101 6 x 103 4 x 102

Chapter 1 1.48

a. b.

186 thousand mi/s 1100 million km/h

186 x 1000 = 1.86 x 105 mi/s 1100 x 1,000,000 = 1.1 x 109 km/h

1.49

6.51 x 10-2 mm, 2.56 x 10-3 in. 1.52

1.50

The decimal point has been moved 22 places to the left. This places 21 zeros to the right of the decimal point and before the numbers 105 g.

1.51

602 000 000 000 000 000 000 000 hydrogen molecules The decimal point has been moved 23 places to the right.

1.52

a. b. c. d. e.

(8.2 x 10-3)(1.1 x 10-2) (2.7 x 102)(5.1 x 104) (3.3 x 10-4)(2.3 x 102) (9.2 x 10-4)(2.1 x 104) (4.3 x 106)(6.1 x 105)

=9.02 x 10-5 =1.377 x 107 =7.59 x 10-2 =1.932 x 101 =2.623 x 1012

=9.0 x 10-5 with significant figures =1.4 x 107 with significant figures =7.6 x 10-2 with significant figures =1.9 x 101 with significant figures =2.6 x 1012 with significant figures

1.53

a. b. c. d. e.

(6.3 x 105)(4.2 x 10-8) (2.8 x 10-3)(1.4 x 10-4) (8.6 x 102)(6.4 x 10-3) (9.1 x 104)(1.4 x 103) (3.7 x 105)(6.1 x 10-3)

=2.646 x 10-2 =3.92 x 10-7 =5.504 x 100 =1.274 x 103 =2.257 x 103

=2.6 x 10-2 with significant figures =3.9 x 10-7 with significant figures =5.5 x 100 with significant figures =1.3 x 108 with significant figures =2.3 x 103 with significant figures

1.54

a. b. c. d.

(144)(0.0876) =(1.44 x 102)(8.76 x 10-2) =1.26144 x 101 (751)(106) =(7.51 x 102)(1.06 x 102) =7.9606 x 104 (0.0422)(0.00119) =(4.22 x 10-2)(1.19 x 10-3) =5.0218 x 10-5 (128,000)(0.0000316)=(1.28 x 105)(3.16 x 10-5) =4.0448 x 100

=1.26 x 101 with SF =7.96 x 104 with SF =5.02 x 10-5 with SF =4.04 x 100 with SF

1.55

a. b. c. d.

(538)(0.154) (600)(524) (22.8)(341) (23.6)(0.047)

=(5.38 x 102)(1.54 x 10-1) =8.2852 x 101 =3.144 x 105 =(6 x 102)(5.24 x 102) 1 2 =(2.28 x 10 )(3.41 x 10 ) =7.7748 x 103 =(2.36 x 101)(4.7 x 10-2) =1.1092 x 100

=8.29 x 101 with SF =3 x 105 with SF =7.77 x 103 with SF =1.1 x 100 with SF

1.56

3.1 x 103 1.2 x 102

=2.583 x 10-5

=2.6 x 10-5 with SF

7.9 x 102 3.6 x 104

=2.194 x 102

=2.2 x 102 with SF

4.7 x 10-1 7.4 x 102

=6.35135 x 10-4

=6.4 x 10-4 with SF

0.00229 3.16

=7.2468354 x 10-4

=7.25 x 10-4 with SF

119 3.8 x 103

=3.131578947 x 10-2

= 3.1 x 10-2 with SF

0.000 000 000 000 000 000 000 105 g

Chapter 1 1.57

1.58

1.59

233 1.67

=1.33532934142 x 102

=1.34 x 102 with SF

6.7 x 103 4.2 x 104

=1.59523809524 x 10-1

=1.6 x 10-1 with SF

8.7 x 10-4 2.3 x 10-2

=3.78260869565 x 10-2

=3.8 x 10-2 with SF

6.8 x 103 2.7 x 10-4

=2.5185185 x 107

=2.5 x 107 with SF

1.8 x 10-2 6.5 x 104

=2.7692307692 x 10-7

=2.8 x 10-7 with SF

(5.3)(0.22) (6.1)(1.1)

=1.7377 x 10-1

=1.7 x 10-1 with SF

(3.8 x 10-4)(1.7 x 10-2) 6.3 x 103

=1.025 x 10-9

=1.0 x 10-9 with SF

=2.6 x 106 with SF

4.8 x 106 (7.4 x 103)(2.5 x 10-4)

=2.59459 x 106

5.6 (0.022)(109)

=2.335279 x 100 =2.3 x 100 with SF

(4.6 x 10-3)(2.3 x 102) (7.4 x 10-4)(9.4 x 10-5)

=1.520989 x 107 =1.5 x 107 with SF

(7.4 x 10-3)(1.3 x 104) (5.5 x 10-2)

=1.749 x 103

=1.7 x 103 with SF

6.4 x 105 (8.8 x 103)(1.9 x 10-4)

=3.82775 x 103

=3.8 x 105 with SF

(6.4 x 10-2)(1.1 x 10-8) (2.7 x 10-4)(3.4 x 10-4)

=7.668845 x 10-3 =7.7 x 10-3 with SF

(963)(1.03) (0.555)(412)

=4.3378 x 100

1.15 (0.12)(0.73)

=1.312785 x 101 =1.3 x 101 with SF

=4.34 x 100 with SF

Chapter 1

SECTION 1.7 SIGNIFICANT FIGURES 1.60

a. b. c. d.

A ruler with a smallest scale marking of 0.1 cm A measuring telescope with a smallest scale marking of 0.1mm A protractor with a smallest scale marking of 1° A tire pressure with a smallest scale marking of 1 lb/in2

0.01 cm 0.01 mm 0.1° 0.1 lb/in2

1.61

a. b. c. d.

A buret with a smallest scale marking of 0.1 mL A graduated cylinder with a smallest scale marking of 1 mL A thermometer with a smallest scale marking of 0.1°C A barometer with a smallest scale marking of 1 torr

0.01 mL 0.1 mL 0.01°C 0.1 torr

1.62

a. b. c. d.

6.0 mL 37.00°C 9.00 s 15.5°

1.63

A length of two and one-half centimeters measured with a measuring telescope with a smallest scale marking of 0.1 mm. An initial reading of exactly 0 for a buret with a smallest scale marking of 0.1 mL. A length of four and one-half centimeters measured with a ruler that has a smallest scale marking of 0.1 cm. An atmospheric pressure of exactly 690 torr measured with a barometer that has a smallest scale marking of 1 torr.

b. c. d. 1.64

2.500 cm 0.00 mL 4.50 cm 690.0 torr

Measured = 5.06 lbs. Exact = 16 potatoes

5.06 lb. = 0.31625 lb. = 0.316 lb. with SF 16 potatoes potato potato

Measured = percentages Exact = 5 players

71.2 % + 66.9% + 74.1% + 80.9% + 63.6% = 71.34% with SF 5 players

Measured = 1pm, 2pm Exact = 19, 24, 17, 31, 40

19 + 24 + 17 + 31 + 40 people = 26.2 people 5 days day

Measured = heights Exact = 5 players

6’9 + 5’8 + 5’6 + 5’1 + 4’11 = 5’7” with SF 5 players

1.66

a. b. c.

0.0400 309 4.006

3 SF (0.04000) 3 SF 4 SF

d. 4.4 x 10-3 e. 1.002 f. 255.02

2 SF 4 SF 5 SF

1.67

a. b. c.

0.040 11.91 2.48 x 102

2 SF 4 SF 3 SF

d. 149.1 e. 10.003 f. 148.67

4 SF 5 SF 5 SF

1.68

a. b.

(3.71)(1.4) (0.0851)(1.2262)

1.65

5.194 0.10434962

= 5.2 with significant figures = 0.104 with significant figures

Chapter 1 c.

(0.1432)(2.81) (0.7762) (3.3 x 104)(3.09 x 10-3) (760.)(2.00) 6.02 x 1020

0.518412780211

= 0.518 with signiﬁcant ﬁgures

101.97 2.52491694352 x 10-18

= 1.0 x 102 with significant figures = 2.52 x 10-18 with significant figures (assuming 0 in 760 is significant)

a. b. c.

(1.21)(3.2) (6.02 x 1023)(0.220) (0.023)(1.1 x 10-3) 100

3.872 1.3244 x 1023 2.53 x 10-7

= 3.9 with significant figures = 1.32 x 1023 with significant figures = 3 x 10-7 with significant figures

(365)(7.00) 60

42.583333

= 4 x 101 with signiﬁcant ﬁgures

(810)(3.1) 8.632 x 10-1

2908.94347

= 2.9 x 103 with signiﬁcant ﬁgures

a. b. c. d.

0.208 + 4.9 + 1.11 228 + 0.999 + 1.02 8.543 – 7.954 (3.2 x 10-2) + (5.5 x 10-1) (Hint: Write in decimal form ﬁrst, then add.) 336.86 – 309.11 21.66 – 0.02387

= 6.218 = 230.019 = 0.589

= 6.2 with significant figures = 2.30 x 102 with significant figures = 0.589 with significant figures

= 0.582

= 0.58 with signiﬁcant ﬁgures

= 27.75 = 21.63613

= 27.75 with significant figures = 21.64 with significant figures

= 7.289 = 8.181 = 1.126

= 7.3 with significant figures = 8.18 with significant figures = 1.126 with significant figures

= 0.355

= 0.355 with signiﬁcant ﬁgures

= 106.1 = 17.7121

= 106.10 with significant figures = 17.71 with significant figures

= 0.004483037867

= 0.00460 with signiﬁcant ﬁgures

= 2.20817413454

= 2.208 with signiﬁcant ﬁgures

= 2.6453333

= 2.65 with signiﬁcant ﬁgures

= - 12.8633592018

= -13 with signiﬁcant ﬁgures

= 3.3902741324

= 3 with signiﬁcant ﬁgures

d. e. 1.69

1.70

e. f. 1.71

a. b. c. d.

e. f. 1.72

2.1 + 5.07 + 0.119 0.051 + 8.11 + 0.02 4.337 – 3.211 (2.93 x 10-1) + (6.2 x 10-2) (Hint: Write in decimal form ﬁrst, then add.) 471.19 – 365.09 17.76 – 0.0479

(0.0267 + 0.00119)(4.626) 28.7794 b. 212.6 – 21.88 86.37 c. 27.99 – 18.07 4.63 – 0.88 d. 18.87 _ 18.07 2.46 0.88 (Hint: do division ﬁrst, then subtract.) e. (8.46 – 2.09)(0.51 + 0.22)

Chapter 1

f. 1.73

1.74

(3.74 + 0.07)(0.16 + 0.2) 12.06 – 11.84 0.271

= 0.811808118081

= 0.81 with signiﬁcant ﬁgures

132.15 – 32.16 = 1.14209023415 87.55 b. (0.0844 + 0.1021)(7.174) = 0.070012768117 19.1101 c. (2.78 – 0.68)(0.42 + 0.4) = 3.66726296959 (1.058 + 0.06)(0.22 + 0.2) d. 27.635 – 21.71 = 1.2852494577 4.97 – 0.36 e. 12.47 _ 203.4 = 0.781167484035 6.97 201.8 (Hint: Do division ﬁrst, then subtract.) f. 19.37 – 18.49 = 1.07055961071 0.822

= 1.142 with signiﬁcant ﬁgures

= 0.07001 with significant figures = 4 with significant figures = 1.29 with significant figures = 0.78 with significant figures

= 1.1 with signiﬁcant ﬁgures

a. Area (A = 1 x w) Black A = 12.00 cm x 10.40 cm – 124.8 cm2 Red A = 20.20 cm x 2.42 cm – 48.884 cm2 = 48.9 cm2 Green A = 3.18 cm x 2.55 cm = 8.109 cm2 = 8.11 cm2 Orange A = 13.22 cm x 0.68 cm – 8.9896 cm2 = 9.0 cm2

Perimeter (P = 2 (l) + 2 (w)) P = 2(12.00 cm) + 2(10.40 cm) = 44.80 cm P = 2 (20.20 cm) + 2(2.42 cm) = 45.24 cm P = 2 (3.18 cm) + 2(2.55 cm) = 11.46 cm P = 2(13.22 cm) + 2(0.68 cm) = 27.80 cm

Width

Length

Area (A = 1 x w) Perimeter (P = 2 (l) + 2 (w)) 2 P = 2(0.1200 m) + 2(0.1040 m) = 0.4480 m Black A = 0.1200 m x 0.1040 m = 0.01248 m P = 2(0.2020 m) + 2(0.0242 m) = 0.4524 m Red A = 0.2020 m x 0.0242 m = 0.0048884 m2 2 = 0.00489 m P = 2(0.0318 m) + 2(0.0255 m) – 0.1146 m Green A = 0.0318 m x 0.0255 m = 8.109 x 10-4 m2 = 8.11 x 10-4 m2 P = 2(0.1322 m) + 2(0.0068 m) = 0.2780 m Orange A = 0.1322 m x 0.0068 m = 8.9896 x 10-4 m2 -4 2 = 9.0 x 10 m c. No, the number of signiﬁcant ﬁgures in the answers remains constant. The numbers of places

Chapter 1 past the decimal are different; however, that could be fixed by rewriting all of the answers in scientific notation.

SECTION 1.8 USING UNITS IN CALCULATIONS: AN INTRODUCTION TO DIMENSIONAL ANALYSIS 1.75

1 kg

2.20 lbs. c. 1.76

1.094 yd

1g 0.035 oz

0.394 in 1 cm

0.015 grain 1 mg

0.0338 ﬂ oz 1 mL

1L 1.057 qt

1m 1.094 yd

1.77 1.78

1.79

1.80

We’ll use the factor-unit method. Given is the factor 80 mg , which equals 40 mg, or 1 mL 2 mL 1 mL 40 mg

Chapter 1 1.81

Using the factor-unit method, two factors will be employed.

1.82

Using the factor-unit method,

= 286 min.

1.83

Table 1.3 shows that 1 kg = 2.20 lb

1.84

SECTION 1.9 CALCULATING PERCENTAGES 1.85 1.86

55 years

65 years $25.73

$467.80

x 100 = 85%

x 100 = 5.500%

1.87

140 lbs−32 lbs

1.88

1.0 day

1.89 1.90

140 lbs

1.4

mg day

x 100 = 77%

x 100 = 71%

2000 Calories �

45 Calories

100 Calories

� = 900 Calories = 9.0 𝑥𝑥 102 Calories with signi�icant �igures

Total = 987.1 mg + 213.3 mg + 99.7 mg + 14.4 mg + 0.1 mg = 1314.6 mg IgG =

IgD =

987.1 mg

1314.6 mg 14.4 mg

1314.6 mg

x 100 = 75.09 %; IgA = x 100 = 1.10%; IgE =

213.3 mg

1314.6 mg 0.1 mg

1314.6 mg

x 100 = 16.23%; IgM =

x 100 = 0.008%

99.7 mg

1314.6 mg

x 100 = 7.58%

Chapter 1

SECTION 1.10 DENSITY 1.91

1.92

1.93

D = m/v D = 2.10 g = 1.56 g/cm3 1.35 cm3

1.94

Volume = (3.98 cm)3 = 63.0 cm3 Density =

1.95

1.96

mass = 718.3 g = 11.4 g volume (3.98 cm)3 cm3

Chapter 1 1.97

Using the factor-unit method,

ADDITIONAL EXERCISES 1.98

1.99

1.100

1.101

Chapter 1

1.102

1.103

CHEMISTRY FOR THOUGHT 1.104

b. c. d.

1.105

To separate wood sawdust and sand, I would add water. The sawdust will float, while the sand will sink. The top layer of water and sawdust can be poured off into a filter. The water will run through the filter leaving the sawdust in the filter. The sawdust can then be allowed to dry. The remainder of the water and sand can be poured off into a filter and the sand can be allowed to dry. To separate sugar and sand, I would add water to dissolve the sugar. I would then filter the mixture to isolate the sand. I would evaporate the water to isolate the sugar. To separate iron filings and sand, I would use a magnet. The iron filings will be attracted to the magnet, while the sand will not be attracted to the magnet. To separate sand soaked with oil, I would pour the mixture through a filter. The oil will go through the filter and leave the sand behind on the filter.

A bathroom mirror becomes foggy when someone takes a hot shower because the steam from the shower condenses on the cold glass of the mirror. This is a physical change because the water molecules are changing phase of matter, but not composition.

1.106

This student should have used the relationship 2.2 lbs = 1 kg to multiply 44.5 kg by 2.2 lbs/kg to ﬁnd a weight of 97.9 lbs. The mistake she made appears to be that she divided 44.5 kg by 2.2 rather than multiplying by it. Consequently, she found a weight of only 20.2 lbs. Since she knows 2.2 lbs = 1 kg, she was expecting the pound value to be larger than the kilogram value and she determined she had made a calculation error.

Chapter 1 1.107

A mercury thermometer cannot be used to measure a temperature that is -45°C. A thermometer ﬁlled with a liquid that has a freezing point below -45°C could be used to measure this temperature.

1.108 Assuming each of the guests eats one serving of oatmeal, 9 cups of dry oatmeal should be prepared. 1.109

All matter is made up of atoms of the elements and therefore contains chemicals.

1.110

The density of the object is only 8.76 g/mL; therefore, it does not have the same density as silver and is not silver.

Chapter 2

Chapter 2: Atomic Structure and the Periodic Table CHAPTER OUTLINE 2.1 The Periodic Table

2.5 Radioactive Nuclei

2.2 Subatomic Particles

2.6 Where Are the Electrons?

2.3 Atomic Symbols

2.7 Trends within the Periodic Table

2.4 Isotopes and Atomic Weights

LEARNING OBJECTIVES/ASSESSMENT When you have completed your study of this chapter, you should be able to: 1.

Locate elements in the periodic table on the basis of their group and period designations. (Section 2.1; Exercises 2.1, 2.7, and 2.9)

Describe the charge, relative mass, and location of the three subatomic particles. (Section 2.2; Exercises 2.11 and 2.13)

Write the atomic symbol (AX) for a given set of subatomic particles. (Section 2.3; Exercise 2.15)

Deﬁne the terms isotope and atomic weight. (Section 2.4; Exercises 2.19 and 2.21)

Write balanced equations for radioactive decay and other nuclear processes. (Section 2.5; Exercises 2.41 and 2.43)

Solve problems using the half-life concept. (Section 2.5; Exercises 2.47 and 2.51)

Describe the applications of radiation in health and medicine. (Section 2.5; Exercises 2.57 and 2.59)

Write correct electron conﬁgurations for each element through atomic number 56. (Section 2.6; Exercises 2.77 and 2.79)

Describe periodic trends in atomic radius, ﬁrst ionization energy, and electronegativity. (Section 2.7; Exercises 2.93, 2.95, and 2.97)

LECTURE HINTS AND SUGGESTIONS 1.

The word “element” has two usages: (1) a homoatomic, pure substance; and (2) a kind of atom. This dual usage confuses the beginning students. It often helps the beginning student for the instructor to distinguish the usage intended in a particular statement. e.g. “There are 118 elements, meaning 118 kinds of atoms.” or “Each kind of atom (element) has a name and a symbol.” or “Water contains the element (kind of atom) oxygen.”

The student will memorize the names and symbols for approximately one-third of the 118 elements to be dealt with those commonly encountered in this course or in daily living. Mentioning both the name and the symbol whenever an element is mentioned in the lecture will aid the student’s memorizing.

While memorization of the names and symbols is important, it should not become the major outcome of this class. Avoid reinforcing the mistaken notion that chemistry is merely learning formulas and equations.

If samples of some of the elements are available, showing them may beneﬁt the discussion of the periodic table as it relates to metals, non-metals, similarity within groups, etc.

Chapter 2 5.

Clearly diﬀerentiate between the radioactive substance and the radiation coming from a radioactive substance. Students often think of these as being interchangeable terms because the similar names: alpha radiation and alpha decay, etc.

Many of the students, for whom this text is written, have some basis for understanding the health eﬀects of X-rays. Others have seen ﬁlm badges used by X-ray technicians. This background knowledge can be helpful in discussing the measurement of radiation from nuclear sources.

Extend the concept of chemical change and chemical properties to the atomic level. Chemical changes cause a change in the constituent particles. The atoms are rearranged to make diﬀerent constituent particles. This rearrangement involves the electrons of the atoms. Thus, atoms with similar chemical properties must have similar electronic arrangements.

Relate the arrangement of the periodic table with groups and periods to the electronic structure of the atom, and the ﬁlling order of the orbitals. This facilitates the understanding of the “outer electrons” determining the chemical combinations.

Solutions to All End-of-Chapter Exercises What follows are more complete explanations/full solutions to the EOC exercises whose answers are published in shorter form at the end of the textbook

SECTION 2.1 THE PERIODIC TABLE 2.1

Group a.

IVA (14)

element number 21

III B (3)

zinc

II B (12)

element number 35

VII A (17)

Group

Period

2.2

2.3

2.4

2.5

Period

element number 27

VIII B (9)

IV A (14)

arsenic

V A (15)

II A (2)

How many elements are located in group VIIB (7) of the periodic table?

How many elements are found in period 5 of the periodic table?

How many total elements are in group IVA (14) and IVB (4) of the periodic table?

How many elements are located in group VIIB (7) of the periodic table?

How many total elements are found in periods 1 and 2 of the periodic table?

How many elements are found in period 5 of the periodic table?

This is a vertical arrangement of elements the periodic table

The chemical properties of the elements repeat in a regular way

group periodic law

Chapter 2 as the atomic numbers increase c.

The chemical properties of elements 11, 19, and 37 demonstrate

periodic law

this principle

2.6

Elements 4 and 12 belong to this arrangement

group

This is a horizontal arrangement of elements in the periodic table

period

Element 11 begins this arrangement in the periodic table

period

The element nitrogen is the ﬁrst member of this arrangement

group

elements 9, 17, 35, and 53 belong to this arrangement

group

2.7

a. Transition metal c. Inner-transition metal e. Representative

b. Representative d. Noble gases

2.8

a. Transition metal c. Noble gases e. Representative

b. Inner-transition metal d. Noble gases

2.9

a. c. e.

Nonmetal Metalloid Metal

b. d.

Metal Metalloid

2.10

a. c. e.

Metal Metal Nonmetal

b. d.

Metalloid Nonmetal

SECTION 2.2 SUBATOMIC PARTICLES 2.11

Charge

Mass (u)

8 protons and 9 neutrons

20 protons and 25 neutrons

52 protons and 78 neutrons

130

Charge

Mass (u)

2.12

2.13

9 protons and 10 neutrons

20 protons and 23 neutrons

47 protons and 60 neutrons

107

The number of protons and electrons are equal in a neutral atom. a.

2.14

10 electrons

18 electrons

50 electrons

The number of protons and electrons are equal in a neutral atom. a.

9 electrons

20 electrons

47 electrons

Chapter 2

SECTION 2.3 ATOMIC SYMBOLS 2.15

Symbol

Name

Belongs to group VIA (16) and period 3

Sulfur

The ﬁrst element (reading down) in group VIB (6)

Chromium

The fourth element (reading left to right) in period 3

Silicon

Belongs to group IB (11) and period 5

Silver

Symbol

Name

2.16 a.

The noble gas belonging to period 4

Krypton

The fourth element (reading down) in group IVA (14)

Tin

Belongs to group VIB (6) and period 5

Molybdenum

The sixth element (reading left to right) in period 6

Neodymium

Electrons

Protons

sulfur

element number 24

Electrons

Protons

2.17

2.18 a.

silicon

element number 74

SECTION 2.4 ISOTOPES AND ATOMIC WEIGHTS 2.19

Protons

Neutrons

Electrons

143

Protons

Neutrons

Electrons

3 2

9 4

235 92

34 16

91 40

131 54

cadmium-110

110 4

cobalt-60

60 2

uranium-235

235 9

silicon-28

argon-40

2.20

2.21

2.22

Chapter 2 c.

strontium-88 Mass Number Atomic Number

2.23 a.

6 protons and 6 neutrons

12 6

8 protons and 9 neutrons

17 8

20 protons and 25 neutrons

45 20

2.24

2.25

2.26

2.27

2.28

2.29

Symbol

Mass Number Atomic Number

Symbol

9 protons and 10 neutrons

19 9

20 protons and 23 neutrons

43 20

47 protons and 60 neutrons

107

107 47

contains 20 electrons and 20 neutrons

40 20

contains 1 electron and 2 neutrons

3 1

a magnesium atom that contains 14 neutrons

26 12

contains 17 electrons and 20 neutrons

a copper atom with a mass number of 65

37 17Cl

H M

65 29

a zinc atom that contains 36 neutrons

66 30

the number of neutrons in the nucleus

22.9898 – 11 = 11.9898 ≈ 12 neutrons

the mass (in u) of the nucleus (to three SF)

23.0 u

the number of neutrons in the nucleus

26.982 – 13 = 13.982 ≈ 14 neutrons

the mass (in amu) of the nucleus (to three SF)

27.0 amu

7.42 % x 6.0151 u + 92.58% x 7.0160 u = 0.0742% x 6.0151 u + 0.9258 x 7.0160 u = 6.94173322 u; 6.942 u with SF or (7.42 x 6.0151 u)+(92.58 x 7.0160 u) 100

= 6.94173322 𝑢𝑢; 6.942 u with SF

The atomic weight listed for lithium in the periodic table is 6.941 u. The two values are very close. 2.30

19.78% x 10.0129 amu + 80.22% x 11.0093 amu = 0.1978 x 10.0129 amu + 0.8022 x 11.0093 amu = 10.81221208 amu; 10.812amu with SF or (19.78 x 10.0129 amu)+(80.22 x 11.0093 amu) 100

= 10.81221208 amu; 10.812 𝑎𝑎𝑎𝑎𝑎𝑎 with SF

The atomic weight listed for boron in the periodic table is 10.81 amu. The two values are

Chapter 2 close to one another. 2.31

92.21% x 27.9769 u + 4.70% x 28.9765 u + 3.09% x 29.9738 u = 0.9221 x 27.9769 u + 0.0470 x 28.9765 u + 0.0309 x 29.9738 u = 28.08558541 u; 28.09 u with SF or (92.21 x 27.9769 u)+(4.70 x 28.9765 u)+(3.09 x 29.9738 u) 100

= 28.08558541 u; 28.09 u with SF

The atomic weight listed for silicon in the periodic table is 28.09 u. The two values are the same. 2.32

69.09% x 62.9298 amu + 30.91% x 64.9278 amu = 0.6909 x 62.9298 amu + 0.3091 x 64.9278 amu = 63.5473818 amu; 63.55 amu with SF or (69.09 x 62.9298 𝑎𝑎𝑎𝑎u)+(30.91 x 64.9278 amu) 100

= 63.5473818 amu; 63.55 amu with SF

The atomic weight listed for copper in the periodic table is 63.55 amu. The two values are the same.

SECTION 2.5 RADIOACTIVE NUCLEI 2.33

Radioactive means a material is able to undergo spontaneous nuclear changes and emit energy in the form of radiation. a. Beta rays are not radioactive, they are a form of radiation. b. Radon is not a stable radioactive element, because by deﬁnition a radioactive element is not stable. It undergoes radioactive decay, but as a noble gas, it is also chemically-inert.

2.34

a. b. c.

mass number = 0: beta, gamma, positron. positive charge: alpha, positron charge = 0: gamma, neutron

2.35

a. b. c.

Those with a negative charge = beta Those with a mass number greater than 0 = alpha, neutron Those that consist of particles = alpha, beta, neutron, positron

2.36

a. b. c.

A beta particle = an electron An alpha particle = 2 protons and 2 neutrons A positron = positive electron

2.37

Radiation which has no mass (and therefore no charge) is able to penetrate matter better than radiation which has mass, whether it is charged or neutral. As the charge and mass increase, radiation is less able to penetrate matter.

2.38 a. b. c. d. e.

An alpha particle is emitted. A beta particle is emitted. An electron is captured. A gamma rat is emitted. A positron is emitted.

Atomic number change decrease by 2 increase by 1 decrease by 1 no change decrease by 1

Mass number change decrease by 4 no change no change no change no change

Chapter 2 2.39

2.40

A tin-117 nuclear

117 50

A nucleus of chromium (Cr) isotope containing 26 neutrons

50 24

A nucleus of element number 20 that contains 24 neutrons

44 20

A nucleus of the element in period 5 group VB(5) with a mass number of 96

96 41

A nucleus of element number 37 with a mass number of 80

80 37

A nucleus of the calcium (Ca) isotope that contains 18 neutrons

38 20

10 4

210 83

15 8

204 82

84 35

2.41

2.42

2.43

2.44

Be → ? + 10 B 5

? = 0−1 β

44 22

Bi →42 α + ?

206 ? = 81 Ti

O → ? + 15 N 7

? = 10 β

Pb → ? + 42α

Ti + 0−1e → ?

? = 44 Sc 21

8 4

Be → ? + 24 He

? = 42α

46 23

V → ? + 46 Ti 22

? = 10 β

200 ? = 80 Hg

149 62

Br → ? + 0−1 β

? = 84 Kr 36

34 ? → 15 P + 0−1 β

34 ? = 14 Si

41 ? + 0−1e →19 Br

? = 41 Ca 20

15 8

O + 10 β → ?

? = 15 N 7

121 50

Sm → ? + 145 Nd 60

? = 42α

Sn →0−1 β + 121 Sb 51

Sn (beta emission)

121 50

55 26

Fe (electron capture)

55 26

55 Fe →−01 e + 25 Mn

22 11

Na (daughter= neon-22)

22 11

22 Na →−01 β + 10 Ne

190 78

Pt →42 α + 186 Os 76

Pt (alpha emission)

190 78

67 28

Ni (beta emission)

67 28

Ni →0−1 β + 67 Cun 29

67 31

Ga (daughter= zinc-67)

67 31

Ga →−01 β + 67 Zn 30

157 63

Eu (beta emission)

157 63

Eu →0−1 β + 157 Gd 64

190 78

Pt (daughter= osmium-86)

190 78

Pt → 42α + 186 Os 76

138 62

Sm (electron capture)

138 62

Sm →−01 e + 138 Pm 61

188 80

Hg (daughter= Au-188)

188 80

Hg →−01 β + 188 Au 79

234 90

Th (beta emission)

234 90

234 Th →0−1 β + 91 Pa

218 85

At (alpha emission)

218 85

214 At →42 α + 83 Bi

2.45

Half-life is the amount of time required for half of a sample to undergo a speciﬁc process.

2.46

Half-life is the amount of time required for half of a sample to undergo a speciﬁc process. For example, in the half-life of a cake is one day, then half of a cake will be eaten by the ﬁrst day, the next day half of the remaining cake (1/4 of the original cake) would be eaten, the following day half of the remaining cake (1/8 of the original cake) would be eaten, etc.

Chapter 2 2.47

99 days in three half-lives. After the ﬁrst half-life, 6.00 mg remains. After the second half-life, 3.00 mg remains. After the third half-life, 1.5 mg remains.

2.48

1 Amount remaining = 9.0 ng   2

2.49

1 1 =  128  2  7

 1 half-life  24 hours    6 hours 

= 0.56 ng

 1 half-life  x years    92 years 

1 1   =  2 2

 1 half-life  x years    92 years 

 1 half-life  7 = x years    92 years   92 years  x years = 7   = 644 years  1 half-life  1980 + 644 = 2624 The archaeologist made the discovery in approximately 2624. or 1 Since the fraction remaining is , the number of half-lives which have passed is 7 because 128 7

1 1 . The t 1/ 2 is 92 years; therefore, the amount of time that has passed is   = 128 2 92 x 7 = 644 years. The discovery was made 644 years after 1980, so the year 2624.

2.50

  1  half-lives   ⇒ log (0.0625) = log     2       1  log (0.0625) = half-lives  log     2   half-lives = 4.00

1 100% - 93.75% = 100%   2

half-lives

 5600 years  4 Time elapsed = 4 half-lives   = 22400 years = 2.24 x 10 years 1 half-life   or 1 remains. This means that 4 half-lives have Since 93.75% of the 14C has decayed, 6.25% or 16 4

1 1 . The is t1/ 2 5600 years; therefore, the amount of time that has passed passed, since   = 16 2 is 5600 x 4 = 2.24 x 104 years.

2.51

1 Amount remaining = 50.0 mg   2 or

 1 half-life  10.0 hours    2.5 hours 

= 3.13 mg

Chapter 2

 10.0  The t 1/ 2 is 2.5 hours. After 10.0 hours, 4 half-lives have passed since   = 4.0.  2.5  4

1 1 At this time the fraction remaining = 1/ 2 x 1/ 2 x 1/ 2= =  x 1/ 2 16 2

 1  The amount remaining is 50.0 mg   = 3.13 mg .  16 

2.52

 1  1 = ⇒ log     2  8    1  1 half-lives  log    = log    2  8  half-lives = 3 1   2

half-lives

 1 = log    8 

 5600 years  4 Time elapsed = 3 half-lives   = 16800 years = 1.68 x 10 years 1 half-life   or 3

1 1 The amount remaining is = , which is 1/ 2 x 1/ 2 x 1/ 2 x 1/ 2 =   ; therefore, 3 half-lives have elapsed. 8 2 The t 1/ 2 is 5600 years; therefore, the amount of time that has passed is 5600 x 3 = 1.68 x 104 years.

Sr →−01 β + 90 Y 39

2.53

90 30

2.54

Long-term, low-level exposure to radiation made lead to genetic mutations because ionizing radiation can produce free radicals in exposed tissues. Short-term exposure to intense radiation destroys tissue rapidly and causes radiation sickness. Both forms of exposure have negative eﬀects on health.

2.55

The rem corresponds to the health eﬀects produced by 1 roentgen of gamma or X-rays regardless of the type of radiation involved. If a person is working in an area where exposure to several types of radiation is possible, then a unit that is independent of the type of radiation is the best unit to use.

2.56

Physical units of radiation indicate the activity of a source of radiation; whereas, biological units of radiation indicate the damage caused by radiation in living tissue. Examples of physical units of radiation include the Curie and the Becquerel. Examples of biological units of radiation include the Sievert, the Rad, the Gray, and the Rem.

2.57

A diagnostic tracer is a radioactive isotope that concentrates in a speciﬁc organ. It should have a short, but reasonable, half-life, have a stable daughter isotope, be non-toxic and give oﬀ radiation that can penetrate out of the tissue and be detected.

2.58

Chapter 2 2.59

A therapeutic radioisotope administered internally should be an alpha or beta emitted, have a half-life long enough to perform the desired radiation therapy, be non-toxic (and have non-toxic, non-radioactive daughter isotopes), and be concentrated by the target issue.

2.60

Radioactive isotopes can be used for diagnostic work. When the radioactive isotope concentrates in a tissue under observation, the location is called a hot spot. When the radioactive isotope is excluded or rejected by a tissue under observation, the location is called a cold spot. Both hot spots and cold spots can be used for diagnostic work.

2.61

1 2.2 mCi = 70. mCi   2 2.2 mCi  1    70. mCi  2  5

 1 half-life  x days    2.7 days 

 1 half-life  x days  

 2.7 days    1 1   =  2 2      1 half-life  5 = x days    2.7 days 

 2.7 days  x days = 5   = 14 days  1 half-life  or The percentage of gold-198 that will remain at the lower activity level is

2.2 mCi x 100 = 3.1% . 70. mCi

1 1 This is approximately   = = 3.125% , which is the amount that will remain after 5 half-lives. 32 2 The is 2.7 days; therefore, the amount of time that must pass is 2.7 x 5 = 14 days.

Cr + −01e →51 V; daughter nucleus = vanadium-51 23

2.62

51 24

2.63

After 10 half-lives, the amount of the radioisotope left is less than 1/1000 of the original amount. This would almost be undetectable. Ten half-lives for 14C = 56,000 years. The amount of 14C left after several millions years would be essentially zero.

2.64

By using water that contains a radioactive isotope of oxygen, the oxygen gas produced could be analyzed to see if it contains the radioactive isotope of oxygen from the water or a nonradioactive isotope of oxygen from the hydrogen peroxide.

2.65

Strontium would be likely to be deposited in bones because it is in group IIA (2), just like calcium (a major component of bone).

SECTION 2.6 WHERE ARE THE ELECTRONS? 2.66

Protons are subatomic particles with a positive charge that are located in the nucleus.

Chapter 2 2.67

According to Bohr theory, an electron in an orbit located farther from the nucleus would have higher energy than an electron in an orbit close to the nucleus.

2.68

a. X-rays have a higher amount of energy. b. The red light from a stoplight has a higher amount of energy.

2.69

a. UV radiation b. Gamma-rays

2.70

2.71

A 2p orbital

2 electrons

A 2p subshell

6 electrons

The second shell

8 electrons

A 3d orbital

2 electrons

A 3d subshell

10 electrons

The third shell

18 electrons

2.72

Four (4) orbitals are found in the second shell: one 2s orbital and three 2p orbitals.

2.73

Sixteen (16) orbitals are found in the fourth shell: one 4s orbital, three 4p orbitals, ﬁve 4d orbitals, and seven 4f orbitals.

2.74

Three (p) orbitals are found in a 4p subshell. The maximum number of electrons that can be located in this subshell is 6 because each of the three orbitals can hold two electrons.

2.75

Five (5) orbitals are found in the 3d subshell. The maximum number of electrons that can be located in this subshell is 14 because each of the seven orbitals can hold two electrons.

2.76 a.

element number 37

titanium

2.77

Electron Conﬁguration

Unpaired Electrons

1s22s22p63s23p64s23d104p65s1

1s 2s 2p 3s 3p

1s 2s 2p 3s 3p 4s 3d 2

1s 2s 2p 3s 3p 2

Electron Conﬁguration

Unpaired Electrons 1

1s 2s 2p 3s 3p 4s

1s 2s 2p 3s 3p 4s 3d

chromium

element number 33

s electrons in magnesium

unpaired electrons in nitrogen

ﬁlled subshells in Al

2.78

2 0

1s22s22p63s23p64s23d104p3

1s22s22p63s23p64s23d2

Electron Conﬁguration

Solutions

1s22s22p63s2

1s22s22p3 1s 2s 2p 3s 3p 2

3 1

Chapter 2 2.79 a.

Electron Conﬁguration

Solutions

1s22s22p63s23p64s23d104p2

unpaired electrons

1s22s22p63s23p4

number desig. = 3

3d electrons in tin

1s22s22p63s23p64s23d104p65s24d105p2

Symbol

Name

total electrons in Ge have a number designation (before the letters) of 4

unpaired p electrons in sulfur, number designation of the

2 electrons,

2.80 a.

Contains only two 2p electrons

carbon

Contains an unpaired 3s electron

sodium

Symbol

Name

Contains one unpaired 5p electron

In or I

indium or iodine

Contains a half-ﬁlled 5s subshell

rubidium

arsenic

[Ar] 4s23d104p3

silicon

[Ne] 3s23p2

An element that

[Ar] 4s23d5

element

[Kr] 5s24d105p5

2.81

2.82

contains 25 electrons 2.83

An element that

number 53 [Ar] 4s23d4

iodine

[Kr] 5s24d105p5

[Ar] 4s23d1

copper

[Ar] 4s23d9

phosphorus

[Ne] 3s23p3

contains 24 electrons b.

element number 21

2.84

sodium

[Ne] 3s1

magnesium

[Ne] 3s2

c. d.

sulfur

[Ne] 3s23p4

aluminum

[Ne] 3s 3p

chlorine

[Ne] 3s23p5

silicon

[Ne] 3s23p2

argon

[Ne] 3s23p6

2.85

Eighteen (18) elements have the symbol [Kr] in their abbreviated electronic conﬁgurations.

2.86

Cesium is the period 6 element with chemical properties most like sodium. Cesium has 1 valence-shell electron. Sodium also has only 1 valence-shell electron.

2.87

The period 5 element with chemical properties most like silicon is tin (Sn). Tin has four valence-shell electrons. Silicon also has four valance-shell electrons.

2.88

element number 54

8 electrons

The ﬁrst element (reading down) in group V A (15)

5 electrons

Chapter 2 c.

4 electrons

the fourth element (reading left to right) in period 3

4 electrons

element number 35

7 electrons

2 electrons

strontium

2 electrons

The second element in group VA (15)

5 electrons

2.90

p area

d area

2.91

p area

c. d area

2.89

s area

SECTION 2.7 TRENDS WITHIN THE PERIODIC TABLE 2.92

I would expect to ﬁnd silver and gold in addition to the copper because these elements are all in the same group on the periodic table. Elements that are in the same group have similar chemical properties; therefore, if copper is part of this ore, then the other elements that are most similar to it are also likely to be part of the ore.

2.93

a. b. c. d.

Mg or Sr Rb or Ca S or Te I or Sn

Sr larger radius Rb Te Sn

2.94

a. b. c. d.

Ga or Se N or Sb O or C Te or S

Ga larger radius Sb C Te

2.95

a. b. c. d.

Mg or Al Ca or Be S or Al Te or O

Mg loses e more easily Ca Al Te

2.96

a. b. c. d.

Li or K C or Sn Mg or S Li or N

K loses e more easily Sn Mg Li

2.97

a. Li or Br b. F or Cl c. Sn or C d. Cl or Al

2.98

a. Electronegativity increases from left to right, so Cl is more electronegative. b. Electronegativity decreases as you go down a group, so Cl is more electronegative than I

Br F C Cl

Chapter 2 c. Nitrogen is more electronegative because electronegativity decreases as you go down a group. d. Boron is more electronegative than Be because electronegativity increases as you go from left to right.

Additional Exercises 2.99

The density of the metallic elements increases from left to right across a period of the periodic table because as the mass slowly increases, the volume rapidly decreases across the period; therefore, the density must increase because a larger mass is divided by a smaller volume.

2.100

2.101

Mercury and bromine share the physical property of being liquids at room temperature. They are not in the same group, though, because their chemical properties differ.

2.102

Chemical properties are dependent on the number of valence electrons an atom contains, not the number of neutrons an atom contains; therefore, the chemical properties of isotopes of the same element are the same because all isotopes of the same element contain the same number of electrons, including valence electrons.

2.103

a. Atoms of different elements contain different numbers of protons. b. Atoms of different isotopes contain different numbers of neutrons, but the same number of protons.

Chemistry for Thought 2.104

Aluminum exists as one isotope; therefore, all atoms have the same number of protons and neutrons as well as the same mass. Nickel exists as several isotopes; therefore, the individual atoms do not have the weighted average atomic mass of 58.69 u.

2.105

The ingestion of non-radioactive potassium iodide will satisfy the body’s need or iodine, blocking the absorption of radioactive iodine by the thyroid gland, the part of the body that is most sensitive.

2.106

While radioactive decay does occur naturally, it is unlikely that lead changes into gold naturally because lead has an atomic number of 82 and hold has an atomic number of 79, which makes a difference of 3. None of the common nuclear decay processes change the atomic number of the parent nucleus by 3.

2.107

Because of their mass charge, alpha particles do not travel far enough to be useful in diagnostics.

2.108

In principle, a radioactive isotope never completely disappears by radioactive decay because only half of the sample decays per half-life, so half of the initial sample remains. In reality, all of a sample will decay because eventually, one atom will remain in a sample and when that one atom undergoes decay the entire atom will undergo decay, not half of the atom.

Chapter 2 2.109

While at ﬁrst sending nuclear waste into outer space might seem like an attractive possibility because it would remove the hazardous materials from the earth, it is unlikely to be the best solution for waste disposal. Unfortunately, successfully launching the materials into space would not be assured, and if the spacecraft used were to explode during or shortly after launch, radioactive material would be scattered and the results could be devastating. In addition, we have no idea what the eventual fate of the radioactive waste would be. The presence of pockets of radioactive materials in the cosmos could have a myriad of unintended eﬀects on our universe.

2.110

a. When plutonium-239 undergoes an alpha emission, two protons and two neutrons are released, and uranium-235 will be formed. When Iodine-131 undergoes a beta emission, it releases a beta particle and is transformed into xenon-131. These are the balanced equations: and b. When iodine-131 is ingested, it is readily absorbed by the body’s thyroid gland. The accumulation of iodine-131 in the thyroid can lead to the irradiation of thyroid tissue and the emission of ionizing radiation, which can increase the risk of developing thyroid cancer or other thyroid disorders.

2.111

238 92

234 U → 42α + 90 Th

234 90

234 Th → −01 β + 91 Pa

234 91

234 Pa → −01 β + 92 U

Chapter 3

Chapter 3: Chemical Bonds: Molecule Formation CHAPTER OUTLINE 3.5 Covalent Bonding 3.6 Naming Binary Covalent Compounds 3.7 Lewis Structures of Polyatomic Ions 3.8 Compounds Containing Polyatomic Ions

3.1 An Introduction to Lewis Structures 3.2 The Formation of Ions 3.3 Ionic Compounds 3.4 Naming Binary Ionic Compounds

LEARNING OBJECTIVES/ASSESSMENT When you have completed your study of this chapter, you should be able to: 1. 2. 3. 4. 5. 6. 7. 8.

Draw correct Lewis dot structures for atoms of the representative elements. (Section 3.1; Exercises 3.1, 3.3, and 3.5) Use electron configurations to determine the number of electrons gained or lost by atoms as they achieve noble gas electron configurations. (Section 3.2; Exercise 3.9) Use electron configurations to determine the series of ions that are isoelectronic with the noble gases. (Section 3.2; Exercise 3.17) Use the octet rule to correctly predict the ions formed during the formation of ionic compounds. (Section 3.3; Exercises 3.21) Write correct formula units for ionic compounds containing a representative metal and a representative nonmetal. (Section 3.3; Exercise 3.23) Determine formula weights for ionic compounds in atomic mass units. (Section 3.3; Exercises 3.27 and 3.28) Correctly name binary ionic compounds. (Section 3.4; Exercise 3.35, 3.39, 3.41) Draw correct Lewis structures for covalent molecules. (Section 3.5; Exercises 3.47, 3.49, and 3.51)

Determine molecular weights for covalent compounds in atomic mass units. (Section 3.5; Exercises 3.53 and 3.55) 10. Correctly name binary covalent compounds. (Section 3.6; Exercise 3.57) 11. Draw correct Lewis structures for polyatomic ions. (Section 3.7; Exercises 3.61 and 3.63) 12. Write correct formulas for ionic compounds containing representative metals and polyatomic ions. (Section 3.8; Exercises 3.65 and 3.69) 13. Correctly name binary ionic compounds containing polyatomic ions. (Section 3.8; Exercises 3.67)

LECTURE HINTS AND SUGGESTIONS 1.

Emphasize how the reactivity of the noble gases is related to the stability of their electronic conﬁgurations. Use several examples of some simple ionic and covalent compounds to show how atoms exchange or share electrons in order to attain noble gas conﬁgurations.

Many students have diﬃculty recognizing the diﬀerence between a covalent molecule such as SO3, sulfur trioxide, and a polyatomic anion, such as SO32-. The importance of putting the charge on the

Chapter 3 ion

should be stressed in this chapter. Also, point out the polyatomic anion is not a compound. It

must be 3.

associated with a positive ion (usually a metal ion) to make a compound.

Be sure that the students understand that there is not really a clear division between covalent and ionic bonds, and that there is a continuous increase in ionic character as the electronegativity diﬀerence between the two bonding atoms increases. Generalizations can be made that ionic compounds are formed from a metal or NH4+ and a non-metal, while covalent compounds are formed from two nonmetals.

