mole key by harry brielmannjr

key

Unit 9: The Mole Table of Contents

3 Neutralization Challenge 4 Perfume Lab 5-9 Mole Lecture 10-18 Mole Worksheets 19-10 How to Ace the mole unit

Ei-Ichi Negishi shared the 2010 Nobel Prize or his discovery of organozinc catalysis reactions. By discovering a reaction similar to the Grignard Reaction, Negishi found that zinc could be inserted catalytically into a carbon-halogen bond, which can then rapidly displace alkyl halides: R-Zn-X + R’-X  R-R’ + Zn + X2.

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H O

H O NaHCO3

H C C

O H

H baking soda _____ g _____ mol

acetic acid _____ g _____ mol

H C C

H O H

H ethyl acetate

water

CO2 carbon dioxide

_____ g

_____ mol

This stamp indicates you performed this reaction successfully

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Example: The combustion of 3 moles of butane (C4H10) with excess oxygen will produce ___ grams of CO2. 3 moles C4H10 x

44 grams CO2

8 moles CO2

x 1 moles CO2

2 moles C4H10

= 528 grams CO2

For you: The combustion of 4 moles of butane (C4H10) will require____ grams of O2. 13 moles O2 4 moles C4H10 x

x 2 moles C4H10

32 grams O2 1 moles O2

= 832 grams O2

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for you: The combustion of 453 grams of butane (C4H10) with excess oxygen will produce ___ grams of H2O. 453 g C4H10 x

1 mole C4H10 x 58 g C4H10

2 moles C4H10 x 10 moles H2O

18 grams H2O

= 703 grams H2O

1 mole H2O

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What is the % composition by mass of water? KEY: assume one mole to make it easy.

H-O-H 1 g/mol

16 g/mol

1 g/mol

Water is 16/18 oxygen by mass: 89% O, 11% H

key C

most answers are rounded to the nearest whole number for simplicity

6.02 x 1023

27 54 1.2 x 1024

22.7

227 6.02 x 1022 (H2)

12 x 6 = 72; 12 x 1 = 12; 6 x 16 = 96; 72 + 12 + 96 = 180 g/mol)

NO2 H

H O2N

NO2 H

C6H15N = 72 + 15 + 14 â&#x20AC;&#x201C; 101 g/mol

key T 

Show the amounts involved in grams when each of these reactions is performed as written; the first is done for you. For simplicity, molar masses are rounded to the nearest whole number.

1. Mn + O2  MnO2 55 g 32 g 87g  137 38 175  24 79 103 103 137 103 

Combine 12 g carbon with 4 g hydrogen.

 7

210

77  264

 74

102

 44

 110

106

116

100

A ď&#x192;

You try some: 1. How many moles of O2 are required to produce 240 moles of H2O? 13 moles O2 240 moles H2O x ___________ = 312 moles H2O 10 moles H2O

Note that this can be done mostly in your head: â&#x20AC;&#x153;13 moles of oxygen are needed for every mole of water, so I need 1.3 x 240 moles of oxygen; and then punch it into your calculator.

2. How many moles of butane (C4H10) are needed to produce 25 moles of water? moles C4H10 = 5 moles H O. 25 moles H2O x 2___________ 2 10 moles H2O

Example: How many grams of O2 are required to produce 9 moles of CO2? 9 moles CO2 x 13 moles O2 x 32 grams O2 = 468 grams O2 8 moles CO2 1 mole O2 You try one: How many grams of CO2 will be produced from 17 moles of O2? 8 moles CO2 44 g CO2 17 moles O2 x ___________ x __________ = 460 g CO2 13 moles O2 1 mole CO2 How many moles of C4H10 are required to produce 100 grams of CO2? moles CO2 1 mole C4H10 = 100 g CO2 x ___________ x __________ 0.57 mol C4H10. 4 moles CO2

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Type 3: Grams to Grams # of Steps: 3

These are three- step problems. We have to convert from the grams given to moles, then from the moles of that substance to the equivalent number of moles of the desired substance, and then finally we convert from the moles of that substance to the equivalent number of grams.

To put it another way, we go from grams to moles, then moles to moles, then moles to grams. Example: For the combustion of hydrogen, ho w many grams of H2O are produced from 40 grams of O2, assuming excess hydrogen? 40 g O2

x ______________

Start with this- it is what you are given

You try one: How many grams of butane (C4H10) are required to produce 25 grams of CO2?

25 g CO2 x ______________

x