Providing a handout summarizing cation and anion charges along with the rules of nomenclature is helpful when introducing ionic and covalent compounds. The example handout titled Naming Ionic and Covalent Compounds on the next two pages could be used as a resource for students to reference.

Chapter 3

Naming Ionic and Covalent Compounds Ionic Compounds Cations 1.

The charge of a cation is the number of electrons lost to achieve a noble gas electron configuration (i.e. excluding transition metals the charge of a cation is the group number of the periodic table.) Group 1 +1 charge Ex. Na+ Group 2 +2 charge Ex. Sr2+

Most transition metals can exist in more than one form. These ions are named with a roman numeral designating the charge.

Name of cations: Element name + ion Examples: Mg2+ Magnesium ion Mn4+ Manganese (IV) ion

There is one polyatomic cation: NH4+ ammonium ion

Anions 1.

The charge of an anion is the number of electrons gained to achieve a noble gas electron configuration Group 16 -2 charge Ex. O2Group 17 -1 charge Ex. Br –

Name of anions: element stem and the suﬃx – ide.

Examples: Cl-

chloride ion

sulﬁde ion

Chapter 3

There are many polyatomic anions:

Naming ionic compounds: 1.

Name = metal + nonmetal stem + -ide Examples:

CaF2

Calcium ﬂuoride

Li2S

Lithium Sulﬁde

Compounds with metals that can form more than one type of ion must indicate which ion is present by a Roman numeral in parentheses following the name of the metal. Examples: NiCl2

Nickel (II) Chloride

Ti2O3

Titanium (III) Oxide

Covalent Compounds Naming Covalent Compounds: 1.

Give the name of the less electronegative element ﬁrst (the element given ﬁrst in the formula)

Give the stem of the name of the more electronegative element next (see Table 3.1) and add the suﬃx -ide.

Indicate the number of each type of atom in the molecule by means of the Greek preﬁxes listed in Table 3.4. a.

the preﬁx mono- is not used when it appears at the beginning of a name.

Examples: NBr3 Nitrogen tribromide Cl2O7 Dichlorine heptoxide

Chapter 3

SOLUTIONS TO ALL END-OF-CHAPTER EXERCISES What follows are more complete explanations/full solutions to the EOC exercises whose answers are published in shorter form at the end of the textbook

SECTION 3.1 AN INTRODUCTION TO LEWIS STRUCTURES 3.1

potassium

aluminum

barium

bromine

iodine

tin

strontium

sulfur

3.3

a. b. c. d.

fluorine element number 37 selenium silicon

3.4

a. germanium

b. cesium

c. indium

d. calcium

a. carbon

b. phosphorus

c. sodium

d. rubidium

3.2

3.5

3.6

[He] 2s22p5 [Kr] 5s1 [Ar] 4s23d104p4 [Ne] 2s22p2

a. lithium [He]2s1

b. chlorine [Ne]3s23p5 c. titanium [Ar] 4s23d2 d. [Ar]4s23d104p3 3.7

Any group IIIA (13) element

Any group VI (16) element

Chapter 3

3.8

Any group IIA (2) element

Any group VA (15) element

SECTION 3.2 THE FORMATION OF IONS 3.9 a.

iodine

element 38 (strontium)

Added electrons 1

Removed electrons 17 (7 not including d electrons)

(6 not including d electrons)

15 (5 not including d electrons)

phosphorus

germanium

Added electrons 4

Removed electrons 14

3.10

(4 not including d electrons)

(7 not including d or f electrons)

element number 49

13 (3 not including d electrons)

calcium

(6 not including d electrons)

3.11

Number of electrons lost/gained

Equation

2 electrons lost

Mg → Mg 2+ + 2e −

silicon

4 electrons lost

Si → Si 4+ + 4e −

element 53

1 electron gained

I + e− → I−

sulfur

2 electrons gained

S + 2e − → S 2 −

Number of electrons lost/gained

Equation

1 electron lost

Cs → Cs + + e −

3.12 a.

oxygen

2 electrons gained

O + 2e − → O 2 −

element number 7

3 electrons gained

N + 3e − → N 3 −

iodine

1 electron gained

I + e− → I−

3.13

Equation

Ion Symbol

A bromine atom that has gained one electron

Br + e → Br

A sodium atom that has lost one electron

Na → Na + + e −

Na +

A sulfur atom that has gained two electrons

S + 2e − → S 2 −

S2−

−

Br −

Chapter 3 3.14

3.15

3.16

3.17

3.18

Equation

Ion Symbol 2−

Se 2 −

A selenium atom that has gained two electrons

Se + 2e → Se

A rubidium atom that has lost one electron

Rb → Rb + + e −

Rb +

An aluminum atom that has lost three electrons

Al → Al 3 + + 3e −

Al3+

E−

−

fluorine

E3−

nitrogen

beryllium

lithium

sulfur

E+

sodium

aluminum

chlorine

E2−

Mg 2 +

neon

N3−

neon

Te 2 −

xenon

Be 2 +

helium

Li +

helium

S2−

argon

xenon

−

3.19

a. Kr: Se2-, Br-, Kr, Rb+, Sr2+ b. Xe: Te2-, I-, Xe, Cs+, Ba2+ c. Rn: Ra2+, Fr+, Rn, At-, Po2-

3.20

Cu+

29 protons, 28 electrons

S2-

16 protons, 18 electrons

Sc3+

21 protons, 18 electrons

−

krypton

SECTION 3.3 IONIC COMPOUNDS 3.21

Cation formation a. b. c.

Ca and S

3.24

3.25

+ 2e

→ 2e

Ca → Ca

Mg and N

Mg + Mg

elements num. 19 & 17

3.22

3.23

K + K → 1e

Anion formation

S + 2e → S

−

N + 3e → N −

−

Ca and Cl

lithium and bromine

Ca → Ca

+ 2e

Mg 3 N 2 KCl

−

Anion formation

Formula

Cl + e → Cl

CaCl 2

−

Li → Li + + e − 2+

−

Br + e − → Br −

LiBr

2−

MgS

elements num. 12 & 16

Br −

2−

CaO

S2−

BaSe

I−

BaI 2

P3−

Ba 3 P2

As 3 −

Ba 3 As 2

XO, X = +2

XBr2, X = +2

Mg → Mg + 2e

CaBr2

S + 2e → S

−

Formula CaS

3−

Cl + 1e → Cl

−

Cation formula 2+

2−

S2− 3−

CaS

Ca 3 N 2

Chapter 3 c.

X3P, X = +1

XI, X = +1

3.26

a. c.

BaX, X = -2 Al2X3, X = -2

b. d.

MgX2, X = -1 Na2X, X = -2

3.27

a. c.

Na2O, 61.98 amu PbS2, 271.32 amu

b. d.

FeO, 71.84 amu AlCl3,133.33 amu

3.28

a. c.

NaBr = 102.89 amu Cu2S = 159.15 amu

b. CaF2 = 78.07 amu d. Li3N = 34.83 amu

SECTION 3.4 NAMING BINARY IONIC COMPOUNDS 3.29

3.30

3.31

3.32

3.33

3.34

3.35

3.36

3.37

binary

H2 S

binary

OF2

binary

MgBr2

binary

H 2 SO 4

not binary

PbO 2

binary

Be 3 N 2

binary

CuCl 2

binary

CaCO 3

not binary

KNO 3

not binary

Ca 2 +

calcium ion

Al 3 +

aluminum ion

potassium ion

rubidium ion

Li +

lithium ion

Ba 2 +

barium ion

magnesium ion

cesium ion

chloride ion

S2−

nitride ion

2−

bromide ion

P3−

phosphide ion

oxide ion

fluoride ion

Cl −

3−

Br −

2−

Na 2 O

sodium oxide

LiF

lithium fluoride

CaCl 2

calcium chloride

Ba 3 N 2

barium nitride

Al 2 S 3

aluminum sulfide

SrS

strontium sulfide

Li 2 O

lithium oxide

CaF2

calcium fluoride

MgO

magnesium oxide

BaCl 2

barium chloride

a. magnesium chloride, MgCl2 b. lithium oxide, Li2O

−

sulfide ion selenide ion

Chapter 3 c. potassium fluoride, KF d. sodium nitride, Na3N 3.38

3.39

3.40

3.41

3.42

3.43

3.44

a. cesium iodide CsI b. calcium sulfide CaS c. aluminum chloride AlCl3 d. beryllium phosphide

Be3P2

CrCl 2 and CrCl 3

chromium (II) chloride and chromium (III) chloride

CoS and Co 2 S 3

cobalt (II) sulfide and cobalt (III) sulfide

FeO and Fe 2 O 3

iron (II) oxide and iron (III) oxide

PbCl 2 and PbCl 4

lead (II) chloride and lead (IV) chloride

PbO and PbO 2

lead(II) oxide and lead(IV) oxide

CuCl and CuCl 2

copper(I) chloride and copper(II) chloride

Au 2 S and Au 2 S 3

gold(I) sulfide and gold(III) sulfide

CoO and Co 2 O 3

cobalt(II) oxide and cobalt(III) oxide

CrCl 2 and CrCl 3

chromous chloride and chromic chloride

CoS and Co 2 S 3

cobaltous sulfide and cobaltic sulfide

FeO and Fe 2 O 3

ferrous oxide and ferric oxide

PbCl 2 and PbCl 4

plumbous chloride and plumbic chloride

PbO and PbO 2

plumbous oxide and plumbic oxide

CuCl and CuCl 2

cuprous chloride and cupric chloride

Au 2 S and Au 2 S 3

aurous sulfide and auric sulfide

CoO and Co 2 O 3

cobaltous oxide and cobaltic oxide

manganese (II) chloride

MnCl 2

iron (II) bromide

FeBr2

iron (III) sulfide

Fe 2 S 3

tin (II) chloride

SnCl 2

chromium (II) oxide

CrO

mercury (I) oxide

Hg 2 O

copper (I) nitride

Cu 3 N

lead (II) oxide

PbO

cobalt (II) sulfide

CoS

platinum (IV) iodide

PtI 4

SECTION 3.5 COVALENT BONDING 3.45

Chapter 3 3.46

3.47

IBr

PH 3 (each H atom is bonded to the P atom)

HClO 2 (the O atoms are each bonded to the Cl, and the H is bonded to one of the O atoms)

3.48

3.49

CH 4 (each H atom is bonded to the C atom)

CO 2 (each O atom is bonded to the C atom)

H 2 Se (each H atom is bonded to the Se atom)

NH 3 (each H atom is bonded to the N atom)

a. CCl2O (Each Cl and O atom is bonded to the C atom)

Chapter 3 b. SiF4 (Each F atom is bonded to the Si atom)

c. PF3 (Each F atom is bonded to the P atom)

d. C2H2 (Each H atom is bonded to one C atom)

3.50

a. Boron does not have a full octet

Sulfur has an expanded octet with 12 electrons around the central atom

Sulfur has an expanded octet with ten electrons around the central atom

d. In SeO3, selenium has 12 electrons around the central atom. 3.51

a. O3 (the O atoms are bonded together, like beads on a string) b. CS2 (each S atom is bonded to the C atom) c. SeO2 (each O atom is bonded to the Se atom)

3.52

a. ClO2 = 67.45 amu

Chapter 3 b. N2O = 44.02 amu c. SO2 = 64.06 amu d. CCl4 = 153.81 amu 3.53

a. sulfur hexafluoride, SF6, 146.05 amu b. silicon tetrachloride, SiCl4, 169.89 amu c. dinitrogen trioxide, N2O3, 76.01 amu d. sulfur trioxide, SO3, 80.06 amu

3.54

a. C3H8 = 44.10 amu b. CO = 28.01 amu

3.55

a. Glucose (C6H12O6), 180.16 amu b. The anesthetic Lidocaine (C14H22N2O), 234.34 amu c. The amino acid methionine (C5H11NO2S), 149.21 amu

SECTION 3.6 NAMING BINARY COVALENT COMPOUNDS 3.56

a. b. c.

phosphorus trichloride dinitrogen monoxide carbon tetrachloride

d. e.

boron trifluoride carbon disulfide

3.57

a. b. c.

silicon dioxide silicon tetrafluoride diboron trioxide

d. e.

nitrogen monoxide carbon tetrabromide

3.58

a. SO2, sulfur dioxide, covalent b. MgBr, magnesium bromide, ionic c. BaS, barium sulfide, ionic d. N2O4, dinitrogen tetroxide, covalent

3.59

a. CS2, carbon disulfide, covalent b. SiF4, silicon tetrafluoride, covalent c. CuO, copper(II) oxide, ionic d. NaF, sodium fluoride, ionic

SECTION 3.7 LEWIS STRUCTURES OF POLYATOMIC IONS 3.60

a. SO32– (Each O atom is bonded to the S atom), sulfite

Chapter 3 b. SO42– (Each O atom is bonded to the S atom), sulfate

3.61

a. NO3– (Each O atom is bonded to the N atom), nitrate

b. NO2– (Each O atom is bonded to the N atom), nitrate

3.62

3.63

NH 4 + (each H atom is bonded to the N atom)

PO 4 3 − (each O atom is bonded to the P atom)

SO 3 2 −

ClO 3 − (each O atom is bonded to the Cl atom)

(each O atom is bonded to the S atom)

CN − 13

Chapter 3

CO32- (each O atom is bonded to the C atom)

SECTION 3.8 COMPOUNDS CONTAINING POLYATOMIC IONS 3.64

3.65

3.66

3.67

Any group I A (1) element and SO 3

Any group I A (1) element and C 2 H 3 O −2

Any metal that forms M 2+ ions and Cr2 O7

Any metal that forms M 3+ ions and PO 4

Any metal that forms M 3+ ions and NO −3

M(NO 3 )3

Any metal that forms M + ions and SO 4 2 −

M 2 SO 4

Any metal that forms M 3+ ions and OH −

M(OH)3

Any metal that forms M 2+ ions and HPO 4 2 −

MHPO 4

calcium and the hypochlorite ion

Ca(ClO)2

calcium hypochlorite

cesium and the nitrite ion

CsNO 2

cesium nitrite

Mg and SO 3

2−

MgSO 3

magnesium sulfite

K and Cr2 O7

2−

K 2 Cr2 O7

potassium dichromate

calcium and carbonate ion

CaCO 3

calcium carbonate

sodium and the sulfate ion

Na 2 SO 4

sodium sulfate

K 3 PO 4

potassium phosphate

Mg(NO 3 )2

magnesium nitrate

3.68

3.69

3.70

2−

K and PO 4 2+

M 2 SO 3 MC 2 H 3 O 2 2−

MCr2 O7

3−

and NO 3 −

MPO 4

MgSO 3

(NH 4 )2 SO 4

Ba(OH)2

LiHCO 3

CaCO 3

KOH

Na 3 PO 4

Na 2 CO 3

Ca(NO 3 )2

NH 4 Cl

a. NaNO3 b. Ca2OH c. MgBr2

Chapter 3 d. K2CO3 3.71

a. CsSO3 should be Cs2SO3 b. LiPO4 should be Li3PO4 c. K(MnO4)2 should be KMnO4 d. AlClO3 should be Al3ClO3

3.72

a. FeSO4, ionic, iron(II) sulfate b. NO, covalent, nitrogen monoxide c. CF4, covalent, carbon tetrachloride

3.73

a. NaHCO3, ionic, sodium bicarbonate, or sodium hydrogen carbonate b. Ca2C2H3O2, ionic, calcium acetate c. KI, ionic, potassium iodide

CHEMISTRY FOR THOUGHT 3.74

Hydrogen is the element with an electronic configuration of 1s1. Nitrogen is the element with the electronic configuration of 1s22s22p3. The molecule that contains 3 hydrogen atoms and 1 nitrogen atom is NH 3 . They hydrogen bond to each other as shown here:

3.75

Magnesium is the element in group IIA (2) and period 3. Fluorine is the element with the electron configuration 1s22s22p5. The formula for a compound made from these elements would be MgF2 , magnesium fluoride.

3.76

Potassium dichromate is K 2 Cr2 O7 .

Potassium chromate is K 2 CrO 4 .

Potassium phosphate is K 3 PO 4 .

Potassium permanganate is KMnO 4 .

The colored compounds of potassium ( K 2 Cr2 K 2 O7 and KMnO 4 ) have a transition metal as part of the polyatomic ion. 3.77

Yes, a reaction similar to the one that occurs between sodium metal and chlorine gas would be expected to occur between potassium metal and fluorine gas because potassium is in the same group as sodium and fluorine is in the same group as chlorine. In fact, any combination of metal from group IA (1) (lithium, sodium, potassium, rubidium, cesium, francium) and nonmetal from group VIIA (17) (fluorine, chlorine, bromine, iodine, astatine) should produce a similar reaction.

Chapter 3 3.78

A negatively charged ion will be larger than a nonmetal atom of the same element, because as the electrons are added to the atom, the electron cloud will increase in size as the nucleus is not able to hold onto the electrons as well as their number increases.

3.79

When a metal changes to form a positively charged metal ion, it loses electrons. The remaining electrons are then pulled closer to the positively charged nucleus, which makes the size of the metal ion smaller than the size of the original metal atom.

3.80

There is one pair of nonbonding electrons present on the nitrogen atom in methamphetamine.

3.81

Neon atoms do not combine to form Ne 2 molecules because each neon atom has a completed valence shell and does not need to form bonds with another atom in order to satisfy its valence.

3.82

There are five nonbonding electron pairs on glycine that are located on the nitrogen and oxygen atoms: two are on the C=O oxygen atom, two are on the OH oxygen atom, and one pair is on the NH2 nitrogen atom.

3.83 3.84

There are four nonbonding electron pairs on propanoic acid that are located on the two oxygen atoms: two are on the C=O oxygen, and two are on the OH oxygen.

Chapter 4

Chapter 4: The Mole and Chemical Reactions CHAPTER OUTLINE 4.1 Avogadro’s Number: The Mole 4.2 The Mole and Chemical Formulas 4.3 Chemical Equations

4.4 The Mole and Chemical Equations 4.5 Reaction Yields 4.6 Types of Reactions

LEARNING OBJECTIVES/ASSESSMENT When you have completed your study of this chapter, you should be able to: 1.

Convert between the number of moles, number of grams, and number of atoms of elements. (Section 4.1; Exercises 4.1 and 4.3)

Convert between the number of moles, number of grams, number of compounds, and number of atoms in compounds. (Section 4.2; Exercise 4.9)

Calculate the percent by mass of atoms in compounds. (Section 4.2; Exercise 4.11)

Write a balanced chemical equation when provided with a chemical reaction. (Section 4.3; Exercises 4.31 and 4.32)

Use the mole concept to perform calculations based on the stoichiometry of balanced chemical equations. (Section 4.4; Exercises 4.35 and 4.37)

Use the mole concept to calculate percent yields. (Section 4.5; Exercises 4.45 and 4.47)

Identify the oxidizing and reducing agents in oxidation–reduction (redox) reactions when given the oxidation states of each element in the reactants and products. (Section 4.6; Exercises 4.50 and 4.51)

Classify reactions ﬁrst as redox or nonredox, then as decomposition, combination, single replacement, double replacement, or combustion. (Section 4.6; Exercises 4.54 and 4.55)

LECTURE HINTS AND SUGGESTIONS 1.

It should be emphasized that the mole is a convenient way of measuring out needed numbers of atoms and molecules in the correct ratios for chemical reactions. Explain that the term “mole” is the same type of term as “dozen,” “pair,” or “gross,” except that it speciﬁes a much larger number of items.

Emphasize the main roles of oxidizing numbers are in recognizing redox reactions and in balancing redox equations. To simplify the problems oxidation numbers are provided and students are simply asked to identify oxidizing and reducing agents, rather than memorize the rules of determining oxidation numbers. 3.

Use some simple redox reactions and relate the oxidation number change with the loss and gain of electrons. For example, in the reaction 2Na + Cl2 → 2NaCl, each sodium atom loses one electron, and its oxidation number goes from 0 to + 1; and each chlorine atom gains an electron, and its oxidation number goes from 0 to -1.

Many double replacement reactions are easy to demonstrate in class. Simply mix two solutions, which will provide two ions that form a precipitate: for example, sodium carbonate and calcium 1

Chapter 4 chloride or barium chloride and sodium sulfate or silver nitrate and potassium chromate. Try to pick examples that have some relevance such as in these cases where the formation of calcium carbonate would illustrate what is meant by hard water and where the insolubility of barium sulfate is related to its usefulness as a radiopaque diagnostic material. The silver chromate is a chemical precursor to modern photography that has been important in neuroscience, as it is used in the “Golgi method” of staining neurons for microscopy. The reaction that produces silver chromate is also highly visible in a large lecture theater because the reaction involves a color change in addition to the formation of a precipitate.

Chapter 4

Solutions To All End-Of-Chapter Questions What follows are more complete explanations/full solutions to the EOC exercises whose answers are published in shorter form at the end of the textbook

SECTION 4.1 AVOGADRO’S NUMBER: THE MOLE 4.1

a. 1.003 x 1023 atoms are in 3.10 grams of phosphorus.

b. 3.207 g of sulfur would contain the same number of atoms as 3.10 grams of phosphorus.

4.2

a. 6.022 x 1022 atoms are in 1.60 grams of oxygen.

b. 1.90 grams of fluorine would contain the same number of atoms as 1.60 grams of oxygen.

4.3

The number of moles of beryllium atoms in a 10.0 g sample of beryllium

The number of lead atoms in a 2.0 mol sample of lead

Chapter 4

4.4

The number of sodium atoms in a 50 g sample of sodium

The mass of grams of one phosphorous atom

The number of grams of aluminum in 1.65 mol of aluminum

The total mass in grams of one-fourth Avogadro’s number of krypton atoms

SECTION 4.2 THE MOLE AND CHEMICAL FORMULAS 4.5

Fructose, C6H12O6 = (6 ´ 12.01 g/mol C) + (12 ´ 1.008 g/mol H) + (6 ´ 16.00 g/mol O) = 180.2 g/mol

4.6

Creatine, C4H9N3O2, (4 x 12.01 g/mol C) + (9 x 1.008 g/mol H) + (3 x 14.00 g/mol N) + (2 x 16.00 g/mol O) = 131.1 g/mol

4.7

Olinvyk, C22H30N2O2S, (22 ´ 12.01 g/mol C) + (30 ´ 1.008 g/mol H) + (2 ´ 14.00 g/mol N) + (2 ´ 16.00 g/mol O) + (1 ´ 32.06 g/mol O) = 386.5 g/mol

4.8

Camzyos, C15H19N3O2, (15 x 12.01 g/mol C) + (19 x 1.008 g/mol H) + (3 x 14.00 g/mol N) + (2 x 16.00 g/mol O) = 273.3 g/mol

4.9

(1 x 31.0 u) + (3 x 1.01 u) = 34.u; 1 mole PH3 = 34.0 g PH3 (1 x 32.1 u) + (2 x 16.0 u) = 1 mole SO2 = 64.1 g SO2

Chapter 4 4.10

(1 x 10.8 u) + (3 x 19.0 u) = 67.8 u; 1 mole BF3 = 67.8 g BF3 (2 x 1.01 u) + (1 x 32.1 u) = 34.1 u; 1 mole H2S = 34.1 g H2S

4.11

a. 2 mol of O atoms

4.12

4.13

4.14

First, let’s use the formula for percent to ﬁnd the mass of iron present.

4.15

72.0 g C x 100 = 40.0% of C in C6 H12 O6 180.0 g C6 H12 O6 24.0 g C x 100 = 52.5% of C in C 2 H6 O 46.0 g C 2 H6 O

4.16

4.04 g H x 100 = 25.2% of H in CH 4 16.04 g CH 4

Chapter 4

6.06 g H x 100 = 20.1% of H in C 2 H6 30.01 g C 2 H6 4.17

Statement 4.

6.02 x 1023 C6H5NO3 molecules contain 36.12 x 1023 C atoms, 30.01 x 1023 H atoms, 6.02 x 1023 N atoms, and 18.06 x 1023 O atoms.

Statement 5.

1 mol C6H5NO3 molecules contain 6 moles of C atoms, 5 moles of H atoms, 1 mole of N atoms, and 3 moles of O atoms.

Statement 6.

139 g of nitrophenol contains 72.0 g of C, 5.05 g of H, 14.0 g of N, and 48.0 g of O.

a. (Statement 6) 139 g of nitrophenol contains 72.0 g of C, 5.05 g of H, 14.0 g of N, and 48.0 g of O

b. (Statement 5) 1 mol of C6H5NO3 molecules contain 6 moles of C atoms, 6 moles of H atoms, 1 mole of N atoms, and 3 mol of O atoms.

c. (Statement 4) 6.02 x 1023 C6H5NO3 molecules contain 36.12 x 1023 C atoms, 30.1 x 1023 H atoms, 6.02 x 1023 N atoms, and 18.06 x 1023 O atoms. 4.18

a. 180 g of fructose contains 72.0 g of C, 12.1 g of H, and 96.0 g of O.

b. 1 mol of C6H12O6 molecules contains 6 mol of C atoms, 12 mol of H atoms, and 6 mol of O atoms.

c. 6.02 x 1023 C6H12O6 molecules contain 36.12 x 1023 C atoms, 72.24 x 1023 H atoms, and 36.12 x 1023 O atoms.

Chapter 4

4.19

Urea (CH4N2O) contains the higher mass percentage of nitrogen as shown below:

4.20

Using the periodic table, the mass of 1 mol of epinepherine can be calculated: 9 C’s = 9 x 12.01 = 108.9 13 H’s = 13 x 1.008 = 13.104 1 N = 1 x 14.01

4.21

Using the periodic table, the mass of 1 mol of cortisone can be calculated: 21 C’s = 21 x 12.01 = 252.21 28 H’s = 28 x 1.008 = 28.224

80.00 or 360.4 g/mol 360.434 Now let’s use the factor-unit method. Step 1: 10.0 mg Step 2: 10.0 mg =

5 O’s = 5 x 16.00 =

Step 3: 10.0 mg x Step 4:

1 g 1000 mg

1g 1000 mg x = mol 1000 mg 360.4 g

1000 mg 360.4 g

= 2.77 x 10 −5 mol

4.22

There are 1759.34 grams in 5.5 moles of chloroquine.

4.23

a. C6H14N2O2

Chapter 4 b. Molar Mass of C6H14N2O2, (6 ´ 12.01 g/mol C) + (14 ´ 1.008 g/mol H) + (2 ´ 14.00 g/mol N) + (2 ´ 16.00 g/mol O) =146.2 g/mol c. 19.2%

d. 0.0171 moles of lysine are in 2.5 g.

4.24

a. C5H8NO4 b. Molar mass of C5H8NO4 = (5 x 12.01 g/mol C) + (8 x 1.008 g/mol H) + (1 x 14.01 g/mol N) + (4 x 16.00 g/mol O) = 146.1 g/mol c. 43.8%

d. Yes, the risk value is 80 mg/kg and if a 9 kg the toddler ingested 5.4 g of glutamate, they will have ingested over the risk value

SECTION 4.3 CHEMICAL EQUATIONS 4.25

Reactants

Products

CH 4 ,O 2

CO 2 ,H 2 O

Al,Cl 2

AlCl 3

CH 4 (g) + 2O 2 (g) → CO 2 (g) + H 2 O(g)

2Al(g) + 3Cl 2 (g) → 2AlCl 2 (g)

methane + water → carbon monoxide + hydrogen

methane, water

copper(II) oxide + hydrogen → copper + water

copper(II) oxide

4.26 a. b.

carbon monoxide, hydrogen copper, water

hydrogen Reactants

Products

H 2 (g) + Cl 2 → (g) 2HCl (g)

H 2 , Cl 2

HCl

2 KClO 3 (s) → 2 KCl (s) + 3 O 2 (g)

KClO 3

KCl, O 2

Chapter 4

magnesium oxide + carbon → magnesium + carbon monoxide

4.27

magnesium oxide, carbon

magnesium carbon monoxide

ethane, oxygen

carbon dioxide,

ethane + oxygen → carbon dioxide + water

Fe (s) + O 2 (g) → Fe 2 O 3 (s) is not consistent with the law of conservation of matter because neither the number of iron atoms nor the number of oxygen atoms are the same on both sides

water

of the reaction. b.

2Na 3 PO 4 (aq) + 3MgCl 2 (aq) Mg 3 (PO 4 )2 (s) + 6NaCl (aq) is consistent with the law of conservation of matter.

3.20 g oxygen + 3.21 g sulfur → 6.41 g sulfur dioxide is consistent with the law of conservation of matter.

Notice, the number of moles of oxygen and sulfur are equal on both sides of the equation.

 1 mole O 2  Reactants: 3.20 g O 2   = 0.100 moles O 2 ;  32.0 g O 2   1 mole S  3.21 g S   = 0.100 moles S  32.1 g S   1 mole SO 2   2 moles O  Products: 6.41 g SO 2    = 0.200 moles O;  64.1 g SO 2   1 mole SO 2   1 mole SO 2   1 mole S  6.41 g SO 2    = 0.100 moles S  64.1 g SO 2   1 mole SO 2  d.

4.28

CH 4 (g) + 2 O 2 (g) → CO 2 (s) + 2 H 2 O (g) is consistent with the law of conservation of matter. ZnS (s) + O 2 (g) → ZnO (s) + SO 2 (g) is not consistent with the law of conservation of matter

because the reactant (left) side of the equation has two moles of oxygen atoms, while the product (right) side of the equation has three moles of oxygen atoms. b.

Cl 2 (aq) + 2 I − (aq) → I 2 (aq) + 2Cl − (aq) is consistent with the law of conservation of matter.

1.50 g oxygen + 1.50 g carbon → 2.80 g carbon monoxide is not consistent with the law of conservation of matter because the mass of the reactants is 3.00 g, while the mass of the products is only 2.80 g.

Notice the number of moles of oxygen and carbon are not equal on both sides of the equation either.

Chapter 4

 1 mole O 2  Reactants: 1.50 g O 2   = 0.0469 moles O 2 ;  32.0 g O 2   1 mole S  1.50 g C   = 0.125 moles C  12.0 g S   1 mole CO   1 mole O  Products: 2.80 g CO    = 0.100 moles O;  28.0 g CO   1 mole CO   1 mole CO   1 mole C  2.80 g CO    = 0.100 moles C  28.0 g CO   1 mole CO  d.

2 C 2 H6 (g) + 7 O 2 → 4 CO 2 (g) + 6 H 2 O (g) is consistent with the law of conversation

of matter.

4.29

4.30

3 Pb(NO 3 )2 (aq) + 2 AlBr3 (aq) → 3 PbBr2( 3) + 2 Al(NO 3 )3 (aq)

K (s) + H 2 O (l) → KOH (aq) + H 2 (g)

CaCO 3 (s) → CaO (s) + CO 2 (g)

Ba(ClO 3 )2 (aq) + H 2 SO 4 (aq) → 2HClO 3 (aq) + BaSO 4 (s)

Ag(s) + Cu(NO 3 )2 (aq) → Cu (s) + AgNO 3 (aq)

2 N 2 O(g) + 3 O 2 (g) → 4 NO 2 (g)

Chapter 4

Mg (s) + O 2 (g) → 2 MgO (s)

H 2 SO 4 (aq) + Ca(OH)2 (aq) → CaSO 4 (s) +2 H 2 O (l)

4.31

Li 3 N (s) → Li (s) + N 2 (g)

2Li 3 N (s) → 6Li (s) + N 2 (g)

4.32

KClO 3 (s) → KCl (s) + O 2 (g)

2 KClO 3 (s) → 2 KCl (s) + 3 O 2 (g)

Chapter 4

SECTION 4.4 THE MOLE AND CHEMICAL EQUATIONS 4.33

4.34

2SO 2 (g) + O 2 (g) → 2SO 3 (g)

4HCl (g) + O 2 (g) → 2Cl 2 (g) + 2 H 2 O (g)

Fe 2 O 3 (s) + 3 C (s) → 2Fe (s) + 3CO (g)

2H 2 O 2 (aq) → 2 H 2 O (l) + O 2 (g)

2 C 3 H6 (g) + 9 O 2 (g) → 6 CO 2 (g) + 6 H 2 O (g)

S (s) + O 2 (g) → SO 2 (g)

Chapter 4 b.

Sr (s) + 2 H 2 O (l) → Sr(OH)2 (s) + H 2 (g)

2 H 2 S (g) + 3 O 2 (g) → 2 H 2 O (g) + 2 SO 2 (g)

4 NH 3 (g) + 5 O 2 (g) → 4 NO(g) + 6 H 2 O (g)

CaO (s) + 3 C (s) → CaC 2 (s) + CO (g)

4.35

For the following equation: 4H2O(l) + SnCl4(s) ® Sn(OH)4(s) + 4HCl(aq) a. How many moles of H2O(l) will react with 1.5 moles of SnCl4(s):

b. 292 grams of HCl will be produced.

Chapter 4 4.36

2 SO 2 + O 2 → 2 SO 3 (g)

Factors:

12.0 x 10 23 SO 2 molecules 6.02 x 10 23 O 2 molecules 12.0 x 10 23 SO 3 molecules 12.0 x 10

; ;

6.02 x 10 23 O 2 molecules 12.0 x 10 23 SO 2 molecules 6.02 x 10 23 O 2 molecules

SO 2 molecules 12.0 x 10

SO 3 molecules

; ;

12.0 x 10 23 SO 2 molecules 12.0 x 10 23 O 2 molecules 12.0 x 10 23 SO 3 molecules 6.02 x 10 23 O 2 molecules

; ;

2 moles SO 3 2 moles SO 2 1 mole O 2 2 moles SO 2 2 moles SO 3 1 mole O 2 ; ; ; ; ; 1 mole O 2 2 moles SO 2 2 moles SO 3 2 moles SO 2 2 moles SO 3 1 mole O 2 160.0 g SO 3 160.0 g SO 3 128 g SO 2 32.0 g O 2 128 g SO 2 32.0 O 2 ; ; ; ; ; 32.0 O 2 128 g SO 2 160.0 g SO 3 128 g SO 2 160.0 g SO 3 32.0 g O 2

This list does not include all possible factors. 4.37

2 SO 2 + O 2 → 2 SO 3 (g)

 128 g SO 2  350. g SO 3   = 280. g SO 2  160. g SO 3  280. g SO 2 must react to produce 350. g SO 2 4.38

CaCO 3 (s) → CaO (s) + CO 2 (g)

 100 g CaCO 3  500. g CaO   = 891.265597148 g CaCO 3  56.1 g CaO  ≈ 891 g CaCO 3 4.39

CaCO 3 (s) → CaO (s) + CO 2 (g)  1 mole CO 2  500. g CaO   = 8.91265597 148 moles CO 2  56.1 g CaO  ≈ 8.91 moles CO 2

Chapter 4 4.40

2 Al (s) + 3 Br2 (l) → AlBr3 (s)

 479 g Br2  50.1 g Al   = 444.4055556 g Br2  54.0 g Al  ≈ 444 g Br2 with SF

4.41

3 Ag 2 S (s) + 2 Al (s) → 6 Ag (s) + Al 2 S 3 (s) a. b.

4.42

 54.0 g Al  0.250 g Ag 2 S   = 0.01814516129 g Al  744 g Ag 2 S  -2 SF  1 mole Al 2 S 3≈ 1.81 x 10 g Al with 0.250 g Ag 2 S  3.36021505 x 10 -4 moles Al 2 S 3 =   744 g Ag 2 S  ≈ 3.36 x 10 -4 moles Al 2 S 3 with SF

TiCl 4 (s) + 2Mg (s) → Ti (s) 2 MgCl 2 (s)  1000 g Ti   48.6 g Mg  1.00 kg Ti    = 1014.61377871 g Mg  1 kg Ti   47.9 g Ti  ≈ 1.01 x 10 3 g Mg with SF

4.43

C6 H12 O6 (aq) + 6 O 2 (aq) → 6 CO 2 (aq) + 6 H 2 O (l) a.

4.44

 108 g H 2 O  1.00 mole glucose   = 108 g H 2 O  1 mole glucose  = 1.08 x 10 2 g H 2 O   192 g O 2 1.00 mole glucose   = 192 g O 2  1 mole glucose  = 1.92 x 10 2 g O 2

C6 H12 O6 (aq) + 8 O 2 (aq) → 6 CO 2 (aq) + 6 H 2 O (l)   256 g O 2 1.00 mol caproic acid   = 256 g O 2  1 mol caproic acid 

SECTION 4.5 REACTION YIELDS 4.45 4.46

14.37 g x100 = 81.88% yield 17.55 g 2 HgO (s) → 2 Hg (l) + O 2 (g)

 402 g Hg  7.22 g HgO   = 6.6876 g Hg  434 g HgO 

5.95 g  402 g Hg  7.22 g Hg    434 g HgO 

x 100 = 89.0% yield

Chapter 4 5.95 g x 100 = 89.0% yield 6.6876 g Hg

4.47

2 Ca (s) + O 2 (g) → + 2 CaO (s)  112 g CaO  2.00 g Ca   = 2.7930 g CaO  80.2 g Ca 

2.26 g

 112 g CaO  2.00 g Ca    80.2 g Ca 

x 100 = 80.9% yield

2.26 g x 100 = 80.9% yield 2.7930 g Ca

SECTION 4.6 TYPES OF REACTIONS 4.48

O.N.

Change

Classiﬁcation

4.49

O.N.

Change

Classiﬁcation

4.50

2 Cu (s) + O 2 (g) → 2 CuO (s)

Cl 2 (aq) + 2 KI (aq) → 2 KCl (aq) + I 2 (aq)

3 MnO 2 (s) + 4 Al (s) → 2 Al 2 O 3 (s) + 3 Mn (s)

Chapter 4

4.51

2 H + (aq) + 3 SO 3 2 − (aq) → 2 NO 3 − (g) + H 2 O (l) + 3 SO 4 2- (aq)

Mg (s) + 2 HCl (aq) → MgCl 2 (aq) + H 2 (g)

4 NO 2 (g) + O 2 (g) → 2 N 2 O 5 (g)

H 2 (g) + Cl 2 (g) → 2 HCl (g)

H 2 O(g) + CH 4 (g) → CO (g) + 3 H 2 (g)

CuO (s) + H 2 (g) → Cu (s) + H 2 O (g)

B2 O 3 (s) + 3 Mg (s) → 2 B (s) + 3 MgO (g)

Fe 2 O 3 (s) + CO (g) → 2 FeO (s) + CO 2 (g)

Chapter 4 f.

4.52

Cr2 O7 2 − (aq) + 2 H + (aq) → 3 Mn 2 + (aq) + Cr 3 + (aq) + 3 MnO 2 (s) + H 2 O (l)

6 NaOH (aq) + 2 Al (s) → 3 H 2 (g) + Na 3 AlO 3 (aq) + heat

The oxidizing agent is NaOH because the oxidation number for the hydrogen changes from +1 to 0. The reducing agent is Al because the oxidation number for the aluminum changes from 0 to +3. 4.53

Yes, using the rules for oxidation numbers, assign oxidation numbers. Rule 1: The O.N. for O2 is O. Rule 4: The O.N. for combined H is +1. Rule 5: The O.N. for combined oxygen is -2. Rule 6: Allows us to calculate the O.N. of C on each side of the equation. In CO2, (O.N. of C) + 2 (O.N. of O) = 0 (O.N. of C) + 2 (-2) = 0 (O.N. of C) + (-4) = 0 O.N. of C - +4 In C6 H12 O6 , 6 (O.N. of C) + 12 (O.N. of H) + 6 (O.N. of O) = 0 Thus, carbon on the left side of the equation goes from O.N. of +4 to 0 (right side). Therefore, carbon has been reduced and is the oxidizing agent. The oxygen on the left side of the equation is going from O.N. of -2 to 0 (in the O2) and stays of -2 (in the C6 H12 O6 ). Therefore, some of the oxygen is being oxidized and is the reducing agent.

4.54 a. nonredox:

decomposition

b. redox: single-replacement c. nonredox: double-replacement d. nonredox: combination e. redox: combination f. redox: combustion g. redox: combustion

4.55

N 2 O 5 (g) + H 2 O (l) → 2HNO 3 (aq)

Chapter 4

4.56

Nonredox decomposition

4.57

Na H CO 3 (aq) + H (aq) → Na (aq) + H 2 O (l) + CO 2 (g) ; This reaction is a nonredox reaction.

+4 − 2

+1 − 2

+4 − 2

4.58

Redox.

4.59

Redox.

4.60

Cl 2 (aq) + H 2 O (l) HOCl (aq) + HCl (aq) ; This is a redox reaction.

4.61

Ca 3 (PO 4 )2 (s) + 4 H 3 PO 4 (aq) → 3 Ca(H 2 PO 4 )2 (s) ; This is a nonredox combination reaction.

4.62

Oxidizing agent: N2

+1 − 2

+5 − 2

+1 − 2 + 2

+1 − 1

+2 + 1 + 5 − 2

+1 + 5 − 2

Reducing agent: H2

ADDITIONAL EXERCISES 4.63

1.0 x 10 9 x 100 = 1.66 x 10 −13 % 6.02 x 10 23

4.64

4.65

D2O : (2 x 2u) + (1 x 16.00 u) = 20 u

4.66

Due to its highly reactive nature and tendency to donate electrons, lithium can participate in redox reactions as a reducing agent.

4.67

I would not expect argon to be involved in a redox reaction because it is a noble gas. Noble gases do not tend to form ions, and in order to participate in a redox reaction, argon would have to form ions.

4.68

    60 60.0 g Fe  = 7.5 g 60 Fe 6.983 g naturally occuring elemental iron   55.9 g naturally occuring    elemental iron  

Chapter 4 4.69

Sodium is the element with an electron configuration of 1s22s22p63s1. Phosphorus is the element with 15 protons in its nucleus.

3 Na (s) + P (s) → Na 3 P (s) 4.70

a. There are approximately 1.479 x 10-5 moles of tetrodotoxin in 2.4 mg of (C11H17N3O8).

b. There are 8.893 x 1018 molecules of tetrodotoxin in 2.4 mg of C11H17N3O8.

CHEMISTRY FOR THOUGHT 4.71

*Many vitamin supplements are candy-like in appearance, which can lead to iron poisoning in small children. Symptoms will appear in a 3-year-old at doses greater than 10 mg/kg. a.

No. If a 10 kg patient ingested 5 multivitamin gummies at 12 mg each, 60 mg of iron was ingested. The threshold for iron poisoning for a 10 kg patient would be 100 mg.

b. A 10 kg pediatric patient would receive 36 g of deferoxamine in 24 hours for the treatment for iron poisoning.

4.72

When a yield of more than 100% occurs for a compound prepared by precipitation from water solutions, it is likely that the “dry” compound is still moist and contains extra water mass.

4.73

The formula of the sample that was decomposed was N2O5, dinitrogen pentoxide.  1 mole O  80.0 g O   = 5.00 moles O  16.0 g O 

4.74

  1 mole N 1.20 x 10 24 atoms N   = 1.99 moles N 23  6.02 x 10 atoms g N  Zinc changing from zinc metal into zinc ions is an oxidation reaction because the zinc atoms are

losing electrons to become cations. This reaction is the source of electrons as shown in the following equation: Zn (s) → Zn2+ (aq) + 2 e-

Chapter 5

Chapter 5: Heating and Changes of State: A Story of Polarity and Intermolecular Forces CHAPTER OUTLINE 5.1 Geometries of Molecules and Polyatomic Ions

5.4 Energy and Properties of Matter

5.2 The Polarity of Covalent Molecules

5.5 Changes of State

5.3 Intermolecular Forces

5.6 Energetics: Changes of State vs. Speciﬁc Heat

LEARNING OBJECTIVES/ASSESSMENT When you have completed your study of this chapter, you should be able to: 1.

Use VSEPR theory to predict the geometries of molecules and polyatomic ions. (Section 5.1; Exercises 5.3 and 5.5)

Use electronegativities to classify the covalent bonds in molecules. (Section 5.2; Exercises 5.20 and 5.21)

Determine whether covalent molecules are polar or nonpolar. (Section 5.2; Exercises 5.22 and 5.23)

Identify the major attractive force between molecules in a given substance. (Section 5.3; Exercises 5.30 and 5.31)

Distinguish phenomena that demonstrate kinetic energy from those that demonstrate potential energy. (Section 5.4; Exercise 5.42)

Identify the states of matter using ﬁve properties: density, shape, compressibility, particle interaction, and molecular movement. (Section 5.4; Exercise 5.41)

Classify changes of state as exothermic or endothermic. (Section 5.5; Exercises 5.45 and 5.47)

Use the factors that aﬀect evaporation and condensation to rank substances in order of increasing vapor pressure. (Section 5.5; Exercise 5.53 and 5.55)

Use the factors that aﬀect boiling and melting to rank substances in order of increasing boiling point and melting point. (Section 5.5; Exercises 5.59 and 5.62)

10. Calculate energy changes that accompany heating, cooling, or changing the state of a substance. (Section 5.6; Exercises 5.69 and 5.71)

LECTURE HINTS AND SUGGESTIONS 1.

Provide students Table 5.1 as a full-page handout when introducing VSEPR theory (see next page.)

Have the students make molecular models using balloons as a 3D visual aid. Stress the idea that the molecular shape is the result of electron pairs spreading out as far from each other as possible. The large magnitude of the heat of fusion of water can be demonstrated by taking equal masses of ice and warm water and stirring them together in a beaker. Show that the temperature produced is 0 °C rather than a temperature which is midway between that of the ice and water as the students might initially guess.

Many of the students, for whom this text is written, have a strong interest in health and medicine. To better understand speciﬁc heat, emphasize the real-world application of this concept with a discussion of therapeutic hypothermia (Example 5.11.)

Chapter 5

VSEPR Theory

Chapter 5

Solutions to All End-of-Chapter Questions What follows are more complete explanations/full solutions to the EOC exercises whose answers are published in shorter form at the end of the textbook

SECTION 5.1 GEOMETRIES OF MOLECULAR AND POLYATOMIC IONS 5.1

a. CH4 (each H atom is bonded to the C atom) four bonding domains

b. SO2 (each O atom is bonded to the S atom) two bonding domains and one nonbonding domain

c. AlCl3 (each Cl atom is bonded to the Al atom) three bonding domains

5.2

a. NH3 (each H atom is bonded to the N atom) has three bonding domains and one nonbonding domains

b. BeCl2 (each Cl atom is bonded to the Be atom) has two bonding domains

c. ClCN (the Cl and N atoms are bonded to the C atom) has two bonding domains

Chapter 5 5.3

a. BF3 (each F atom is around the B atom) trigonal planar

b. PH3 (each H atom is around the P atom) trigonal pyramidal

c. SCl2 (each Cl atom is bonded to the S atom) bent

5.4

H 2 S (each H atom is bonded to the S atom) Lewis structure

VSEPR

PCl 3 (each Cl atom is bonded to the P atom) Lewis structure

3D Structure

VSEPR

OF2 (each F atom is bonded to the O atom) Lewis structure

3D Structure

VSEPR

Chapter 5 5.5

O 3 (the O atoms are bonded together, like beads on a string) Lewis structure

3D Structure

VSEPR

3D Structure

VSEPR

NO 2 − (each O is bonded to N) Lewis structure

VSEPR

SO 3 (each O atom is bonded to the S atom) Lewis structure

5.6

3D Structure

PH 3 (each H atom is bonded to the P atom) Lewis structure

VSEPR

SeO 2 (each O atom is bonded to the Se atom) Lewis structure

3D Structure

VSEPR

ClO 3 − (each O is bonded to Cl) Lewis structure

3D Structure

VSEPR

Chapter 5

CO 3 2 − (each O is bonded to C) Lewis structure

3D Structure

VSEPR trigonal planar with C in the middle

H 3 O + (each H is bonded to O. Note the positive charge; compare with NH 4 + ) Lewis structure

5.7

3D Structure

VSEPR

PO 3 3 − (each O is bonded to P) Lewis structure

VSEPR

NH 2 − (each H is bonded to N) Lewis structure

3D Structure

VSEPR

BeCl 4 2 − (each Cl is bonded to Be) Lewis structure

3D Structure

VSEPR

Chapter 5

ClO 4 − (each O is bonded to Cl) Lewis structure

5.8

Trigonal pyramidal

5.9

Trigonal planar

5.10

Bent

5.11

Trigonal planar

3D Structure

VSEPR

Chapter 5

SECTION 5.2 THE POLARITY OF COVALENT MOLECULES 5.12

a. O (EN = 3.5) and S (EN = 2.5) are in the same group, but O is above S, so O is more electronegative b. Electronegativity increases as one goes from left to right and top to bottom in the periodic table, which means F (EN = 4.0) is more electronegative than Si (EN = 2.1).

5.13

a. F and Cl, F is more electronegative b. C and N, N is more electronegative

5.14

a. N-F is more polar than N-C as F is more electronegative than C, which makes the bond more polar (ENC = 2.5, ENN = 3.0, ENF = 4.0). b. Si-Cl is more polar than Si-S because Cl is more electronegative than S, which makes the bond more polar (ENSi = 2.1, ENCl = 3.0, ENS = 2.5)

5.15

a. C-O or C-S, C-O is more polar b. Br-Cl or Br-F, Br-F is more polar

5.16

because N (EN = 3.0) is more electronegative than H (EN = 2.1)

because F (EN = 4.0) is more electronegative than Cl (EN = 3.0)

because F (EN = 4.0) is more electronegative than C (EN = 2.5)

5.17

a. b. c.

5.18

5.19

Chapter 5

5.20 a.

LiBr

Calculation 2.8 – 1.0 = 1.8

Classiﬁcation ionic

HCl

3.0 – 2.1 = 0.9

polar covalent

PH 3 (each H is bonded to P)

2.1 – 2.1 = 0.0

nonpolar covalent

SO 2 (each O is bonded to S)

3.5 – 2.5 = 1.0

polar covalent

CsF

4.0 – 0.7 = 3.3

ionic

Calculation

Classiﬁcation

5.21 a.

MgI 2 (each I is bonded to Mg)

2.5 – 1.2 = 1.4

polar covalent

NCl 3 (each Cl is bonded to N)

3.0 – 3.0 = 0.0

nonpolar covalent

H 2 S (each H is bonded to S)

2.5 – 2.1 = 0.4

polar covalent

d. e.

RbF SrO

4.0 – 0.8 = 3.2 3.5 – 1.0 = 2.5

ionic ionic

magnesium and chlorine

Calculation 3.0 – 1.2 = 1.8

Classiﬁcation polar covalent

5.22

5.23

5.24 a.

Chapter 5 b. c.

carbon and hydrogen phosphorus and hydrogen

2.5 – 2.1 = 0.4 2.1 – 2.1 = 0.0

nonpolar covalent nonpolar covalent

C and Br aluminum and chlorine nitrogen and oxygen

Calculation 2.8 – 2.5 = 0.3 3.0 – 1.5 = 1.5 3.5 – 3.0 = 0.5

Classification polar covalent polar covalent polar covalent

5.25 a. b. c.

5.26

5.27

5.28

a. SOCl2 (S atom in the middle) SOCl2 is polar

Chapter 5

b. NF3 (N atom in the middle) NF3 is polar

c. CH4 (C atom in the middle) CH4 is nonpolar

5.29

a. AlCl3 (Al atom in the middle) AlCl3 is nonpolar

b. OF2 (O atom in the middle) OF2 is polar

c. H2S (S atom in the middle) H2S is nonpolar

SECTION 5.3 INTERMOLECULAR FORCES 5.30

a. CF4 London dispersion forces b. CS2 London dispersion forces c. HCl Dipole-dipole

Chapter 5 5.31

a. CH3OH hydrogen bonding b. CH3CH2CH2CH3 London dispersion forces c. CH2Cl2 London dispersion forces

5.32

The alcohol has higher melting and boiling points than the ether. The forces that hold the alcohol molecules together must be stronger and harder to break than the forces that hold the ether molecules together.

5.33

CH4 is nonpolar because its symmetric tetrahedral shape cancels out the weak individual bond dipoles, while CH3Cl is polar due to the presence of a polar C-Cl bond that creates an imbalance in electron distribution in the tetrahedral molecular shape.

5.34

London dispersion forces < Dipole-dipole forces < Hydrogen bonding < Ion-dipole interactions.

5.35

Hydrogen bonding is the most important intermolecular force vitamin C experiences with water. In addition to hydrogen bonding, dipole-dipole interactions are present due to the polar functional groups as well as London dispersion forces.

5.36

A single water molecule can form a maximum of four hydrogen bonds. This is because each water molecule has two hydrogen atoms that can donate a hydrogen bond and two lone pairs of electrons that can accept a hydrogen bond.

5.37

In methylamine, the nitrogen atom has a lone pair of electrons available for hydrogen bonding. This lone pair can form a hydrogen bond with the hydrogen atom of another methylamine molecule. Since methylamine has two hydrogen atoms bonded to the nitrogen atom, it can potentially form two hydrogen bonds with neighboring methylamine molecules.

SECTION 5.4 ENERGY AND PROPERTIES OF MATTER 5.38

The molecules of a liquid possess kinetic and potential energy. The kinetic energy is not great enough to overcome the attractive forces between the molecules; therefore, the molecules are able to ﬂow together into the shape of the container, but the volume of the liquid remains constant.

Solid and liquids are composed of molecules with considerable attractive forces between the molecules. These attractive forces bring the molecules close together and cause them to be diﬃcult to compress further.

Gases are composed of molecules with high kinetic energy and low potential energy. The molecules are small compared to the amount of space occupied by the gas. The molecules strike each other and the walls of the container; however, all of the collisions are elastic and no net energy is lost from the system. Since the molecules are in constant motion, the same average number of gas molecules will strike the walls of the container at any given time and the elastic collisions result in uniform pressure on the walls of its container.

Chapter 5 5.39

The distance between gas molecules is much greater than between molecules of solids and liquids; hence, the mass of a volume of gas is much less than the mass of an equal volume of either a solid or liquid. Consequently, gases have low densities.

The diﬀerence between a solid and a liquid is that the molecules of a solid are organized, whereas the molecules of a liquid are random. The intermolecular distances are similar, and the amount of space occupied by a particular mass of the substance remains very much the same; hence, the densities of liquids and solids are very similar.

The average molecular speed increases as the temperature rises resulting in the tendency to overcome the intermolecular attractive forces; hence, the molecules tend to be further apart with an increase in temperature. Consequently, solids, liquids, and gases all expand when heated.

5.40

Liquids and gases are similar in their fluidity, ability to mix, and response to temperature and pressure changes. However, liquids have stronger intermolecular forces, a fixed volume, and are denser, while gases have weaker intermolecular forces, are more compressible, and have greater diffusion rates.

5.41

a. Phosgene is a gas that is shipped in a compressed steel cylinder. This is because gases have a low density and are easily compressed, making them ideal for shipping in cylinders. It is a colorless gas that is heavier than air. b. Benzocaine is a solid that has a high boiling point, low compressibility, and a density heavier than water. This means that it is difficult to turn into a gas, and it is not easily compressed.

5.42

Kinetic energy is the energy of motion. The particles in a solid have the lowest kinetic energy because the particles are moving the least in this phase of matter. The particles in a liquid have higher kinetic energy than a solid, as well as lower kinetic energy than a gas. The particles in a gas have the highest kinetic energy because the particles are moving the most in this phase of matter. The potential energy of the three phases of matter is easiest to compare during a phase transition. When a solid melts into a liquid, the temperature does not change, even though energy is added to the solid; therefore, both the liquid and the solid particles have the same average kinetic energy at the melting point and the liquid must have higher potential energy than the solid. Similarly, when a liquid evaporates into a gas, the temperature does not change, even though energy is added to the liquid; therefore, both the liquid and the gas particles have the same average kinetic energy at the melting point and the gas must have higher potential energy than the liquid.

SECTION 5.5 CHANGES OF STATE 5.43

a. b.

This state is characterized by the lowest density of the three. This state is characterized by an indeﬁnite shape and a high density. 13

gaseous liquid

Chapter 5

5.44

c. d.

In this state, disruptive forces prevail over cohesive forces. In this state, cohesive forces are most dominant.

a. b. c.

Temperature changes influence the volume of this state substantially. gaseous In this state, constituent particles are less free to move about than in other states. solid Pressure changes influence the volume of this state more than that of the other gaseous two states. This state is characterized by an indefinite shape and a low density. gaseous

(l) → (s); remove heat (s) → (g); add heat (l) → (g); add heat

gaseous solid

5.45

a. b. c.

Freezing Sublimation Vaporization

5.46

a. b. c.

Condensation Liquefaction Boiling

5.47

a. Frost appearing on a car window after a cold evening is an exothermic process. This is because the water vapor in the air condenses to form ice, releasing heat in the process. The heat is released because the molecules in the water vapor lose energy as they come closer together to form ice.

(g) → (l); remove heat (g) → (l); remove heat (l) → (g); add heat

exothermic endothermic endothermic exothermic exothermic endothermic

b. Coffee is freeze-dried before shipping is an endothermic process. This is because the water in the coffee is removed by sublimation, which requires energy. Sublimation is the process of going directly from a solid to a gas without going through the liquid state. The energy required for sublimation comes from the coffee, which cools down as the water is removed. 5.48

a. Rubbing alcohol on skin before an injection quickly disappears due to an endothermic process. Evaporation requires energy, and breaking intermolecular forces needs energy. The rubbing alcohol evaporates, cooling down the skin. b. Frost on food in the freezer is an exothermic process. Water vapor in the air condenses and turns into ice, releasing heat as molecules lose energy and come closer together.

5.49

a. A cold pack is used to reduce swelling on a muscle from an injury. This is an endothermic process because the cold pack absorbs heat from the surrounding tissue, which helps to reduce the swelling. b. A heat pack is used to relax a muscle cramp. This is an exothermic process. This is because the heat pack releases heat to the surrounding tissue, which helps to relax the muscle.

5.50

a. An ice cube left melts after being left out on the table. This is an endothermic process. This is because the ice cube absorbs heat from the surrounding environment, which melts it.

Chapter 5 b. Cellular metabolism breaks chemical bonds and generates heat. This is an exothermic process. This is because the breaking of chemical bonds releases energy, which is then used to power the cell's activities. 5.51

(lowest vapor pressure) heptane < hexane < pentane < butane (highest vapor pressure) All of the compounds are nonpolar liquid hydrocarbons; thus, they all experience the same type of intermolecular forces (dispersion forces) which increase as the molecular weight increases. Vapor pressure is low for substances with high intermolecular forces and vapor pressure is high for substances with low intermolecular forces. Therefore, the substance with the lowest vapor pressure will have the highest molecular weight and the substance with the lowest vapor pressure will have the lowest molecular weight.

5.52

Liquid water was heated to become steam (endothermic process), and then the steam was cooled to become liquid water again (exothermic process).

5.53

CH3Cl < CH2Cl2 < CCl4 < CH4

5.54

Methylene chloride is a volatile liquid. When it was sprayed in the mouth, the methylene chloride absorbed heat from the tissue and evaporated. The tissue became cold and was anesthetized.

5.55

Water has a higher vapor pressure than hydrogen sulﬁde. This is because water molecules have stronger intermolecular forces compared to hydrogen sulﬁde molecules, leading to a higher vapor pressure for water at the same temperature.

5.56

Water and ethylene glycol diﬀer in their boiling points. Water has a boiling point of 100°C, while ethylene glycol has a boiling point that is higher than 100°C. To determine the identity of the two boiling liquids, use a thermometer to measure the boiling point.

5.57

As pressure on a liquid increases, the boiling point also increases. The temperature inside an autoclave would be 121°C.

5.58

Water (H2O) has a lower vapor pressure than 1-propanol (CH3CHOHCH3). This is because the hydrogen bonding intermolecular forces in water are stronger than in 1-propanol, which makes it more difficult for the water molecules to escape from the liquid state and enter the gas state.

5.59

Octane (CH3CH2CH2CH2CH2CH2CH2CH3) will have a higher boiling point than pentane (CH3CH2CH2CH2CH3). This is because octane has a longer carbon chain than pentane, which means that the octane molecules have stronger London dispersion forces. Stronger intermolecular forces make it more difficult for the octane molecules to escape from the liquid state and enter the gas state, which means that octane has a higher boiling point.

5.60

C5H12 < C6H14 < C8H18 < C10H22 This is because the boiling point increases as the number of carbon atoms in the molecule increases. This is because the longer carbon chain results in stronger London dispersion forces, which leads to a higher boiling point.

Chapter 5 5.61

CF4 < CCl4 < CBr4 < CI4 The boiling point of substances is determined by intermolecular forces. CF4, CCl4, CBr4, and CI4 are all nonpolar, but CCl4, CBr4, and CI4 have more electrons and stronger London dispersion forces. CI4 has the most electrons and the strongest dispersion forces among the four compounds.

5.62

Mg(OH)2 will have a higher melting point than palmitic acid CH3(CH2)14COOH because Mg(OH)2 is an ionic compound, while palmitic acid is a molecular compound. Ionic compounds have stronger intermolecular forces than molecular compounds and therefore have higher melting points.

5.63

Calcium carbonate (CaCO3) will have the higher melting point than cetyl palmitate (C32H64O2) because CaCO3 is an ionic compound, while cetyl palmitate is a molecular compound. Ionic compounds have stronger intermolecular forces than molecular compounds, and therefore have higher melting points.

5.64

To obtain pure solid iodine from a mixture of solid iodine and sand, heat the mixture until the iodine sublimes and provide a cold surface above the mixture so the iodine can be deposited as a solid. The sand will not sublime and the solid on the cold surface will be pure iodine.

5.65

Dry ice (solid CO2) will sublime gaseous CO2 without leaving any liquid behind, keeping the samples dry. Water ice will melt to leave behind liquid water, which may destroy the samples.

5.66

When the ice and water mixture reached a constant temperature of 0.0°C, the system shared not only the same temperature throughout, but also the same vapor pressure. The vapor pressure of both the water and the ice is 4.58 torr.

SECTION 5.6 ENERGETICS: CHANGES OF STATE VS SPECIFIC HEAT 5.67

a. 1 kilocalorie is equal to 4.184 kJ. So, a protein bar that produces 280 kilocalories when burned produces 280 x 4.184 = 1171.5 kJ. b. Two apples that produce 144 kilocalories when burned produce 144 x 4.184 = 602.5 kJ.

5.68

a. An IV of dextrose that produces 170 kilocalories when burned produces 170 x 4.184 kJ/kcal = 711.3 kJ = 710 jK. b. One packet of electrolyte powder that produces 25 kilocalories when burned produces 25 x 4.184 = 104.6 kJ = 1.0 x 102 kJ.

5.69

Heat needed = 2.5 g × 80 cal/g = 200 cal

5.70

Heat needed = 66 g x 540 cal/g = 35640 cal = 36 kcal

Chapter 5 5.71

210. g of copper from 40°C to 95°C

heat = (210. g)(0.093

cal )(95 C − 40 C)  g C

= 1.1 x 10 3 cal b.

150. g of mercury from 120°C to 300°C

heat = (150. g)(0.033

cal )(300 C − 120 C) g C

= 8.9 x 10 2 cal c.

2.50 x 103 g of helium gas from 250°C to 900°C

heat = (2.50 x 10 3 g)(1.25

cal )(900 C − 250 C) g C

= 2.0 x 10 6 cal 5.72

50. g of aluminum from 25°C to 55°C

heat = (50. g)(0.24

cal )(55 C − 25 C)  g C

= 3.6 x 10 2 cal b.

2.50 x 103 g of ethylene glycol from 80°C to 85°C

heat = (2.50 x 10 3 g)(0.57

cal )(85 C − 80 C) g C

= 7 x 10 3 cal c.

500. g of steam from 110°C to 120°C

heat = (500. g)(0.48

cal )(120 C − 110 C) g C

= 2.4 x 10 3 cal

5.73

CaCl 2 6H 2 O: melting point = 30.2 C ‫ ﻠ‬heat of fusion = 40.7 cal/g

 1000 g   40.7 cal  heat = (1000. kg)    ‫ككككككككككككككككككك‬ g  1 kg    = 4.07 x 107 cal

b.‫ك‬LiNO 3 3H 2 O: melting point = 29.9 C, heat of fusion = 70.7 cal/g  1000 g   70.7 cal  heat = (1000. kg)    ‫ككككككككككككككككككك‬ g  1 kg    = 7.07 x 107 cal

Chapter 5

c. Na 2 SO 4 10H 2 O: melting point = 32.4 C, heat of fusion – 57.1 cal/g  1000 g   57.1 cal  heat = (1000. kg)    ‫ككككككككككككككككككك‬ g  1 kg    7 = 5.71 x 10 cal

5.74

In a solar heat storage system, solids are melted, and the energy is stored in the liquid form until the heat is released upon re-solidiﬁcation. The melting point of K2SO4 is 1069˚C which is much too high to be useful. The melting temperature must be low enough to easily maintain the compound in the liquid state for storage on cloudy days.

5.75

2550. cal + 40000. cal + 50000. cal + 270000. cal + 4800. cal = 367350. cal ⇒ 3.67 x 105 cal

5.76

 1000 g   38.6 cal  4 (2.00 kg)    = 7.72 x 10 cal 1 kg 1 g   

ADDITIONAL EXERCISES 5.77

As the sevoflurane is heated, the kinetic energy of the molecules increases, thus increasing the disruptive forces. At the boiling point, the disruptive forces overcome the cohesive forces, and the molecules become gaseous, escaping the liquid.

5.78

If one atom of oxygen reacted with two atoms of nitrogen to form a molecule, the formula of the molecule would be N 2 O . The electronegativity difference between nitrogen and oxygen is 0.5; therefore, the bond between nitrogen and oxygen is covalent. The name of N 2 O is dinitrogen monoxide.

5.79

Boiling water on the summit of Mount Everest might not make it safe to drink because the boiling point of water on the summit of Mount Everest is only 76.5°C. This might not be a high enough temperature to sterilize the water.

Chapter 5 5.80

The heat from the burner does not increase the temperature of the water-containing cup to the ignition temperature because the cup is the same temperature as the water. The energy from the burner must ﬁrst heat the water to the boiling point and completely boil the water to dryness before the cup reaches a temperature above the boiling point of water. If a sample of water were heated to the boiling point in a glass beaker using a single burner and then a second burner were added, the temperature of the boiling water would stay the same. The boiling point of water is not dependent on the amount of energy added to the water. The time required to reach the boiling point of water may decrease when using two burners rather than just one.

5.81

If four 250-mL bottles of water were placed in an ice chest ﬁlled with crushed ice, 2.1 x 102 g of ice would be needed to cool the bottles of water form 23°C to 5°C, as shown by the calculation below:

 250. mL   1 g   1 cal   4 bottles  − 5 C)     (23 C =   1 bottle  1 mL   g ⋅ C 

 1 cal    80. cal   (x)   + (x)    (5 C − 0. C) g g C ⋅    

 85. cal    g 

‫كككككككككككككككككككككككككككككككككككككككككككككككككككك‬18000. cal = (x) 

‫ككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككككك‬ ‫ك‬ ‫كككك‬x = 211.76 g ⇒ 2.1 x 10 2 g ice 5.82

Lead has the smallest speciﬁc heat value, which indicates a greater change in temperature when equal amounts of energy are added to metals at the same mass.

5.83

If 43.4 kJ of heat is released when all ice melts at 0°C, 43.4 kJ = (mass of water ) x 334 J/g is rearranged to

5.84

The heat required = 0.0688 kg x 3.32 kJ/kg•˚C x (183˚C-175˚C) = 1.83 kJ

5.85

The heat released from 1 mg of protein can be calculated by the following:

Chapter 6

Chapter 6: Gases and Solutions CHAPTER OUTLINE 6.1 Properties of Gases

6.5 Solutions and Solubility

6.2 Pressure, Temperature and Volume Relationships

6.6 Electrolytes and Net Ionic Equations

6.3 The Ideal Gas Law

6.8 Solution Preparation

6.4 Dalton’s Law

6.9 Osmosis and Dialysis

6.7 Solution Concentrations

LEARNING OBJECTIVES/ASSESSMENT When you have completed your study of this chapter, you should be able to: 1.

Calculate the kinetic energy of moving particles. (Section 6.1; Exercise 6.2)

Convert pressure and temperature values into various units. (Section 6.1; Exercise 6.5)

Do calculations based on Boyle’s law, Charles’s law, and the combined gas law. (Section 6.2; Exercises 6.11, 6.13, and 6.15)

Do calculations based on the ideal gas law. (Section 6.3; Exercises 6.31, 6.33, and 6.35)

Do calculations based on Dalton’s law. (Section 6.4; Exercises 6.44 and 6.45)

Apply your knowledge of intermolecular forces to determine the solubility of substances in a given solvent. (Section 6.5; Exercises 6.49)

Write molecular equations in total ionic and net ionic forms. (Section 6.6; Exercises 6.61 and 6.62)

Calculate solution concentrations in units of molarity, weight/weight percent, weight/volume percent, and volume/volume percent. (Section 6.7; Exercises 6.65, 6.73, and 6.79)

Describe how to prepare solutions of speciﬁc concentration by diluting concentrated solutions. (Section 6.8; Exercises 6.91 and 6.93)

10. Describe the process of osmosis. (Section 6.9; Exercise 6.95) 11. Describe the role of osmosis and diﬀusion in dialysis. (Section 6.9; Exercises 6.98)

LECTURE HINTS AND SUGGESTIONS 1.

The individual gas laws are important from a historical point of view, however, the ideal gas law PV=nRT can be used to solve most problems, thus your emphasis should be on this equation more than the individual laws. Moreover the major focus on the gas laws should be relating the kinetic molecular theory to the physical behavior of gases (how the laws relate to common and everyday phenomenon)

Combining Table 6.2 and Study Tools 6.1 together on a one-page handout (see next page) creates a

Treat Avogadro's volume, 22.4 L/mol at STP, as another conversion factor that is always available for

useful resource for students to reference when solving gas law problems. use when using dimensional analysis to solve problems. 4.

The solubility of sugar, cooking oil and rubbing alcohol in water can be demonstrated. The diﬀerence between the terms soluble, insoluble and miscible should be illustrated. Explain the general statement "like dissolves like" and explain in terms of chemical structure and the attractive forces between molecules. 1

Chapter 6

Units of Pressure and Gas Laws

Chapter 6

Solutions to All End-of-Chapter Questions What follows are more complete explanations/full solutions to the EOC exercises whose answers are published in shorter form at the end of the textbook.

SECTION 6.1 PROPERTIES OF GASES 6.1

The kinetic energy of the halfback is greater than the kinetic energy of the tackle as shown in the calculations below. The tackle will be pushed back because he has less kinetic energy.

KE =

1 mv 2 2 2

 1 m 3 KE halfback = ( 81.8 kg )  8.0  = 2.6 x 10 2 s   2

1  m 2 KE tackle = ( 118.2 kg )  3.0  = 5.3 x 10 2 s  

6.2

N 2 O kinetic energy = 3.14 x 10 KE =

kg  m 2 s2

g cm 2 10 s2

1 44 u  4 cm  mv 2 =  3.78 x 10  2 2  s 

Sevoflurane kinetic energy = 3.13 x 10 KE =

kg  m 2

g cm 2 10 s2

1 200.1 u 44 u  4 cm  mv 2 =  1.77 x 10  2 2 s  

6.3

A gas law is a mathematical relationship that describes the behavior of gases as they are mixed, as they are subjected to pressure or temperature changes, or allowed to diﬀuse.

6.4

atm

 1atm  28.6 in. Hg   = 0.957 atm  29.9 in. Hg 

torr

 1atm   760 torr  28.6 in. Hg    = 727 torr 29.9 in. Hg    1 atm 

psi

 1atm   14.7 psi  28.6 in. Hg    = 14.1 psi  29.9 in. Hg   1 atm 

bars

 1.01bars  28.6 in. Hg   = 0.966 bars  29.9 in. Hg 

Chapter 6

6.5

6.6

6.7

atm

 1atm  210. psi   = 14.3 atm  14.7 psi 

bars

 1atm   1.01 bars  210. psi    = 14.4 bars  14.7 psi   1 atm 

mm Hg

 1atm  760 mm Hg  4 210. psi    = 1.09 x 10 mm Hg   14.7 psi   1 atm

in. Hg

 1atm  29.9 in. Hg  210. psi    = 427 in. Hg  14.7 psi   1 atm 

The kinetic energy of the hydrogen will be greater than the kinetic energy of the helium as shown in the calculations below.

1 mv 2 2 1 KEH2 = (2.02 u)(2v)2 = 4.04 v 2 u 2 1 KEHe = (4.00 u)(v)2 = 2.00 v 2 u 2 KE =

6.8

6.9

K = 273 +  C = 273 + (-268.9  C) = 4.1 K



K = 273 +  C - 273 + 63.7  C = 336.7 K

The melting point of gold, 1337.4K, to degrees in Celsius

1337.4 K – 273 = 1064°C

The melting point of tungsten, 3410°C to kelvins.

3410°C + 273 = 3683 K

The melting point of tin, 505 K, to degrees Celsius.

505 K -273 = 232°C

C = K - 273 = 14.1 K - 273 = 258.9  C

Chapter 6

SECTION 6.2 PRESSURE, TEMPERATURE, AND VOLUME RELATIONSHIPS 6.10

Gas A:

(1.50 atm)(2.00 L) (Pf )(3.00 L) = ; 300. K 450. K

Gas B:

(2.35 atm)(1.97 L) (1.09 atm)(Pf ) = 293. K 310. K

Pf = 1.50 atm Gas C:

Vf = 4.49 L

(9.86 atm)(11.7 L) (5.14 atm)(9.90 L) = 500. K Tf

Tf = 221 K

6.11

(690. torr)(200. mL) (760 torr)(Vf ) ⇒ Vf = 166 mL = 26.0 C + 273 0.0 C + 273

6.12

(610. torr)(200. mL) (760 torr)(Vf ) = ⇒ Vf = 138 mL 45.0 C + 273 0.0 C + 273

6.13

(Pf )(0.50 L) (1.00 atm)(3.00 L) = ⇒ Pf = 7.1 atm  0.0 C + 273 30.0 C + 273

6.14

(Pf )(0.75 L) (1.00 atm)(2.50 L) = ⇒ Pf = 3.70 atm  0.0 C + 273 30.0 C + 273

6.15

(14.7 psi) (1.00 atm) = Vi = 1 atm

(65.0 psi)(1.00 L) ⇒ Vi = 4.42 L

6.16

(14.7 psi) (1.00 atm) = Vi = 1 atm

(32.0 psi)(14.5 L) ⇒ Vi = 31.6 L

6.17

Constant Temp →

V2 =

6.18

760 torr (1000. mL) 620 torr

Constant Temp →

V2 =

P1 V1

P1 V1 T1

2000 psi (5.0 L) 14.0 psi

P2 V2 T2

→ 760 torr (1000. mL) = 620 torr (V2 )

= 1.23 x 10 3 mL

P2 V2 T2

→ 2000 psi (5.0 L) = 14.0 psi (V2 )

= 714 L

Chapter 6

6.19

Constant Temp →

T1 = 4 C = 277 K 

P1 V1 T1

P2 V2 T2

T2 = 37 C = 310 K 

300 mL x mL 300 mL (310 K) = → = 336 mL 277 K 310 K 277 K

6.20

Vf 3.8 L = ⇒ Vf 4.6 L   20. C + 273 85 C + 273

6.21

If the gas pressure remains constant, the equation

P1 V1 P2 V2 V V = reduces to 1 = 2 . T1 T2 T1 T2

Remember the T must be in Kelvins. K = °C = 273 K = 16°C + 273 = 289 K for T1 K = 37°C + 273 = 310 K for T2

V1 = 1.00 L; V2 = ?

V1 V2 V2 100 = = = T1 T2 289 K 310 K Cross multiplying gives (1.00 L)(310 K) = (289 K)(V2) Dividing both sides by 289 gives V2 = 1.07 L

6.22

V1 2.0 L = ⇒ V1 = 2.5 L  120. C + 273 40. C + 273

6.23

P1 V1 P2 V2 = T1 T2

T1 = 20 C = 293 K

T2 = 37  C = 310 K

2500 torr (180 mL) 94 torr (x mL) 2500 torr (180 mL)(310 K) = 506 mL = → 293 K 310 K 940 torr (293) K

6.24

P1 V1 P2 V2 = T1 T2

T1 = 37  C = 310 K

T2 = 22 C = 295 K

210 K Pa (x mL) 94.6 K Pa (7.64 mL) = 310 K 295 K 94.6 K Pa (7.64 mL)(310 K) = 3.62 mL 210.0 K Pa (295 K)

Chapter 6

6.25

P1 V1 P2 V2 = T1 T2

T1 = 0.0 C = 283 K

T2 = 37  C = 310 K

3.25 atm (6.25 L) 250 atm (x L) = 283 K 310 K 3.25 atm (6.25 L)(310K) = 89.0 L 0.250 atm + (283 K)

 1 atm  (400. torr)  (V ) 760 torr  f = ⇒ Vf 1.4 x 10 4 ft 3 5 C + 273

6.26

(0.98 atm) (8000. ft 3 ) = 23 C + 273

6.27

(1.50 L) 2(1.50 L) Tf 333 C = ⇒ =  30 C + 273 Tf + 273

6.28

P1 V1 P2 V2 at constant temperature gives P1 V1 = P2 V2 = T1 T2

1.00 atm (4.00 L) = 3.00 atm (x L) 1.00 atm (4.00 L) = 1.33 L 3.00 atm 6.29

 1000 mL  3 (Pi )(250. mL) = (700. torr)(2.00 L)   ⇒ Pi = 5.60 x 10 torr 1 L  

6.30

 760 torr  (760 torr)(2.00 L) = (4.00 atm)   (Vf ) ⇒ Vf = 0.500 L  1 atm 

density =

g 2.50 g mass ⇒ = 5.00 L volume 0.500 L

SECTION 6.3 THE IDEAL GAS LAW 6.31

 1L  L ⋅ atm PV = nRT ⇒ P ⇒ (400. mL)  )((20.0 + 273) K)  = (2.00 mol) (0.0821 ⋅K 1000 mL mol   L ⋅ atm (2.00 mol)(0.0821 )((20.0 + 273) K) ⋅K mol = 120. atm V= (400. mL)

PV = nRT ⇒ (3.00 atm) V= (0.525 mol)(0.0821 V=

L ⋅ atm )((20.0 + 273) K) mol ⋅ K

L ⋅ atm )((15.0 + 273) K) mol ⋅ K = 4.14 L (3.00 atm)

(0.525 mol)(0.0821

Chapter 6

L ⋅ atm  1 atm  PV = nRT ⇒ (300. torr)  )T  (2.25 L) = (0.100 mol)(0.0821 ⋅K 760 torr mol    1 atm  (300. torr)  (2.25 L) torr   760 = T= 108 K - 273 = − 165 C L ⋅ atm (0.100 mol)(0.0821 ) mol ⋅ K

6.32

6.33

PV = nRT ⇒ P(0.750 L) = (0.156 mol)(0.0821 P=

L ⋅ atm )((27 + 273)K) mol ⋅ K

L ⋅ atm )((27 + 273)K) mol ⋅ K = 5.12 atm (0.750 L)

(0.156 mol) (0.0821

6.34

6.35

PV = nRT ⇒ P(0.890 atm) V = (8.75 g)( (8.75 g)( V=

1 mole L ⋅ atm )(0.0821 )((35.0 + 273)K) 32.00 g O 2 mol ⋅ K

1 mole L ⋅ atm )(0.0821 )((35.0 + 273)K) 32.00 g O 2 mol ⋅ K = 7.77 L (0.890 atm)

Chapter 6 6.36

6.37

PV = nRT ⇒ P(400. mL)( (12.0 g)( P=

1 mole 1 mole L ⋅ atm ) = (12.0 g)( )(0.0821 )((35.0 + 273)K) 1000 mL 44.01 g CO 2 mol ⋅ K

1 mole L ⋅ atm )(0.0821 )((35.0 + 273)K) 44.0 g CO 2 mol ⋅ K = 17.2 atm 1 mole (400. mL)( ) 1000 mL

6.38

6.39

L ⋅ atm )(273K) mol ⋅ K  6.02 x 10 23 molecules  0.446 moles  =  1 mole  

PV = nRT ⇒ (1 atm)(10.0 L) = n(0.0821

(1 atm)(10.0 L) n= = L ⋅ atm (0.0821 )(273K) mol ⋅ K

mRT 6.40 = MW = P/v

6.41

L atm 29.7 g (0.0821 278 K) mol K = 1 atm (3.96 L)

168 g/mol

Kevin equation 6.11 will be helpful.

PV =

mRT mw

Substitution into the key equation 6.11 gives:

(1.00 atm)(1.00 L) =

L atm )(308 K) mol K 28.6 g/mol

m (0.0821

2.69 x 10 23 molecules

Chapter 6 Multiplying both sides of the equation by 28.6 g/mol gives: L atm (28.6 g/mol)(1.00 atm)(1.00 L) = m (0.0821 308 K) mol K Dividing by (0.0821

L atm mol K

) (308 K) isolates m

g    28.6  (1.00 atm)(1.00 L) mol   m= = 1.13 g L atm    0.0821 mol K  (308 K)  

6.42

The gas is CO2 (MW = 44.0 g/mole) according to the calculation below: PV = nRT

6.43

The gas is Ar (MW = 39.9 g/mole) according to the calculation below: PV = nRT

SECTION 6.4 DALTON’S LAW 6.44

PT = PO2 + PHe 2100. torr = PO2 + 1680. torr PO2 = 420. torr

part 1680. torr x 100% He = x 100 whole 2100. torr

%O 2 =

6.45

420. torr 80% He x 100 = 2100. torr 20% O 2

effusion rate A = effusion rate B

molecular mass B molecular mass A

PO2 + PH2 O = PT ⇒ PO2 + 22 torr = 720 torr ⇒ PO2 = 698 torr

Chapter 6

SECTION 6.5 SOLUTIONS AND SOLUBILITY 6.46 a.

Solvent

Solutes

water

sodium hypochlorite

isopropyl alcohol

water

hydrogen peroxide

Liquid laundry bleach: sodium hypochlorite 5.25%, inert ingredient 94.75%

Rubbing alcohol: isopropyl alcohol 70%

Hydrogen peroxide: 3% hydrogen peroxide

After-shave: SD alcohol,

SD alcohol

water, glycerin, fragrance,

water, glycerin, fragrance, menthol, benzophenone-1,

menthol, benzophenone-1,

coloring

6.47

Solvent a.

Antiseptic mouthwash: alcohol 25%,

water

Solutes alcohol, thymol, eucalyptol,

thymol, eucalyptol, methyl, salicylate,

methyl salicylate, menthol,

benzoic acid, boric acid

benzoic acid, boric acid,

Paregoric: alcohol 45%, opium 0.4%

Baby oil: mineral oil, lanolin

Distilled vinegar: acetic acid 5%

water

alcohol, opium

mineral oil*

lanolin

water

acetic acid

*Mineral oil is nonpolar; therefore, polar water will not mix with mineral oil.

6.48

25 mL of cooking oil and 25 mL of vinegar—the resulting mixture is cloudy and gradually separates into two layers.

25 mL of water and 10 mL of rubbing alcohol—the resulting mixture is clear and colorless.

Soluble

25 mL of chloroform and 1 g of rooﬁng tar—the resulting mixture is clear but dark brown in color.

6.49

Immiscible

Soluble

25 mL of water and 1 g of salt—the resulting mixture is clear and colorless.

Soluble

Chapter 6

25 mL of water and 1 g of solid silver chloride—the resulting mixture is cloudy and solid settles out.

Insoluble

25 mL of water and 5 mL of mineral oil—the resulting mixture is cloudy and gradually separates into two layers.

6.50

A solution to which a small piece of solute is added, and it dissolves.

b. c. 6.51

Immiscible

Unsaturated

A solution to which a small piece of solute is added, and much more solute comes out of solution.

Supersaturated

The ﬁnal solution resulting from the process in part b

Saturated

Unsaturated. At 0°C, 100 g of water is saturated with 70.6 g of ammonium sulfate. 35.8 g is less than 70.6 g, so the water is unsaturated.

6.52

Miscible means that two liquids can be mixed in any concentration; there are no solubility limits.

6.53

The hydrated ion is surrounded by water molecules. For the Cl- ion, the hydrogen atoms in the water molecule will be oriented toward the Cl- ion. A nonhydrated ion is not surrounded by water molecules.

6.54

This solution could become supersaturated by slowly lowering the temperature of the solution or by allowing some of the solvent to evaporate. The solution must be handled very gently as a supersaturated solution is unstable.

6.55

soluble in water b.

HCl soluble in water

c. soluble in water

N≡N

soluble in benzene

Chapter 6 6.56

soluble in benzene

Ne soluble in benzene

c. soluble in water d.

soluble in benzene 6.57

A solid would dissolve faster (1) at a higher temperature; (2) if the solid were crushed into smaller particles; or (3) is the solution were stirred.

6.58

Insoluble. The structure indicates that halothane is a nonpolar material. Therefore, halothane would not be soluble in a polar solvent, such as water.

SECTION 6.6 ELECTROLYTES AND NET IONIC EQUATIONS 6.59

6.60

6.61

Na 2 SO 4

2Na + , SO 4 2 −

NaOH

Na + , OH −

CaCl 2

Ca 2 + , 2Cl −

KNO 3

K + , NO 3 −

(NH 4 )3 PO 4

3NH 4 + , PO 4 3 −

NaMnO 4

Na + , MnO 4 −

LiNO 3

Li + , NO −3

KOH

+ − K OH Mg, 2+ , 2Br-

Na 2 HPO 4

2 Na + , HPO 4 2 −

MgBr2

(NH 4 )2 SO 4

Ca(ClO 3 )2

H 2 O (l) + Na 2 SO 3 (aq) + SO 2 (aq) → + 2 NaHSO 3 (aq)

Ca , 2 ClO 3

−

2NH4+, SO42-

Chapter 6

6.62

Cu (s) + 8 HNO 3 (aq) → + 3 Cu(NO 3 )2 (aq) + 2 NO (g) + 4 H 2 O (l)

2 HCl (aq) + CaO (s) → CaCl 2 (aq) + H 2 O (l)

CaCO 3 (s) + 2 HCl (aq) → CaCl 2 (aq) + CO 2 (aq) + H 2 O (l)

MnO 2 (s) + 4 HCl (aq) → MnCl 2 (aq) + Cl 2 (aq) + 2 H 2 O (l)

2 AgNO 3 (aq) + Cu (s) → Cu(NO 3 )2 (aq) + Ag (s)

SO 2 (aq) + H 2 O (l) → H 2 SO 3 (aq)

CuSO 4 (aq) + Zn (s) → ZnSO 4 (aq)

Chapter 6

2 Kbr (aq) + 2 H 2 SO 4 (aq) → Br2 (aq) + SO 2 (aq) + K 2 SO 4 (aq) + 2H 2 O (l)

AgNO 3 (aq) + NaOH (aq) → AgOH (s) + NaNO 3 (aq)

BaCO 3 (s) + 2 HNO 3 (aq) → Ba(NO 3 )2 (aq) + CO 2 (g) + H 2 O (l)

N 2 O 5 (aq) + H 2 O (l) → + 2 HNO 3 (aq)

SECTION 6.7 SOLUTION CONCENTRATIONS 6.63

2.50 L of solution that contains 0.860 mol of solute.

0.860 mol = 0.344 M 2.50 L 400. mL of solution that contains 0.304 mol of solute.

 1L  400. mL   = 0.400 L  1000 mL  M=

moles 0.304 mol 0.760 M ⇒ = L 0.400 L or

0.304 mol = 0.760 M  1L  400. mL    1000 mL 

0.210 mol of solute is put into a container and enough distilled water is added to give 250. mL of solution.

Chapter 6

 1L  250. mL   = 0.250 L  1000 mL  M=

moles 0.120 mol ⇒ = 0.480 M L 0.250 L or

0.120 mol = 0.480 M  1L  250. mL    1000 mL 

6.64

1.25 L of solution contains 0.455 mol of solute.

0.445 mol = 0.364 M 1.25 L 250. mL of solution contains 0.215 mol of solute.  1L  250. mL   = 0.250 L  1000 mL 

moles 0.215 mol ⇒ = 0.860 M L 0.250 L or 0.215 mol = 0.860 M  1L  250 mL    1000 mL 

0.175 mol of solute is put into a container and enough distilled water is added to give 100. mL of solution.  1L  100. mL   = 0.100 L  1000 mL 

moles 0.175 mol ⇒ = 1.75 M L 0.100 L or 0.175 mol = 1.75 M  1L  100. mL    1000 mL 

6.65

A sample of solid NaSO 4 is weighing 0.140 g is dissolved in enough water to make 10.0 mL of solution.

Chapter 6

A 4.50 g sample of glucose ( C6 H12 O6 ) is dissolved in enough water to give 150. mL of solution.

A 43.5 g sample of K 2 SO 4 is dissolved in a quantity of water, and the solution is stirred well. A 25.0 mL sample of the resulting solution is evaporated to dryness and leaves behind a 2.18 g of solid K 2 SO 4 .

6.66

A sample of solid KBr weighing 11.9 g is put in enough distilled water to give 200. mL of solution.

 1 mole  11.9 g   = 0.100 mol  119.00 g   1L  200. mL   = 0.200 L  1000 mL 

moles 0.100 mol ⇒ = 0.500 M L 0.200 L or

 1 mole  11.9 g    119.00 g  = 0.500 M  1L  200. mL    1000 mL  b.

A 14.2 g sample of solid Na 2 SO 4 is dissolved in enough water to give 500. mL of solution.  1 mole  14.2 g   = 0.0100 M  142.00 g 

Chapter 6  1L  500. mL   = 0.500 L  1000 mL 

moles 0.100 mol ⇒ = 0.200 M L 0.500 L or

 1 mole  14.2 g    142.00 g  = 0.200 M  1L  500. mL    1000 mL  c.

A 10.0 mL sample of solution is evaporated to dryness and leaves 0.29 g of solid residue that is identiﬁed as Li 2 SO 4 .

 1 mole  0.29 g   = 0.0026 moles  109.9 g   1L  10.0 mL   = 0.0100 L  1000 mL 

moles 0.0026 mol ⇒ = 0.26 M L 0.100 L or

 1 mole  0.29 g    109.95 g  = 0.26 M  1L  10.0 mL    1000 mL  6.67

6.68

278 mmol 1 mol x = 0.348 mol dextrose 1L 1000 mmol

1.25 L x

200. mL x

0.500 mol x

How many moles of solute are contained in 1.75 L of 0.215 M solution?

1L 154 mmol 1 mol x x = 0.0308 mol NaCl 1000 mL 1L 1000 mmol 1000 mmol 1L x = 3.25 L NaCl solution 1 mol 154 mmol

 0.215 moles  1.75 L   = 0.376 moles 1L  

How many moles of solute are contained in 250. mL of 0.300 M solution? 1L    0.300 moles  250. mL    = 0.0750 moles 1000 mL 1L   

How many mL of 0.350 M solution contains 0.200 mol of solute?

Chapter 6

   1000 mL  1L 0.200 mol   = 571 mL   0.350 moles   1 L 

6.69

6.70

How many grams of solid would be left behind if 20.0 mL of 0.550 M KCl solution was evaporated to dryness?

   0.550 moles   74.6 g  1L 20.0 mL   = 0.821 g   1L  1000 mL     1 mole  b.

How many liters of 0.315 M HNO 3 solution is needed to provide 0.0410 mol of HNO 3 ?   1L 0.0410 mol   = 0.130 L 0.315 moles  

How many mL of 1.21 M NH 4 NO 3 solution contains 50.0g of solute?

 1 mole     1000 mL  1L 50.0 g     = 517 mL 80.0 g 1.21 moles 1L    

6.71

How many grams of solid AgNO 3 will be needed to prepare 200. mL of a 0.200 M solution?

 1L  0.200 moles   170. g  200. mL   = 6.80 g   1L  1000 mL    1 mole  b.

How many grams of vitamin C ( C6 H 5 O6 ) would be contained in 25.0 mL of a 1.00 M solution?

 1L  1.00 moles   176. g  25.0 mL   = 4.40 g   1L  1000 mL   1 mole  c.

How many moles of HCl are contained in 250. mL of a 6.0 M solution?

 1L  6.0 moles  250. mL    = 1.5 moles 1L  1000 mL  

Chapter 6 6.72

5.3 g of sugar in 100. mL of water

 1.00 g  100. mL   = 100. g water  1 mL  solution mass = 5.3 g sugar + 100. g water = 105.3 g (without SF) %(w/w)=

5.3 g solute mass x 100 ⇒ x 100 = 5.0%(w/w) solution mass 105.3 g

or 5.3 g sugar  1.00 g  5.3 g sugar + 100. mL    1 mL 

x 100 = 5.0%(w/w)

5.3 g of any solute and 100. mL of water

 1.00 g  100. mL   = 100. g water  1 mL  solution mass = 5.3 g solute + 100. g water = 105.3 g (without SF) %(w/w)=

5.3 g solute mass x 100 = 5.0%(w/w) x 100 ⇒ solution mass 105.3 g

or 5.3 g solute  1.00 g  5.3 g solute + 100. mL    1 mL 

x 100 = 5.0%(w/w)

5.3 g of any solute and 100. g of any solvent. 5.3 g solute x 100 = 5.0%(w/w) 5.3 g solute + 100. g solvent

6.73

6.5 g of table salt and 100. mL of water

 1.00 g  100. mL   = 100. g water  1 mL  solution mass = 6.5 g table salt + 100. g water = 106.5 g (without SF)

%(w/w)=

6.5 g solute mass x 100 ⇒ x 100 = 6.1%(w/w) solution mass 106.5 g or

6.5 g table salt  1.00 g  6.5 g table salt + 100. mL    1 mL 

x 100 = 6.1%(w/w)

Chapter 6

6.5 g of any solute and 100. mL of water/

 1.00 g  100. mL   = 100. g water  1 mL  solution mass = 6.5 g solute + 100. g water = 106.5 g (without SF)

%(w/w)=

6.5 g solute mass x 100 ⇒ x 100 = 6.1%(w/w) solution mass 106.5 g or 6.5 g solute

 1.00 g  6.5 g solute + 100. mL    1 mL 

x 100 = 6.1%(w/w)

6.5 g of any solute and 100. g of any solvent

6.5 g solute x 100 = 6.1%(w/w) 6.5 g solute + 100. g solvent

6.74

20.0 g of salt is dissolved in 250. mL of water

 1.00 g  250. mL   = 250. g water  1 mL  solution mass = 20.0 g salt + 250. g water = 270.0 g (without SF) %(w/w)=

20.0 g solute mass x 100 ⇒ x 100 = 7.41%(w/w) solution mass 270.0 g

or 20.0 g salt  1.00 g  20.0 g salt + 250. mL    1 mL 

x 100 = 7.41%(w/w)

0.100 mol of solid glucose ( C6 H12 O6 ) is dissolved in 100. mL of water

 53.49 g  mass solute = 0.100 mol   = 18.0 g  1 mole   1.00 g  mass of water = 100. mL   = 100. g  1 mL  mass of solution = 18.0 g + 100. g = 118.0 g %(w/w)=

18.0 g mass of solute x 100 = 15.3%(w/w) x 100 ⇒ mass of solution 118.0 g

Chapter 6

 180. g  0.100 mol    1 mole  x 100 = 15.3%(w/w)  1.00 g   180. g  0.100 mol   + 100. mL water    1 mole   1 mL  c.

120 g of solid is dissolved in 100. mL of water

 1.00 g  mass of water = 100. mL   = 100. g  1 mL  mass of solution = 120.0 g + 100. G = 220.0 g %(w/w)=

120.0 g mass of solute x 100 ⇒ x 100 = 54.5%(w/w) mass of solution 220.0 g

or 120.0 g solute  1.00 g  120.0 g solute + 100. mL water    1 mL 

x 100 = 54.5%(w/w)

10.0 mL of ethyl alcohol (density = 0.789 g/mL is mixed with 10.0 mL of water

 0.789 g  mass of ethyl alcohol = 0.00 mL ethyl alcohol   = 7.89 g  1 mL   1.00 g  mass of water = 100. mL   = 100. g  1 mL 

mass of solution = 7.89 g ethyl alcohol + 10.0 g water = 17.89 g (without SF) %(w/w)=

7.89 g mass of solute x 100 = 44.1%(w/w) x 100 ⇒ mass of solution 17.89 g

 7.89 g  10.0 mL ethyl alcohol    1 mL  x 100 = 44.1%(w/w)  7.89 g   1.00 g  10.0 mL ethyl alcohol  + 10.0 mL water     1 mL   1 mL 

6.75

5.20 g of CaCl 2 is dissolved in 125 mL of water.

 1.00 g  125 mL water   = 125 g  1 mL  solution mass = 5.20 g CaCl 2 + 125 g water = 130.20 g (without SF)

%(w/w)=

5.20 g solute mass x 100 ⇒ x 100 = 3.99%(w/w) solution mass 130.20 g or

Chapter 6 5.20 g CaCl 2  1.00 g  5.20 g CaCl 2 + 125 mL water    1 mL 

x 100 = 3.99%(w/w)

0.200 mol of solid KBr is dissolved in 200. mL of water

 119 g  mass solute = 0.200 mol   = 23.8 g  1 mole   1.00 g  mass of water = 200. mL   = 200. g  1 mL  mass of solution = 23.8 g + 100. g = 223.8 g

%(w/w)=

23.8 g mass of solute x 100 = 10.6%(w/w) x 100 ⇒ mass of solution 223.8 g or

 119 g  0.200 mol    1 mole  x 100 = 10.6%(w/w)  119 g   1.00 g  0.200 mol   + 200. mL water    1 mole   1 mL 

50.0 g of solid is dissolved in 250. mL of water

 1.00 g  mass of water = 250. mL   = 250. g  1 mL  mass of solution = 50.0 g + 250. g = 300. g

%(w/w)=

50.0 g mass of solute x 100 ⇒ x 100 = 16.7%(w/w) mass of solution 300. g or 50.0 g solute

 1.00 g  50.0 g solute + 250. mL water    1 mL 

x 100 = 16.7%(w/w)

10.0 mL of ethyl alcohol (density = 0.789 g/mL is mixed with 10.0 mL of ethylene glycol (density 1.11 g/mL)

 0.789 g  mass of ethyl alcohol = 10.0 mL   = 7.89 g  1 mL   1.11 g  mass of ethylene glycol = 10.0 mL ethylene glycol   = 11.1 g  1 mL  mass of solution = 7.89 g ethyl alcohol + 11.1 g ethylene glycol = 18.99 g (without SF)

%(w/w)=

7.89 g mass of solute x 100 ⇒ x 100 = 41.5%(w/w) mass of solution 18.99 g or

Chapter 6

 0.789 g  10.0 mL ethyl alcohol    1 mL  x 100 = 41.5%(w/w)  0.789 g   1.11 g  10.0 mL ethyl alcohol   + 10.0 mL ethylene glycol    1 mL   1 mL 

6.76

20.0 g of solute is dissolved in enough water to give 150. mL of solution. The density of the resulting solution is 1.20 g/mL.

 1.20 g  mass of solution = 150.0 mL solution   = 180. g  1 mL  20.0 g mass of solute %(w/w)= x 100 ⇒ x 100 = 11.1%(w/w) mass of solution 180. g or 20.0 g solute  1.20 g  150. mL solution    1 mL 

x 100 = 11.1%(w/w)

A 10.0 mL solution sample with a density of 1.10 g/mL leaves 1.18 g of solid residue when evaporated.  1.10 g  mass of solution = 10.0 mL solution   = 11.0 g  1 mL  1.18 g mass of solute %(w/w)= x 100 ⇒ x 100 = 10.7%(w/w) mass of solution 11.0 g

or 1.18 g solute  1.10 g  10.0 mL solution    1 mL 

x 100 = 7.48%(w/w)

A 25.0 g sample of solution on evaporation leaves 1.87 g residue of MgCl 2 . 1.87 g MgCl 2 x 100 = 7.48%(w/w) 25.0 g solution

6.77

424 g of solute is dissolved in enough water to give 1.00 L of solution. The density of the resulting solution is 1180. g/mL.

 1 mL   1.18 g  mass of solution = 1.00 mL solution  -3    = 1180. g  10 L   1 mL  %(w/w)=

424 g mass of solute x 100 ⇒ x 100 = 35.9%(w/w) mass of solution 1180 g or

Chapter 6 424 g solute  1 mL   1.18 g  1.00 L solution  -3     10 L   1 mL 

x 100 = 35.9%(w/w)

A 50.0 mL solution sample with a density of 0.898 g/mL leaves 12.6 g of solid residue when evaporated.

 0.898 g  mass of solution = 50.0 mL solution   = 44.9 g  1 mL 

%(w/w)=

12.6 g mass of solute x 100 ⇒ x 100 = 28.1%(w/w) mass of solution 44.9 g or

12.6 g solute  0.898 g  50.0 mL solution    1 mL 

x 100 = 28.1%(w/w)

A 25.0 g sample of solution on evaporation leaves a 2.32 g residue of NH 4 Cl . 2.32 g NH 4 Cl x 100 = 9.28%(w/w) 25.0 g solution

6.78

6.79

200. mL of solution contains 15 mL of alcohol. 15.0 mL alcohol x 100 = 7.5%(v/v) 200. mL solution

200. mL of solution contains 15 mL of any soluble liquid solute. 15.0 mL alcohol x 100 = 7.5%(v/v) 200. mL solution

8.0 ﬂuid ounces of oil is added to 2.0 gallons (256 ﬂuid ounces) of gasoline. 8.0 fl. oz. oil x 100 = 3.0%(v/v) 8.0 fl. oz. oil + 256 fl. oz. gasoline

A solution of alcohol and water is separated by distillation. A 200 mL solution sample gives 85.9 mL of alcohol. 85.9 mL alcohol x 100 = 43.0%(v/v) 200 mL solution

250. mL of solution contains 15.0 mL of acetone. 15.0 mL alcohol x 100 = 6.00%(v/v) 250.mL solution

250. mL of solution contains 15.0 mL of any soluble liquid solute. 15.0 mL solute x 100 = 6.00%(v/v) 250.mL solution

Chapter 6

6.80

1.0 quart of acetic acid is put into a 5 gallon container, and enough water is added to ﬁll the container. 1.0 quart acetic acid x 100 = 5%(v/v)  4 quarts  5 gallons    1 gallon 

A solution of acetone and water is separated by distillation. A 300. mL solution sample gives 109 mL of acetone. 109 mL alcohol x 100 = 36.3%(v/v) 300.mL solution

 1000 mL  (5.0 L)(0.50% (v/v)) = (5.0 L)(0.0050) = 0.025 L   = 25 mL  1L 

Note: This number may seem small when compared with the volume of a shot of alcohol. (Typically, 1 shot is 1.5 ﬂ. oz. or 44 mL.) This question assumed that the drink was pure ethanol and that it was completely absorbed into the bloodstream. These assumptions do not take into account that most alcoholic beverages are not pure ethanol (many spirits are approximately 40% alcohol by volume) and that the alcohol is distributed throughout the body (not just the bloodstream). 6.81

1.8 mg  10 −3 g   1 mL     x 100 = 0.0023%(v/v) 100.mL  1 mg   0.79 g 

6.82

7.50 g x 100 = 5.00%(w/v) dextrose 150.0 mL

7.50 g x 100 = 5.00%(w/v) 150.0 mL

3.15 g x 100 = 0.900%(w/v) saline 350.0 mL

26.5 of solute is dissolved in 200. mL of water to give a solution with a density of 1.10 g/mL.

6.83

26.5 g solute   1.00 g   1 mL   26.5 g + 200. mL water      1 mL    1.10 g   b.

x 100 = 12.9%(w/v)

A 30.0 mL solution sample on evaporation leaves a solid residue of 0.38 g. 0.38 g solute x 100 = 1.3%(w/v) 30.0 mL solution

On analysis for total protein, a blood serum sample of 15.0 mL is found to contain 1.15 g of total protein.

Chapter 6 1.15 g solute x 100 = 7.67%(w/v) 15.0 mL solution

6.84

6.85

65 g KBr x 100 = 39%(w/w) 65 g KBr + 100. g H 2 O

M = moles solute/liters of solution. Thus, we need to ﬁrst calculate both the moles of phenol and the volume in liters. 1L = 0.100 L solution 100 mL x 1000. mL The molecular weight of C6 H 5 O6 is: 6 C’s = 12.01 = 72.06 6 H’s = 6 x 1.008 = 6.048 1 O = 16.00 = 16.00 94.108 or 94.11 g/mol Use the factor-unit method to ﬁnd the number of moles of phenol.

1.4 g x

1 mol 94.11 g

= 0.015 phenol

The last step is to use the equation for molarity.

moles 0.015 mol M= = = liters 0.100 L

0.15 M

SECTION 6.8 SOLUTION PREPARATION 6.86

 1 L   0.150 moles Na 2 SO 4   142.05 g Na 2 SO 4  200. mL   = 4.26 g Na 2 SO 4   1L  1000 mL     1 mole Na 2 SO 4  I would mass 4.26 Na2SO4 and add it to a 200. mL volumetric ﬂask. I would add enough water to dissolve the Na2SO4, then add water up to the mark on the volumetric ﬂask, cap, and shake to ensure the solution is homogeneous.

 1 L   0.250 moles   189.41 g Zn(NO 3 )2  250. mL   = 11.8 g Zn(NO 3 )2    1L  1000 mL     1 mole Zn(NO 3 )2  I would mass 11.8 g Zn(NO3)2 and add it to a 250. mL volumetric ﬂask. I would add enough water to dissolve the Zn(NO3)2, then add water up to the mark on the volumetric ﬂask, cap, and shake to ensure the solution is homogeneous. b.

150. g x 2.25% (w/w) = 150. g x 0.0225 = 3.38 g NaCl  1 mL  150. mL − 3.38 g = 146.62 g H 2 O  146.62 mL H 2 O ≈ 147 mL H 2 O =  1.00 H 2 O 

Chapter 6 I would mass 3.38 g NaCl and add it to a 250. mL Erlenmeyer flask. I would measure 147.mL of water and add enough of it to the Erlenmeyer flask to dissolve the salt, then I would add the rest of the water and swirl the flask to ensure that the solution is homogeneous. d. 125. mL x 0.75%(w/v) = 0.938 KCl ≈ 0.94 g KCl I would add 0.94 g KCl to a 125. mL volumetric flask. I would add enough water to dissolve the KCl, then dilute the mixture to the mark on the volumetric flask. I would then cap the flask and shake it to ensure the solution is homogeneous.

6.87

 1 L  2.00 moles NaOH   40.0 g NaOH  250. mL   = 20.0 g NaOH    1000 mL  1 L solution   1 mole NaOH  I would weigh 20.0 g NaOH and add it to a 250. mL volumetric ﬂask. I would add enough water to dissolve the NaOH, then add water up to the mark on the volumetric ﬂask, cap, and shake to ensure the solution is homogeneous. a.

 40.0 mL alcohol  500. mL   = 200. mL alcohol  100. mL solution  I would pipette 200. mL of alcohol and add it to a 500. mL volumetric ﬂask. I would add water up to the mark on the volumetric ﬂask, cap, and shake to ensure the solution is homogeneous. b.

 15.0 g glycerol  100. mL   = 15.0 g glycerol  100. mL solution 

 15.0 g glycerol   1 mL glycerol  100. mL   = 11.9 g glycerol   100. mL solution  1.26 g glycerol  I would either weigh 15.0 g glycerol on a balance or pipette 11.9 mL glycerol into a 100. mL volumetric ﬂask. I would add up to the mark on the volumetric ﬂask, cap, and shake to ensure the solution is homogeneous.  1 g solution   0.89 g NaCl  50. mL   = 0.45 g NaCl   1 mL solution  100. g solution  (Note: The assumption made in the calculation is that the density of the 0.89%(w/w) solution is approximately 1 g/mL.) I would add 0.45 NaCl to a 250 mL beaker. I would add enough water to make approximately 50. mL of solution. I would stir the mixture to dissolve the NaCl and ensure the solution is homogeneous. d.

6.88

The equation for molarity is

moles liters

Chapter 6

Thus, moles NaCl = 0.40 mol NaCl. To ﬁnd the mass in grams, let’s use the factor-unit method. From the periodic table the molecular weight of NaCl is 58.44 g/mol.

250. mL of solution is 0.250 L M is given as 0.16 M moles moles NaCl M= = 0.16 M = liters 0.250 L 0.400 mol NaCl x

6.89

6.90

6.91

58.44 g 1 mol NaCl

= 2.3 g NaCl

58.44 g

0.400 mol NaCl x

250. mL x

NaCl = 58.44 g/mol

120. mL x

 1 L   1.75 moles Li 2 CO 3   73.89 g Li 2 CO 3  250. mL   = 32.3 g Li 2 CO 3    1 L solution  1000 mL     1 mole Li 2 CO 3 

 1 L   3.50 moles NH 3  200. mL   = 0.700 moles NH 3   1000 mL   1 L solution 

250. mL x 12.5%(v/v) = 250. mL x 0.125 = 31.3 mL

50.0 mL x 4.20%(w/v) = 50.0 mL x 0.0420 = 2.10 g CaCl 2

(0.500 M)(200. mL) = (6.00 M)(Vc )

1 mol NaCl

= 2.3 g NaCl

1L 0.278 mol x = 0.0695 mol dextrose 1000. mL 1L

1L 0.154 mol 58.44 g x x = 1.08 g NaCl 1000. mL 1L 1 mol

Vc = 16.7 mL I would add 150 mL of water to a 200. mL volumetric ﬂask, then add 16.7 mL of 6.0 M HCl to the ﬂask, let it cool, and dilute to the mark with water. (Always add acid to water, not the reverse!) b.

(2.00 M)(50. mL) = (6.00 M)(Vc ) Vc = 16.7 mL

I would add 30. mL of water to a 50. mL volumetric ﬂask, then add 16.7 mL of 6.0 M H2SO4 to the

Chapter 6 ﬂask, let it cool, and dilute to the mark with water. (Always add acid to water, not the reverse!) c.

(0.89%(w/v))(100. mL) = (5.0%(w/v))(Vc ) Vc = 17.8 mL

I would add 17.8 mL of 5.0% (w/v) NaCl to a 100. mL volumetric ﬂask, then dilute to the mark with water. I would be sure to shake well. d.

(5.00%(v/v))(250. mL) = (20.5%(v/v))(Vc ) Vc = 61.0 mL

I would add 61.0 mL of 20.5%(v/v) acetone to a 250. mL volumetric ﬂask, then dilute to the mark with water. I would be sure to shake well.

6.92

(6.00 M)(5.00 L) = (18.0 M)(Vc ) Vc = 1.67 L

I would add 3.00 L of water to a 5.00 L volumetric ﬂask, then add 1.67 L of 18.0 M H2SO4 to the ﬂask, let it cool, and dilute to the mark with water. (Always add acid to water, not the reverse!) b.

(0.500 M)(250.0 mL) = (3.00 M)(Vc ) Vc = 41.7 mL

I would add 41.7 mL of 3.00 M CaCl2 to a 250. mL volumetric ﬂask, then dilute to the mark with water. I would be sure to shake well. c.

(1.50%(w/v))(200.0 mL) = (10.0%(w/v))(Vc ) Vc = 30.0 mL

I would add 30.0 mL of 10.0%(w/v) KBr to a 200. mL volumetric ﬂask, then dilute to the mark with water. I would be sure to shake well. (10.0%(v/v))(500. mL) = (50.0%(v/v))(Vc ) Vc = 100 mL

I would add 100 mL of 50.0%(v/v) alcohol to a 500. mL

volumetric ﬂask, then dilute to the mark with water. I would be sure to shake well.

6.93

40.0 mL

100. mL

(25.0 mL)(0.412 M) = (40.00 mL)(Cd ) Cd = 0.258 M

(25.0 mL)(0.412 M) = (100. mL)(Cd ) Cd = 0.103 M

Chapter 6

6.94

1.10 L

350. mL

1.50 L

200. mL

500. mL

700. mL

 1000 mL  (25.0 mL)(0.412 M) = (1.10 L)   (Cd )  1L  Cd = 0.00936 M

(25.0 mL)(0.412 M) = (350. mL)(Cd ) Cd = 0.0294 M

 1000 mL  (50. mL)(0.195 M) = (1.50 L)   (Cd )  1L  Cd = 0.00650 M (50. mL)(0.195 M) = (200. mL)(Cd ) Cd = 0.0488 M

(50. mL)(0.195 M) = (500. mL)(Cd ) Cd = 0.0195 M

(50. mL)(0.195 M) = (700. mL)(Cd ) Cd = 0.0139 M

SECTION 6.9 OSMOSIS AND DIALYSIS 6.95

With osmosis, pore size is smaller than other small molecules so only water would migrate into the bag. With dialysis, pore size is larger than the other small molecules so both the water and the other small molecules would migrate out of the bag.

6.96

The water will ﬂow from the 5.00% sugar solution into the 10.0% sugar solution because the 5.00% sugar solution contains more solvent (water) than the 10.0% sugar solution does. The 10.0% sugar solution will become diluted as osmosis takes place. Allowed enough time, the two solutions will eventually have the same concentration.

6.97

A 1.00 quart (0.946 L) bottle would contain 0.0422 moles of gas at STP, as shown by the calculation below.

 1 mole gas  0.946 L   = 0.0422 moles gas  22.4 L  6.98

The hydrated sodium and chloride ions will pass through the dialyzing membrane, but the starch (colloid) will not.

The urea will pass through the dialyzing membrane because it is a small organic molecule, but the starch (colloid) will not.

Chapter 6 c.

The hydrated potassium and chloride ions as well as the glucose molecules will pass through the dialyzing membrane, but the albumin (colloid) will not.

ADDITIONAL EXERCISES 6.99

PV = nRT

 29 moles gas   L ⋅ atm  P(1.0L) = (10 g)    0.0821 mol ⋅ K  (750 K) 908 g C H O N  3 5 9 3   P = 20 atm 6.100

2 H 2 (g) + O 2 (g) → 2 H 2 O (l)

 1 L O2  2.31 L H 2   = 1.16 L O 2  2 L H2  This reaction is classiﬁed as both a combination reaction and a redox reaction.

Chapter 7

Chapter 7: Chemical Equilibrium: Acids, Bases, and Buﬀers CHAPTER OUTLINE 7.1 Principles of Chemical Reactions 7.2 Equilibrium

7.5 The Strengths of Acids and Bases

7.3 Acid–Base Equilibria

7.7 Maintaining pH: The Balancing Act

7.6 Acid–Base Titrations

7.4 The pH Concept

LEARNING OBJECTIVES/ASSESSMENT When you have completed your study of this chapter, you should be able to: 1.

Use energy diagrams to represent and interpret the energy relationships for reactions. (Section 7.1; Exercises 7.1 and 7.2)

Explain how factors such as reactant concentrations, temperature, and catalysts inﬂuence reaction rates. (Section 7.1; Exercises 7.11 and 7.12)

Write equilibrium expressions based on reaction equations. (Section 7.2; Exercises 7.15 and 7.17)

Use Le Châtelier’s principle to predict the inﬂuence of changes in concentration and reaction temperature on the position of equilibrium for reactions. (Section 7.2; Exercises 7.23 and 7.27)

Write reaction equations that illustrate Brønsted-Lowry acid–base behavior. (Section 7.3; Exercises 7.35, 7.37, and 7.43)

Name acids. (Section 7.3; Exercises 7.47 and 7.49)

7. Classify solutions as acidic, basic, or neutral based on hydroxide ion or hydronium ion concentrations. (Section 7.4; Exercises 7.51 and 7.52) 8. Calculate the pH of solutions given hydroxide ion or hydronium ion concentrations. (Section 7.4; Exercise 7.57) 9. Use equilibrium constants to rank the strengths of acids and bases. (Section 7.5; Exercises 7.63 and 7.64) 10. Write reaction expressions for the neutralization reactions between acids and bases. (Section 7.6; Exercises 7.72 and 7.73) 11. Perform titration calculations using balanced acid–base neutralization expressions. (Section 7.6; Exercises 7.77 and 7.78) 12. Write reaction expressions to illustrate how buﬀers work. (Section 7.7; Exercise 7.84) 13. Determine the pH of buﬀers. (Section 7.7; Exercises 7.78 and 7.89) 14. Describe respiratory and urinary control of blood pH. (Section 7.7; Exercises 7.90, 7.96, and 7.97)

LECTURE HINTS AND SUGGESTIONS 1.

An example of equilibrium of biological signiﬁcance is the interaction of oxygen with hemoglobin (introduced in Example 7.2.) One can build on this example in class by discussing how both carbon monoxide and oxygen can compete with one another for the same hemoglobin molecules. Carbon monoxide poisoning and its antidote, oxygen-enriched air, can be used to show Le Châtelier’s principle of a shift in equilibrium.

Chapter 7 2.

Hand out some pH paper in class and have the students test various household products or foods at home for their degree of acidity or basicity. This information could then be discussed in class after a few days. The pH of some household products, such as cleaning agents, could be related to their function. Various foods, e.g., fruit juices or dairy products should show certain characteristic pH values. Some of these materials could be brought to class or the laboratory and the pH determined more accurately with a pH meter.

Have the students look up ﬁrst aid procedures for dealing with accidental ingestion or skin contact with strong acids and bases. Note any trends in the type of ﬁrst aid measures or the types of antidotes that might be useful with certain acids and bases.

A pH meter with a large visible readout could be brought to class and the pH of various types of solutions could be demonstrated to the students. The hydrolysis of a number of different salts to produce acidic or basic solutions could be easily shown. The ability of buffer solutions to resist a change in pH brought about by the addition of strong acids or bases could also be demonstrated as well as the limited capacity of any buffer.

When describing buffer solutions, be sure that the student understands how a buffer works by reacting with strong acids or bases. Write the equations for these reactions and explain why the pH change is much less in the presence of the buffer, using the concepts of equilibrium and Le Châtelier’s principle already developed. The biologically significant bicarbonate buffer is an example you can use to demonstrate this idea (Section 7.7A.)

Provide Chapter 7 Key equations, Table 7.1, and Figure 7.21 on a full-page handout for students to reference while they work through calculations involving equilibrium, Le Châtelier’s Principle, pH and buﬀers (see next page).

You can expand on the concept of urinary control of blood pH using a urinalysis kit (available from Sargent Welch, Rochester, NY 14692-9012.) These kits supply synthetic urine samples, which can be analyzed for pH, sugar, and protein.

Chapter 7

Equilibrium and Le Châtelier’s Principle 𝑎𝑎𝑎𝑎 + 𝑏𝑏𝑏𝑏 ⇄ 𝑐𝑐𝑐𝑐 + 𝑑𝑑𝑑𝑑

(7.1)

[𝐶𝐶]𝑐𝑐 [𝐷𝐷]𝑑𝑑 𝐾𝐾eq = [𝐴𝐴]𝑎𝑎 [𝐵𝐵]𝑏𝑏

pH and Buﬀer Equations 𝐾𝐾𝑤𝑤 = [𝐻𝐻3 𝑂𝑂+ ][𝑂𝑂𝐻𝐻 − ] = (1.0 × 10−7 𝑀𝑀)(1.0 × 10−7 𝑀𝑀) = 1.0 × 10−14 (𝑀𝑀)2 𝑝𝑝𝑝𝑝 = − log[𝐻𝐻3 𝑂𝑂+ ] [𝐻𝐻3 𝑂𝑂+ ] = 10−𝑝𝑝𝑝𝑝

(7.2) (7.3) (7.4)

HA(aq) + H2 O(l) ⇄ H3 O+ (aq) + A− (aq) [𝐻𝐻3 𝑂𝑂+ ][𝐴𝐴− ] 𝐾𝐾𝑎𝑎 = [𝐻𝐻𝐻𝐻] [A− ] pH = p𝐾𝐾a + log [HA]

(7.5)

Figure 7.21 Diagram providing the mathematical relationships between pH, [H3O+], and [OH-].

(7.6) 3

Chapter 7

Solutions To All End-of-Chapter Questions What follows are more complete explanations/full solutions to the EOC exercises whose answers are published in shorter form at the end of the textbook.

SECTION 7.1 PRINCIPLES OF CHEMICAL REACTIONS 7.1

Exothermic (exergonic) reaction with activation energy

Exothermic (exergonic) reaction without activation energy

Both of these energy diagrams have the same average energy of the reactants, average energy of products, and energy difference between reactants and products. The main difference between these two energy diagrams is that the first diagram has an activation energy and the second diagram does not. 7.2

Endothermic (endergonic) reaction with activation energy

Endothermic (endergonic) reaction without activation energy

Chapter 7 7.3

A catalyst lowers the activation energy for a reaction whether the reaction is exergonic or endergonic, but it does not change the energy level of the reactants or the products. 7.4

As temperature increases, the molecules move faster. The faster the molecules move, the more likely they are to collide. As the temperature increases, the molecules gain energy. The more energy the molecules have, the more they are likely to react once they collide.

7.5

Water is decomposed into nonspontaneous hydrogen and oxygen gas by passing electricity through the liquid.

Electricity is required for the reaction to occur. Without a continuous ﬂow of electrical current, the reaction will stop.

An explosive detonates being struck by a falling rock.

spontaneous

Once the explosive is struck, the explosive reacts without any additional sources of energy.

A coating of magnesium oxide forms on a clean piece of magnesium exposed to air.

spontaneous

This reaction occurs without any source of energy.

A light bulb emits light when an electric current is passed through it.

nonspontaneous

Continuous electric current is required to light the bulb.

A cube of sugar dissolves in a cup of hot coﬀee.

spontaneous

This process occurs without any additional energy.

The space shuttle leaves nonspontaneous its pad and goes into orbit.

Rocket engines must continually operate to push the shuttle into orbit

The fuel in a booster rocket of the space shuttle burns.

Once the fuel is ignited, it will continue to burn. No additional energy has to be provided.

7.6

spontaneous

Chapter 7

7.7

7.8

Water boils at 100°C and 1 atm pressure.

nonspontaneous

Heat must be continually supplied to maintain boiling.

Water temperature increases to 100°C at 1 atm pressure.

nonspontaneous

Increasing the temperature of water requires a continual supply of energy.

Your bedroom becomes orderly.

nonspontaneous

A room will not become orderly on its own. Cleaning requires energy.

Any combustion process

exergonic

The reaction produces heat and is spontaneous once begun.

Perspiration evaporating from the skin

endergonic

The process requires energy to change a liquid to a gas. The energy is taken from the skin as long as the body temperature is higher than the temperature of the surroundings, which cools the skin as the process occurs.

Melted lead solidifying

exergonic

The process releases energy as a liquid changes to a solid.

An explosive detonating

exergonic

The reaction releases heat and the reaction continues without additional energy input once it has been activated.

An automobile being pushed up a slight hill (from point of view of the of the automobile)

endergonic

The car receives energy from the person pushing, and the process would not occur spontaneously without continued energy input.

An automobile being exothermic pushed up a slight hill (from POV of the one pushing)

The person pushing the car gives energy to the car.

Ice melting (from POV of the ice)

endothermic

Melting ice requires energy.

Ice melting (from POV surroundings of the ice)

exothermic

The surroundings release heat into the ice.

Chapter 7

7.9

7.10

Steam condensing to liquid water (from POV of the steam)

exothermic

Heat must be released from the stem.

Steam condensing to endothermic liquid water (from POV of surroundings of the steam)

the diﬀusion of ink from a drop placed in a pan of quiet, undisturbed water

The changing diameter of the ink drop or the changing color of the water could be measured.

the loss of water from a pan of boiling water

The changing volume or depth of the water could be measured.

the growth of a corn plant

The changing height of the corn plant could be measured.

the melting of a block of ice

The changing height of the block, the changing mass of the block, or the increasing volume of liquid formed could be measured.

the setting (hardening) of concrete

The ability of an object to penetrate or mark the surface of the concrete could be measured.

the burning of a candle

The changing height of the candle or the changing mass of the candle could be measured.

Heat must be absorbed by the surroundings.

7.11

A reaction is started by mixing reactants. As time passes, the rate decreases because the concentration of reactants decreases and the number of reaction-producing collisions also decreases.

7.12

To speed up a reaction, I might (1) heat the reactants in order to increase the energy of the reactants and the frequency of collision, (2) stir the reactants in order to increase the frequency of collision, (3) increase the surface area of the reactants to increase the number of collisions, and (4) add a catalyst to lower the activation energy for the reaction. Only three of these steps are needed for a complete answer.

Chapter 7

SECTION 7.2 EQUILIBRIUM 7.13

2 CO

colorless gas

2 CO 2



colorless gas

The pressure of the gas mixture will stop changing once equilibrium is reached. b.

LiOH

colorless solid

CO 2



colorless gas

LiHCO 3

colorless solid

The gas pressure and the amount of colorless solid will stop changing once equilibrium. is reached c.

7.14

checking account checks to pay bills paycheck → → The bank balance is constant (usually zero) when equilibrium is established.

colorless gas

2 HI



violet gas

colorless gas

The color of the gas mixture will stop changing once equilibrium is reached. b.

solid sugar

water

 

sugar solution

The amount of solid sugar will become constant once the mixture has reached equilibrium. c.

colorless gas

 

2 O2

colorless gas

2 NO 2

red-brown gas

Both the color and the pressure of the gas mixture will stop changing once equilibrium is reached.

7.15

3 O2  2 O3

O3  K eq =   3 O 2 

COCl 2  CO + Cl 2

Cl 2  CO  K eq =  COCl 2 

CS 2 + 4 H 2  CH 4 + 2 H 2 S

CH 4  H 2 S  K eq =  4 H 2  CS 2 

SO 3 

2 SO 2 + O 2  2 SO 3

K eq =

CO + H 2 O  CO 2 + H 2

CO 2  H 2  K eq =  H 2 O  CO 

O 2  SO 2 

Chapter 7

7.16

CO 2 

2 CO + O 2  2 CO 2

K eq =

N 2 O 4  2 NO 2

 NO 2  K eq =   N 2 O 4 

2 C 2 H6 + 7 O 2  4 CO 2 + 6 H 2 O

CO 2  H 2 O  K eq =  2 7 C 2 H6  O 2 

2 NOCl  4 CO 2 + 6 H 2 O

 NO 2  Cl 2  K eq =  2  NOCl 

2 Cl 2 O 5  O 2 + 4 ClO 2

O 2  ClO 2  K eq =    2 Cl 2 O 5 

CO  O 2  2

7.17

Ni 2 + + 6 NH 3  Ni(NH 3 )62 +

 Ni(NH 3 )62 +   K eq =  6 2+  Ni   NH 3   

Sn 2 + + 2 Fe 3 +  Sn 4 + + 2 Fe 2 +

Sn 4 +  Fe 2 +    K eq =  2 2+ 3+ Sn  Fe    

F2 + 2 Cl  2 F + Cl 2

F −  Cl 2  K eq =   2 F2  Cl − 

7.18

−

Fe(CN)63 −   K eq =  3 + − Fe  CN    

 Ag(NH 3 )2+   K eq =  2  Ag +   NH 3   

K eq =

 AuCl 4−     Au 3 +  Cl −    

Chapter 7 3

7.19

PH 3  F2  K eq =  3 HF  PF3 

O 2   NH 3  K eq =   4 6  NO 2  H 2 O 

O 2  ClO 2  K eq =    2 Cl 2 O 5 

 N 2  H 2 O  K eq =   2 2  NO  H 2 

CO 2  H 2 O  K eq =  2 CH 4  O 2 

CH 4  H 2 O  K eq =  3 H 2  CO 

O2  K eq =   2 O 3 

 NH 3  O 2  K eq =  4 6  NO 2  H 2 O 

a. b.

K = 2.1 x 10 −6 K = 0.15

[reactants] larger than [products] [reactants] larger than [products]

c. d.

K = 1.2 x 10 8 K = 0.00036

[reactants] smaller than [products] [reactants] larger than [products]

K = 5.9

[reactants] smaller than [products]

3 HF + PF3  PH 3 + 3 F2

4 NO 2 + 6 H 2 O  7 O 2 + 4 NH 3

2 Cl 2 O 5  O 2 + 4 ClO 2

7.20

2 NO + 2 H 2  N 2 + 2 H 2 O

7.21

7.22

b. c. d.

7.23

[reactants] smaller than [products]

−4

[reactants] larger than [products] [reactants] larger than [products]

K = 3.3 x 10

K = 2.7 x 10 K = 0.0000558

heat

Co 2 + (aq)

pink

4 Cl − (aq)



colorless

CoCl 24 − (aq) blue

The equilibrium mixture is heated. The equilibrium will shift to the right and the reaction will be less pink and more blue. The reaction will generate less heat and the container will become cooler. b.

heat

Co 2 + (aq)

pink

4 Cl − (aq) colorless

Co 2 + is added to the equilibrium mixture. 10



CoCl 24 − (aq) blue

Chapter 7 The equilibrium will shift to the right with the same results as in part a. c.

Fe 3 + (aq)

brown

colorless

6 SCN − (aq)



Fe(SCN)63 − (aq) red

SCN is added to the equilibrium mixture. The equilibrium will shift to the right and the reaction will be less brown and more red. −

Pb 2 + (aq)

2 Cl − (aq)

colorless

PbCl 2 (s)



colorless

heat

white solid

Pb is added to the equilibrium mixture. The reaction will shift to the right and more white solid will form. The reaction will generate heat and the container will become warmer. e.

C2 H4

colorless gas



violet gas

C2 H4 I 2

heat

colorless gas

A catalyst is added to the equilibrium mixture. The catalyst does not have any eﬀect on the equilibrium. The intensity of the color will not change. No other physical changes will be observed.

7.24

Cu 2 + (aq)

blue

4 NH 3 (aq)



Cu(NH 3 )24 + (aq)

dark purple

colorless

Some NH3 is added to the equilibrium mixture. The equilibrium will shift to the right and the mixture will become less blue and more purple. b.

Pb 2 + (aq)

colorless

2 Cl − (aq)



colorless

PbCl 2 (s)

heat

white solid

The equilibrium mixture is cooled. The equilibrium will shift to the right and more precipitate will form. Heat will also be generated and the temperature of the container will increase. c.

C2 H4

colorless gas



violet gas

C2 H4 I 2

heat

colorless gas

Some C 2 H 4 I 2 is removed from the equilibrium mixture. The equilibrium will shift to the right and the mixture will become less violet and more heat will be produced. d.

C2 H4

colorless gas



violet gas

C2 H4 I 2

heat

colorless gas

The equilibrium mixture is cooled. The equilibrium will shift to the right and the mixture will become less violet and more heat will be produced.

Chapter 7

heat

4 NO 2

6 H2 O

brown gas



colorless gas

7 O2

colorless gas

4 NH 3

colorless gas

A catalyst is added, and NH3 is added to the equilibrium mixture. The catalyst does not have any eﬀect on the equilibrium; however, adding NH3 shifts the equilibrium to the left and this produces heat, lowers the pressure because there will be fewer moles of gas present, and increases the brown color of the equilibrium mixture.

7.25

H + (aq) + HCO −3 (aq)  H 2 O(l) + CO 2 (g); HCO −3 is added

The equilibrium will shift to the right. The amounts of H + and HCO 3− will decrease, while the amounts of H 2 O and CO 2 will increase. The concentration of the water will not be signiﬁcantly changed because this reaction is occurring in an aqueous environment. b.

CO 2 (g) + H 2 O(l)  H 2 CO 3 (aq) + heat; CO 2 is removed.

The equilibrium will shift to the left. The amount of CO 2 and H 2 O will increase, while the amount of H 2 CO 3 and the heat decrease. The concentration of the water will not be signiﬁcantly changed because this reaction is occurring in an aqueous environment. c.

CO 2 (g) + H 2 O(l)  H 2 CO 3 (aq) + heat; the system is cooled.

The equilibrium will shift to the right. The amount of CO 2 and H 2 O will decrease, while the amount of H 2 CO 3 and heat increase. The concentration of the water will not be signiﬁcantly changed because this reaction is occurring in an aqueous environment. 7.26

LiOH(s) + CO 2 (g)  LiHCO 3 (s) + heat; CO 2 is removed.

The equilibrium will shift to the left. The amount of LiHCO 3 will decrease, the amount of LiOH will increase and the concentration of CO 2 will increase. Heat will be used and the container will be cooler to the touch. b.

NaHCO 3 (s) + heat  Na 2 O(s) + 2 CO 2 (g) + H 2 O(g); the system is cooled.

The equilibrium will shift to the left. The concentration of CO 2 and H 2 O will decrease as will the amount of Na 2 O and the amount of NaHCO 3 will increase. Heat will be generated and the container will feel warmer. c.

CaCO 3 (s) + heat  CaO(s) + CO 2 (g); the system is cooled.

The equilibrium will shift to the left. The concentration of CO 2 will decrease as will the amount of CaO and the amount of CaCO 3 will increase. Heat will be generated and the container will feel warmer.

Chapter 7

2 HBr (g) heat  H 2 (g) + Br2 (g)

7.27 a.

Some H 2 is added

to the left

The temperature is increased

to the right

Some Br2 is removed

to the right

d. A catalyst is added e. Some HBr is removed f. The temperature is decreased, some HBr is removed

N 2 (g) + 3 H 2 (g)  2 NH 3 (g) + heat

7.28

7.29

Some N 2 is added

to the right

The temperature is increased

to the left

Some NH 3 is removed

to the right

Some H 2 is removed

to the left

e. f.

A catalyst is added The temperature is increase, some H2 is removed

no shift to the left

K eqc = 0.0245

PCl 5 (g)  PCl 3 (g) + Cl 2 (g) PCl 3  Cl 2  K eq =  PCl 5  0.250 M  0.250 M  0.0245 =  PCl 5  PCl 5  =2.55 M

7.30

no shift to the left to the left

K eqc = 50.5

H 2 + I 2  2 HI 2

HI  K eq =   H 2  I 2  2

0.500 M  50.5 =  0.050  I 2  I 2  =0.099 M

7.31

 Br2  Cl 2  0.26 M  0.26 M  = K eq =    2  = 0.47 2  BrCl  0.38 M 

7.32

 NO  Cl 2  0.92 M  0.20 M  = K eq =  = 0.099 2 2  NOCl  1.31 M 

Chapter 7

SECTION 7.3 ACID-BASE EQUILIBRIA 7.33

7.34

HI(aq) → H + (aq) + I − (aq)

HBrO

HBrO(aq) → H + (aq) + BrO − (aq)

HCN

HCN(aq) → H + (aq) + CN − (aq)

HClO 2

HClO 2 (aq) → H + (aq) + ClO −2 (aq)

HBrO 2

HBrO 2 (aq) → H + (aq) + BrO −2 (aq)

HS −

HS − (aq) → H + (aq) + S 2 − (aq)

HBr

HBr(aq) → H + (aq) + Br − (aq)

HC 2 H 3 O 2

HC 2 H 3 O 2 (aq) → H + (aq) + C 2 H 3 O −2 (aq)

7.35

7.38

Brønsted Bases H 3 O + , Br −

HBr(aq) + H 2 O(l)  H 3 O (aq) + Br (aq)

HBr, H 3 O

H 2 O(l) + N −3 (aq)  HN 3 (aq) + OH − (aq)

H 2 O, HN 3

N −3 ,OH −

H 2 S(aq) + H 2 O(l)  H 3 O + (aq) + HS − (aq)

H2 S , H3 O+

H 2 O , HS −

SO 23 − (aq) + H 2 O(l)  HSO 3− (aq) + OH − (aq)

H 2 O , HSO −3

SO 23 − , OH −

HCN(aq) + H 2 O(l)  H 3 O + (aq) + CN − (aq)

HCN, H 3 O +

H 3 O + , CN −

−

7.36

7.37

Brønsted Acids a.

Brønsted Acids

Brønsted Bases

HC 2 O (aq) + H 2 O(l)  H 3 O (aq) + C 2 O (aq)

HC 2 O , H 3 O

H 2 O , C 2 O 24 −

HNO 2 (aq) + H 2 O(l)  H 3 O + (aq) + NO 2 (aq)

HNO 2 , H 3 O +

H 2 O , NO 2

PO 34 − (aq) + H 2 O(l)  HPO 24 − (aq) + OH − (aq)

H 2 O , HPO 24 −

PO 34 − , OH −

H 2 SO 3 (aq) + H 2 O(l)  HSO −3 (aq) + H 3 O + (aq)

H 2 SO 3 , H 3 O +

H 2 O , HSO −3

F − (aq) + H 2 O(l)  HF(aq) + OH − (aq)

H 2 O , HF

F − , OH −

HI(aq) + H 2 O(l)  I − (aq) + H 3 O + (aq)

HBrO(aq) + H 2 O(l)  BrO − (aq) + H 3 O + (aq)

HCN(aq) + H 2 O(l)  CN − (aq) + H 3 O + (aq)

HSe − (aq) + H 2 O(l)  Se 2 − (aq) + H 3 O + (aq)

HF(aq) + H 2 O(l)  F − (aq) + H 3 O 3 (aq)

HClO 3 (aq) + H 2 O(l)  ClO 3− (aq) + H 3 O + (aq)

− 4

2− 4

− 4

Chapter 7 c.

HClO(aq) + H 2 O(l)  ClO − (aq) + H 3 O + (aq)

HS − (aq) + H 2 O(l)  S 2 − (aq) + H 3 O + (aq)

HSO (aq) + H 2 O(l)  SO (aq) + H 3 O (aq)

SO 23 −

HPO 24 − (aq) + H 2 O(l)  PO 34 − (aq) + H 3 O + (aq)

PO 34 −

HClO 3 (aq) + H 2 O(l)  ClO 3− (aq) + H 3 O + (aq)

ClO −3

CH 3 NH +3 (aq) + H 2 O(l)  CH 3 NH 2 (aq) + H 3 O + (aq)

H 2 C 2 O 4 (aq) + H 2 O(l)  HC 2 O (aq) + H 3 O (aq)

HSO (aq) + H 2 O(l)  SO (aq) + H 3 O (aq)

CH 3 NH +3 (aq) + H 2 O(l)  CH 3 NH 2 (aq) + H 3 O + (aq)

HClO 4 (aq) + H 2 O(l)  ClO (aq) + H 3 O (aq)

NH 4+ (aq) + H 2 O(l)  NH 3 (aq) + H 3 O + (aq)

NH 3

HCl(aq) + H 2 O(l)  Cl − (aq) + H 3 O + (aq)

Cl −

NH (aq) + H 2 O(l)  NH 3 (aq) + OH (aq)

NH 3

CO 23 − (aq) + H 2 O(l)  HCO −3 (aq) + OH − (aq)

HCO −3

OH − (aq) + H 2 O(l)  H 2 O(aq) + OH − (aq)

(CH 3 )2 NH(aq) + H 2 O(l)  (CH 3 )2 NH (aq) + OH (aq)

NO 2− (aq) + H 2 O(l)  HNO 2 (aq) + OH − (aq)

HCO (aq) + H 2 O(l)  H 2 CO 3 (aq) + OH (aq)

H 2 CO 3

S 2 − (aq) + H 2 O(l)  HS − (aq) + OH − (aq)

HS −

HS − (aq) + H 2 O(l)  H 2 S(aq) + OH − (aq)

H2 S

HC 2 O 4− (aq) + H 2 O(l)  H 2 C 2 O 4 (aq) + OH − (aq)

H2 C2 O4

HN 2 O (aq) + H 2 O(l)  H 2 N 2 O 2 (aq) + OH (aq)

H2 N2 O2

HI(aq) + ? → H 3 O (aq) + I (aq)

H 2 O (l)

NH 3 (l) + ? → NH (aq) + NH (aq)

NH 3 (l)

H 2 C 2 O 4 (aq) + H 2 O(l) → ? + HC 2 O −4 (aq)

H 3 O + (aq)

7.39

Conjugate Base 2− 3

− 3

− 4

CH 3 NH 2 HC 2 O 4−

7.40

Conjugate Base 2− 4

− 4

SO 24 −

− 4

CH 3 NH 2 ClO −4

7.41

Conjugate Base − 2

−

H2 O

+ 2

−

(CH 3 )2 NH 2+

HNO 2

7.42

Conjugate Acid − 3

−

− 2

−

7.43

Missing Formula +

+ 4

−

− 2

Chapter 7 d.

H 2 N 2 O 2 (aq) + H 2 O(l) → H 3 O + (aq) + ?

HN 2 O 2− (aq)

? + H 2 O(l) → H 3 O + (aq) + CO 32 − (aq)

HCO 3− (aq)

H 2 AsO (aq) + NH 3 (aq) → NH (aq) + ?

HAsO 24 − (aq)

? + H 2 O(l) → C6 H 5 NH 3+ (aq) + OH − (aq)

C6 H 5 NH 2 (aq)

7.44

Missing Formula

2−

H 2 O (l)

S (aq) + ? → HS (aq) + OH (aq)

? + HBr(aq) → (CH 3 )2 NH (aq) + Br (aq)

(CH 3 )2 NH (aq)

CH 3 NH 2 (aq) + HCl(aq) → ? + Cl −

CH 3 NH 3+ (aq)

−

+ 2

−

Acid

Base

Equation

HOCl

H2 O

HOCl(aq) + H 2 O(l) → H 3 O + (aq) + OCl − (aq)

HClO 4

NH 3

HClO 4 (aq) + NH 3 (aq) → NH 4+ (aq) + ClO 4_ (aq)

H2 O

NH −2

H 2 O(l) + NH 2− (aq) → NH 3 (aq) + OH − (aq)

H2 O

OCl −

H 2 O(l) + OCl − (aq) → HOCl(aq) + OH − (aq)

HC 2 O 4−

H2 O

HC 2 O 4− (aq) + H 2 O(l) → H 3 O + (aq) + C 2 O 42 − (aq)

Acid

Base

HPO

2− 4

NH 3

H 2 PO 4 (aq) + NH 3 (aq) → NH +4 (aq) + HPO 24 − (aq)

HS 2 O 3−

OCl −

HS 2 O −3 (aq) + OCl − (aq) → HOCl(aq) + S 2 O 32 − (aq)

H2 O

ClO −4

H 2 O(l) + ClO −4 (aq) → HClO 4 (aq) + OH − (aq)

H2 O

NH 3

H 2 O(l) + NH 3 (aq) → OH − (aq) + NH +4 (aq)

H 2 Se(aq) hydroselenic acid

H 2 SO 4 sulfuric acid

HClO 3

HNO 3 nitric acid

H 2 Te(aq) hydrotelluric acid

H 2 SO 3 sulfurous acid

HClO

HNO 2 nitrous acid

7.46 a.

7.48

+ 4

7.45

7.47

− 4

H3 O

− 2

Equation +

H 3 O (aq) + NH (aq) → NH 3 (aq) + H 2 O(l) − 2

chloric acid

hypochlorous acid

7.49

H 3 C6 H 5 O7 = citric acid

7.50

H 2 C 4 H 4 O 4 = succinic acid

Chapter 7

SECTION 7.4 THE PH CONCEPT 7.51

a. b. c. d.

Basic Acidic Acidic Basic

7.52

a. b. c. d.

Acidic Basic Acidic Basic

7.53

Kw 1.0 x 10 -14 (mol/L)2 = = 1.0 x 10-8 M a. [OH − ]= + 1.0 x 10 -6 mol/L [H ] Kw 1.4 x 10 -14 (mol/L)2 2.1 x 10 -11 M = = b. [OH − ]= + -14 [H ] 4.8 x 10 mol/L Kw 1.0 x 10 -14 (mol/L)2 = = 2.0 x 10-12 M c. [OH − ]= + 0.0051 mol/L [H ]

Kw 1.04 x 10 -14 (mol/L)2 1.0 x 10-7 M = = d. [OH − ]= + [H ] 1.0 x 10-7 mol/L Kw 1.0 x 10 -14 (mol/L)2 1.6 x 10-6 M = = e. [OH − ]= + [H ] 6.3 x 10-9 mol/L

7.54

Kw 1.0 x 10 -14 (mol/L)2 1.0 x 10 -7 M = = a. [OH − ]= + [H ] 1.0 x 10 -7 mol/L Kw 1.0 x 10 -14 (mol/L)2 7.7 x 10 -11 M = = b. [OH − ]= + [H ] 1.3 x 10 -4 mol/L Kw 1.0 x 10 -14 (mol/L)2 = = 1.6 x 10-12 M c. [OH − ]= + 0.0063 mol/L [H ] Kw 1.0 x 10 -14 (mol/L)2 = = 1.0 x 10-9 M d. [OH − ]= + 1.0 x 10-5 mol/L [H ] Kw 1.0 x 10 -14 (mol/L)2 2.0 x 10 -6 M = = e. [OH − ]= + [H ] 5.0 x 10 -9 mol/L

7.55

Kw 1.0 x 10 -14 (mol/L)2 [H 3 O + ]= 1.0 x 10 -8 M = a. = [OH − ] 1.0 x 10 -6 mol/L Kw 1.0 x 10 -14 (mol/L)2 [H 3 O + ]= 2.0 x 10-8 M = b. = [OH − ] 5.0 x 10 -7 mol/L Kw 1.0 x 10 -14 (mol/L)2 [H 3 O + ]= 3.8 x 10-6 M = c. = [OH − ] 2.6 x 10 -9 mol/L

Chapter 7

Kw 1.0 x 10 -14 (mol/L)2 [H 3 O + ]= = 7.7 x 10-4 M d. = [OH − ] 1.3 x 10 -11 mol/L Kw 1.0 x 10 -14 (mol/L)2 [H 3 O + ]= 1.0 x 10 -7 M = e. = [OH − ] 1.0 x 10 -7 mol/L

7.56

Kw 1.0 x 10 -14 (mol/L)2 = 1.0 x 10 -3 M [H 3 O + ]= a. = [OH − ] 1.0 x 10 -11 mol/L Kw 1.0 x 10 -14 (mol/L)2 [H 3 O + ]= = 7.7 x 10-5 M b. = [OH − ] 1.3 x 10 -10 mol/L

Kw 1.0 x 10 -14 (mol/L)2 = [H 3 O + ]= 1.0 x 10 -7 M c. = [OH − ] 1.0 x 10 -7 mol/L Kw 1.0 x 10 -14 (mol/L)2 = [H 3 O + ]= 3.1 x 10 -8 M d. = − [OH ] 3.2 x 10 -7 mol/L Kw 1.0 x 10 -14 (mol/L)2 [H 3 O + ]= 2.0 x 10 -4 M = e. = − -11 [OH ] 5.0 x 10 mol/L

7.57

a. b. c. d. e.

Blood, pH = 7.41 Gastric juice, pH 1.60 Urine, pH = 5.93 Saliva, pH = 6.85 Pancreatic juice, pH = 7.85

[H+] = 10-7.41 = 3.9 x 10-8 M [H+] = 10-1.60 = 2.5 x 10-2 M [H+] = 10-5.93 = 1.2 x 10-6 M [H+] = 10-6.85 = 1.4 x 10-7 M [H+] = 10-7.85 = 1.4 x 10-8 M

basic acidic acidic acidic basic

7.58

a. b. c. d. e.

Bile, pH = 8.05 Vaginal ﬂuid, pH = 3.93 Semen, pH = 7.38 Cerebrospinal ﬂuid, pH = 7.40 Perspiration, pH = 6.23

[H+] = 10-8.05 = 8.9 x 10-9 M [H+] = 10-3.93 = 1.2 x 10-4 M [H+] = 10-7.38 = 4.2 x 10-8 M [H+] = 10-7.40 = 4.0 x 10-8 M [H+] = 10-6.23 = 5.9 x 10-7 M

basic acidic basic basic acidic

7.59

[H 3 O + ] = 1.0 x 10 -ph = 1.0 x 10 -7.5 = 3.2 x 10 -8 M

[H 3 O + ] = 1.0 x 10 -ph = 1.0 x 10 -8.5 = 3.2 x 10 -9 M

[H 3 O + ] = 1.0 x 10 -ph =

3.2 x 10 -8 M (for pH 7.5) 3.2 x 10 -9 M (for pH 8.5)

pH 7.5 is 10 times as acidic as pH 8.5

7.60

[H 3 O + ] = 1.0 x 10 -ph = 1.0 x 10 -7.2 = 6.3 x 10 -8 M

[H 3 O + ] = 1.0 x 10 -ph = 1.0 x 10 -7.8 = 1.6 x 10 -8 M

[H 3 O + ] = 1.0 x 10 -ph =

6.3 x 10 -8 M (for pH 7.2) 1.6 x 10 -8 M (for pH 7.8)

Chapter 7 pH 7.2 is 3.9 times as acidic as pH 7.8

7.61

7.62

pH = 9.00

[H+] = 10-9.00 = 1.0 x 10-9 M

[OH − ]=

10 -14 = 1.0 x 10 -5 M 10 -9

pH = 6.27

[H+] = 10-6.27 = 5.4 x 10-7 M

[OH − ]=

10 -14 = 1.9 x 10 -8 M -7 5.4 x 10

pH = 3.10

[H+] = 10-3.10 = 8.0 x 10-4 M

[OH − ]=

10 -14 = 1.3 x 10 -11 M 8.0 x 10 -4

pH = 3.95

[H+] = 10-3.95 = 1.1 x 10-4 M

[OH − ]=

10 -14 = 8.9 x 10 -11 M 10 -3.95

pH = 4.00

[H+] = 10-4.00 = 1.0 x 10-4 M

[OH − ]=

10 -14 = 1.0 x 10 -10 M 10 -4.00

pH = 11.86

[H+] = 10-11.86 = 1.4 x 10-12 M

[OH − ]=

10 -14 = 7.2 x 10 -3 M 10 -11.86

SECTION 7.5 THE STRENGTHS OF ACIDS AND BASES 7.63

a. (weakest) acid B < acid A < acid C , acid D (strongest) The smaller the Ka, the weaker the acid. The larger the Ka, the stronger the acid. b. (weakest) base D < base C < base A < base B (strongest) The smaller the Ka, the weaker the acid and the stronger the conjugate base. The larger the Ka, the stronger the acid and the weaker the conjugate base.

7.64

a. (weakest) acid B < acid A < acid C < acid D (strongest) The smaller the Ka, the weaker the acid, The larger the Ka, the stronger the acid. b. (weakest) base D < base C < base A < base B (strongest) The smaller the Ka, the weaker the acid and the stronger the conjugate base. The larger the Ka, the stronger the acid and the weaker the conjugate base

7.65

HBrO

HBrO(aq)  H + (aq) + BrO − (aq)

Ka =

[H + ][BrO − ] [HBrO]

H 2 SO 3

H 2 SO 3 (aq)  H + (aq) + HSO 3− (aq)

Ka =

[H + ][HSO −3 ] [H 2 SO 3 ]

HSO 3−

HSO 3− (aq)  H + (aq) + SO 32 − (aq)

Ka =

[H + ][SO 23 − ] [HSO 3− ]

Chapter 7

7.66

H 2 Se

H 2 Se(aq)  H + (aq) + HSe − (aq)

Ka =

[H + ][HSe − ] [H 2 Se]

H 3 AsO 4

H 3 AsO 4 (aq)  H + (aq) + H 2 AsO 4− (aq)

Ka =

[H + ][H 2 AsO −4 ] [H 3 AsO 4 ]

HSe −

HSe − (aq)  H + (aq) + Se 2 − (aq)

Ka =

H 2 BO 3−

H 2 BO 3− (aq)  H + (aq) + BO 23 − (aq)

Ka =

HBO 23 −

HBO 32 − (aq)  H + (aq) + BO 33 − (aq)

Ka =

HAsO 24 −

HAsO 42 − (aq)  H + (aq) + AsO 34 − (aq)

Ka =

HClO

HClO(aq)  H + (aq) + ClO + (aq)

Ka =

[H + ][Se 2 − ] [HSe − ]

[H + ][HBO 23 − ] [H 2 BO −3 ] [H + ][BO 33 − ] [HBO 23 − ] [H + ][AsO 34 − ] [HAsO 24 − ] [H + ][ClO − ] [HClO]

7.67

(weakest acid) HC < HA < HB (strongest acid) All of these solutions have the same concentration of acid; therefore, each pH value reﬂects the H+ concentration and the degree of dissociation of these acids—the higher the degree of dissociation, the lower the pH value and the stronger the acid.

7.68

The 0.05 M HCl is a dilute strong acid solution. If someone wanted this solution instead, they should use the term “dilute.” The 20% acetic acid solution is the weak acid solution. It is a weak acid solution because acetic acid does not completely dissociate.

7.69

H 3 AsO 4 (aq) + H 2 O(l)  H 3 O + (aq) + H 2 AsO −4 (aq) H 2 AsO −4 (aq) + H 2 O(l)  H 3 O + (aq) + HAsO 24 − (aq)

HAsO 24 − (aq) + H 2 O(l)  H 3 O + (aq) + AsO 43 − (aq)

The strongest Brønsted base corresponds to the weakest Brønsted acid; therefore, AsO43- is the strongest Brønsted base. The weakest Brønsted base corresponds to the strongest Brønsted acid; therefore, H2AsO4- is the weakest Brønsted base.

SECTION 7.6 ACID-BASE TITRATIONS 7.70

M*: 3 RbOH(aq) + H 3 PO 4 (aq) → Rb 3 PO 4 (aq) + 3 H 2 O(l) TIE**: 3 Rb + (aq) + 3 OH − (aq) + 3 H + (aq) + PO 24 − (aq) → 3 Rb + (aq) + PO 34 − (aq) + 3 H 2 O(l) NIE***: OH − (aq) + H + (aq) → H 2 O(l)

Chapter 7

M*: 2 RbOH(aq) + H 2 C 2 O 4 (aq) → Rb 2 C 2 O 4 (aq) + 2 H 2 O(l) TIE**: 2 Rb + (aq) + 2 OH − (aq) + 2H + + C 2 O 24 − (aq) → 2 Rb + (aq) + C 2 O 24 − (aq) + 2 H 2 O(l) NIE***: OH − (aq) + H + (aq) → H 2 O(l)

M*: RbOH(aq) + HC 2 H 3 O 2 (aq) → RbC 2 H 3 O 2 (aq) + H 2 O(l) TIE**: Rb + (aq) + OH − (aq) + H + (aq) + C 2 H 3 O −2 (aq) → Rb + (aq) + C 2 H 3 O −2 (aq) + H 2 O(l) NIE***: OH − (aq) + H + (aq) → H 2 O(l)

7.71

M*: RbOH(aq) + HCl(aq) → RbCl(aq) + H 2 O(l) TIE**: Rb + (aq) + OH − (aq) + H + (aq) + Cl − (aq) → Rb + (aq) + Cl − (aq) + H 2 O(l) NIE***: OH − (aq) + H + (aq) → H 2 O(l)

M*: RbOH(aq) + HNO 3 (aq) → RbNO 3 (aq) + H 2 O(l) TIE**: Rb + (aq) + OH − (aq) + H + (aq) + NO 3− (aq) → Rb + (aq) + NO 3− (aq) + H 2 O(l) NIE***: OH − (aq) + H + (aq) → H 2 O(l)

M*: 2 RbOH(aq) + H 2 SO 4 (aq) → Rb 2 SO 4 (aq) + 2 H 2 O(l) TIE**: 2 Rb + (aq) + 2 OH − (aq) + 2 H + (aq) + SO 24 − (aq) → 2 Rb + (aq) + SO 24 − (aq) + H 2 O(l) NIE***: OH − (aq) + H + (aq) → H 2 O(l)

7.72

Al(OH)3 + 3 HCl → AlCl 3 + 3 H 2 O

7.73

NaHCO 3 + HCl → NaCl + H 2 O + CO 2

7.74

M*: 2 KOH(aq) + H 3 PO 4 (aq) → K 2 HPO 4 (aq) + H 2 O(l) TIE**: 2 K + (aq) + 2 OH − (aq) + 2 H + (aq) + HPO 42 − (aq) → 2 K + (aq) + HPO 42 − (aq) + 2 H 2 O(l) NIE***: OH − (aq) + H + (aq) → H 2 O(l)

M*: 3 KOH(aq) + H 3 PO 4 (aq) → K 3 PO 4 + 3 H 2 O(l) TIE**: 3 K + (aq) + 3 OH − (aq) + 3 H + (aq) + PO 34 − (aq) → 3 K + (aq) + PO 34 − + 3 H 2 O(l) NIE***: OH − (aq) + H + (aq) → H 2 O(l)

M*: KOH(aq) + H 2 C 2 O 4 (aq) → KHC 2 O 4 + H 2 O(l) TIE**: K + (aq) + OH − (aq) + H + (aq) + HC 2 O 4− (aq) → K + (aq) + HC 2 O 4− + H 2 O(l) NIE***: OH − (aq) + H + (aq) → H 2 O(l)

Chapter 7 7.75

Molecular: KOH(aq) + H 2 CO 3 (aq) → KHCO 3 + H 2 O(l) Total ionic: K + (aq) + OH − (aq) + H + (aq) + HCO −3 (aq) → K + (aq) + HCO −3 + H 2 O(l) Net Ionic: OH − (aq) + H + (aq) → H 2 O(l)

Molecular: 2 KOH(aq) + H 2 CO 3 (aq) → K 2 CO 3 (aq) + 2 H 2 O(l) Total ionic: 2 K + (aq) + 2 OH − (aq) + 2 H + (aq) + CO 32 − (aq) → 2 K + (aq) + CO 23 − (aq) + 2 H 2 O(l) Net ionic: 2 OH − (aq) + 2 H + (aq) → 2 H 2 O(l)

Molecular: KOH(aq) + H 3 PO 4 (aq) → KHPO 4 + H 2 O(l) Total ionic: K + (aq) + OH − (aq) + H + (aq) + H 2 PO 4− (aq) → K + (aq) + HPO −4 + H 2 O(l) Net ionic: OH − (aq) + H + (aq) → H 2 O(l)

7.76

HBr + NaOH → H 2 O + NaBr

 1 L  0.400 moles HBr  1 mole NaOH  250. mL     = 0.100 moles NaBr 1L  1000 mL   1 mole HBr  b.

HClO 4 + NaOH → H 2 O + NaClO 4

 1 L   0.300 moles HClO 4   1 mole NaOH  750. mL   = 0.225 moles HClO 4   1L  1000 mL     1 mole HClO 4 

7.77

HCl + NaOH → H 2 O + NaCl

 0.250 moles HCl  1 mole NaOH  1.00 L    = 0.250 moles NaOH 1L   1 mole HCl  b.

HNO 3 + NaOH → H 2 O + NaNO 3  1 L   0.300 moles HNO 3   1 mole NaOH  500. mL   = 0.150 moles NaOH    1L  1000 mL    1 mole HNO 3 

7.78

H 2 C 2 O 4 + 2 NaOH → 2 H 2 O + Na 2 C 2 O 4

 1 L  0.1891 moles NaOH   1 mole H 2 C 2 O 4  43.88 mL     1L  1000 mL    2 mole NaOH  = 0.1660 M H 2 C 2 O 4  1L  25.00 mL    1000 mL 

Chapter 7 7.79

HCl + NaOH → NaCl + H 2 O  1 L  0.0210 moles NaOH  1 mole HCl  26.4 mL    1 mole NaOH  1L  1000 mL    = 0.0222 M HCl  1L  25.00 mL    1000 mL 

7.80

H 2 SO 4 + 2 NaOH → 2 H 2 O + Na 2 SO 4

 1 L  1.17 moles NaOH   1 mole H 2 SO 4  29.88 mL     2 moles NaOH  1L  1000 mL    = 3.50 M H SO 2 4  1L  5.00 mL    1000 mL  b.

HC 2 H 3 O 2 + KOH → 2 H 2 O + KC 2 H 3 O 2

 1 L  0.250 moles KOH   1 mole HC 2 H 3 O 2  35.62 mL     1 mole KOH  1L  1000 mL    = 0.891 M HC H O 2 3 2  1L  10.00 mL    1000 mL  c.

HCl + NaOH → H 2 O + NaCl

 1 L  6.00 moles NaOH  1 mole HCl  20.63 mL    1 mole NaOH  1L  = 12.4 M HCl  1000 mL    1L  10.00 mL    1000 mL  7.81

HI + NaOH → H 2 O + NaI

 1 L  0.250 moles NaOH  1 mole HI  27.15 mL    1 mole NaOH  1L  = 0.272 M HI  1000 mL    1L  25.00 mL    1000 mL  b.

H 2 SO 4 + 2 KOH → 2 H 2 O + K 2 SO 4

 1 L   0.109 moles NaOH   1 mole H 2 SO 4  11.12 mL     2 moles KOH  1L  1000 mL     = 0.0303 M H SO 2 4  1L  20.00 mL    1000 mL 

HCl + NaOH → H 2 O + NaCl

Chapter 7

 1 L  0.250 moles NaOH  1 mole HI  18.40 mL    1 mole NaOH  1L  1000 mL    = 0.0184 M HCl  1L  25.00 mL    1000 mL 

SECTION 7.7 MAINTAINING PH: THE BALANCING ACT 7.82

One component reacts with acid, and one component reacts with base.

7.83

NH 3 (aq) + H + (aq)  NH +4 (aq)

NH +4 (aq) + OH − (aq)  NH 3 (aq)

Yes, a mixture of ammonia and ammonium chloride could behave as a buﬀer when dissolved in water. The ammonia would react with acid to produce ammonium ions and the ammonium ions would react with base to produce ammonia. 7.84

HPO 24 − (aq) + H 3 O + (aq)  H 2 PO −4 (aq) + H 2 O(l) H 2 PO 24 − (aq) + OH − (aq)  HPO 24 − (aq) + H 2 O(l)

(1) = 3.74 (1)

7.85

pH = 3.74 + log

7.86

pH = 3.85 + log

(0.1) = 3.85 (0.1)

pH = 3.85 + log

(1) = 3.85 (1)

The solution in part b has greater buﬀer capacity than the solution in part a because the higher concentration of the buﬀer components will allow it to react with larger added amounts of acid or base.

7.87

Nitrous acid and its conjugate base would make the best buffer with a pH of 3.00. The pK a for nitrous acid is 3.33, which is the closest to 3.00 of the choices given. A buffer has the greatest buffering capacity if the pH it is required to maintain is equal to its pK a .

7.88

pH = 4.74 + log

(0.25) = 4.54 (0.40)

pH = 7.21 + log

(0.40) = 7.81 (0.10)

Chapter 7

7.89

(0.20) = 6.12 (1.50)

pH = 7.00 + log

pH = 12.66 + log

pH = 3.33 + log

pH = 10.25 + log

(0.52) = 12.86 (0.33)

(0.065) = 3.68 (0.029) (0.15) = 9.73 (0.50)

7.90

Three systems that cooperate to maintain blood pH in an appropriate narrow range are blood buﬀer, respiratory, and urinary. The blood buﬀer system provides the most rapid response to changes in blood pH.

7.91

To increase the pH of a buﬀer, more conjugate base needs to be added. To raise the pH of a citric acid-citrate buﬀer from 3.20 to 3.35, add more sodium citrate to the solution.

7.92

No. When the blood buﬀering systems become taxed, the respiratory and urinary systems act to help maintain blood pH.

7.93

Acidosis is a condition of abnormally low blood pH. Alkalosis is a condition of abnormally high blood pH.

7.94

Hb − + H 3 O +  HHb + H 2 O HHb + OH −  Hb − + H 2 O

7.95

A stronger acid reacts with HCO3- to produce H3CO3, which is a weaker acid. The weaker acid has only a slight impact on blood pH.

7.96

Hyperventilation is a symptom of low pH in the body (acidosis)

7.97

The kidneys play an important role in controlling blood pH by controlling the concentration of H+ ions excreted in the urine. Excretion of more H+ ions causes a decrease in the urine pH with a corresponding increase in the blood pH.

7.98

When kidneys function to control blood pH, the concentration of H+ in the blood decreases. As a result, the CO2 in the blood is converted to bicarbonate ions in the blood, in order to maintain the equilibria: H2O + CO2 + H2CO3 ⇆ H+ + HCO3- . Removal of H+ ions causes more carbonic acid to form HCO3-. The source of carbonic acid is CO2. Overall CO2 is converted to HCO3-.

7.99

The pH increases.

7.100

The buﬀer system in urine that keeps the urine pH from dropping lower than 6 is the phosphate buﬀer system.

Chapter 7 7.101

As hydrogen ions are removed from the blood by the kidneys, sodium ions enter the bloodstream. This ionic shift maintains electron charge balance when the kidneys function to increase the pH of blood.

ADDITIONAL EXERCISES 7.102

HA(aq) + H 2 O(l)  H 3 O + (aq) + A − (aq) H 3 O +  = 2.63% x 0.150 M = 0.003945   ph = − log H 3 O +  = − log(0.003945) = 2.404

7.103

If the concentration of H+ ions, in the urine exceeds the urine’s buﬀering capacity, the pH of the urine will drop below pH 6 to produce very acidic urine.

7.104

Ordinary water may be the most eﬀective “sports drink” because the body needs to be hydrated in order to maintain ﬂuid and electrolyte balance in the body. Water comprises a large percentage of the composition of the human body and drinking water can help restore natural balance within the body during and after exercise.

7.105

Phenolphthalein would not be a useful indicator to differentiate between two solutions with pH values of 5 and 7 because it is clear at both of those pH values. The color change in phenolphthalein occurs between pH 8 and 9. Methyl red changes from red to orange around pH 6 and would be a useful indicator to differentiate between solutions with pH values of 5 and 7. Bromothymol blue changes from yellow to blue around pH 7 and would be a useful indicator to differentiate between solutions with pH vales of 5 and 7.

7.106

Smoking is dangerous in the presence of oxygen gas. The abundance of oxygen (an oxidizing agent) would increase the reaction rate for a redox reaction occurring between any reducing agent and the oxygen gas. A lit cigarette or even a small amount of ash containing an ember could provide the activation energy needed for an explosive redox reaction.

7.107

liquid (l) + CO 2 (g)  carbonated beverage (l) While under pressure, the carbon dioxide remains dissolved in the carbonated beverage. The layer of carbon dioxide above the carbonated beverage in the sealed container escapes once the pressure is released; therefore, the reaction shifts to the left as more CO2 is liberated as bubbles in an attempt to reestablish equilibrium. Once the pressure is released, some of the carbon dioxide bubbles out of the solution because the reaction shifts to the left.

Chapter 7 7.108

Cl − (aq) + H 3 O + (aq)  HCl(aq) + H 2 O(l)

The chloride ion is a Brønsted base because it is a proton acceptor. The hydronium ion is a Brønsted acid because it is a proton donor. An alternative way to look at this is reaction is that the chloride ion has a pair of electrons that it can donate to one of the hydrogen atoms in the hydronium ion in order to form a covalent bond and HCl. The electron pair donor (chloride ion) is acting as a base. The electron pair acceptor (hydronium ion) is acting as an acid.

7.109

HbCO + O 2  Hb O 2 + CO By administering pure oxygen to victims of CO poisoning, the equilibrium shifts to the right so that the CO is ﬂushed from the hemoglobin and replaced with oxygen.

Chapter 8

Chapter 8: Introduction to Organic Chemistry: Hydrocarbons CHAPTER OUTLINE 8.1 Organic Chemistry: The Story of Carbon 8.2 Representations of Organic Molecules 8.3 Functional Groups

8.7 Alkenes and Alkynes 8.8 Reactions of Hydrocarbons 8.9 Physical and Chemical Properties of

8.4 Isomers and Conformations 8.5 Classiﬁcation and Naming of Alkanes

Hydrocarbons 8.10 Aromatic Compounds and Benzene

8.6 Cycloalkanes

LEARNING OBJECTIVES/ASSESSMENT When you have completed your study of this chapter, you should be able to: 1.

Distinguish between organic and inorganic chemicals. (Section 8.1; Exercises 8.4, 8.5, and 8.9)

Write condensed, expanded, and skeletal structural representations for compounds. (Section 8.2; Exercises 8.17 and 8.23)

Identify functional groups within larger molecules. (Section 8.3; Exercises 8.30 and 8.31)

Use structural formulas to identify compounds that are isomers of each other. (Section 8.4; Exercises 8.37 and 8.39)

Assign IUPAC names to alkanes. (Section 8.5; Exercise 8.45)

Draw structural formulas for alkanes when given the IUPAC name. (Section 8.5; Exercise 8.47)

Assign IUPAC names to cycloalkanes. (Section 8.6; Exercise 8.53)

Write the IUPAC names of alkenes and alkynes from their molecular structures. (Section 8.7; Exercise 8.69)

Write chemical equations for addition reactions of alkenes. (Section 8.8; Exercises 8.89)

10. Use Markovnikov’s rule to predict the major products of certain alkene addition reactions. (Section 8.8; Exercises 8.85 and 8.87) 11. Write chemical equations for addition polymerization. (Section 8.8; Exercises 8.91 and 8.92) 12. Describe the key physical properties of alkanes, alkenes, and alkynes. (Section 8.9; Exercises 8.94 and 8.95) 13. Name aromatic compounds. (Section 8.10; Exercise 8.105 and 8.107) 14. Draw structural formulas for aromatic compounds. (Section 8.10; Exercises 8.109 and 8.111)

LECTURE HINTS AND SUGGESTIONS 1.

When introducing structural representation make sure to provide examples explaining why the diﬀerent representations are useful. The expanded form shows every bond and is useful when learning that carbon needs four bonds to achieve an octet. The condensed form provides less detail; however, it still shows the general attachment of the compound and saves space. Even though the condensed form saves space, it is completely impractical when representing something like a fatty acid tail, or something larger like a DNA molecule. In the case of complex biomolecules, it is essential to represent structures in the skeletal form, as the condensed and expanded representations would be too chaotic.

Chapter 8 2.

Emphasize the usefulness of classifying organic compounds according to functional groups. Illustrate how compounds with the same functional group will participate in similar chemical reactions. Additionally, compounds with the same functional groups will often exhibit similar physical characteristics (melting points, boiling points, and solubility.)

Explain that unsaturated hydrocarbons contain carbon-carbon multiple bonds and since carbon can only have four bonds, the carbons involved in these bonds have lost hydrogens, thus the compound is no longer saturated with hydrogens.

Use molecular models in class to illustrate the diﬀerent types of isomerism. Students are easily confused as to when structures are equivalent or non-equivalent. Models are incredibly useful at this point to illustrate the diﬀerences. Two models which represent equivalent rather than isomeric structures can be shown to superimpose upon turning or upon rotation about single bonds.

A full-page image of Table 8.2 is a useful handout for students to reference when learning pertinent chemical functional groups (see next page.)

Figures 8.4 and 8.6, Tables 8.4-8.6, and an alkane nomenclature example can be combined to create a double-sided handout introducing students to structural representations and organic molecule naming (see next three pages.)

Chapter 8

Functional Groups

Chapter 8

Structural Representations Expanded structural representation: A structural molecular formula showing all the covalent bonds. Condensed structural representation: A structural molecular formula showing the general arrangement of atoms but without all the covalent bonds. Skeletal structural representation: A structural molecular formula showing carbon atoms as points, with hydrogens bonded to carbon assumed. Other atoms (e.g., nitrogen, oxygen, and chlorine) are shown explicitly, as well as hydrogens bonded to these noncarbon atoms.

Naming Alkanes 1.

Number the longest chain in the direction that give the substituents the lowest numbers.

Name the longest linear carbon chain.

Chapter 8 3.

Locate and name the attached groups.

Combine the longest chain and indicate the number and position of attached groups a to give the full IUPAC name (list attached groups in alphabetical order.) Example: Provide the IUPAC name of the following alkane

Number the longest chain:

Name the longest linear carbon chain. hexane

Locate and name the attached groups. The one-carbon group at position 2 is a methyl group. The two-carbon group at position 4 is an ethyl group.

Combine the longest chain and indicate the number and position of attached groups a to give the full IUPAC name (list attached groups in alphabetical order.) 4-ethyl-2-methylhexane

Chapter 8

Solutions to All End-of-Chapter Questions What follows are more complete explanations/full solutions to the EOC exercises whose answers are published in shorter form at the end of the textbook.

SECTION 8.1 ORGANIC CHEMISTRY: THE STORY OF CARBON 8.1

The term, organic compound, originally referred to those compounds found in living, “organic” matter.

8.2

Fruits and vegetables, the family pet, plastics, sugar, cotton, and wood are a few of many items composed of organic compounds.

8.3

Wöhler heated a compound that was commonly accepted as an inorganic compound. It had not been associated with the vital force that characterized organic material. Upon heating, it formed urea, an organic substance found in many animals. This experiment caused serious questions about the vital force theory.

8.4

All organic compounds contain carbon atoms.

8.5

KBr

inorganic

H2 O

inorganic

c. d.

H—C≡C—H LiOH

organic inorganic

8.6

a. b. c. d.

inorganic organic inorganic inorganic

8.7

Covalent bonding is the most prevalent type of bonding in organic compounds.

8.8

The majority of all compounds that are insoluble in water are organic because organic compounds are often nonpolar and nonpolar compounds are insoluble in polar water.

8.9

a. b. c.

A liquid that readily burns is most likely an organic compound because organic compounds can exist in any of the states of matter and are often flammable. A white solid with a melting point of 735 °C is most likely an inorganic compound because inorganic compounds are usually high melting point solids. A liquid that floats on the surface of water and does not dissolve is most likely an organic compound because organic compounds can exist in any of the states of matter and often have low water solubility. A compound that exists as a gas at room temperature and ignites easily is most likely an organic compound because organic compounds can exist in any of the states of matter and are often flammable. A solid substance that melts at °65 C is most likely an organic compound because organic compounds can exist as low melting point solids. 6

Chapter 8 8.10

a. b. c. d.

A relatively low melting point is characteristic of organic compounds. Solubility in water and lack of ﬂammability is typical of inorganic compounds. Conductivity of water solutions is typical for ionic substances (inorganic compounds). Inorganics often have high melting points.

SECTION 8.2 REPRESENTATIONS OF ORGANIC MOLECULES 8.11

8.12

Chapter 8 8.13

8.14

a. b. c. d.

Correct Incorrect, the center carbon has five bonds. Incorrect, the carbon on the right has five bonds. Incorrect, the second carbon from the left has only three bonds.

8.15

8.16

Chapter 8

CH 3 CH 2 CH 2 − NH 2

condensed structural formula

CH 3 − CH 2 − CH 2 − NH 2

condensed structural formula

C3 H9 N

molecular formula

8.18

a. b. c. d.

Expanded structural formula Condensed structural formula Molecular formula Condensed structural formula

8.19

Expanded

Condensed

Expanded

Condensed

Expanded

Condensed

8.17

a. b.

8.20

Chapter 8

8.21

8.22

Expanded

Condensed

Expanded

Condensed

Expanded

Chapter 8 8.23

8.24

8.25

8.26

Chapter 8

SECTION 8.3 FUNCTIONAL GROUPS 8.27

Unsaturated hydrocarbons contain one or more multiple bonds. Saturated hydrocarbons contain only single bonds.

8.28

a. b. c. d.

saturated unsaturated unsaturated unsaturated

8.29

a. b. c. d.

saturated unsaturated saturated unsaturated

8.30

Name the four functional groups circled in the antibiotic cephalexin: a. carboxylic acid b. amide c. amine d. aromatic ring

8.31

Name the four functional groups circles in the steroid levonorgestrel: a. ketone b. alkene c. hydroxyl group (alcohol) d. alkyne

Chapter 8

SECTION 8.4 ISOMERS AND CONFORMATIONS 8.32

Numerous organic compounds exist because each carbon atom can form four covalent bonds, including bonds to other carbon atoms, and carbon containing molecules can exhibit isomerism. Carbon has the unique capacity to form extremely long chains which are stable. Since the same number of carbon atoms can link together to form both straight and branched chains, the number of possibilities rapidly increases as the number of carbon atoms increases.

8.33

When a central carbon atom bonds to four other atoms, a tetrahedral molecular geometry exists.

8.34

When a central carbon atom bonds to three other atoms, a trigonal planar molecular geometry exists.

8.35

carbon hydrogen oxygen

8.36

not isomers (different molecular formulas— C 4 H8 and C 4 H10 )

isomers (same molecular formula— C 5 H12 , but different structures)

not isomers (different molecular formulas— C 4 H10 O and C 4 H8 O )

isomers (same molecular formula— C 3 H6 O , but different structures)

8.37

a. b. c. d.

Yes, structural isomers. Yes, structural isomers. No, they differ in number of hydrogens. No, they differ in number of hydrogens.

8.38

The isomers are (a), (b), (e), and (f)—all have the molecular formula (C3H6O2). e. C 3 H6 O 2 a. C 3 H6 O 2

4 bonds 1 bond 2 bonds

nitrogen bromine

C3 H8 O 2

C3 H6 O 2

C4 H8 O

C3 H8 O 2

8.39

a. b. c. d.

branched branched normal normal

8.40

a. b. c. d.

branched (3 C longest chain) normal (5 C longest chain) normal (4 C longest chain) normal (5 C longest chain)

3 1

Chapter 8 8.41

8.42

same compound ( C 4 H10 normal)

same compound ( C6 H14 normal)

isomers ( C 4 H10 branched and normal)

same compound ( C 5 H12 branched)

a. b. c. d.

the same compound, 2-methylbutane the same compound, 2-methylbutane structural isomers the same compound, butane

SECTION 8.5 CLASSIFICATION AND NAMING OF ALKANES 8.43

isopropyl

butyl

isobutyl

methyl

8.44

a. b. c. d.

propyl ethyl sec-butyl t-butyl

8.45

3-methylpentane

Chapter 8

8.46

3,4,6-trimethylpentane

3-ethyl-2,3-dimethylpentane

4,7-diethyl-5-methyldecane

3-ethyl-5-methylnonane

2-methylbutane

2,2-dimethylpropane

Chapter 8

8.47

2,3-dimethylpentane

4,6,6-triethyl-2,7-dimethylnonane

5-sec-butyl-2,4-dimethylnonane

3-ethylpentane

2,2-dimethylbutane

Chapter 8

8.48

4-ethyl-3,3-dimethyl-5-propyldecane

5-sec-butyldecane

2,2,4-trimethylpentane

4-isopropyloctane

3,3-diethylhexane

5-t-butyl-2-methylnonane

Chapter 8

8.49 pentane 8.50

2-methylbutane

2,2-dimethylpropane

Diﬀerent conformations of an alkane are not considered structural isomers because the bonding order is not changed in the diﬀerent conformations, which means the conformations correspond to the same molecule. Two structures would be structural isomers only if bonds were broken and remade to convert one into the other.

SECTION 8.6 CYCLOALKANES 8.51

a. b. c.

two one zero

8.52

a. b. c.

two one zero

8.53

cyclopentane

1,2-dimethylcyclobutane

1,1-dimethylcyclohexane

Chapter 8

8.54

8.55

1,2,3-trimethylcyclobutane

1-ethyl-1-1-methylcyclopentane

t-butylcyclohexane

2-chloro-1,3-dimethylcyclopentane

1-ethyl-2-propylcyclopropane

ethylcyclobutane

1,1,2,5-tetramethylcyclohexane

Chapter 8

8.56

8.57

8.58

1-butyl-3-isopropylcyclopentane

1-ethyl-1-propylcyclopentane

1,2-dimethylcyclopropane

same compound ( C6 H12 ; 1,2-dimethylcyclobutane)

structural isomers ( C8 H16 ; 1-ethyl-methylcyclopentane & 1-ethyl-3-methylcyclopentane)

structural isomers ( C 5 H10 ; ethylcyclopropane & 1,2-dimethylcyclopropane)

structural isomers ( C8 H16 ; 1,1-dimethylcyclohexane & 1,2-dimethylcyclohexane)

isopropylcyclobutane

ethylcyclopropane

cyclopentane

Chapter 8

1,1-dimethylcyclopropane

methylcyclobutane

1,2-dimethylcyclopropane 8.59

Structural isomers diﬀer in the order the atoms are bonded together. Geometric isomers have the same atomic bonding order but diﬀer in the way the branches are oriented in space. For geometric isomers to be possible, the structure must have some feature which prevents free rotation about the bond in question. This occurs in cycloalkanes.

8.60

trans-1-ethyl-2-methylcyclopropane

cis-1-bromo-2-chlorocyclopentane

trans-1-methyl-2-propylcyclobutane

trans-1,3-dimethylcyclohexane

SECTION 8.7 ALKENES AND ALKYNES 8.61

An unsaturated hydrocarbon is a hydrocarbon containing one or more multiple bonds.

Chapter 8 8.62

An alkene is a hydrocarbon that contains at least one carbon-carbon double bond. An alkyne is a hydrocarbon that contains at least one carbon-carbon triple bond. An aromatic hydrocarbon is a compound that contains a benzene ring or other similar feature.

8.63

a. b. c.

CH3—CH2—CH3 CH3CH=CHCH3

saturated unsaturated unsaturated

alkene alkyne

unsaturated

alkene

saturated

8.64

a. b. c. d.

unsaturated, alkene unsaturated, alkene unsaturated, alkene not unsaturated

8.65

a. b. c. d.

Can exist in nature. Cannot exist, because two carbons each have ﬁve bonds. Can exist in nature. Cannot exist, because carbon 2 has ﬁve bonds.

8.66

A B C

Br H — CH 3

— CH 3

A B C D

H F F Cl (Note that C and D can be reversed)

8.67

8.68‫ككك‬a.

2-methyl-3-hexene

6-methyl-2,4-heptadiene

Chapter 8

8.69

cyclopentene

2-pentyne

2,5-dimethyl-1,5-hexadiene

3-isopropyl-5-methylcyclohexene

3-ethylcyclohexene

2-butene CH3CH=CHCH3

3-ethyl-2-pentene

4,4-dimethyl-2-hexyne

4-methylcyclopentene

‫ككككككككككككككككككك‬

Chapter 8

6-bromo-2-methyl-3-heptyne

1-ethyl-2,3-dimethylcyclopropene

‫ككككككككككك‬

8.70

8.71

8.72

6-methyl-1,4-heptadiene

4-methylcyclohexane

2-ethyl-1,4-pentadiene

3-ethyl-2-hexene

3-methyl-1,3-pentadiene

C 5 H8 ; alkyne

1-pentyne

4-isopropyl-3,3-dimethyl-1,5-octadiene

3,4-dimethyl-1-pentene

‫ك‬

2-pentyne

3-methyl-1-butyne

Chapter 8

C 5 H8 ; diene 1,2-pentadiene

2,3-pentadiene

1,3 pentadiene

1,4-pentadiene

3-methyl-1,2-butadiene

8.73

C 5 H8 ; cyclic alkene

1-ethylcyclopropene

3-ethylcycloproene

1,3-dimethylcyclopropene

3,3-dimethylcyclopropene

3-methylcyclobutene

cyclopentene

1-butyne

2-butyne

1,3-butadiene

1,2-butadiene

1,2-dimethylcyclopropene

1-methylcyclobutene

Chapter 8 8.74

Carbon atoms in a double bond have a trigonal planar geometry where the two atoms attached to the double bonded caron atom as well as the other carbon atom in the double bond are in the same plane, separated by bond angles of 120°.

8.75

Structural isomers have a diﬀerent order of linkage of atoms. Geometric isomers have the same order of linkages of atoms; however, the three-dimensional structures are unique because of restricted rotation as a result of a ring or a double bond.

8.76

a. CH 3CH 2CH 2CH 2CH=CH2; no geometric isomers b.

cis-3-hexene

trans-3-hexene

8.77

; no geometric isomers

CH 2 =CHCH 3 ; no geometric isomers

cis-2-bromo-butene

trans-2-bromo-2-butene

Chapter 8

cis-1-chloro-propene

8.78

trans-1-chloro-propene

cis-3-hexene b.

trans-3-heptene

8.79

cis-4-methyl-2-pentene 8.80

a. b. c. d.

1,1,2,2-tetrachloroethene cis-2-pentene trans-1-chloropropene 2-methyl-2-butene

8.81

a. b. c. d.

cis-1,2-dibromo-1,2-dichloroethene trans-2-bromo-2-butene trans-3-hexene propene

trans-5-methyl-2-heptene

Chapter 8 8.82

Geometric isomerism is not possible in alkynes because the geometry around a triple bond is linear. Each carbon atom in the triple bond only has one other attached group, unlike each carbon atom in a double bond which has two attached groups.

8.83

Acetylene is the simplest alkyne and it is used in torches for welding steel and in making plastics and synthetic fibers.

8.84

The geometry of a triple bond is linear.

SECTION 8.8 REACTIONS OF HYDROCARBONS 8.85

8.86

Markovnikov’s rule states that when a heteroatomic compound containing hydrogen is added to a multiple bond, the hydrogen will attach to the carbon atom in the multiple bonds that is directly bonded to more hydrogen atoms. The following reaction is an example of this rule:

Using a Pt catalyst: CH2=CHCH2CH3 + H2 → CH3CH2CH2CH3

8.87

8.88

Chapter 8 b.

8.89

8.90

a. H 2, Pt catalyst b. H 2 O, H2SO4

Chapter 8

8.91

8.92

a. -CH2CH2-CH2CH2-CH2CH2b.

c‫كﻟ‬

d‫كﻟ‬

8.93

A monomer is the starting material for a polymer. It can be a small molecule. A polymer is a large molecule made up of repeating units (often thousands of repeating units). An addition polymer is a polymer formed by the reaction of monomers that contained multiple bonds to form the repeating units of a polymer. A copolymer is a polymer formed by the reaction of at least two diﬀerent types of monomers.

Chapter 8

SECTION 8.9 PHYSICAL AND CHEMICAL PROPERTIES OF HYDROCARBONS 8.94‫ك‬

a. b. c. d.

Decane is a liquid at room temperature. Decane is not soluble in water. Decane is soluble in hexane. Decane is less dense than water.

8.95

a. b.

Propane has the higher molecular weight and the higher boiling point. Hexane has the higher molecular weight and the higher boiling point.

8.96

a. b.

hexane decane

8.97

a. b. c.

liquid gas liquid

8.98

a. b. c.

gas solid liquid

8.99

Alkanes of low molecular weight have lower melting and boiling points than water because alkanes only experience dispersion forces, which are much weaker than the hydrogen bonding that water experiences. Weaker intermolecular forces result in lower melting and boiling points.

8.100

SECTION 8.10 AROMATIC COMPOUNDS AND BENZENE 8.101

The circle within the hexagon represents the evenly distributed electrons in the pi lobes of the six carbon atoms in the benzene ring in the structural formula for benzene.

8.102

Aromatic means a molecule contains a benzene ring or one of its structural relatives.

Chapter 8 8.103

Limonene does not contain a benzene ring; therefore, it is not considered aromatic. Aromaticity reﬂects structure, not the physical property of a fragrance.

8.104

a. 1,3,5-trimethylbenzene b. 1,4-diethylbenzene or p-diethylbenzene

8.105

ethylbenzene

8.106

a. m-bromophenol b. p-ethlylaniline

8.107

1-bromo-3-ethylbenzene or m-bromoethylbenzene

m-bromotoluene p-ethylaniline 8.108

a. o-ethylphenol

Chapter 8

b. m-chlorobenzoic acid

c. 3-methyl-3-phenylpentane

8.109

p-chloroaniline

8.110

m-ethylphenol

a. 2-chlorophenol or o-chlorophenol

2-phenyl-1-pentene

Chapter 8 b. 3-chlorophenol or m-chlorophenol

c. 4-chlorophenol or p-chlorophenol

8.111

8.112

Benzene does not readily undergo addition reactions characteristic of other unsaturated compounds because the delocalized pi cloud of the benzene ring makes the ring so stable that addition reactions do not occur. An addition reaction would result in the loss of one of the double bonds, and consequently, disrupt the delocalized π system. This is not favored because it results in a loss of stability. Benzene, therefore, undergoes substitution reactions instead of addition reactions.

8.113

Aromatic hydrocarbons are nonpolar molecules that are insoluble in water and soluble in nonpolar solvents. They are also less dense than water.

8.114

a. b. c. d.

A solvent A vitamin An essential amino acid Starting material for dyes

benzene, toluene riboﬂavin phenylalanine aniline

Chapter 8

CHEMISTRY FOR THOUGHT 8.115

Cyclohexene readily undergoes addition reactions. Benzene resists addition reactions and favors substitution reactions. Both benzene and cyclohexene can undergo combustion.

8.116

The study of organic compounds might be important to someone interested in the health or life sciences because living organisms are composed of mostly organic compounds and most medications are also organic compounds.

8.117

Very few aqueous solutions of organic compounds conduct electricity because very few organic compounds dissolve in water, and of those that do, very few dissociate into ions (ions are needed to conduct electricity through the solution).

8.118

The low melting point of aspirin indicates that the molecules in aspirin have weak forces between the molecules. (Note: the forces are dispersion forces.)

8.119

If a semi-truck loaded with cyclohexane overturns during a rainstorm, spilling its contents over the road embankment and the rain continues, then the cyclohexane will ﬂoat on top of the standing water and run oﬀ along the water drainage route. Some of the cyclohexane will probably also evaporate because it has weak intermolecular forces.

8.120

Alkanes do not mix with water and are less dense than water; therefore, when an oil spill occurs, the oil floats on top of the water. Alkanes are relatively unreactive; therefore, short of burning the oil, a chemical reaction cannot be performed to “neutralize” the oil spill. Burning the oil is not a good idea because of the wildlife that quickly becomes disabled by contact with the oil and because crude oil often contains compounds (other than the pure hydrocarbons) that may produce toxic products when burned. Birds that come into contact with the oil must be cleaned with soap before they are able to fly again. A bird will try to clean the oil off its own feathers and ingest the oil, which does damage to its internal organs. Sea otters and killer whales as well as small organisms at the bottom of the food chain are also impacted by contact with oil.

8.121

1-isopropyl-4-methylcyclohexene (menthene) 8.122

Propene does not exhibit geometric isomerism because one of the carbon atoms in the double bond is also bonded to two hydrogen atoms. In order to exhibit geometric isomerism, both carbon atoms in the double bond must also be bonded to two unique groups.

8.123

Alkynes have slightly higher boiling points and densities than structurally equivalent alkanes because the linear alkyne molecules can pack more closely together than the alkane molecules.

Chapter 8 This close packing allows for stronger dispersion forces, which cause the alkynes to have higher boiling points and densities than structurally equivalent alkanes.

Chapter 9

Chapter 9: Alcohols, Ethers, and Amines CHAPTER OUTLINE 9.6 The Nomenclature and Classiﬁcation of Amines 9.7 Properties of Amines 9.8 Biologically Signiﬁcant Amines

9.1 Nomenclature and Classiﬁcation of Alcohols 9.2 Physical Properties of Alcohols 9.3 Reactions of Alcohols 9.4 Applications of Alcohols 9.5 Ethers

LEARNING OBJECTIVES/ASSESSMENT When you have completed your study of this chapter, you should be able to: 1.

Name alcohols using common names and the IUPAC naming system. (Section 9.1; Exercises 9.3, 9.4, and 9.7)

Classify alcohols as primary, secondary, or tertiary on the basis of their structural formulas. (Section 9.1; Exercises 9.13 and 9.15)

Explain how hydrogen bonding inﬂuences the physical properties of alcohols. (Section 9.2; Exercises 9.19 and 9.21)

Predict the products when alcohol reacts with diﬀerent acids and bases. (Section 9.3; Exercises 9.25 and 9.31)

Write chemical equations for alcohol dehydration and oxidation reactions. (Section 9.3; Exercises 9.33 and 9.35)

List uses for three speciﬁc alcohols. (Section 9.4; Exercises 9.42 and 9.43)

Name ethers by using common names. (Section 9.5; Exercises 9.47 and 9.49)

Describe the key physical and chemical properties of ethers. (Section 9.5; Exercises 9.53 and 9.55)

Assign common names to simple amines. (Section 9.6; Exercises 9.57)

10. Classify amines as primary, secondary, or tertiary based on their structural formulas. (Section 9.6; Exercises 9.61 and 9.63) 11. Predict the products when amines react with different acids and bases. (Section 9.7; Exercises 9.67) 12. Explain how hydrogen bonding influences the physical properties of amines. (Section 9.7; Exercises 9.71, 9.73, and 9.79) 13. Describe the uses of three different crucial biological amines. (Section 9.8; Exercises 9.83, 9.87, and 9.89)

LECTURE HINTS AND SUGGESTIONS 1.

Have the students investigate the ingredients of commercial products that contain alcohol and phenols. In class, relate the function of the compound to its molecular structure and its chemical and physical properties.

Point out that amines are derivatives of ammonia, where some, or all, of the hydrogen-nitrogen bonds have been replaced with carbon-nitrogen bonds. Show the similarities between the common names of amines and alcohols.

Chapter 9 3.

Some basic principles of "drug design" may be explained using nitrogen-containing drugs from this chapter. To better understand the real-world application of this concept, discuss medicines such as carvedilol (Example 9.10,) phenylephrine (Example 9.11,) and morphine (Figure 9.31.)

Many of the students for whom this text is written, have a strong interest in health and biological systems. To better understand the biological roles of amines, emphasize the structure of neurotransmitters: acetylcholine, norepinephrine, dopamine, and serotonin (Figure 9.35.)

Solutions to All End-of-Chapter Questions What follows are more complete explanations/full solutions to the EOC exercises whose answers are published in shorter form at the end of the book.

SECTION 9.1 NOMENCLATURE AND CLASSIFICATION OF ALCOHOLS 9.1

Alcohol: R—OH

Phenol:

where the R and the OH is a hydrocarbon group 9.2

9.3

9.4

9.5

2-propanol

3-methyl-3-pentanol

2,2-dimethylcyclopentanol

1,2,4-hexanetriol

1- propanol

3-phenyl-1-propanol

2-chloro-3-methyl-1-butanol

2-chloro-5-ethylcyclopentanol

methyl alcohol

isopropyl alcohol

ethyl alcohol

ethylene glycol

2-methyl-1-butanol 2-bromo-3-methyl-3-pentanol

1-methylcyclopentanol

Chapter 9 9.6

2-methyl-2-pentanol

9.7

1,3-butanendiol

2-isopropylphenol o-isopropylphenol

9.8

1-ethyl-cyclopentanol

2,3-dimethylphenol

2-ethylphenol o-ethylphenol 2-ethyl-4,6-dimethylphenol 9.9

m-methylphenol 9.10

2,3-dicholorophenol b.

p-chlorophenol

2,5-diisopropylphenol

Chapter 9 9.11

A primary alcohol has a hydroxy group bonded to a carbon atom that is only bonded to one other carbon atom. A secondary alcohol has a hydroxy group bonded to a carbon atom that is bonded to two other carbon atoms. A tertiary alcohol has a hydroxy group bonded to a carbon that is bonded to three other carbon atoms.

9.12

9.13

Tertiary

Secondary

Tertiary

Primary

Secondary

9.14

9.15 1-butanol/1° 9.16

2-butanol/2°

2-methyl-2-propanol/3°

2-methyl-1-propanol/1°

Both compounds are aromatic but only the structure on the left has the –OH attached to the benzene ring making it a phenol.

SECTION 9.2 PHYSICAL PROPERTIES OF ALCOHOLS 9.17

a. b.

1-propanol < 1-butanol < 1-pentanol pentane < 2-butanol < 1,2-butanediol

Chapter 9 9.18

9.19

(lowest boiling point) methanol < ethanol < 1-propanol (highest boiling point) All three molecules experience both dispersion forces and hydrogen bonding. Since they all hydrogen bond to the same extent, their boiling points differ according to the strength of their dispersion forces. Dispersion forces are stronger for larger molecules.

(lowest boiling point) butane < 1-propanol < ethylene glycol (highest boiling point) These three compounds have comparable molar masses, so their dispersion forces are approximately equal. Butane experiences only dispersion forces, and therefore, has the lowest boiling point. 1-Propanol has only one hydrogen bonding site per molecule, while ethylene glycol has two. Ethylene glycol can, therefore, hydrogen bond more strongly than 1-propanol and has the higher boiling point.

2,3,4-pentanetriol should be more soluble in water because it has more alcohol groups capable of forming hydrogen bonds with water molecules.

9.20

3-hexanol for the same reason as a.

1-butanol should be more soluble in water because it has fewer carbons.

butane or 2-butanol

2-butanol will be more soluble in water than butane because 2-butanol will hydrogen bond with the water molecules, while nonpolar butane will not.

2-propanol or 2-pentanol

2-propanol will be more soluble in water than 2-pentanol because while both compounds will hydrogen bond with water, the 2-propanol has a shorter carbon chain and lower molecular weight than the 2-pentanol, which increases water solubility.

2-butanol or 2,3-butanediol

2-,3-butanediol will be more soluble in water than 2-butanol because it contains two hydroxy groups, while the 2-butanol only contains one hydroxy group. The two hydroxy groups increase the hydrogen bonding interactions with water, which increases the solubility of the compound in water.

Chapter 9 9.21

9.22

SECTION 9.3 REACTIONS OF ALCOHOLS 9.23

The use of glycerol in lotions helps retain water and keep the skin moist because glycerol hydrogen bonds to water molecules, which keeps the skin hydrated.

9.24

9.25

a. b.

9.26

Reactant

Product

Chapter 9 9.27

Reactant

Product

9.28

CH 3 − O − CH 3

CH 3 CH 2 CH 2 CH 2 − O − CH 2 CH 2 CH 2 CH 3

9.29

Reactant a.

Product

CH 3 CH 2 CH 2 − O − CH 2 CH 2 CH 3

CH 3 CH 2 CH 2 − OH

9.30

9.31

Chapter 9 9.32

CH3CH=CHCH3

CH 3 CH 2 CH 2 − O − CH 2 CH 2 CH 3

9.33

There are two possible answers: and

CH 3 CH 2 CH 2 CH 2 − OH

CH 3 CH 2 − OH

9.34

b. c.

CH3-OH

9.35

9.36

secondary

primary

secondary

primary

secondary

primary

Chapter 9 9.37

Reactant

Product

9.38

Reactant

Product

Reactant

9.39 a.

Chapter 9

SECTION 9.4 IMPORTANT ALCOHOLS 9.40

If in the laboratory methanol is accidentally spilled on one’s clothing, it would be a serious mistake to just let it evaporate because the methanol in contact with the skin will be absorbed through the skin. Methanol is toxic for humans and can cause blindness and death. The methanol-infused clothing needs to be changed immediately.

9.41

Glycerol can be added to the ﬁlling of chocolate cordials because it is nontoxic and it increases the moisture in foods.

9.42

9.43

a moistening agent in many cosmetics

1,2,3-propanetriol (glycerin, glycerol)

the solvent in solutions called tinctures

ethanol (ethyl alcohol)

automobile antifreeze

1,2-ethanediol (ethylene glycol)

rubbing alcohol

2-propanol (isopropyl alcohol)

menthol

ethanol

methanol

SECTION 9.5 ETHERS 9.44

R—O—R’ where the R and the R’ are hydrocarbon groups

9.45

o-phenylphenol

a disinfectant for cleaning walls

2-benzyl-4-chlorophenol

9.46

an antiseptic found in some mouth washes

4-hexylresorcinol

an antioxidant used to prevent rancidity in foods

BHA, BHT

CH 3 − O − CH 2 CH 3 ethyl methyl ether

phenyl propyl ether

di-sec-butyl-ether

Chapter 9 9.47

CH 3 − O − CH 2 CH 2 CH 2 CH 3

butyl methyl ether ethyl isopropyl ether

diphenyl ether

9.48

CH3—O—CH2CH2CH3

methyl propyl ether b. isopropyl phenyl ether cyclopentyl ethyl ether

9.49

CH3CH2—O—CH2CH2CH3 1-ethoxypropane

b. 2-ethoxypropane d. c. 1,2-dimethoxycyclopentane ethoxybenzene

9.50

ethyl methyl ether

butyl phenyl

Chapter 9 9.51

methyl isopropyl ether

1,2-diethoxycyclopentane

phenyl propyl ether

2-phenoxy-2-butene

2-methoxypentane

9.52

For compounds of comparable molecular weight. Alcohols are more soluble than ethers, which are more soluble than alkanes. 9.53

CH3CH2—OH> CH3CH2—O—CH3> CH3CH2CH2CH3 All of the compounds have approximately the same molecular mass. All compounds experience dispersion forces, but since these three have similar molecular masses, the strength of their dispersion forces is similar. The major diﬀerences between these molecules are their key functional groups and the strength of their interparticle forces. The 1-propanol will have the strongest hydrogen bonds to water because of its hydroxy group; therefore, 1-propanol is the most soluble compound. The ethyl methyl ether (methoxyethane) can also hydrogen bond to water; therefore, it is more soluble than the butane, but less soluble than the 1-propanol. The butane cannot form hydrogen bonds with water; therefore, it is the least soluble.

9.54

The order is the same as in the answer to Exercise 9.52. Alcohols have the highest boiling point due to hydrogen bonding. Pure ethers exhibit no hydrogen bonding. Ethers are slightly higher boiling than alkanes due to the oxygen atom: alcohol, ether, alkane.

9.55

CH3CH2CH2—OH> CH3CH2—O—CH3> CH3CH2CH2CH3 All of the compounds have approximately the same molecular mass. All compounds experience dispersion forces, but since these three have similar molecular masses, the strength of their dispersion forces is similar. The major diﬀerences between these molecules are their key functional groups and the strength of their interparticle forces. The 1-propanol will have the strongest intermolecular forces (hydrogen bonds in addition to the dispersion forces); therefore,

Chapter 9 1-propanol has the highest boiling point. The ethyl methyl ether (methoxyethane) has the next strongest intermolecular forces (dipolar forces in addition to the dispersion forces); therefore, methoxyethane has a lower boiling point than 1-propanol. The butane has the weakest intermolecular forces (only dispersion forces); therefore, the butane has a lower boiling point than methoxyethane. 9.56

Diethyl ether is hazardous to use as an anesthetic or as a solvent in the laboratory because it is very volatile and ﬂammable.

SECTION 9.6 THE NOMENCLATURE AND CLASSIFICATION OF AMINES 9.57

9.58

a. cyclohexyl amine b. N-cyclohexylethylamine c. sec-butylamine

9.59

Chapter 9

9.60

9.61

Primary amines have one carbon atom bonded to the nitrogen atom. Secondary amines have two carbon atoms directly bonded to the nitrogen atom. Tertiary amines have three carbon atoms directly bonded to the nitrogen atom.

9.62

Primary amines have the formula: Secondary amines have the formula: Tertiary amines have the formula:

9.63

9.64

Chapter 9

SECTION 9.7 PROPERTIES OF AMINES 9.65

CH3—NH2 is a Brønsted base because it can accept a proton from a proton donor. The lone pair of electrons on the nitrogen atom is available to form a bond to hydrogen ions.

9.66 9.67

9.68

CH 3

OH −

N H2

CH 3

9.69

Amine salts are more soluble in water, and therefore in blood, than their corresponding amines; consequently, the amine salts are the preferred form for drugs.

9.70

The structure of a quaternary ammonium salt diﬀers from the structure of a salt of a tertiary amine because the nitrogen atom of a quaternary ammonium salt is bonded directly to four carbon atoms, while the nitrogen atom of a salt of a tertiary amine is bonded directly to three carbon atoms and one hydrogen atom.

9.71

All low molecular weight amines are water soluble because all amines can hydrogen bond with water. Low molecular weight amines have small aliphatic portions, thus the hydrogen bonds to water are strong enough to allow the amines to dissolve.

9.72

The boiling points of amines are lower than those of the corresponding alcohols because the hydrogen bonds formed between nitrogen and hydrogen are weaker than the hydrogen bonds formed between oxygen and hydrogen.

Chapter 9 9.73

The boiling points of tertiary amines are lower than the boiling points of primary and secondary amines because tertiary amines cannot hydrogen bond to each other, while primary and secondary amines can. The weaker intermolecular forces in tertiary amines allow them to boil at a lower temperature.

9.74

9.75

9.76

9.77

alkene < amine < carboxylic acid

9.78

alkane < amine < alcohol

Chapter 9 9.79

9.80

because it has fewer hydrophobic carbon atoms (lower molecular weight)

As a primary amine, it can exhibit greater hydrogen bonding with water molecules.

CH 3 CH 2 NH 2 CH 3 Has a lower molecular weight and fewer hydrophobic carbon atoms.

SECTION 9.8 BIOLOGICALLY SIGNIFICANT AMINES 9.81

A neuron consists of a bulbous body called the soma attached to a long stemlike projection called an axon. Numerous short extensions called dendrites are attached to the large bulbous end of the soma, and filaments called synaptic terminals are attached to the end of the axon. A small gap called a synapse exists between the synaptic terminals of the axon and the dendrites of the next neuron.

9.82

The gap between neurons is called a synapse.

9.83

Two amino acids that are starting materials for the synthesis of neurotransmitters are tyrosine (produces norepinephrine) and tryptophan (produces serotonin).

9.84

The two amines often associated with the biochemical theory of mental illness are norepinephrine and serotonin.

9.85

During nerve impulse transmission, neurotransmitters carry the nerve impulses from one nerve cell to another.

9.86

Four neurotransmitters important in the central nervous system are acetylcholine, norepinephrine, dopamine, and serotonin.

9.87

Epinephrine is the fight-or-flight hormone because it is released in response to pain, anger, or fear and provides glucose for a sudden burst of energy.

9.88

Alkaloids are derived from plants.

9.89

Alkaloids are weakly basic because they are amines and amines are weakly basic.

9.90

a. b. c.

found in cola drinks used to reduce saliva flow during surgery present in tobacco

caffeine atropine nicotine

Chapter 9 d. a cough suppressant e. used to treat malaria f. an effective painkiller

codeine quinine morphine

9.91 This reaction is exothermic in the forward direction; therefore, at the high temperature of 180 °C, the reaction equilibrium shifts to the left to use the excess heat and the dehydration reaction is favored.

ADDITIONAL EXERCISES 9.92

R—O—H + X − → R— O − + HX R—O—H + HA → R— OH 2 + A +

9.93

−

The physiological eﬀects of amphetamines on the body include raising the glucose level in the blood and increasing pulse rate and blood pressure. Abuse of amphetamines can cause long periods of sleeplessness, weight loss, and paranoia.

CHEMISTRY FOR THOUGHT 9.94

The three products are: diethyl ether

ethyl propyl ether

dipropyl ether Three products form because the two types of alcohol molecules can react with a molecule of the same type to produce either diethyl ether or dipropyl ether or a molecule of a different type to produce ethyl propyl ether.

9.95

9.96

Skin takes on a dry appearance after diethyl ether is spilled on the skin because diethyl ether, a volatile solvent, removes natural skin oils by dissolving them and then evaporating, taking these oils with it.

9.97

It might be more practical for chemical suppliers to ship amines as amine salts because amine salts are solids, while amines are often liquids. It is easier to ship a solid than a liquid.

Chapter 9 9.98

Hydrochloric acid is used to prepare an amine hydrochloride salt.

9.99

DNA and RNA contain components called bases because the components are amines.

9.100

9.101

Chapter 9 9.102

Alkaloids can be extracted from plant tissues using dilute hydrochloric acid because the acid reacts with the amines to produce water-soluble amine salts.

Chapter 10: Carbonyl and Carboxyl Compounds CHAPTER OUTLINE 10.1 10.2

The Nomenclature of Aldehydes and Ketones Properties of Aldehydes and Ketones

10.3

Reactions of Aldehydes and Ketones

10.4

The Nomenclature of Carboxylic Acids

10.5

Properties of Carboxylic Acids

10.6

The Nomenclature of Esters and Amides

10.7

Properties of Esters and Amides

10.8

Formation and Reactions of Esters and Amides

and Carboxylate Salts

LEARNING OBJECTIVES/ASSESSMENT When you have completed your study of this chapter you should be able to: 1.

Classify carbonyl compounds as aldehydes or ketones. (Section 10.1; Exercises 10.1 and 10.7)

Identify speciﬁc uses for aldehydes and ketones. (Section 10.1; Exercises 10.3)

Assign IUPAC names to aldehydes and ketones. (Section 10.1; Exercises 10.9 and 10.11)

Explain how intermolecular forces inﬂuence the physical properties of aldehydes and ketones. (Section 10.2; Exercises 10.21 and 10.22)

Predict the products of the hydrogenation, oxidation, reduction, and hydration reactions of aldehydes and ketones. (Section 10.3; Exercises 10.28 and 10.29)

Assign IUPAC names to carboxylic acids. (Section 10.4; Exercises 10.30 and 10.31)

Assign common and IUPAC names to carboxylate salts. (Section 10.4; Exercises 10.34 and 10.35)

Describe uses for carboxylate salts. (Section 10.4; Exercises 10.40 and 10.41)

Explain how hydrogen bonding aﬀects the physical properties of carboxylic acids. (Section 10.5; Exercises 10.49, 10.51, and 10.53)

10. Write key acid–base reactions of carboxylic acids. (Section 10.5; Exercises 10.59 and 10.61) 11. Assign common and IUPAC names to esters and amides. (Section 10.6; Exercises 10.63, 10.65, and 10.73) 12. Explain how hydrogen bonding aﬀects the physical properties of esters and amides. (Section 10.7; Exercises 10.77 and 10.79) 13. Write reactions for the formation and hydrolysis of esters and amides. (Section 10.8; Exercises 10.87 and 10.88) 14. Write key reactions for saponiﬁcation. (Section 10.8; Exercises 10.97 and 10.98)

LECTURE HINTS AND SUGGESTIONS 1.

As with most of the chapters on organic compounds, one should resist spending too much time on nomenclature. Try to explain the rules of giving a few simple examples and compare the systematic names to the common names. Show the similarities to the rules of naming alcohols.

When teaching Organic Chemistry students often ask, “why do I have to learn this?” To answer this question and help students connect with the material better emphasize the uses of common aldehydes and ketones. Examples include but are not limited to the following:

Formaldehyde is a biological preservative and used in embalming ﬂuid.

Acetone is an ingredient in the common household item nail polish remover. It is also

Benzaldehyde, vanillin, and cinnamaldehyde are all common ﬂavoring agent.

medically relevant as it is present in urine in diabetics. d. Glucose and fructose are common sugars at the core of carbohydrate nutrition. 3.

Begin by brieﬂy describing the general structure of carboxylic acids and their acidity. Then show several examples of carboxylic acids, starting with formic acid and working up by adding additional carbon atoms. Mention their physical characteristics as you compare the common name to the systematic name. a.

Formic Acid from the Latin formica, meaning ant. It was ﬁrst isolated by the distillation of ants and found in the venom of insects. Formic acid is a corrosive liquid.

Acetic Acid from the Latin acetum, meaning vinegar. Vinegar is about 5% acetic acid in water. Acetic acid is formed during the bacterial oxidation of wine.

Propionic Acid from the Latin pion, meaning fat. The calcium salt is used as a mold inhibiter in bread.

d. Butyric Acid from the Latin butyrum, meaning butter. The compound is responsible for the odor of rancid butter. e.

Benzoic Acid. Sodium benzoate is used as a preservative in carbonated soda.

Oxalic Acid is a toxic substance found in the leaves of rhubarb. It is used as a cleaning agent for metals.

Explain esters as condensation products derived from alcohols and carboxylic acids. Describe the diﬀerent fruity odors which are produced by simple esters (e.g., methyl butanoate-apple; ethyl butanoate-strawberry; butyl butanoate-pineapple).

Discuss the medical uses of carboxylic acids and esters. a.

Aspirin-analgesic, anti-inﬂammatory, antipyretic

Ibuprofen – analgesic

Describe amides as condensation products between amines and carboxylic acids.

Discuss the most biologically relevant amides, mainly peptide bonds in proteins and important amides in medicine. a.

Ampicillin is a broad-spectrum antibiotic used to treat bacterial infections in patients.

Thiopental is used for intravenous anesthesia.

Amobarbital is a medicine to treat insomnia.

Solutions to All End-of-Chapter Questions What follows are more complete explanations/full solutions to the EOC exercises whose answers are published in shorter form at the end of the book.

SECTION 10.1 THE NOMENCLATURE OF ALDEHYDES AND KETONES 10.1

The structural diﬀerence between an aldehyde and a ketone is that an aldehyde has the carbonyl group on a terminal carbon atom, which is also bonded to a hydrogen atom, while a ketone has the carbonyl group on an interior carbon atom, which is also bonded to two other carbon atoms.

10.2

10.3

The ending –one indicates the presence of a ketone group.

10.4

b and d

10.5

a and c

10.6

10.7

10.8

3-methylbutanal

4-methylheptanal

2-phenylethanal

butanal

10.09

10.10

10.11

10.12

10.13

10.14

3-methylpentanal

2-methylpropanal

3-phenylbutanal

2,2-dimethylpropanal

2-pentanone

3-pentanone

2,4-dimethyl-3-hexanone

2-methyl-4-phenyl-3-pentanone

2-heptanone

2,3,5-trimethyl-4-heptanone

4-methylcyclohexanone

3-methyl-2-butanone

10.15

SECTION 10.2 PROPERTIES OF ALDEHYDES AND KETONES 10.16

Hydrogen bonding does not occur between molecules of aldehydes or ketones because aldehydes and ketones do not contain an oxygen atom bonded directly to a hydrogen atom.

10.17

The acetone dissolves the remaining water in the glassware and the bulk of the mixture is discarded. The remaining traces of acetone evaporate quickly because it is much more volatile than water.

10.18

c < b < a

10.19

a < c < b

10.20

10.21

10.22

b < c < a

10.23

The order of increasing boiling point is (c) methylcyclopentane, (a) cyclopentanone, (b) cyclopentanol. The boiling points increase with the strength of the intermolecular forces each molecule experiences. Methylcyclopentane experiences only dispersion forces between its molecules. Cyclopentanone is polar and experiences both dispersion and dipolar forces between its molecules. Cyclopentanol has a hydroxy group and experiences hydrogen bonding in addition to dispersion forces between its molecules; therefore, cyclopentanol has the strongest intermolecular forces and the highest boiling point.

SECTION 10.3 REACTIONS OF ALDEHYDES AND KETONES 10.24

10.25

10.26

10.27

a. b.

Oxidation of a secondary alcohol produces a ketone. Oxidation of a primary alcohol produces an aldehyde that can be further oxidized to a carboxylic acid. c. Hydrogenation of a ketone produces a secondary alcohol. d. Hydrogenation of an aldehyde produces a primary alcohol. e. Hydrolysis of an acetal produces an aldehyde and two alcohol molecules. f. Hydrolysis of a ketal produces a ketone and two alcohol molecules.

10.28

10.29

SECTION 10.4 THE NOMENCLATURE OF CARBOXYLIC ACIDS AND CARBOXYLATE SALTS 10.30

10.31

c. butanoic acid

10.32

10.33

10.34

10.35

10.36

10.37

10.38

The sodium salt of valeric acid

sodium valerate

The magnesium salt of lactic acid

magnesium lactate

The potassium salt of citric acid

potassium citrate

10.39

10.40

10.41

potassium acetate

sodium benzoate

magnesium oxalate

sodium stearate

calcium propanoate

citric acid and sodium citrate

sodium benzoate

zinc-10-undecylenate

acetic acid

sodium benzoate

citric acid and sodium citrate

10.42

b c d

10.43

b c d

10.44

The structural features of a fatty acid are the carboxylic acid functional group and a long hydrocarbon tail. They are called fatty acids because they were originally isolated from fats.

10.45

Acetic acid is responsible for the sour or tart taste of Italian salad dressing.

10.46

The carboxylic acid present in sour milk and sauerkraut is lactic acid.

SECTION 10.5 PROPERTIES OF CARBOXYLIC ACIDS 10.47

c < a < d < b

10.48

(b) hexane < (a) pentanal < (c) 1-pentanol < (d) butanoic acid The order of increasing boiling point is alkane, aldehyde, alcohol, carboxylic acid. All of these compounds experience approximately equal dispersion forces since their molar masses are similar. Hexane experiences only dispersion forces, which are weaker than the combination of dispersion forces and dipolar forces experienced by pentanal. Both 1-pentanol and butanoic acid form hydrogen bonds, but butanoic acid has two sites for hydrogen bonding per molecule compared to only one site per molecule for 1-pentanol. This results in a greater degree of hydrogen bonding and a higher boiling point.

10.49

10.50

10.51

Acetic acid, sodium acetate, and sodium caprate are all soluble in water, but capric acid is not. Acetic acid is a small carboxylic acid that hydrogen bonds with water strongly enough to dissolve in water. Sodium acetate and sodium caprate are both ionic compounds that dissolve in water because of the ionic charges present in the compounds. Capric acid is not soluble in water because it contains a large hydrophobic carbon chain that repels water more than the hydrogen bonding of the carboxylate group can attract water.

10.52

The carboxylic acid functional group allows caproic acid to be soluble in water. The solubility is limited by the aliphatic portion of the acid because it is hydrophobic.

10.53

(a) hexane < (c) 3-pentanone < (b) 2-pentanol < (d) pentanoic acid Alkanes are the least soluble in water because they are nonpolar and cannot experience either dipolar forces or hydrogen bonds with water. Ketones are more soluble in water than alkanes because they are polar and can experience dipolar forces and hydrogen bonding with water. Alcohols are more soluble in water than ketones because they have a hydroxy group that has stronger hydrogen bonds with water than the carboxyl group of the ketones. Carboxylic acids are more soluble in water than alcohols because they have two sites for hydrogen bonding per molecule, which results in a greater degree of hydrogen bonding to the water molecules.

10.54

pentane < ethoxyethane < 1-butanol < propanoic acid Alkanes are the least soluble in water because they are nonpolar and cannot experience either dipolar forces or hydrogen bonds with water. Ethers are more soluble in water than alkanes because they are polar and can experience dipolar forces and hydrogen bonding with water. Alcohols are more soluble in water than ethers because they have a hydroxy group that has stronger hydrogen bonds with water than the oxygen atom in the ether. Carboxylic acids are more soluble in water than alcohols because they have two sites for hydrogen bonding per molecule, which results in a greater degree of hydrogen bonding to the water molecules.

10.55

10.56

10.57

Within the cell, lactic acid will dissociate to form the lactate ion and H+ because body fluids have a pH of 7.4 and lactic acid is a weak acid, which will dissociate in basic pH environments to form its conjugate base. The pKa of lactic acid is 3.85, which is much lower than the pH of most body fluids, so at physiological pH (which ranges from slightly acidic to slightly basic), the acid will exist primarily as its conjugate base.

10.58 At a pH of 12, the propanoate ion is the predominant form because the OH- ions remove the H3O+ ions from solution and shift the equilibrium to the right. At a pH of 2, the propanoic acid is the predominant form because the excess H3O+ ions shift the equilibrium to the left.

10.59

10.60

10.61

10.62

SECTION 10.6 THE NOMENCLATURE OF ESTERS AND AMIDES 10.63

10.64

10.65

10.66

10.67

10.68

10.69

nonanal

ethyl acetate

methyl propionate

propyl acetate

ethyl propanoate

propyl pentanoate

methyl propionate

propyl propanoate

butyl butanoate

ethyl butanoate

propyl butanoate

10.70

ethyl ethanoate

propyl ethanoate

butyl ethanoate

10.71

b, c, d

10.72

c and d

10.73

10.74

10.75

10.76

SECTION 10.7 PROPERTIES OF ESTERS AND AMIDES 10.77

Amides can form hydrogen bonds with water molecules.

10.78

N-ethylpropanamide because as a secondary amine, it can exhibit greater hydrogen bonding than a tertiary amine.

10.79

10.80

10.81

The boiling points of disubstituted amides are often lower than those of unsubstituted amides because the disubstituted amides are unable to form hydrogen bonds between their molecules, while the unsubstituted amides are able to form such hydrogen bonds.

10.82

Esters have a lower boiling point than carboxylic acids of similar molecular weight due to the difference in their intermolecular forces and molecular structure. Esters can only engage in weaker London dispersion forces and dipole-dipole interactions. Carboxylic acids contain a carboxyl functional group (-COOH) which allows them to form stronger intermolecular hydrogen bonds, which require more energy to break and transition from the liquid to the gas phase. As a result, carboxylic acids have higher boiling points.

SECTION 10.8 FORMATION AND REACTIONS OF ESTERS AND AMIDES 10.83

not an ester

b. c. d.

ester not an ester not an ester

10.84

10.85

10.86

10.87

10.88

10.89

10.90

10.91

10.92

10.93

10.94

10.95

10.96

10.97

10.98

10.99

10.100

ADDITIONAL EXERCISES 10.101 The most reduces carbon atom is part of the alkane, the least reduced carbon atom is part of the aldehyde. The alcohol carbon is less reduced than the alkane, but more reduced than the aldehyde. 10.102 A carboxylic acid solution will react with a sodium bicarbonate solution to produce carbon dioxide bubbles, while an alcohol solution will not react with sodium bicarbonate. 10.103 Formaldehyde can be prepared in the form of a 37% solution in water because it is a small, polar molecule that is water soluble. Decanal has 10 carbon atoms, which results in a mostly nonpolar molecule that is not water soluble; therefore, decanal cannot be prepared as a 37% solution in water. 10.104 Butanone is the preferable IUPAC name for this structure because 2-butanone is repetitive in nature. The only possible location of the carbonyl carbon atom in the four-carbon parent chain is carbon 2; therefore, butanone is a sufficient name for this structure.

10.105 Aldehydes are easily oxidized to carboxylic acids; therefore, they are not used in many consumer products. 10.106 The chemical structures of the aldehydes should be the same; however, the vanilla extract from an orchid is likely to contain other natural compounds in addition to the vanillin, which may enhance the flavor of the vanilla extract. Imitation vanilla extract is typically less expensive. 10.107 Fingernail polish remover evaporates fairly quickly when used because the acetone has weak intermolecular forces and a high vapor pressure. 10.108 Ester formation and ester hydrolysis are exactly the same reaction only written in reverse order. The presence (or absence) of heat as well as the concentration of reactants and products determines which direction the reaction proceeds and what actually forms. 10.109 Citric acid is viewed as a “safe” food additive because it is nontoxic. It is an acid that can be found in citrus fruit. The acidity of citric acid is not a concern because the body has many buffer systems in place to control pH within the body, although too large a concentration of citric acid in a particular food (such as sour candy) can cause gastric discomfort.

10.110 a.

10.111 a.

10.112 Ester hydrolysis is the reversible reaction in which water reacts with an ester in the presence of an acid catalyst to produce a carboxylic acid and an alcohol. Saponiﬁcation is the basic cleavage of an ester linkage that produces the salt of a carboxylic acid and an alcohol.

Chapter 11

Chapter 11: Carbohydrates CHAPTER OUTLINE 11.1

Carbohydrate Structure and Function

11.5

Important Monosaccharides

11.2

Carbohydrate Stereochemistry

11.6

Properties of Monosaccharides

Fischer Projections, Cyclization, and

11.7

Disaccharides

Haworth Projections

11.8

Polysaccharides

11.3 11.4

Classiﬁcation of Monosaccharides

LEARNING OBJECTIVES/ASSESSMENT When you have completed your study of this chapter, you should be able to: 1.

Describe the four major functions of carbohydrates in living organisms. (Section 11.1; Exercise 11.1)

Classify carbohydrates as monosaccharides, disaccharides, or polysaccharides. (Section 11.1; Exercise 11.5)

Identify chiral carbon atoms in molecules. (Section 11.2; Exercises 11.9 and 11.10)

Draw pairs of enantiomers using wedges and dashes. (Section 11.2; Exercises 11.17 and 11.18)

Use Fischer projections to represent D- and L-enantiomers. (Section 11.3; Exercises 11.21,11.24, and 11.26)

Identify the following functional groups: hemiacetal, hemiketal, acetal, and ketal. (Section 11.3; Exercises 11.37 and 11.39)

Draw Haworth projections of carbohydrate anomers. (Section 11.3; Exercises 11.45 and 11.46)

Classify monosaccharides as aldoses or ketoses. (Section 11.4; Exercise 11.51)

Classify monosaccharides according to the number of carbon atoms they contain. (Section 11.4; Exercise 11.53)

10. Describe uses of four important monosaccharides. (Section 11.5; Exercises 11.57 and 11.58) 11. Describe the purpose of Benedict’s test. (Section 11.6; Exercise 11.68) 12. Identify glycosidic linkages in carbohydrate derivatives. (Section 11.6; Exercise 11.71) 13. Describe the uses of three important disaccharides. (Section 11.7; Exercise 11.76) 14. Describe the linkage in disaccharides. (Section 11.7; Exercises 11.77, 11.81, and 11.83) 15. Write reactions for the condensation and hydrolysis of disaccharides. (Section 11.7; Exercise 11.89) 16. Describe the uses of three important polysaccharides. (Section 11.8; Exercise 11.91) 17. Describe the linkage in polysaccharides. (Section 11.8; Exercises 11.92, 11.93, and 11.94)

LECTURE HINTS AND SUGGESTIONS 1.

Describe sugars as polyfunctional alcohols which can contain aldehyde or ketone groups. Explain that many sugars are composed of two or more simple sugars linked together (disaccharides and polysaccharides). A useful way to show this to students is to provide a handout (next few pages) showing the progression from monosaccharides to disaccharides to polysaccharides.

Chapter 11 2.

Use everyday objects to demonstrate chirality. Hold up your two hands and show how they have a mirror image relationship. Explain that the right hand is a mirror image of the left hand. Show that they are similar but not identical. Be sure to emphasize the enantiomers (e.g., right and lefthanded) are real and not mirror images (they only have a mirror image relationship). Sometimes students get confused and think we are speaking about images and not real objects (molecules). Emphasize that everything has a mirror image and that we are speaking only of a special type of structural relationship between two real objects (molecules).

When teaching Fischer Projections students usually grasp that the chiral carbon is represented by the intersection of two lines where the horizontal bonds are coming toward you and vertical bonds are going away from you, however, it is less common for students to remember that the carbonyl end is placed at the top and the whole conﬁrmation must ﬁrst be eclipsed in this representation. Make sure to emphasize this by building the sugar in the extended form with a molecular model and then manipulate it so that it is fully eclipsed in front of the class so students can see the bonds coming out at them in 3D space and observe the carbonyl at the top.

Explain the diﬀerence between capital "D and L" and lower case "d and 1" to prevent confusion with material that they may encounter in other courses. Explain that diﬀerent systems are used for expressing stereochemical relationships and that the D/L system is historically the earliest. It is a system useful for talking about carbohydrates and amino acids because their relationship to D/L glyceraldehyde is fairly easy to establish.

The concept of reducing sugars can be illustrated by the simple test for glucose in the blood and urine using Benedict's reagent. The test kits are readily available and can be easily used as a demonstration in the classroom.

Solutions to All End-of-Chapter Questions What follows are more complete explanations/full solutions to the EOC exercises whose answers are published in shorter form at the end of the textbook

SECTION 11.1 CARBOHYDRATE STRUCTURE AND FUNCTION 11.1

The four important roles of carbohydrates in living organisms are: (1) to provide energy through their oxidation, (2) to supply carbon for the synthesis of cell components, (3) to serve as a stored form of chemical energy, and (4) to form a part of the structural elements of some cells and tissues.

11.2

11.3

cellulose—structural material in plants

sucrose, table sugar– an energy source in our diet.

glycogen—form of stored energy in animals

starch—form of stored energy in pants

Monosaccharides are the simplest carbohydrates. Disaccharides contain two monosaccharides.

Chapter 11 Polysaccharides contain many monosaccharides. 11.4

table sugar—carbohydrate, disaccharide

11.5

11.6

starch—carbohydrate, polysaccharide

glycogen—carbohydrate, polysaccharide

monosaccharide, carbohydrate

polysaccharide, carbohydrate

disaccharide, carbohydrate

polysaccharide, carbohydrate

Carbohydrates are polyhydroxy aldehydes and ketones, or substances that yield such compounds on hydrolysis. In other words, carbohydrates contain alcohol, aldehyde, and ketone functional groups or will contain those functional groups upon hydrolysis.

SECTION 11.2 CARBOHYDRATE STEREOCHEMISTRY 11.7

Carbon atom 1 is not chiral because it is only bonded to three other atoms. Carbon atom 3 is not chiral because it is bonded to two hydrogen atoms. Chiral carbon atoms are bonded to four unique groups.

11.8

The chiral carbon atom is marked with an asterisk. The four attached groups are: (1) a hydrogen (2) an amino group (3) a methyl group (4) a

11.9

yes

group

Chapter 11 11.10

yes

11.11

3 chiral carbons, 8 stereoisomers

11.12

2 chiral carbons, 4 stereoisomers

11.13

11.14

11.15

11.16

no chiral carbon atoms; no enantiomers

1 chiral carbon atom; a pair of enantiomers

Chapter 11

b. 1 chiral carbon atom; 2 stereoisomers; a pair of enantiomers

c. 1 chiral carbon atom; 2 stereoisomers; a pair of enantiomers

11.17

First, identify the chiral carbon, indicated by the asterisk (*). The stereochemistry at the chiral carbon can be represented by drawing the substituent to the main carbon chain either dashed or wedged. (a) yields

and

as its two enantiomers.

and

as its two enantiomers.

(b) yields

11.18 First, identify the chiral carbon, indicated by the asterisk(*). The stereochemistry at the chiral carbon can be represented by drawing the substituent to the main carbon chain either dashed or wedged. a. yield

11.19

and

yields

and

as its two enantiomers.

With multiple chiral centers, the enantiomer switches the stereochemistry at each chiral center from wedge to dash and vice versa.

Chapter 11

(a) The enantiomer

of would be

(b) The enantiomer of

11.20

would be

With multiple chiral centers, the enantiomer switches the stereochemistry at each chiral center from wedge to dash and vice versa. a. The enantiomer of

b. The enantiomer of

would be

SECTION 11.3 FISCHER PROJECTIONS, CYCLIZATION, AND HAWORTH PROJECTIONS 11.21

11.22

Chapter 11 11.23

The intersection of the lines represents a chiral carbon atom. The H and the OH groups (groups on horizontal bonds) stick out in front of the plane of the paper and the CHO and CH2CH3 groups (groups on vertical bonds) stick out behind the plane of the paper.

11.24

11.25

11.26

11.27

Chapter 11 11.28

The study of chiral molecules is important in biochemistry because living organisms consist largely of chiral substances. The stereochemistry of a biological compound plays a key role in determining which receptors with which it will interact, and enzymes will be able to catalyze its reactions, as well as what its physiological eﬀect on an organism will be.

11.29

11.30

11.31

same

rotate plane-polarized light in diﬀerent directions.

same

rotate plane-polarized light in diﬀerent directions.

same

Optically active molecules contain at least one chiral carbon atom, have nonsuperimposable mirror images, and rotate the plane of polarized light.

11.32

False. The D and L signify the spatial relationships at chiral carbons relative to the reference compound, glyceraldehyde. Both D and L compounds can rotate either dextrorotatory or levorotatory, they do not signify the direction of how plane polarized light is rotated by the molecule.

11.33

b, c

11.34

11.35

a, d

11.36

Identify which of the following structures are acetals: The general structure of an acetal is a carbon attached to two –OR groups, a R group, and a hydrogen.

is not an acetal as the carbon is not

is an acetal.

attached to hydrogen.

Chapter 11

is an acetal.

11.37

11.38

11.39

11.40

11.41

Chapter 11 11.42

11.43

(a) The ketal within the following molecule is highlighted in blue.

(b) The ketal within the following molecule is highlighted in blue.

(c ) None of the listed functional groups is present in the following molecule as there is only one oxygen present.

11.44

a. The hemiacetal within the following molecule is highlighted in blue.

b. The hemiacetal within the following molecule is highlighted in blue.

Chapter 11

11.45

11.46

Identify each of the following as an α or a β form and draw the structural formula of the other anomer: Recall the three steps used to draw anomers: STEP 1: Draw the pyranose ring with the oxygen atom to the back. STEP 2: Put the anomeric carbon on the right side of the ring. Number the ring starting at the right side. Position number 1 will be the anomeric carbon. STEP 3: The terminal -CH2OH group (position 6) should always be shown above the ring for d-monosaccharides as shown below.

Since the -OH group at position 1 is in the up direction the anomer drawn is the β form. To draw the α form the –OH group at position 1 should be drawn in the down direction as shown here:

Since the -OH group at position 1 is in the down direction the anomer drawn is the α form. To draw the β form the –OH group at position 1 should be drawn in the up direction as shown here:

Chapter 11

11.47

11.48

11.49

A pyranose ring refers to a carbohydrate which has a six-membered ring containing an oxygen atom. A furanose ring refers to a carbohydrate which has a ﬁve-membered ring containing an oxygen atom.

SECTION 11.4 CLASSIFICATION OF MONOSACCHARIDES 11.50

a. b. c.

ketose aldose aldose

11.51

a. b. c.

ketose aldose ketose

11.52

a. b. c.

pentose hexose tetrose

11.53

a. b. c.

tetrose hexose pentose

Chapter 11

11.54

11.55

11.56

SECTION 11.5 IMPORTANT MONOSACCHARIDES 11.57

Ribose and deoxyribose are monosaccharides used in the synthesis of nucleic acids.

11.58

D-glucose can be injected directly into the bloodstream to serve as an energy source because glucose is the sugar transported by the blood to body tissues to satisfy energy requirements. Glucose has a high solubility in aqueous solutions; it dissolves readily in water and in blood;

Chapter 11 therefore, it poses no risks of toxicity itself and will not form dangerous solid occlusions in the blood vessels. 11.59

Fructose can be a low-calorie sweetener because it is sweeter, gram for gram, than the other common sugars; consequently, less fructose is needed to obtain the same degree of sweetness.

11.60

Fructose is a ketohexose.

11.61

Glucose and galactose have similar structures. The only diﬀerence is the orientation of the hydroxy group attached to carbon 4.

11.62

a. The natural sources of glucose are honey and fruits, as well as sucrose, which is a disaccharide containing glucose. b. The natural source of fructose is fruit c. The natural source of galactose is milk which contains the disaccharide lactose.

SECTION 11.6 PROPERTIES OF MONOSACCHARIDES 11.63

Certain carbohydrates are called sugar because they taste sweet.

11.64

the arrows indicate the sites on the molecule where water can hydrogen bond to glucose.

11.65

When a sugar fails to react with Cu2+, it is a nonreducing sugar.

11.66

The cyclic compound β-D-galactose can react with Cu2+ and be classified as a reducing sugar because as it undergoes mutarotation in a solution, the ring opens, an aldehyde is formed, and the compound can react with Benedict’s reagent, which is an oxidizing agent.

11.67

All reducing sugars.

Chapter 11 11.68

All will give a positive Benedict’s test.

11.69

11.70

11.71

Chapter 11 11.72

11.73

a. b.

CH3CH2OH CH3CH2CH2OH

11.74

a. b.

CH3CH2CH2OH CH3CH2OH

SECTION 11.7 DISACCHARIDES 11.75

11.76

glucose, fructose

galactose, glucose

glucose

The most common household sugar

sucrose

Formed during the digestion of starch

maltose

An ingredient of human milk

lactose

Found in germinating grain

maltose

Hydrolyzes when cooked with acidic foods to give invert sugar

sucrose

Found in high concentrations in sugar cane

sucrose

11.77

Glycosidic linkages are broken when disaccharides are hydrolyzed to monosaccharides.

11.78

The hemiacetal group of a lactose molecule is able to react with Benedict’s reagent because the ring containing the hemiacetal group is not “locked”. In solution the hemiacetal group can undergo mutarotation, opening the ring into an open-chain aldose that can react with Benedict’s reagent.

11.79

11.80

glucose, glucose

galactose, fructose

galactose, glucose

glucose, fructose

glucose, glucose

galactose, glucose

Chapter 11 11.81

11.82

11.83

11.84

11.85

11.86

1→ 4

1→ 5

1→ 6

1→ 4

1→ 6

1→ 4

yes

11.87

Chapter 11 11.88 Lactose is a reducing sugar because it contains a hemiacetal group.

Sucrose is not a reducing sugar because it does not contain a hemiacetal or a hemiketal group.

11.89

a. b. c.

When melibiose is hydrolyzed, glucose and galactose are the monosaccharides produced. Melibiose is a reducing sugar because the glucose ring contains a hemiacetal group. The glycosidic linkages in melibiose is an α (1 → 6) linkage.

11.90

In the case of melibiose the anomeric carbon on glucose contains and intact -OH group that can be oxidized by an oxidizing agent. In other words melibiose can reduce an oxidizing agent, and is thus a reducing sugar.

SECTION 11.8 POLYSACCHARIDES 11.91

a. b. c. d. e.

The unbranched polysaccharide in starch A polysaccharide widely used in textile fiber The most abundant polysaccharide in starch The primary constituent of paper A storage form of carbohydrates in animals

11.92

a. b. c. d.

amylose, cellulose amylose cellulose glycogen

11.93

a. b.

Contains both α(1→4) and α(1→6) glycosidic linkages Is composed of glucose monosaccharide units

Contains acetal linkages between monosaccharides units Is composed of highly branched molecular chains

amylose cellulose amylopectin cellulose glycogen

amylopectin, glycogen amylose, amylopectin, glycogen, cellulose amylose, amylopectin, glycogen, cellulose amylopectin, glycogen

Chapter 11 11.94

a. b. c.

11.95

Both are comprised of glucose units with α(1→4) and α(1→6) linkages. Glycogen contains more branching. Both are comprised of glucose units linkages which amylose lacks. Amylose is a straight chain, while amylopectin is branched. Both are comprised of linear chains of glucose units, linked 1→4. In amylose, the linkages are alpha, while in cellulose they are beta.

Celery is a good snack for people on a diet because it contains large amounts of cellulose, which the human body cannot hydrolyze. Cellulose passes unchanged through the digestive tract without contributing to the caloric value of food. Celery can fill a stomach without providing many calories. It is also an excellent source of dietary fiber which combats constipation and aids in the elimination of body wastes, flushing the body of toxins.

ADDITIONAL EXERCISES 11.96

Maltose (C12H22O11) has approximately twice the mass per molecule of glucose (C6H12O6); therefore, if two 10% (w/v) solutions are made with maltose and glucose as solutes, the maltose solution would contain roughly half the number of molecules contained in the glucose solution. Neither of the solutes dissociates in water since both are molecular. The boiling point of a solution increases with the number of solute particles in solution; therefore, the glucose solution would contain more molecules of solute and would have the higher boiling point.

11.97

One mole of D-glucose only reacts with one mole of methanol to form an acetal because the pyranose ring already contains one –OR group.

11.98.

11.99

 1 mole sucrose   1 mole glucose   180 g glucose  100 g sucrose    = 52.6 g glucose   342 g sucrose   1 mole sucrose   1 mole glucose  Glucose is a solid at room temperature because it is able to form many hydrogen bonds per molecule (the ratio of oxygen to carbon is one to one) and it has a high molecular weight. The combination of a high degree of hydrogen bonding with fairly strong dispersion forces causes it to be a solid at room temperature. Hexanal and 1-hexanol each contain one oxygen atom per six carbon atoms and a relatively low molecular weight. Hexanal cannot hydrogen bond amongst its own molecules. It primarily experiences dipolar forces between its molecules. Since this is a much weaker intermolecular force than the hydrogen bond, it will have a lower boiling and melting point than glucose. 1-Hexanol can hydrogen bond but makes fewer hydrogen bonds per molecule than glucose. This, combined with its weaker dispersion forces, causes it to be a liquid at room temperature.

Chapter 11

CHEMISTRY FOR THOUGHT 11.100.

2.80 x 10 5 u  1 molecule linked glucose  3   = 1.64 x 10 glucose molecules 1 molecule starch  171 u linked glucose  Note: The molecular mass of glucose is 180 u, but when glucose links together to form a molecule of starch, 1 molecule of water (18.0 u) is lost for every two glucose units or 0.5 molecules of water per glucose unit; therefore, the molecular mass of a linked glucose unit is 180-9.0 = 171 u.

11.101 Genetically altering human intestinal bacteria so they could hydrolyze the β linkages of cellulose would allow humans to digest this plant material for use as an energy source. This genetic change would compromise the ability of humans to eat plant matter for roughage, which is needed to provide bulk, stimulate contraction of the intestines, and aid in the passage of food through the digestive system. Since a wide variety of foods are currently available for energy and cellulose plays an important role in digestion, this genetic modification would not be beneficial for most humans. 11.102 Aspartame (Nutrasweet) contains calories and yet is used in diet drinks. A drink can contain aspartame as a sweetener and yet be low in calories because the amount needed to sweeten a diet drink is so small that very few calories are added to the drink. 11.103 The open-chain form of glucose constitutes only a small fraction of any glucose sample, yet when Cu2+ is used to oxidize the open-chain form, nearly all the glucose in a sample reacts. The reason this occurs is that the open-chain form of glucose is in equilibrium with the ring forms of glucose and as the open-chain form of glucose reacts with the Cu2+ it is essentially removed from the equilibrium; consequently, in an attempt to reestablish equilibrium, the ring forms of glucose undergo mutarotation to produce the open-chain form of glucose, which in turn reacts with the Cu2+ and causes more of the ring forms of glucose to undergo mutarotation. This cycle continues until nearly all of the glucose in a sample reacts with the Cu2+. This is an example of Le Châtelier’s principle. 11.104 Foods that would be expected to give a positive starch test include bread, crackers, pasta, and rice. 11.105 When lactose in cow’s milk is hydrolyzed in a child’s digestive system, 1 molecule of galactose and 1 molecule of glucose are produced for each molecule of lactose. 11.106 Amylose is a straight-chain glucose polymer similar to cellulose. Paper manufactured with amylose instead of cellulose would have less longevity because amylose forms loose spiral structures unlike cellulose which forms extended straight chains that can be aligned side by side to form well-organized, water-insoluble fibers in which the hydroxy groups form numerous hydrogen bonds with the neighboring chains that confer rigidity and strength to the overall structure of the paper.

Chapter 11 11.107 Unlike the coiled form of amylose which reacts with iodine by incorporating the iodine molecules into its structure to produce a positive bluish-black iodine test, amylopectin has a branched structure that cannot incorporate iodine into its structure, and consequently, when amylopectin is tested in its pure form it has a negative iodine test.

Chapter 11

Chapter 11: Carbohydrates Monosaccharides

Disaccharides

Chapter 11 Polysaccharides

Amylose

Amylopectin

Cellulose

Chapter 12

Chapter 12: Amino Acids, Proteins, and Enzymes CHAPTER OUTLINE 12.1 12.2 12.3 12.4 12.5

12.6 12.7 12.8 12.9

Amino Acid Structure and Classiﬁcation Reactions of Amino Acids Protein Structure Protein Functions Classes of Enzymes

Enzyme Activity Factors Aﬀecting Enzyme Activity Enzyme Inhibition and Regulation Enzymes in Medical Applications and Disease

LEARNING OBJECTIVES/ASSESSMENT When you have completed your study of this chapter, you should be able to: 1.

Identify the characteristic parts and stereochemistry of alpha-amino acids. (Section 12.1; Exercises 12.1 and 12.2.)

Draw structural formulas to illustrate the various ionic forms assumed by amino acids. (Section 12.2; Exercises 12.7 and 12.8.)

Write reactions to represent the formation of peptides and the oxidation of cysteine. (Section 12.2: Exercises 12.14 and 12.15.)

Explain what is meant by the primary structure of proteins. (Section 12.3; Exercise 12.21.)

Describe the role of hydrogen bonding in the secondary structure of proteins. (Section 12.3; Exercises 12.25 and 12.27.)

Describe the role of side-chain interactions in the tertiary structure of proteins. (Section 12.3; Exercises 12.28-12.31.)

Explain what is meant by the quaternary structure of proteins. (Section 12.3; Exercises 12.32-12.35.)

Describe four biological functions of proteins. (Section 12.4; Exercises 12.36 and 12.38.)

Describe three reactions that enzymes catalyze. (Section 12.5; Exercises 12.41 and 12.43.)

10. Describe enzyme activity in terms of eﬃcacy and mechanism of action. (Section 12.6; Exercises 12.45 and 12.48.) 11. Identify four factors that aﬀect enzyme activity. (Section 12.7; Exercises 12.56, 12.59, and 12.60.) 12. Compare the mechanisms of competitive and noncompetitive enzyme inhibition and allosteric regulation. (Section 12.8; Exercise 12.63.) 13. Name three enzymes that can be used in diagnostic testing to identify the presence of disease. (Section 12.9; Exercises 12.68 and 12.69.)

LECTURE HINTS AND SUGGESTIONS 1.

Explain that the amino acids are the monomer units for the formation of peptide or protein polymers. Provide Table 12.1 (next page) as a full-page handout to help students learn the amino acids and help classify them into groups based on chemical characteristics. Line up a few amino acids on the whiteboard and show how they condense in a head to tail fashion forming amide (peptide) bonds.

Chapter 12

The zwitterion nature of amino acids could be easily demonstrated to the entire class by using an aqueous solution of an amino acid as a buﬀering agent. By using a pH meter with a large readable scale, the ability of an amino acid solution to resist a change in pH brought about by the addition of a strong acid or base could be easily demonstrated. Explain how the zwitterion form of an amino acid could react with either a strong acid or a strong base by losing or gaining an H+.

The oxidation of the water-soluble cysteine to the water insoluble cysteine is an easy demonstration. Make a fairly concentrated solution of cysteine in water and bubble air through it or simply allow it to sit until the next class. As the thiol amino acid is oxidized to the disulfide, a fine white precipitate appears. Relate the formation of these disulfide bonds to the formation of disulfide bridges in the quaternary structure of proteins.

Describe enzymes as catalysts for biological reactions. Draw a sketch of a model of an enzyme on the whiteboard showing the active site as a crevice. Explain that when biomolecules enter and bind to this site a subsequent chemical reaction is made easier because the organic functional groups in the active site participate directly in the making and breaking of chemical bonds.

Many enzyme inhibitors are useful drugs: a) Sulfa drugs inhibit the enzyme which incorporates PABA into folic acid. b) Penicillin inhibits the aminoacylase which is involved in bacterial cell wall synthesis. c) Methotrexate, an anti-cancer agent, inhibits dihydrofolic reductase. d) AZT inhibits viral reverse transcriptase in AIDS.

Chapter 12

Solutions for All End-of-Chapter Questions What follows are more complete explanations/full solutions to the EOC exercises whose answers are published in shorter form at the end of the textbook

SECTION 12.1 AMINO ACID STRUCTURE AND CLASSIFICATION 12.1

All amino acids contain an amine and a carboxylic acid key functional group.

12.2

12.3

12.4

Using Table 12.1 the amino acids below can be classiﬁed as follows: a. threonine – neutral and polar b. aspartate – acidic c. serine – neutral and polar d. phenylalanine – neutral and nonpolar

Chapter 12 12.5

SECTION 12.2 REACTIONS OF AMINO ACIDS 12.6

A zwitterion is a dipolar ion that carries both a positive and a negative charge as a result of an internal acid-base reaction in an amino acid molecule.

12.7

12.8

12.9

The isoelectric point is the characteristic solution pH at which an amino acid has a net charge of 0. At this point, both the amino group and the carboxylic acid termini are ionized, and the amino acid exists as a true zwitterion.

Chapter 12 12.10

12.11

12.12

12.13

12.14

Chapter 12

12.15

12.16

12.17

glycine – serine – threonine

12.18

alanine – cysteine – valine

12.19

Chapter 12

12.20

SECTION 12.3 PROTEIN STRUCTURE 12.21

The primary structure of proteins is the order of the amino acid residues in a protein.

12.22

Even though two polypeptides have the same amino acid composition, the sequence of amino acids may be different. A different order of amino acids would give rise to a different threedimensional structure and thus different functions.

12.23

Covalent bonding, speciﬁcally peptide (amide) bonds between the amino acid residues, accounts for the primary structure of proteins.

12.24

12.25

Alpha secondary structures in proteins are helices formed when hydrogen bonds form between carbonyl oxygen atoms and amide hydrogen atoms in adjacent turns of the helical backbone, 4 residues away in the chain. Beta secondary structures in proteins are pleated sheets that form when several protein chains lie side by side and are held in position by hydrogen bonds between the amide carbonyl oxygen atoms of one chain and the amide hydrogen atoms of an adjacent chain.

12.26

The formation of secondary structures (α-helix and the β-pleated sheet) arises from hydrogen bonding between amide groups and is independent from the primary structure.

12.27

Hydrogen bonding between amide hydrogen atoms and carbonyl oxygen atoms of amino acid residues in the backbone of the protein or polypeptide chain is the most important type of

Chapter 12 bonding for maintaining a speciﬁc secondary conﬁguration. 12.28

The formation of salt bridges can only occur if the side chain is charged. The amino acids with side-chain groups that can form salt bridges are histidine, lysine, arginine, aspartate, and glutamate.

12.29

The hydrogen bonding in the tertiary structure of a protein occurs between the side chains on the amino acid residues. The hydrogen bonding in the secondary structure of a protein occurs between the atoms in the backbone of the protein.

12.30

c, d

12.31

tyrosine and glutamine

hydrogen bonds

aspartate and lysine

salt bridge

leucine and isoleucine

hydrophobic interactions

phenylalanine and valine

hydrophobic interactions

12.32

Quaternary protein structure is the arrangement of subunits that form a larger protein.

12.33

Hydrophobic forces, hydrogen bonds, disulﬁde bridges, and salt bridges between subunits give rise to quaternary structure.

12.34

The quaternary structure of hemoglobin is four subunits (two alpha chains containing 141 amino acid residues each and two beta chains containing 146 residues each) that each contain a heme group (a planar ring structure centered around an iron atom) located in crevices near the exterior of the molecule. The molecule is nearly spherical with the four subunits held together rather tightly by hydrophobic forces.

12.35

The term subunit describes a polypeptide chain that has its own primary, secondary, and tertiary structure that combines with other polypeptide chains to form the quaternary structure of a protein.

SECTION 12.4 PROTEIN FUNCTIONS 12.36

The role of enzymes in the body is to catalyze chemical reactions.

12.37

Using Table 12.3 the proteins below can be classiﬁed as follows: a. insulin – regulation b. rhodopsin – nerve impulse transmission c. antibodies – protection

Chapter 12 12.38

a. b. c.

structure transport catalysis

12.39

Using Table 12.3 the proteins below can be classiﬁed as follows: a.

myosin – movement

serum albumin – transport

thyrotropin – regulatory (see text before table)

SECTION 12.5 CLASSES OF ENZYMES 12.40

a. b. c. d.

hydrolase transferase lyase ligase

12.41

a. b. c. d.

isomerase oxidoreductase hydrolase transferase

a. b. c. d.

Enzyme decarboxylase phosphatase peptidase esterase

12.42

12.43

Reaction Catalyzed remove of carboxyl group from compounds hydrolysis of phosphate ester linkages hydrolysis of peptide linkages formation of ester linkages

a. dehydrolase— removes water molecules b. dehydrogenase—removes hydrogen from molecules c. oxygenase—adds oxygen to molecules d. carboxylase—adds carboxyl groups to molecules

SECTION 12.6 ENZYME ACTIVITY 12.44

Enzyme catalysis of a reaction is superior to normal laboratory conditions because enzymes are speciﬁc in the type of reaction they catalyze and enzyme activity can be regulated by the cell.

12.45

The relationship between an enzyme and the energy of activation for a reaction is that an enzyme reduces the energy of activation for a reaction.

12.46

Many types of enzymes are needed because enzymes can only catalyze speciﬁc reactions and a body requires many reactions to occur in order to maintain life. 10

Chapter 12 12.47

Enzyme specificity means that an enzyme is quite specific both in the type of reaction it catalyzes and the substrate it will bind (either a particular substance or a specific class of compound).

12.48

Enzyme activity in an experiment is observed by any method that allows the rate of product formation or reactant usage to be determined. The disappearance or appearance of a characteristic color is an example.

12.49

An enzyme international unit is the quantity of enzyme that catalyzes the conversion of 1 micromole of substrate per minute. This is useful in medical diagnosis because a specific enzyme activity of a patient can be measured and compared to normal activity.

12.50

The equation shows that the enzyme (E) and substrate (S) establish an equilibrium with the enzyme—substrate complex (ES). This is a reversible reaction. The enzyme—substrate complex can break apart into the enzyme and substrate or react to produce the enzyme and the product (P). The reaction to produce the product is not a reversible reaction.

12.51

The substrate binds to the active site of an enzyme to form the enzyme—substrate complex.

12.52

CH3CH2—OH and HO—

CH3CH2—COO- and Fe2+

CH3CH2CH2CH2CH3 and

—OOC-CH2—N+H3 and –COO-

12.53

through attraction of oppositely charged ions. b.

CH3CH2—NH2 and HO— through hydrogen bonding.

CH3CH=CHCH3 and through hydrophobic interactions.

12.54

Covalent bonds are strong bonds, too strong to permit the eventual release of the substrate.

12.55

The lock-and-key theory states that the active site of the enzyme is rigid (like a lock) and the substrate acts (like a key) by entering the active site. The induced ﬁt model states that the active site of the enzyme accommodates the substrate and is not rigid, but has a shape that becomes complementary to that of the substrate only after the substrate is bound.

Chapter 12

SECTION 12.7 FACTORS AFFECTING ENZYME ACTIVITY 12.56

The rate of an enzyme-catalyzed reaction reaches a maximum when the substrate concentration is raised beyond the saturation point. At this point the enzyme becomes the limiting reagent and the substrate is in excess.

12.57

Hypothermia will decrease enzyme activity in the body because the enzymes will no longer be in their ideal temperature range for biological activity.

12.58

With more enzyme present, the Vmax will be higher than it should be.

12.59

Reaction rates generally increase with increasing temperature. However, at temperatures too high, enzyme systems can be denatured and reaction rates slow down.

12.60

Enzyme structures are unique, so enzyme properties are also unique.

12.61

The pH of enzymes is maintained near 7.0 to prevent the enzymes from denaturing. Enzymes are proteins that can lose their three-dimensional structure if the pH is drastically changed.

SECTION 12.8 ENZYME INHIBITION AND REGULATION 12.62

Irreversible enzyme inhibition involves the formation of a covalent bond between an irreversible inhibitor and the enzyme which renders the enzyme inactive. This is not reversible because the bond between enzyme and inhibitor is extremely strong. Reversible enzyme inhibition involves the reversible binding, which may or may not be covalent, of a reversible inhibitor to an enzyme. An equilibrium is established; therefore, the inhibitor can be removed from the enzyme by shifting the equilibrium.

12.63

In competitive enzyme inhibition, an inhibitor binds to the active site of an enzyme and prevents the substrate from binding to the active site. In noncompetitive enzyme inhibition, an inhibitor binds to the enzyme at a location other than the active site and changes the shape of the active site, thus preventing the substrate from binding to the active site.

12.64

Chapter 12

12.65

12.66

12.67

reversible—competitive

reversible—noncompetitive

irreversible—reversible noncompetitive

reversible—competitive

irreversible

SECTION 12.9 ENZYMES IN MEDICAL APPLICATIONS AND DISEASE 12.68

A medical professional might run a diagnostic test detecting speciﬁc enzymes in the blood serum in the following scenarios:

12.69

A patient has experienced an injury causing major tissue/cellular damage.

The medical professional is testing a patient for the presence of cancer.

Diagnosis of diseases of the heart, liver, pancreas, prostate, and bones

Table 12.8 provides the following reasons a medical professional may run a test to detect enzymes in blood serum.

1. 2. 3. 4. 5. 6. 7. 8. 9.

Acid phosphatase may indicate prostate cancer. Alanine transaminase (ALT) could indicate a hepatitis infection. Alkaline phosphatase (ALP) might mean a patient has liver or bone disease. Amylase may indicate disease of the pancreas. Aspartate transaminase (AST) is used to detect heart attacks or hepatitis. Creatine kinase (CK) is a sign of heart attack. Lactate dehydrogenase (LDH) might mean a heart attach or liver damage. Lipase assays are used to diagnose acute pancreatitis. Lysozyme could indicate monocytic leukemia.

Chapter 12

ADDITIONAL EXERCISES 12.70

12.71

The two Ka values of alanine reﬂect the ionization of the carboxylic acid group and the protonated amine group, respectively, on the molecule. The carboxylic acid group has the Ka value of 5.0 x 10-3, since it is the stronger acid, and the –NH3+ group has the Ka value of 2.0 x 10-10. Glutamate would have 3 Ka values, since it has 3 ionizable groups.

CHEMISTRY FOR THOUGHT 12.72

The ethyl ester of alanine melts 200 degrees below the melting point of alanine because the ethyl ester of alanine is unable to form a zwitterion, and consequently, it is unable to make ionic bonds to other molecules of the same type. The weak interparticle forces between the molecules of the ethyl ester of alanine correspond to a low melting point, while the strong interparticle forces between the alanine zwitterions correspond to a melting point more like an ionic compound than an organic compound.

Chapter 13

Chapter 13: Nucleic Acids and Protein Synthesis CHAPTER OUTLINE 13.1

The Flow of Genetic Information

13.2

The Components of Nucleic Acids

13.6

Transcription: RNA Synthesis from

13.3

The Structure of DNA

13.7

The Genetic Code and Protein Synthesis

13.4

DNA Replication

13.8

Viruses

13.5

Ribonucleic Acid (RNA)

13.9

Recombinant DNA

DNA

LEARNING OBJECTIVES/ASSESSMENT When you have completed your study of this chapter, you should be able to: 1.

Describe the central dogma of molecular biology. (Section 13.1; Exercises 13.3 and 13.4)

Identify the components of nucleotides. (Section 13.2; Exercise 13.7)

Classify the sugars and bases of nucleotides. (Section 13.2; Exercises 13.6, 13.8, and 13.9)

Describe the structure of DNA. (Section 13.3; Exercises 13.19 and 13.20)

Write the complementary base strand of a DNA sequence. (Section 13.3; Exercises 13.27 and 13.28)

Outline the process of DNA replication. (Section 13.4; Exercises 13.32 and 13.35)

Describe three diﬀerences between DNA and RNA. (Section 13.5; Exercises 13.45 and 13.46)

Describe the function of the three main types of cellular RNA. (Section 13.5; Exercise 13.47)

Describe the process by which RNA is synthesized from DNA in cells. (Section 13.6; Exercise 13.51)

10. Explain how the genetic code functions in the ﬂow of genetic information. (Section 13.7; Exercises 13.58 and 13.61) 11. Outline the process by which proteins are synthesized in cells. (Section 13.7; Exercises 13.69, 13.70, and 13.73) 12. Describe the downstream eﬀects of genetic mutations in a living organism. (Section 13.7; Exercises 13.74 and 13.75) 13. Describe three viruses and the diseases they cause in human patients. (Section 13.8; Exercises 13.79 and 13.81) 14. Explain the process of immunization using vaccines. (Section 13.8; Exercise 13.81) 15. Describe the technology used in molecular cloning. (Section 13.9; Exercises 13.87 and 13.89) 16. Describe one biomedical application of CRISPR. (Section 13.9; Exercises 13.93 and 13.94)

LECTURE HINTS AND SUGGESTIONS 1.

Stress that all of the information that controls the cell is stored in the nucleus in the form of DNA. Follow the ﬂow of this information from the DNA to the messenger RNA to the synthesis of proteins. Show that all of this is a result of interactions between base pairs which ﬁt together and bind with just the right geometry and orientation. Relate all of this to intermolecular forces such as hydrogen bonding and organic chemistry between the functional groups.

Chapter 13 2.

If you want to teach with 3D models, plastic snap-lock DNA model kits are available from Sargent-Welch, Buﬀalo Grove, IL. These kits include workbooks and instructions to explore the structural characteristics of nucleic acids and their role in protein synthesis.

Figure 13.7 and Figure 13.10 can be combined to create a full-page handout that is useful in describing the structure of DNA molecules (see Chapter 13: DNA handout below). Depending on the level of detail you want to teach these two ﬁgures show students the structure of all four DNA nitrogenous bases, the phosphodiester linkage in the DNA backbone, and the hydrogen bonding between complementary bases in the DNA double helix.

Most students will have a reasonable understanding of messenger RNA (mRNA) from high school; however, fewer students will know about the structure and role of transfer RNA (tRNA) and ribosomal RNA (rRNA) in protein translation. Figures 13.20 and 13.29 can be combined to create a full-page handout (see handout Chapter 13: tRNA and Polypeptide Elongation) that walks the students through tRNA structure and shows them how tRNA and the ribosome translate a polypeptide chain. Figure 13.29 shows a cartoon of the ribosome and allows you to elaborate and explain that the ribosome is a protein-RNA complex consisting of 65% rRNA (which makes up 80% of the total RNA in the cell.)

An interactive activity useful in teaching students the genetic code is to take a simple DNA sequence and have students transcribe it into mRNA and translate it into a polypeptide chain. Start by providing the “Ch. 13: Genetic Code and Protein Translation Handout”, then use it to walk students through Example 13.4 and 13.5. After students have learned the concepts then have them test their knowledge by transcribing and translating the following DNA template sequence: 5’ CAC-ATC-GGG-CCT-TGT-CAT 3’ Solution: mRNA Sequence: 5’ AUG-ACA-AGG-CCC-GAU-GUG 3’ Protein Sequence: Met–Thr–Arg–Pro–Asp–Val

The COVID-19 pandemic will still be fresh in the minds of your students. This provides an opportunity to explain virology and vaccines in the context of a real-life experience for them. Explain the diﬀerences between DNA, RNA, and retroviruses and emphasize that COVID-19 is an RNA virus. Explain vaccines in general terms and then use this opportunity to teach students what a mRNA vaccine is, and how it is diﬀerent from other vaccines.

Students are in general more interested in recent technologies. In addition to covering molecular cloning touch on the CRISPR (clustered regularly interspaced short palindromic repeat) technique and discuss the implications it presents in terms of treating diseases such as sickle cell anemia. Also, if you are looking to incorporate lab activities, experiment kits to explore nucleic acids are available from Sargent-Welch, Buﬀalo Grove, IL. These cover some topics such as natural genetic engineering, where antibiotic resistance is transferred from one strain of bacteria to another, and plasmid insertion into bacterial DNA.

Chapter 13

Chapter 13: DNA Handout

Chapter 13

Chapter 13: tRNA and Polypeptide Elongation

Chapter 13

Ch. 13: Genetic Code and Protein Translation Handout

Chapter 13

Solutions to All End-of-Chapter Questions What follows are more complete answers/solutions to the EOC exercises whose answers are published in shorter form at the end of the textbook.

SECTION 13.1 THE FLOW OF GENETIC INFORMATION 13.1

The nucleus is the principle location for DNA within the eukaryotic cell.

13.2

RNA is found primarily in the cytoplasm of eukaryotic cells, though it may be found in other locations, too.

13.3

The central dogma of molecular biology is the well-established process by which genetic information stored in DNA molecules is expressed in the structure of synthesized proteins.

13.4

Transcription is the transfer of necessary information from a DNA molecule onto a molecule of messenger RNA. Translation is the decoding of the messenger RNA into an amino acid sequence, resulting in the synthesis of a specific protein.

13.5

A gene is the basic unit of physical and functional hereditary characteristics. It is made up of DNA and passed from parent to child.

SECTION 13.2 THE COMPONENTS OF NUCLEIC ACIDS 13.6

The carbohydrate, or sugar, that is part of nucleic acid structure diﬀers between DNA and RNA. In DNA, the sugar is 2-deoxyribose. In RNA, the sugar is ribose.

13.7

The three components of nucleotides are a ribose or deoxyribose unit, a phosphate group, and a pyrimidine or purine base.

13.8

13.9

guanine

purine

cytosine

pyrimidine

adenine

purine

thymine

pyrimidine

uracil

pyrimidine

The bases found in DNA are thymine, cytosine, adenine, and guanine. The bases found in RNA are uracil, cytosine, adenine, and guanine.

13.10

13.11

deoxyribose

adenine, guanine, cytosine, uracil

phosphate

ribose

adenine, guanine, cytosine, thymine

Chapter 13 c.

phosphate

false

true

false

true

false

13.12

13.13

13.14

13.15

SECTION 13.3 THE STRUCTURE OF DNA 13.16

Two DNA molecules that contain the same number of nucleotides may diﬀer in the order as well as relative numbers of the bases.

Chapter 13 13.17

The 3’ end of the DNA segment AGTCAT is T. The 5’ end of the DNA segment AGTCAT is A.

13.18

The data obtained from the chemical analysis of DNA that supported the idea of complementary base pairing in DNA proposed by Watson and Crick was that in all DNA the percentages of adenine and thymine were always equal to each other as were the percentages of guanine and cytosine.

13.19

The revolutionary three-dimensional structure proposed by Watson and Crick was the DNA double helix. This was a revelation as the double helix allowed for the copying of genetic information, a key feature of the study of genetics.

13.20

The double helix is established because the nucleotide base pairs hydrogen bond with one another, making up the “rungs” of the double helix “ladder”. Due to their complimentary structures, adenine hydrogen bonds with thymine and cytosine hydrogen bonds with guanine. See Figure 13.10.

13.21

13.22

13.23

CAGTAG 15 hydrogen bonds

TTGACA 14 hydrogen bonds

GCATGC 16 hydrogen bonds

TATGGC

15 hydrogen bonds

If each GC pair is held together by three hydrogen bonds and each AT pair is held together by two hydrogen bonds, strand A and its complementary strand would be held by 15 hydrogen bonds, while strand B and its complementary stand have only 14 hydrogen bonds.

13.24

Strand A and its complementary strand exhibit 16 hydrogen bonds, while strand B and its complementary strand have just 15 hydrogen bonds.

13.25

13.26

13.27

Thymine is 43% because it must match the amount of adenine.

Guanine is 7%. The total of A and T is 86%. The remaining 14% is split equally.

Cytosine is 7%.

14%

36%

14%

5’ ATGCATC 3’ original strand 3’ TACGTAG 5’ complementary strand

13.28

5’ GATTCA 3’ original strand 3’ CTAAGT 5’ complementary strand 8

Chapter 13 13.29

For the two strands, the amount of A = T and the amount of G = C. Thus, the complementary strand has 19% T, 34% G, 28% C, and 19% A.

13.30

For the two strands, the amount of A = T and the amount of G = C. Thus, the complementary strand has 21% T, 29% G, 36% C, and 14% A.

SECTION 13.4 DNA REPLICATION 13.31

A chromosome is a structure within a cell that contains one molecule of DNA coiled about small, basic proteins called histones. Each human cell contains 46 chromosomes. The approximate number of genes in a human cell is 25,000.

13.32

Semiconservative replication means a replication process in which DNA molecules contain one strand from the parent and a new strand that is complementary to the strand from the parent.

13.33

A replication fork is the point along the DNA helix at which the DNA helix unwinds for replication to begin.

13.34

The function of the enzyme helicase in the replication of DNA is to catalyze the separation and unwinding of the nucleic acid strands at a speciﬁc point along the DNA helix.

13.35

The steps for DNA replication are: (1) unwinding the double helix, (2) synthesizing DNA segments, and (3) closing the nicks.

13.36

The enzymes involved in DNA replication are helicase, DNA polymerase, and DNA ligase.

13.37

The new DNA strand is formed from the 5’ end to the 3’ end.

13.38

The synthesis of a DNA daughter strand growing toward a replication fork grows continuously. A daughter strand growing away from a replication fork has Okazaki fragments separated by nicks.

Chapter 13 13.39

13.40

GTACGT CATGCA

13.41

The origin of Okazaki fragments is that when a daughter strand grows away from the replication fork it will be separated by a gap from the next replication fork.

13.42

Using this diagram will help. Remember to write the answers in the 5’ to 3’ direction. Daughter Q

Daughter R

5’

3’

5’

3’

5’

3’

5’

New

Parent

New

Parent

Strand Strand a.

CCAGTT

AACTGG

Strand Strand

Chapter 13 13.43

Using this diagram will help. Remember to write the answers in the 5’ to 3’ direction. Daughter Q 5’

3’

5’

3’

5’

3’

5’

New

Parent

New

Parent

Strand Strand

13.44

Daughter R

ATACGT

ACGTAT

Strand Strand

Response c is correct. The repeating nucleotide units of a DNA strand are linked together by 3’ to 5’ phosphodiester linkages.

SECTION 13.5 RIBONUCLEIC ACID (RNA) 13.45

The sugar in the sugar phosphate backbone of RNA contains a hydroxyl group on the 2’ carbon, while in DNA it does not. RNA contains ribose and DNA contains deoxyribose.

13.46

DNA has two intertwined complementary strands joined by hydrogen bonds in a double helix. RNA has a single strand that has some regions of double-helical structure where complementary bases are joined by hydrogen bonds and some regions without hydrogen bonding that are loops.

13.47

Messenger RNA functions as a carrier of genetic information from the DNA of the cell nucleus directly to the cytoplasm. It contains 75-3000 nucleotides. Ribosomal RNA is located in the cytoplasm of the cell in organelles called ribosomes which serve as the sites of protein synthesis. Ribosomal RNA may contain 120, 1700, or 3700 nucleotides. Transfer RNA delivers amino acids to the site of protein synthesis. Strands of transfer RNA contain 73-93 nucleotides

13.48

13.49

rRNA

mRNA

tRNA

The two important regions of the tRNA molecule are the anticodon that binds to the mRNA and the 3’ end that binds to an amino acid.

Chapter 13

13.50

SECTION 13.6 TRANSCRIPTION: RNA SYNTHESIS FROM DNA 13.51

The DNA strands unwind, exposing the sequence to be transcribed. The mRNA forms on the DNA template, after which it is released. Finally, the DNA strands rewind.

13.52

5’ GCCATATTG 3’ DNA template 3’ CGGUAUAAC 5’ mRNA

13.53

DNA strand 5’ ATTGACTCG 3’ hnRNA

3' TAACTGAGC 5'

hRNA written 5' → 3' is CGAGTCAAT 13.54

RNA strand 5' UUGCGAAUC 3' DNA strand 3' AACGCTTAG 5' Written 5' → 3' DNA strand is GATTCGCAA

13.55

RNA strand 5' AUGCGUACG 3' DNA strand 3' TACGCATGC 5' Written 5' → 3' DNA strand is CGTACGCAT Once the introns are removed, the exon TACG remains.

13.56

DNA 5' TACG 3' mRNA 3' AUGC 5' Written 5' → 3' is CGUA Once the introns are removed, the exon TTACGCAT remains.

13.57

DNA 5' TTACGCAT 3' mRNA 3' AAUGCGUA 5' Written 5' → 3' is AUGCGUAA

Chapter 13

SECTION 13.7 THE GENETIC CODE AND PROTEIN SYNTHESIS 13.58

A codon is a sequence of three nucleotide bases that represents a “code word” on the mRNA molecules. Each codon provides genetic code information for a speciﬁc amino acid.

13.59

13.60

UAU

5’ AUA 3’

tyrosine

CAU

5’ AUG 3’

histidine

UCA

5’ UGA 3’

serine

UCU

5’ AGA 3’

serine

5’ UAC 3’ valine

5’ GGG 3’ proline

5’ GUG 3’ histidine

5’ UGG 3’ proline

13.61

cysteine—alanine—tyrosine—methionine—lysine

13.62

glutamic acid—aspartic acid—leucine—glycine—glycine

13.63

The base T is not found in mRNA

13.64

All codons are three-letters.

13.65

Researchers made a synthetic mRNA strand which was composed only of uracil bases. They incubated the synthetic mRNA with ribosomes, amino acids, tRNAs, and the appropriate enzymes for protein synthesis. The polypeptide that was synthesized was composed only of phenylalanine; therefore, UUU was determined to be the codon for phenylalanine.

13.66

False

Each codon is composed of four bases. (Correction: not four bases, three bases)

True

Some amino acids are represented by more than one codon.

False

All codons represent an amino acid. (Correction: Most, not all; some codons represent stop signals)

False

Each living species is thought to have its own unique genetic code. (Correction: share the same genetic code, not have its own unique genetic code.)

True

The codon AUG at the beginning of a sequence is a signal for protein synthesis to begin at that codon.

False

It is not known whether or not the code contains stop signals for protein synthesis. (Correction: UAG, UAA, and UGA are the known stop signals for protein synthesis.)

Chapter 13 13.67

The β-chain of hemoglobin is a protein that contains 146 amino acid residues. The minimum number of nucleotides on the mRNA that is coded for this protein is 444 bases.

 1 codon   3 bases   3 bases   3 bases  146 residues    + 1 stop codon   + 1 initiation codon   = 444 bases 1 residue 1 codon 1 codon       1 codon 

13.68

A polysome is the conﬁguration of several ribosomes simultaneously translating the same mRNA

13.69

To synthesize a protein, DNA is transcribed into mRNA in the nucleus. mRNA leaves the nucleus and aligns on a ribosome. As each codon is read on the mRNA, a tRNA unit with an attached amino acid hydrogen bonds with a codon. The amino acids are linked together to form a continuous peptide strand. This process continues until the ribosome complex reaches a stop codon on the mRNA. Hydrolysis at the stop codon releases the protein.

13.70

The stages of protein synthesis are: (1) initiation of the polypeptide chain, (2) elongation of the chain, and (3) termination of polypeptide synthesis.

13.71 13.72

Protein synthesis begins with the N-terminal amino acid. The A site is the aminoacyl site located on the mRNA-ribosome complex next to the P site where an incoming tRNA carrying the next amino acid will bond. The P site is the peptidyl site and it is the location on the ribosome that the initiating codon must occupy for the initiation process to begin.

13.73

Identify the peptide to form, written N-terminal end to C-terminal end: Gly-Leu-Pro-Cys Identify the codons for this peptide, including the initiation and stop codons: fMet = AUG Gly = GGA, GGC, GGG, or GGU Leu = CUA, CUC, CUG, CUU, UUA, or UUG Pro = CCA, CCC, CCG, or CCU Cys = UGC or UGU Stop = UAG, UAA, or UGA

Rewrite the peptide to include the initiation and stop codons: fMet-Gly-Leu-Pro-Cys-Stop Write the sequence of mRNA 5’ to 3’ that corresponds to the peptide: mRNA 5’ AUG-GGA-CUA-CCA-UGC-UAG 3’ Write the sequence of DNA 3’ to 5’ that corresponds to the mRNA needed: DNA

3’ TAC-CCT-GAT-GGT-ACG-ATC 5’

Remember by convention, nucleotide sequences are always read from 5’ to 3’.

Chapter 13 A segment of DNA is unwound to reveal CTA-GCA-TGG-TAG-TCC-CAT, which is transcribed into mRNA as AUG-GGA-CUA-CCA-UGC-UAG. The mRNA leaves the nucleus and attaches to the small ribosomal subunit, so the codon AUG is at the ribosomal P site. A tRNA with an attached N-formylmethionine forms hydrogen bonds with the codon. A large ribosomal subunit binds to the small subunit and completes the initiation complex. The codon GGA is at the A site and a tRNA brings a glycine molecule that forms a peptide bond to the N-formylmethionine. The whole ribosome moves one codon along the mRNA toward the 3’ end, the empty tRNA is released from the P site, and the tRNA attached to the A site moves to the P site. The CUA codon moves to the A site and a tRNA brings a leucine molecule that forms a peptide bond to the glycine residue. The empty tRNA is released from the P site, and the tRNA attached to the A site moves to the P site. The CCA codon moves to the A site and a tRNA brings a proline molecule that forms a peptide bond to the leucine residue. The empty tRNA is released from the P site, and the tRNA attached to the A site moves to the P site. The UGC codon moves to the A site and a tRNA brings a cysteine molecule that forms a peptide bond to the proline residue. The empty tRNA is released from the P site, and the tRNA attached to the A site moves to the P site. The UAG codon moves to the A site and a termination factor binds to the stop codon. It catalyzes the hydrolysis of the completed polypeptide chain from the ﬁnal tRNA. The “empty” ribosome dissociates and can then bind to another strand of mRNA to once again begin the process of protein synthesis. The N-formylmethionine is then cleaved from the peptide to leave the tetrapeptide Gly-Leu-Pro-Cys. 13.74

A genetic mutation is a change in the genetic code that results from an incorrect sequence of bases on a DNA molecule.

13.75

Genetic mutations can harm an organism by producing genetic diseases whenever an important protein (or enzyme) is not correctly synthesized.

Genetic mutations can help an organism by making it more capable of surviving in its environment.

13.76

The result of a genetic mutation that causes the mRNA sequence GCC to be replaced by CCC is that instead of the amino acid alanine, the polypeptide formed will contain the 15

Chapter 13 amino acid proline. 13.77

13.78

glycine—tyrosine—serine—arginine

glycine—tyrosine—serine—isoleucine

serine—glycine—isoleucine—tryptophan

threonine—glycine—isoleucine—tryptophan

serine—glycine—isoleucine—tryptophan

SECTION 13.8 VIRUSES 13.79

Viruses consist of genetic material in the form of nucleic acids, either DNA or RNA that are normally encapsulated in a protein coat.

13.80

Viruses “hijack” the cell machinery by forcing the host organism to replicate the genetic material, translate viral proteins in order to produce new virions. This allows the virus to propagate and multiply.

13.81

RNA: One prominent example of RNA viruses are the family of coronaviruses that cause such diseases as SARS, MERS, and COVID. DNA: One example of a DNA virus is the adenovirus, which causes symptoms similar to the common cold. Respiratory symptoms may be more severe in children. Retrovirus: One example of a retrovirus is HIV, which can lead to acquired immunodeﬁciency Syndrome (AIDS).

13.82

Prominent differences between an RNA virus, DNA virus, and retrovirus the way their genetic information is stored and expressed. An RNA virus contains 1 or 2 strands RNA genome that serves as the direct template for viral protein translation. DNA viruses contain single or double stranded DNA genomes that are first transcribed into RNA in the nucleus, and then translated into viral proteins in the cytoplasm of the host cell. Retroviruses also contain RNA genomes, however, the genetic information is first reverse transcribed into DNA, then integrated into the host cell genome, before undergoing transcription and translation into viral proteins.

13.83

Reverse transcriptase is the enzyme that converts viral genomic RNA into DNA, whereas integrase is responsible for splicing the viral DNA into the host cell genome.

13.84

The reverse transcriptase enzyme is highly error prone and introduces many DNA mutations when the genomic RNA is converted into DNA. This characteristic is a major contributor to the increased genetic variability and resulting rapid evolution of retroviruses.

Chapter 13 13.85

The broad idea of vaccines is to prepare the body to mount a defense upon exposure to a pathogen. By priming the immune system with some inactive or incomplete form of the pathogen, the body can develop antibodies to ward oﬀ the progression of disease if exposed to it again in the future.

13.86

Vaccine immunology can work in diﬀerent ways depending on the vaccine technology employed. Most commonly employed vaccines use inactivated (attenuated) forms of the pathogen to inoculate patients. This inactivated form does not cause the patient to progress in terms of disease, but instead activates the patient’s immune system to build a defense if exposed to the pathogen in the future. Newer mRNA vaccine technology has recently been used in some COVID vaccines. These vaccines inoculate patients with the genetic material of a part of the target virus, instead of the fully inactivated form. After inoculation, the immune system is then prompted to produce antibodies to match the genetic code of the vaccine, which should then target the pathogen upon exposure.

SECTION 13.9 RECOMBINANT DNA 13.87

Restriction enzymes are protective enzymes found in some bacteria that catalyzes the cleaving of all but a few speciﬁc types of DNA. They are important in the formation of recombinant DNA because they take DNA apart and reduce it to fragments of known size and nucleotide sequence. DNA ligase joins the breaks in the two DNA strands in order to return the plasmid to a circular piece of double-stranded DNA that has both human and bacterial DNA.

13.88

Recombinant DNA contains segments of DNA from two diﬀerent organisms.

13.89

Plasmids are circular, double-stranded DNA found in the cytoplasm of bacterial cells. They are used to get bacteria to synthesize a new protein that they normally do not synthesize by splicing the plasmid with a restriction enzyme, inserting a gene from another organism, and using DNA ligase to reform the circular plasmid. The desired protein can then be synthesized by the bacteria.

13.90

Substances likely to be produced on a large scale by genetic engineering are: human insulin (needed by diabetics allergic to insulin from pigs or cattle), human growth hormone (treatment of dwarﬁsm), interferon (needed to ﬁght viral infections), hepatitis B vaccine (protection against hepatitis), malaria vaccine (protection against malaria), and tissue plasminogen activator (needed to promote the dissolving of blood clots).

13.91

CRISPR stands for clustered regularly interspaced short palindromic repeat.

13.92

CRISPR refers to a series of repeating patterns of DNA nucleotides discovered in bacteria. These repeating patterns in bacterial DNA originated as a primitive immune system to recognize invading viruses.

Chapter 13 13.93

RNA transcribed from the DNA repeating patterns will bind to an enzyme, Cas 9, that cuts apart nucleotide strands. When Cas 9 is combined with a guide RNA, it will bind a complementary invading DNA or RNA sequence and chop it up, thereby preventing viral invasion. Interestingly, if the CRISPR-Cas 9 system is used to target human DNA, it triggers the body’s repair mechanism and it is possible to edit very speciﬁc regions of DNA.

13.94 Since inceS the CRISPR-Cas 9 system can be used to target and edit very speciﬁc regions of human DNA it could be used as potential cures for cystic ﬁbrosis and sickle-cell disease.

ADDITIONAL EXERCISES 13.95

Heating a double-stranded DNA fragment to 94-96 °C when performing PCR causes the DNA to unravel into single strands because the heat causes vibrations within the DNA molecule. The vibrations at this temperature are so intense that the hydrogen bonds that typically stabilize DNA are destroyed, which allows the DNA complementary strands to separate.

13.96

In general, hydrolysis reactions are used to break nucleotides polymers apart as is the case when introns are excised from pre-mRNA, and condensation reactions are used to join nucleotides as is the case when exons are spliced together to forms mature mRNA.

13.97

A segment of DNA has an original code of –ACA–. This segment was mutated to a –AAA– After the mutation occurred, the protein that was coded for was no longer able to maintain its tertiary structure. The mRNA codon from DNA –ACA– is –UGU–, while the mRNA codon from DNA – AAA– is –UUU–. The codon –UGU– codes for cysteine, which can make disulfide bonds to another cysteine residue in the protein. The codon –UUU– codes for phenylalanine, which cannot make disulfide bonds in the protein. The loss of tertiary structure is a result of losing the ability to make a disulfide bond to this amino acid residue because a phenylalanine residue is now in place of a cysteine residue.

CHEMISTRY FOR THOUGHT 13.98

Some benefits of genetic engineering include improved availability of vaccines, enzymes, and hormones for humans and improved production of agricultural food products. Some concerns about genetic engineering include the long-term health effects of overusing these enzymes and the effects of the agricultural hormone residues on humans.

13.99

Guanine and cytosine are able to form three hydrogen bonds each, while adenine and thymine are able to form two hydrogen bonds each. The DNA sample with a greater percentage of guanine-cytosine base pairs will have greater attractive forces holding the double helix together because the guanine-cytosine base pairs are held together with more hydrogen bonds than the adenine-thymine base pairs.

Chapter 13 13.100 AZT, a drug used to fight HIV, is believed to act as an enzyme inhibitor. The type of enzyme inhibition most likely caused by this drug is competitive inhibition. AZT is similar to the substrate of the virus enzyme and thus most likely would be a competitor for the enzyme. 13.101 We are fortunate that the genetic code is degenerate, that is, several codons code for the same amino acid. If this were not the case and there were only one codon per amino acid and 44 stop signals, then most mutations would result in the premature termination of protein synthesis. Because protein synthesis is vital to the continuation of life, we are fortunate that mutations have a more limited effect because most amino acids have several corresponding codons and there are only three stop signals. 13.102 DNA specifies only the primary structure of a protein. The three-dimensional protein structure develops as a result of backbone and side chain interactions as well as interactions between the protein and its environment. These interactions occur as a natural consequence of the primary structure of the protein. 13.103 a.

It would take one DNA polymerase molecule approximately 6 years to replicate the genetic material in one cell.

  1 hour   1 day   1 year  1 min 3 x 10 9 nucleotides   = 5.7 years    1000 nucleotides    60 min   24 hours  365.25 days  ≈ 6 years b.

To replicate the cell’s genetic material in 10 minutes, 3 x 105 molecules of DNA polymerase would be needed.

3 x 10 9 nucleotides  1 min x molecule  5   = 3 x 10 molecules 10 min 1000 nucleotides  

Chapter 14

Chapter 14: Lipids CHAPTER OUTLINE 14.1

The Classiﬁcation of Lipids

14.2

Fatty Acids

14.3

The Structure of Fats and Oils

14.4

Chemical Properties and Reactions of Fats and Oils

14.5

Waxes

14.6

Phosphoglycerides

14.7

Sphingolipids

14.8

Biological Membranes

14.9

Steroids and Hormones

LEARNING OBJECTIVES/ASSESSMENT When you have completed your study of this chapter, you should be able to: 1.

List two of the major functions of lipids. (Section 14.1; Exercise 14.2)

Classify lipids as saponiﬁable or nonsaponiﬁable. (Section 14.1; Exercises 14.4 and 14.5)

Describe four general characteristics of fatty acids. (Section 14.2; Exercises 14.8 and 14.9)

Draw structural formulas of triglycerides given the formulas of the component parts. (Section 14.3; Exercises 14.20 and 14.22)

Describe the structural similarities and diﬀerences of fats and oils. (Section 14.3; Exercise 14.24)

Write key reactions for fats and oils. (Section 14.4; Exercises 14.31, 14.33, and 14.35)

Compare and contrast the structures of fats and waxes. (Section 14.5; Exercises 14.44)

Draw structural formulas for phosphoglycerides. (Section 14.6; Exercises 14.49 and 14.50)

Describe two uses for phosphoglycerides. (Section 14.6; Exercises 14.56 and 14.57)

10. Draw structural formulas for sphingolipids. (Section 14.7; Exercises 14.58 and 14.59) 11. Describe two uses for sphingolipids. (Section 14.7; Exercises 14.63, 14.64, and 14.68) 12. Describe the major features of cell membrane structure. (Section 14.8; Exercise 14.73) 13. Identify the structural characteristic that is typical of steroids. (Section 14.9; Exercise 14.76) 14. List ﬁve important groups of steroids in the body. (Section 14.9; Exercises 14.81, 14.82, 14.85, and 14.92) LECTURE HINTS AND SUGGESTIONS 1.

Explain that lipids, unlike most other types of biological compounds, are distinguished by a common physical property (oil solubility). Because of this, lipids can represent a wide variety of organic compounds with diﬀerent functional groups.

The wide variety of lipids can sometimes contribute to confusion in understanding the structural characteristics of the different classes. The handout titled “Chapter 14: Lipids” (see the next two pages) first introduces the classification of lipids based on if they are saponifiable or nonsaponifiable. It then shows a side-by-side structural comparison of the six different groups of lipids to help students understand how to properly classify lipid examples.

Chapter 18 Chapter 14: Lipids

Waxes:

Triglycerides:

Phosphoglycerides:

Chapter 18

Sphingolipids:

Glycolipids (subset of Sphingolipids)

Steroids:

Prostaglandins:

Chapter 18

When introducing fats and fatty acids, many common industrial processes such as the manufacture of soaps or margarine can be discussed. This chapter is an appropriate time to explain how a soap-like molecule can stabilize an emulsion and function as a cleaning agent. It is also an appropriate time to introduce hard and soft water as well as the function of water softeners.

Many health topics broadcast on the news involve lipid materials: anabolic steroids, saturated fat, gallstones (ultrasound surgery), "good" and "bad" cholesterol, fatty acids as nutrients, prostaglandin in medicine, birth control pills, liposomes for drug delivery, etc. These would make excellent topics for group discussion posts or student presentations.

Chapter 18

Solutions to All End-of-Chapter Questions What follows are more complete answers/solutions to the EOC exercises whose answers are published in shorter form at the end of the textbook.

SECTION 14.1 THE CLASSIFICATION OF LIPIDS 14.1

The basis for deciding if a substance is a lipid is whether the molecule from a living organism is soluble only in nonpolar solvents and insoluble in water.

14.2

Lipids are a form of stored energy and a structural component of the human body. Some hormones in the human body are lipids.

14.3 14.4

14.5

An ester is the functional group common to all saponiﬁable lipids. a.

A steroid

nonsaponiﬁable

A wax

saponiﬁable

A phosphoglyceride

saponiﬁable

A simple wax

saponiﬁable

nonsaponiﬁable

14.6

a and c

14.7

a, b, and d

SECTION 14.2 FATTY ACIDS 14.8

14.9

Four structural characteristics exhibited by most fatty acids are: (1)

they are usually straight-chain carboxylic acids

(2)

the sizes of most common fatty acids range from 10 to 20 carbon atoms

Chapter 18 (3)

fatty acids usually have an even number of carbon atoms

(4)

fatty acids can be saturated or unsaturated

14.10

A micelle is a spherical cluster of fatty acid ions in which the nonpolar chains extend toward the interior of the structure away from water and the polar carboxylate groups face outward in contact with the water. The nonpolar chains in a micelle are held together by weak dispersion forces.

14.11

Insoluble lipids form complexes with proteins to form lipoprotein aggregates (chylomicrons) These complexes pass into the lymph system and then into the bloodstream. They are modified by the liver into smaller lipoprotein particles, which allows most lipids to be transported to various parts of the body by the bloodstream.

14.12

14.13

Two essential fatty acids are linolenic acid and linoleic acid. The human body cannot synthesize these acids; therefore, they must be obtained through diet.

14.14

a. b. c. d.

CH 3 (CH 2 )14 COOH CH 3 (CH 2 )4 CH=CH 2 CH=CH(CH 2 )7 COOH CH 3 (CH14 H 24 )COOH

14.15

a. b. c. d.

unsaturated, liquid unsaturated, liquid saturated, liquid saturated, solid

14.16

Unsaturated fatty acids have lower melting points than saturated fatty acids because they contain

CH 3 (CH10 H 20 )COOH

saturated unsaturated unsaturated saturated

solid liquid liquid solid

double bonds which form kinks and prevent the unsaturated fatty acids from packing together as eﬀectively as the saturated fatty acids. The greater separation between the molecules causes the intermolecular forces in the unsaturated fatty acids to be weaker than the intermolecular forces in the saturated fatty acids. Weaker intermolecular forces result in lower melting points. 14.17

polyunsaturated, 5 double bonds, omega-3-fatty acid

saturated

monounsaturated

14.18 a. b.

saturated polysaturated, 4 double bonds, omega-6 fatty acid

Chapter 18 c.

monounsaturated

polyunsaturated, 6 double bonds, omega-3 fatty acid

14.19 The structural feature of a fatty acid responsible for the name omega-3 fatty acid is that the last double bond is three carbons from the methyl end of the chain: CH3CH2CH=CH….COOH

SECTION 14.3 THE STRUCTURES OF FATS AND OILS 14.20

14.21

14.22

Chapter 18

14.23

14.24

Fats and oils are both triglycerides that contain a glycerol backbone and three fatty acid chains. Fats contain more saturated than unsaturated fatty acids and oils contain more unsaturated than saturated fatty acids. This causes fats to be solid at room temperature, while oils are liquid under the same conditions.

14.25

The order of increasing percentage of unsaturated fatty acids is: butter, beef fat, chicken fat, corn oil, sunﬂower oil.

14.26

Triglycerides from animal sources (other than ﬁsh) tend to have more saturated fatty acids. Triglycerides from plants or ﬁsh tend to have more unsaturated fatty acids.

14.27

Triglyceride B should have the lower melting point because it contains a higher percentage of unsaturated fatty acids than Triglyceride A.

14.28

Triglyceride B has a greater percentage of unsaturated fatty acid (oleic) and the lowest melting point.

14.29

The amount of saturated fat in the diet is a health concern because it is a risk factor inﬂuencing the development of chronic disease. Saturated fat raises blood cholesterol levels, which is a recognized risk factor in the development of coronary heart disease, the leading cause of death in Americans every year.

SECTION 14.4 CHEMICAL PROPERTIES AND REACTIONS OF FATS AND OILS 14.30

The process used to prepare a number of useful products such as margarines and cooking shortenings from vegetable oils is hydrogenation. In this process some of the double bonds in the unsaturated fatty acids are hydrogenated by reaction with hydrogen gas in the presence of a catalyst. Since fewer double bonds are present, the melting point of the mixture increases and the vegetable oils become more solid in consistency.

Chapter 18

14.31

14.32

14.33

14.34

Chapter 18

14.35

14.36

14.37

ﬁve

14.38

two

14.39

Chapter 18

14.40

Chapter 18 14.41

Hydrogenation of vegetable oils is of great commercial importance because it increases in the melting point of the fat or oil and allows the product to be used as margarine and cooking shortenings.

14.42

a. b. c. d.

Acid hydrolysis of a fat → glycerol + 3 fatty acids (more saturated than unsaturated)

Acid hydrolysis of an oil → glycerol + 3 fatty acids (more unsaturated than saturated)

Saponiﬁcation of a fat → glycerol + 3 salts of fatty acids (soaps)

Saponiﬁcation of an oil → glycerol + 3 salts of fatty acids (soaps)

SECTION 14.5 WAXES

14.43 14.44

Waxes and fats are esters of long-chain fatty acids; however, waxes do not contain glycerol backbones and fats do. Waxes contain one fatty acid chain, while fats contain three fatty acid chains.

14.45

14.46

14.47

14.48

Waxes are protective coatings on feathers, fur, skin, leaves, and fruits. They also occur in secretions of the sebaceous glands to keep skin soft and prevent dehydration.

Chapter 18

SECTION 14.6 PHOSPHOGLYCERIDES 14.49

Phosphoglycerides contain glycerol and 2 fatty acids as well as phosphoric acid and an aminoalcohol. Triglycerides contain glycerol and 3 fatty acids.

14.50

Glycerol

Fatty Acid Fatty Acid Phosphoric Acid Aminoalcohol

14.51

14.52

This is a cephalin because it contains serine.

Chapter 18

14.53

14.54

Lecithins are phosphoglycerides that contain the aminoalcohol choline. Cephalins are phosphoglycerides that contain ethanolamine or serine.

14.55 Lecithin is an important commercial emulsifying agent in certain food products because the phosphate-choline area of the molecule is charged and strongly hydrophilic, while the rest of the molecule is hydrophobic.

14.56

Lecithins are important structural components in cell membranes as well as emulsifying micelleforming agents.

14.57

Cephalins are found in most cell membranes and are especially abundant in brain tissue.

SECTION 14.7 SPINGOLIPIDS 14.58

Sphingolipids include sphingomyelins and glycolipids. Both of these subclasses contain the sphingosine backbone and a fatty acid.

14.59

Sphingosine

Chapter 18

Fatty Acid Phospholipid Aminoalcohol

Fatty Acid Carbohydrate

Sphingolipids have a sphingosine backbone, 1 fatty acid, and either phosphoric acid and an aminoalcohol or a carbohydrate. Phosphoglycerides contain a glycerol backbone, 2 fatty acids, phosphoric acid, and an aminoalcohol.

14.60

14.61

Chapter 18 14.62

Sphingomyelins and glycolipids both have a sphingosine backbone and 1 fatty acid. Sphingomyelins have phosphoric acid and an aminoalcohol, while glycolipids have a carbohydrate.

14.63

Another name for glycolipids is cerebrosides. These compounds are abundant in brain tissue.

14.64

14.65

14.66

A person with blood type A has a glycolipid on the surface of their blood cells called antigen A and a person with blood type B has a diﬀerent glycolipid on their cells called antigen B (see Figure 14.19). Antigen B has galactose in place of N-Acetylgalactosamine (GalNAc) when compared to antigen A. Because an individual with blood type A only expresses the A antigen, they will develop antibodies to the B antigen and cannot receive blood from an individual with blood type B or AB because their immune system will attack the donated blood cells.

14.67

The glycolipids attached to the surface of blood cells are called antigens. A person’s blood type indicates the types of antigens their blood expresses. For instance, blood type AB will express both glycolipid antigens A and B. Interestingly, all the blood antigens share the same structure through the sugar fucose (Fuc), and it is only the addition of GalNAc or Gal, that result in antigen A or B, respectively. Since antigen O shares the same root structure as A and B, it will not trigger

Chapter 18 the immune system to produce antibodies to it. Essentially the body will recognize it as “self” and will not mount an immune response to it. 14.68

Three diseases caused by abnormal metabolism and accumulation of sphingolipids are Tay-Sachs, Gaucher’s, and Niemann-Pick diseases.

SECTION 14.8 BIOLOGICAL MEMBRANES 14.69

The membrane in a prokaryotic cell is around the outside of the cell. The membranes in a eukaryotic cell are around the outside of the cell and around some of the organelles. The external cell membrane acts as a selective barrier between the living cell and its environment, and the internal membranes surround some organelles, creating cellular compartments that have separate organization and functions.

14.70

The three classes of lipids found in membranes are phosphoglycerides, sphingomyelins, and steroids (cholesterol).

14.71

Cholesterol (hydrophobic) is found internally in the hydrophobic portion of the cell membrane.

14.72

The polarity of the phosphoglycerides contributes to their function of forming cell membranes by allowing a lipid bilayer to form with hydrophilic groups oriented to the outside and hydrophobic groups oriented to the inside.

14.73

The ﬂuid-mosaic model contains lipids organized in a bilayer in such a way that the hydrophilic heads are pointed toward the outside of the bilayer and the hydrophobic tails are pointed toward the inside of the bilayer. Some proteins ﬂoat in the lipid bilayer and other proteins extend completely through the bilayer. The lipid molecules move freely laterally within the bilayer.

14.74

Lipid nanoparticles can encase the medicine making it more soluble in the aqueous environment that is the body. Increased solubility results in increased concentrations of the drug entering the body’s circulation allowing more eﬀective delivery of medicines to target tissues in human patients.

14.75

Lipid nanoparticles are used as the delivery system for the mRNA COVID-19 vaccines. The COVID-19 vaccine contains fragments of mRNA that code for the spike protein of the SARS-CoV2 virus, so that the body can develop antibodies to the viral proteins without ever being infected. On their own the viral mRNA fragments would be degraded by various enzymes in the body, however, by enveloping the mRNA within lipid nanoparticles, these nucleic acids strands are suﬃciently protected to reach their target.

Chapter 18

SECTION 14.9 STEROIDS AND HORMONES 14.76

All steroids contain a 4 fused ring system composed of 3 six membered rings and 1 ﬁve membered ring arranged as shown in the structure to the left.

14.77

Testosterone and progesterone both contain the fused ring

system

characteristic of steroids as well as a ketone functional group, a double bond, and two methyl groups in the same locations. Testosterone only has one additional functional group, a hydroxy group. Progesterone has a carboxyl group bonded to a methyl group in the position that testosterone has the hydroxy group.

14.78

Prostaglandins are characterized as cyclic compounds synthesized from the 20-carbon unsaturated fatty acid arachidonic acid. They contain a carboxylic acid group, various ring substituents, and side-chain double bonds.

14.79

At portion 3 on the ﬁve-membered ring, both testosterone and methandrostenolone have an –OH group, while progesterone does not.

14.80

Bile salts aid in the digestion of lipids by emulsifying lipids and breaking the large globules into many smaller droplets in order to increase the surface area available for hydrolysis reactions.

14.81

Important groups of compounds the body synthesizes from cholesterol are bile salts, male and female sex hormones, vitamin D, and the adrenocorticoid hormones. Cholesterol is an essential component of cell membranes.

14.82

With a large hydrophobic region and an ionic region, sodium glycocholate functions like a soap, emulsifying cholesterol.

14.83

The symptoms that might indicate the presence of gallstones in the gallbladder are excruciating pain, nausea, yellowing of the skin, and gray-colored stool.

14.84

The major component in most gallstones is cholesterol.

14.85

A hormone is a chemical messenger secreted by speciﬁc glands and carried by the blood to a target tissue, where it triggers a particular response. Hormones are eﬀective in very small amounts.

14.86

The two groups of adrenocorticoid hormones are mineralocorticoids and glucocorticoids. Mineralocorticoids regulate the concentration of ions in body ﬂuids. An example of a mineralocorticoid is aldosterone which regulates the retention of sodium and chloride ions in

Chapter 18 urine formation. Glucocorticoids regulate carbohydrate metabolism. An example of a glucocorticoid is cortisol which helps to increase glucose and glycogen concentrations in the body. 14.87

The primary male sex hormone is testosterone. The three principal female sex hormones are estradiol, estrone, and progesterone.

14.88

Athletes use anabolic steroids to promote muscular development without excessive masculinization. The side eﬀects of anabolic steroid use range from acne to deadly liver tumors. In males, anabolic steroid use can cause testicular atrophy, a decrease in sperm count, and temporary infertility.

14.89

The estrogens are involved in egg development in the ovaries. Progesterone causes changes in the wall of the uterus to prepare it to accept a fertilized egg and maintain the resulting pregnancy.

14.90

It is suggested that some people restrict cholesterol intake in their diet because high levels of cholesterol in the blood are linked to atherosclerosis (hardening of the arteries).

14.91

The starting material for prostaglandin synthesis is arachidonic acid.

14.92

Body processes regulated in part by prostaglandins are menstruation, conception, uterine contractions during childbirth, blood clotting, inﬂammation, and fever, to name only a few.

14.93

Hormones are synthesized in certain glands, while prostaglandins are synthesized within cells not associated with glands.

14.94

Therapeutic uses of prostaglandins include inducing labor, therapeutic abortion, treating asthma, inhibiting gastric secretions, and treating peptic ulcers.

ADDITIONAL EXERCISES 14.95

Unsaturated fatty acids are susceptible to oxidation, which causes rancidity in food products. The amount of oxidative rancidity could be decreased by vacuum sealing the food (removing the oxygen-containing air from around the food during packaging), keeping the product chilled (decreases the rate of reaction), or adding an antioxidant like BHA or BHT.

14.96

Complex lipids are more predominant in cell membranes than simple lipids because the cell membranes have a lipid bilayer which is composed of lipids with polar head groups and nonpolar tails. Phospholipid molecules have a structure that is more conducive to forming a lipid bilayer than a fat, oil, or wax. Simple lipids do not form as eﬀective a barrier as the complex

Chapter 18 lipids. 14.97

Unsaturated fatty acids have lower melting points than saturated fatty acids because the presence of one or more double bonds in the hydrocarbon chain introduces a definite bend in the chain. This prevents the molecules from packing together as effectively as a saturated chain, decreasing the intermolecular interactions and decreasing the effectiveness of the intermolecular forces. Compound (b) will have a lower melting point than compound (a).

CHEMISTRY FOR THOUGHT 14.98

Gasoline is soluble in nonpolar solvents, but it is not classiﬁed as a lipid because it does not come from a living organism.

14.99

The structure of cellular membranes is such that ruptures are closed naturally. The molecular forces that cause the closing to occur are dispersion forces between the nonpolar “tails” in the lipid bilayer as well as dipolar forces between the polar “heads” in the lipid bilayer and hydrogen bonding between the polar “heads” in the lipid bilayer and the water in the surrounding ﬂuids.

14.100 Seeds are rich in oils because as a new plant begins to grow from the seed it needs energy from the oils (a low density, but calorie rich source) since the plant cannot photosynthesize until it has grown leaves. 14.101 A soap has a fatty acid tail and a polar head group that has a negative charge, which has a positive counterion. A lecithin has two fatty acid tails and a polar head group that contains both a negative charge (phosphate group) and a positive charge (choline). These compounds are similar in their structure because they both contain fatty acids as well as a highly charged polar head group.

Chapter 15: Nutrition and Metabolism CHAPTER OUTLINE 15.1 15.2 15.3 15.4

The Basics of Nutrition An Overview of Metabolism ATP: The Energy Currency of Cells Carbohydrate Metabolism and Blood Glucose

15.5 15.6 15.7 15.8

An Overview of Cellular Respiration Glycogen Metabolism and Gluconeogenesis Lipid Metabolism and Biosynthesis Amino Acid Metabolism and Biosynthesis

LEARNING OBJECTIVES/ASSESSMENT When you have completed your study of this chapter, you should be able to: 1.

Describe the primary functions in the body of carbohydrates, lipids, proteins, vitamins, and minerals. (Section 15.1; Exercises 15.10, 15.25, and 15.31)

Deﬁne metabolism, anabolism, and catabolism. (Section 15.2; Exercises 15.32, 15.33, and 15.34)

Outline the three stages in the extraction of energy from food. (Section 15.2; Exercise 15.35)

Explain how ATP plays a central role in the production and use of cellular energy. (Section 15.3; Exercises 15.39, 15.43, and 15.45)

Identify the monosaccharide products of the digestion of carbohydrates. (Section 15.4; Exercise 15.46)

Explain the importance to the body of maintaining proper blood sugar levels. (Section 15.4; Exercises

Describe the process and cellular regulation of glycolysis and the citric acid cycle. (Section 15.5;

15.48 and 15.50) Exercises 15.54, 15.57, 15.60, 15.63, and 15.66) 8.

Describe in detail the process of oxidative phosphorylation and how it results in ATP production. (Section 15.5; Exercises 15.71, 15.73, and 15.74)

Explain the importance of glycogenesis, glycogenolysis, and gluconeogenesis. (Section 15.6; Exercises 15.75, and 15.83)

10. Describe how hormones regulate carbohydrate metabolism. (Section 15.6; Exercises 15.85 and 15.86) 11. Describe the metabolic pathways by which glycerol and fatty acids are catabolized. (Section 15.7; Exercises 15.88, 15.92, and 15.94) 12. Name the three ketone bodies. (Section 15.7; Exercise 15.96) 13. List the conditions that cause the overproduction of ketone bodies. (Section 15.7; Exercise 15.97 and 15.103) 14. Describe the pathway for fatty acid synthesis. (Section 15.7; Exercises 15.104 and 15.105) 15. Describe the source and function of the body’s amino acid pool. (Section 15.8; Exercises 15.109 and 15.110) 16. Explain the chemical fate of the nitrogen atoms and carbon skeleton of amino acids. (Section 15.8; Exercises 15.111, 15.112, and 15.115) 17. Explain the relationship between intermediates of carbohydrate metabolism and the synthesis of nonessential amino acids. (Section 15.8; Exercise 15.125)

Chapter 15

LECTURE HINTS AND SUGGESTIONS 1.

An interesting discussion post could be to investigate from a nutritional viewpoint some of the current popular diets or weight loss programs. Students could examine each program regarding its content of vitamins and minerals and compare them to the DV values. Also, it would be useful to note whether each diet supplies recommended minimum amounts of protein, fats, and carbohydrates.

Food and nutrient analysis kits are available for student use from Sargent-Welch. By using some simple test solutions and procedures, students can analyze foods for vitamins, carbohydrates, proteins, and fats.

A kit for learning about cell respiration (Cell Respiration Made Easy Kit) is available from Carolina.com. It consists of plastic pieces that stick onto large demonstration boards. A teacher's manual and student worksheets are provided for exercises in learning about biochemical pathways such as the citric acid cycle and glycolysis. Carolina also has numerous kits that allow students to investigate cellular respiration vs fermentation using yeast in a laboratory setting.

Depending on the student audience and the time constraints of the course, it may be advisable to condense the material on the biochemical pathways. Introductory students may be a bit overwhelmed at ﬁrst by all of the complex structural formulas. Stress the overall functions and ATP output of each of the major pathways that makeup cellular respiration, glycolysis, pyruvate oxidation, citric acid cycle and oxidative phosphorylation. By emphasizing function and importance and de-emphasizing organic structures, the material may become more understandable to these introductory students. Emphasize ﬁgures that depict the “big picture” of each pathway (See handout Chapter 15: Cellular Respiration.)

When teaching gluconeogenesis explain that seven out of the eleven reactions in the pathway utilize glycolysis enzymes that can catalyze the reverse reaction. This means the body only needs four new enzymes to produce glucose from pyruvate. To emphasize these enzymes and show that glycolysis and gluconeogenesis are opposing pathways that help the body breakdown or produce glucose based on need in real time, provide students with Figure 15.28 as a full-page image (see handout Chapter 15: Glycolysis and Gluconeogenesis.) Have them use one color to highlight the enzymes shared by both pathways and then use two distinct colors to separately highlight enzymes that are speciﬁc to either glycolysis or gluconeogenesis.

Chapter 15

Chapter 15: Cellular Respiration Overview:

Figure 15.9 The three stages in the extraction of energy from food. 3

Chapter 15

Glycolysis (Steps and Regulation)

Figures 15.17 A summary of glycolysis

Figure 15.18 The regulation of the glycolysis pathway

Chapter 15

Pyruvate Oxidation

Citric Acid Cycle (Steps and Regulation)

Figure 15.21 The citric acid cycle 5

Chapter 15

Oxidative Phosphorylation

Figure 15.23 Oxidative phosphorylation

Chapter 15

Chapter 15: Glycolysis and Gluconeogenesis

Figure 15.28 The reactions of glycolysis and gluconeogenesis are opposing pathways. 7

Chapter 15

Solutions for All End-of-Chapter Questions What follows are more complete answers/solutions to the EOC exercises whose answers are published in shorter form at the end of the textbook.

Section 15.1 The Basics of Nutrition 15.1 15.2

The principle component of dietary ﬁber is cellulose. Macronutrients are nutrients needed by the body in gram quantities every day. Micronutrients are nutrients needed by the body in milligram or microgram quantities every day.

15.3

10% =

250 mg x 100 total sodium

(10)(total sodium) = (250 mg)(100) Total sodium = 2500 mg

15.4

2500 mg x

15.5

8% =

1 tsp 4,250 mg

= 0.59 tsp

23 g x 100 total carbohydrate

(8)(total carbohydrate) = (23g)(100) Total carbohydrate = 287.5 g or 300 g for one significant figure

15.6

300 g carbohydrate x

1 potato 26 g carbohydrate

= 11.54 potatoes

or 10 potatoes if 300 is considered to have just one significant figure. 15.7

6% =

1.5 g x 100 total fiber

(6)(total fiber) = (1.5 g)(100) total fiber = 25 g or 30 g to one significant figure.

Chapter 15

15.8

30 g fiber =

1 potato = 15 potatoes 2 g fiber

or 20 potatoes to one significant figure. 15.9

Fiber is needed in the diet to prevent or relieve constipation by absorbing water and softening the stool for easier elimination.

15.10

Carbohydrates are used for energy and materials for the synthesis of cell and tissue components. Lipids are a concentrated source of energy; they store fat-soluble vitamins and help carry them through the body, and provide essential fatty acids. Lipids are also an essential component of cell membranes. Proteins are used to make new tissues, maintain and repair cells, synthesize enzymes, hormones, and other nitrogen-containing compounds in the body, and provide energy.

15.11

15.12

15.13

15.14

15.15

15.16

carbohydrates (simple)

sugar (many other correct answers)

carbohydrates (complex)

potato (many other correct answers)

lipid

butter (many other correct answers)

protein

steak (many other correct answers)

carbohydrates

proteins

lipids

proteins

proteins, lipids

carbohydrates

protein

carbohydrate

lipids

lipid

protein

carbohydrate

potato chips

carbohydrates, lipids, proteins

buttered toast

carbohydrates, lipids, proteins

plain toast with jam

carbohydrates, proteins

cheese sandwich

carbohydrates, lipids, proteins

Chapter 15 15.17

15.18

proteins, lipids

lipids, proteins, carbohydrates

carbohydrates

carbohydrates, protein, lipids

Linoleic acid is called an essential fatty acid for humans because humans cannot synthesize this compound. It must be obtained from the diet.

15.19

The daily values are 65 g of total fat; 20 g of saturated fat; 300 g of total carbohydrates; and 25 g of ﬁber for a 2,000 calorie diet.

15.20

Note: The signiﬁcant ﬁgures from the calorie per gram conversion were ignored in these calculations. 15.21

15.22

tocopherol

fat-soluble

niacin

water-soluble

folic acid

water-soluble

retinol

fat-soluble

water-soluble

15.23

water-soluble

15.24

fat-soluble

15.25

Eight of the water-soluble vitamins function as coenzymes

15.26

Large doses of fat-soluble vitamins are potentially dangerous because excess fat-soluble vitamins will be stored in the body’s fat reserves and stay in the body. Fat-soluble vitamins can be harmful in large quantities; however, large doses of water-soluble vitamins will be excreted and any eﬀect due to a higher concentration will be only temporary.

Chapter 15 15.27

15.28

15.29

night blindness

Vitamin A (retinol)

blood clotting

Vitamin K

calcium and phosphorus use in forming bones and teeth

Vitamin D (calciferol)

preventing oxidation of fatty acids

Vitamin E (tocopherol)

scurvy

Vitamin C (ascorbic acid)

beriberi

Vitamin B1 (thiam)

pernicious anemia

Vitamin B12 (cobalamin)

pellagra

niacin

A vitamin is an organic micronutrient that the body cannot produce in the amounts needed for good health. A mineral is a metal or a nonmetal used in the body in the form of ions or compounds.

15.30

A major mineral is a mineral found in the body in quantities greater than 5 g. A minor mineral is a mineral found in the body in quantities less than 5 g.

15.31

Trace minerals are components of vitamins, enzymes, hormones, and specialized proteins.

Section 15.2 AN OVERVIEW OF METABOLISM 15.32

A metabolic pathway is a sequence of reactions used to produce one ultimate product or accomplish one overall process. The product(s) of one reaction is(are) used up in the next reaction in the sequence so that all the reactions are essential to the overall process.

15.33

Catabolism includes all breakdown processes. Anabolism includes all synthesis processes.

15.34

Catabolism is energy producing; anabolism is an energy-consuming process.

15.35

The three stages in the extraction of energy from food are digestion, production of acetyl CoA, and the common catabolic pathway.

15.36

15.37

formation of ATP

stage III

digestion of fuel molecules

stage I

consumption of O2

stage III

generation of acetyl COA

stage II

stage III

stage II 11

Chapter 15

15.38

stage III

stage I

The main purpose of the catabolic pathway is to produce ATP.

Section 15.3 ATP: THE ENERGY CURRENCY OF CELLS 15.39

ATP is involved in both anabolic and catabolic processes.

15.40

ATP represents adenosine triphosphate. ADP represents adenosine diphosphate. AMP represents adenosine monophosphate.

15.41

15.42

ΔG°’ represents the free energy change for a reaction under normal physiological conditions (normal body temperature and the appropriate pH and concentration for the tissue involved) in kcal/mol. A negative ΔG°’ value indicates an exergonic reaction. A positive ΔG°’ value indicates an endergonic reaction.

15.43

The triphosphate end of ATP is the high-energy portion of the ATP molecule. The energy is stored in the bonds within this portion of the ATP molecule.

15.44

Anabolic processes require energy and should have a +∆G0’. Catabolic processes liberate energy and thus should have a -∆G0’.

Chapter 15 15.45

Mitochondria serve as the site for most ATP synthesis in the cells.

Section 15.4 Carbohydrate Metabolism and Blood Glucose 15.46

starch

glucose

lactose

glucose, galactose

sucrose

fructose, glucose

maltose

glucose

15.47

The digestion of carbohydrates involves hydrolysis.

15.48

Blood sugar level is the concentration of glucose in the blood. Normal fasting level is the concentration of glucose in the blood after an 8-12 hour fast.

15.49 15.50

The normal fasting level of glucose is 70-110 mg/100 mL. a.

hypoglycemia

blood sugar level is below normal fasting level

hyperglycemia

blood sugar level is above normal fasting level

renal threshold

blood sugar level is above 108 mg/100 mL and sugar is not completely reabsorbed from the urine by the kidneys

d. glucosuria

blood sugar level is above the renal threshold and glucose appears in the urine.

15.51

Molarity = moles/L

Moles =

Volume is 100.0 mL = 0.100 L Mass is 100.0 mg or 0.100 g

0.100 g = 0.000555 or 5.55 x 10 -4 180.2 g

5.55 x 10 -4 moles = 5.55 x 10 -3 M 0.100 L

15.52

Insulin could be used to lower higher levels of blood glucose (hyperglycemia)

15.53

The liver responds to the increase in blood glucose that follows a meal by removing glucose from the bloodstream and converting it to glycogen or triglycerides for storage. When blood glucose levels are low, the liver responds by converting stored glycogen to glucose and by synthesizing new glucose from noncarbohydrate substances.

Chapter 15

Section 15.5 AN OVERVIEW OF CELLULAR RESPIRATION 15.54

The main purpose of glycolysis is to provide energy for the body from carbohydrates.

15.55

The starting material for glycolysis is glucose. The major products of glycolysis are pyruvate and ATP.

15.56

Glycolysis occurs in the cellular cytoplasm.

15.57

Two steps in glycolysis require ATP and two steps produce ATP. The glycolysis of one molecule of glucose results in the net production of 2 molecules of ATP.

15.58

The oxidizing agent in glycolysis is

15.59

Feedback inhibition is a process in which the product of a reaction (or series of reactions) inhibits the reaction or an earlier step in a series of reactions.

15.60

A high concentration of glucose-6-phosphate inhibits glycolysis because hexokinase, the enzyme that catalyzes the conversion of glucose to glucose 6-phosphate, is inhibited by glucose 6-phosphate. This feedback inhibition decreases the rate of glycolysis when a high concentration of glucose 6-phosphate is present.

15.61

Fructose enters the glycolysis pathway as dihydroxyacetone phosphate and glyceraldehydes-3-phosphate. Galactose is metabolized to glucose and enters the glycolysis pathway in the form of glucose-6-phosphate.

15.62

Pyruvate reacts with coenzyme A and NAD+ inside the mitochondria to produce Acetyl-CoA, NADH and CO2 during stage 2 of cellular respiration. Equation 15.5 shows that the pyruvate dehydrogenase complex is the enzyme responsible for catalyzing this reaction.

15.63

The primary function of the citric acid cycle in ATP production is to meet the cellular needs for ATP by generating the reduced coenzymes NADH and FADH2. The other vital role served by the citric acid cycle is the production of intermediates for biosynthesis.

15.64

a. b c.

15.65

The citric acid cycle is dependent upon a supply of NAD+ and FAD, which in turn, are dependent upon a supply of oxygen.

15.66

NAD +

the fuel needed by the cycle the form in which carbon atoms leave the cycle the form in which hydrogen atoms and electrons leave the cycle

acetyl CoA CO2 NADH, FADH2

The enzymes at the three control points of the citric acid cycle are citrate synthetase (step 1), isocitrate dehydrogenase (step 3), and the -ketoglutarate dehydrogenase complex (step 4).

Chapter 15 b.

The citric acid cycle produces both NADH and ATP. If the cell has high levels of NADH and ATP, then in all likelihood, the energy needs of the cell are or will be adequately met. Therefore, when the NADH levels are high, NADH serves as an inhibitor to slow the production of NADH and FADH2. When ATP levels are high, the entry of acetyl CoA into the citric acid cycle is reduced. Energy production is lowered because the requirements of the cell are already being met. If the cell has low levels of NADH and ATP, the cell’s energy requirements are probably not being met. If levels of ATP are low, levels of ADP and usually high. High ADP levels activate the cycle and stimulate the production of more ATP. Similarly, if the levels of NADH are low, this compound is not available to inhibit the control point enzymes and the cycle is stimulated to produce more NADH and ATP.

Two of the control point enzymes are activated by ADP. This beneﬁts the cell because a high level of ADP indicates a low level of ATP, so the cell can respond immediately to low ATP levels. If the cycle were simply stimulated by a feedback mechanism involving only high or low levels of ATP, the cellular response to low levels of ATP (and low energy) would be slower. Since ADP stimulates the cycle, the cell can respond immediately to its presence rather than waiting for the level of ATP to fall and its activity as an inhibitor to be reversed. A second beneﬁt occurs because ADP is converted directly into ATP, the substance which stimulates the pathway is the direct substrate of that pathway, and conversion between the two compounds is, therefore, quick.

15.67

The primary function of the electron transport chain is to reduce molecular oxygen to water.

15.68

a. within the cytoplasm of the cell b. within mitochondria c. within mitochondria d. within mitochondria

15.69

NADH and FADH2 bring electrons to the electron transport chain.

15.70

The cytochromes in the electron transport chain function to pass electrons along the chain to oxygen atoms and to provide H+ ions to the oxygen, resulting in the formation of water.

15.71

Oxygen acts as the electron acceptor that combines with two H+ ions to form water. This reaction is often summarized as: 2H+ + 2e- + ½O2 → H2O Note: The reaction cannot really involve ½ a molecule of O2, but the reaction is balanced in this way to make it easier to relate the individual substances involved in the reaction to a single molecule of glucose or a single unit of acetyl CoA. The overall reaction is given as: 4H+ + 4e- + O2 → 2 H2O

Chapter 15 15.72

Iron deficiency might impact the electron transport chain, thus reducing the utilization of oxygen and the production of energy.

15.73

According to the chemiosmosis, ATP synthesis results from the ﬂow of H+ particles, across the inner mitochondrial membrane through the ATP synthase protein.

15.74

Glycolysis which converts 1 glucose → 2 pyruvate produces 2 NADH and 2 ATP. Oxidation of the 2 pyruvates to 2 acetyl CoA produces 2 more NADH. When the 2 acetyl CoA are reacted in the citric acid cycle they produce 2 FADH2, 6 NADH, and 2 GTP. The combined 10 NADH and 2 FADH2 produced up until this point are used in oxidative phosphorylation to produce a total of 28 more ATP. Thus 2 ATP (from glycolysis) + 2 GTP (from citric acid cycle) + 28 ATP (from oxidative phosphorylation) = 32 ATP equivalents.

Section 15.6 GLYCOGEN METABOLISM AND GLUCONEOGENESIS 15.75

Glycogenesis is the process by which glucose is converted to glycogen. Glycogenolysis is the process by which glycogen is broken back down into glucose. These are reverse processes.

15.76

Fructose and galactose from digested food enter the glycolysis pathway. The entry point for fructose is dihydroxyacetone phosphate or glyceraldehydes-3-phosphate. The entry point for galactose is glucose which enters in the form of glucose-6-phosphate.

15.77

The high-energy compound involved in the conversion of glucose to glycogen is uridine triphosphate (UTP).

15.78

It is important that the liver store glycogen because one of the functions of the liver is to maintain a relatively constant level of blood glucose. When the blood sugar levels are low, the liver can break down glycogen to release glucose. When the blood sugar levels are high, the liver can synthesize glycogen.

15.79

When glycogen enters the glycolysis pathway a number of different processes are required. The α(1→4) linkages between glucose units are broken first, releasing glucose-1-phosphate. Next the chain branches are broken by cleavage of α(1→6) linkages, and the unbranched chain can then be hydrolyzed. During debranching glucose-1-phosphate is converted to glucose-6-phosphate, which enters the glycolysis pathway.

15.80

The principal site for gluconeogenesis is the liver.

15.81

After fasting for three days, the glycogen stores are depleted and the carbon skeletons of lactate, glycerol (derived from the hydrolysis of fats), and certain amino acids are used to synthesize pyruvate, which is then converted to glucose. This process is called gluconeogenesis. 16

Chapter 15 15.82

Lactate, glycerol, and certain amino acids undergo gluconeogenesis. Amino acids are derived from proteins, glycerol is derived from lipids, and lactate has a variety of sources, including pyruvate, which is produced from carbohydrates.

15.83

In the absence of adequate carbohydrates, gluconeogenesis would be elevated by synthesizing glucose from lactate, certain amino acids, and glycerol.

15.84

The Cori cycle is the metabolic pathway in which lactate produced in the muscles is ultimately converted to glucose in the liver. As lactate levels increase in the muscle tissue, the lactate diﬀuses out of the tissue into the blood, where it is transported to the liver. An enzyme converts the lactate back into pyruvate, which is then converted to glucose by gluconeogenesis, and the glucose re-enters the blood. The glucose goes to the muscle tissue via the blood. Muscle activity quickly uses up stored ATP and more energy must be produced by the breakdown of glycogen. Under normal aerobic conditions glycogenolysis produces glucose, which is further broken down into pyruvate that feeds into the citric acid cycle as acetyl CoA. The lactate pathway does not operate until oxygen supplies are limited, such as during intense muscular activity, and the anaerobic pathway that produces lactate is activated. The Cori cycle provides a mechanism for energy to be generated under anaerobic conditions and returns an energy source (glucose) to the muscles.

15.85

The β-cells of the pancreas release insulin. The α-cells of the pancreas release glucagon.

15.86

Glucagon increases the blood sugar levels by enhancing the breakdown of glycogen in the liver, while insulin reduces the blood sugar level by enhancing the absorption of glucose from the blood into cells of active tissue. Glucagon inhibits glycogen formation, while insulin increases the rate of glycogen formation.

Section 15.7 LIPID METABOLISM AND BIOSYNTHESIS 15.87

Epinephrine stimulates fat mobilization.

15.88

Fat mobilization is the process whereby epinephrine stimulates the hydrolysis of triglycerides (in adipose tissue) into fatty acids and glycerol, which enter the bloodstream.

15.89

Resting muscles and liver cells use fatty acids rather than glucose to satisfy their energy needs.

15.90

Glycerol enters the glycolysis pathway as dihydroxyacetone phosphate.

15.91

cytoplasm 17

Chapter 15 15.92

The two fates of glycerol after it has been converted to an intermediate of glycolysis are to provide energy to cells or be converted to glucose.

15.93

Fatty acid catabolism occurs in the mitochondria.

15.94

A fatty acid is prepared for catabolism by a conversion into fatty acyl CoA. This activation occurs outside the mitochondria in the cytoplasm of the cell.

15.95

15.96

Ketone bodies are the compounds synthesized within the liver from excess acetyl CoA. They are acetoacetate, β-hydroxylbutyrate, and acetone.

15.97

a. Excessive ketone bodies may form during starvation because fatty acids from stored fats become the body’s primary energy source. With a minimum amount of cellular glucose available, the level of glycolysis decreases and a reduced amount of oxaloacetate is synthesized. Also, oxaloacetate is used for gluconeogenesis to a greater-than-normal extent as the cells react to make their own glucose. The lack of oxaloacetate reduces the activity of the citric acid cycle. Consequently, more acetyl CoA is produced by fatty acid oxidation than can be processed through the citric acid cycle. As the concentration of acetyl CoA builds up, the excess is converted within the liver to ketone bodies. b. Excessive ketone bodies may form in patients with diabetes mellitus because of an imbalance in carbohydrate and lipid metabolism. Blood glucose reaches hyperglycemic levels, but a deficiency of insulin prevents the glucose from entering tissue cells in sufficient amounts to meet cellular energy needs. The resulting increase in fatty acid metabolism leads to excessive production of acetyl CoA and a substantial increase in the level of ketone bodies in the blood.

15.98

Ketone bodies are formed in the liver. The brain, heart, and skeletal muscles can use these compounds to meet energy needs.

15.99

With a low-carbohydrate diet, there is increased fatty acid oxidation. There is a decreased synthesis of oxaleacetate, slowing down the citric acid cycle. Acetyl CoA, the fuel for the citric acid cycle, then builds up in concentration.

15.100 In ketoacidosis, the kidneys excrete excessive amounts of water in response to the low blood pH.

Chapter 15 15.101 β-hydroxybutyrate 15.102 Acid-base balance can be restored by using bicarbonate to neutralize the acids. 15.103 a.

ketonemia—an elevated level of ketone bodies in the blood.

ketonuria – the presence of ketone bodies in the urine

acetone breath – a condition in which acetone can be detected in the breath

ketosis – a condition in which ketonemia, ketonuria, and acetone breath exist together

15.104 Acetyl CoA supplies the carbon atoms for fatty acid synthesis. 15.105 The energy for fatty acid synthesis is provided by ATP and NADPH. 15.106 There is a transport system which helps to move citrate from the mitochondria to the cytoplasm. In other words, there are molecules which assist the passage of citrate through the membrane. 15.107 The liver can convert glucose to fatty acids via acetyl CoA; however, it does not have the enzyme that can catalyze the conversion of acetyl CoA to pyruvate which is needed for gluconeogenesis. 15.108 Two polyunsaturated fatty acids, linoleic, and linolenic, cannot be synthesized by the body and yet are required for good health.

Section 15.8 AMINO ACID METABOLISM AND BIOSYNTHESIS 15.109 Amino acids serve the vital functions of providing building blocks for the synthesis of proteins in the body, being used to synthesize other nitrogen-containing biomolecules, and providing energy. 15.110 The sources of amino acids in the pool are digested dietary proteins, turnover of body proteins, and synthesis in the liver. 15.111 Nitrogen atoms end up in molecules of urea. 15.112 Other than proteins synthesis, amino acids are used for the production of purine and pyrimidine bases of nucleic acids, heme structures for hemoglobin and myoglobin, choline and ethanolamine for phospholipids, and neurotransmitters such as acetylcholine and dopamine. 15.113 a.

transamination

An amino group is removed from an amino acid and donated to an α-keto acid

deamination

Ammonium ion is produced 19

Chapter 15 c.

transamination

New amino acids are synthesized from other amino acids

A keto acid is produced from an amino acid

deamination and transamination

15.114 Urea is synthesized in the liver. 15.115 The source of the carbon atom in urea is a bicarbonate ion. The sources of the two nitrogen atoms in urea are an ammonium ion and aspartate.

15.116

15.117 Glucogenic amino acids are amino acids whose carbon skeletons can be converted metabolically to an intermediate used in the synthesis of glucose. Ketogenic amino acids are amino acids whose carbon skeletons can be converted metabolically to acetyl CoA or acetoacetyl CoA. 15.118 The amino acids isoleucine, phenylalanine, tryptophan, and tyrosine can be both glucogenic and ketogenic. 15.119 The five locations at which the carbon skeletons of amino acids enter the citric acid cycle are: (1) oxaloacetate, (2) citrate, (3) α-ketoglutarate, (4) succinyl CoA, and (5) fumarate. 15.120 The lack of phenylalanine hydroxylase causes PKU. 15.121 a.

aspartate → oxaloacetate

glucogenic

leucine → acetyl CoA

ketogenic

tyrosine → acetoacetyl CoA

ketogenic

15.122 During fasting or starvation, glucogenic amino acids are used to synthesize glucose through the process of gluconeogenesis after 12-18 hours without food. 15.123 Essential amino acids are amino acids which must be obtained from the diet because unlike nonessential amino acids, they cannot be synthesized within the liver in the amounts needed by the body. 15.124 The essential amino acid that makes tyrosine nonessential is phenylalanine. 15.125 glycolysis intermediates, citric acid cycle intermediates, and phenylalanine 20

Chapter 15 15.126 The amino acids synthesized from glutamate are proline, glutamine, and arginine.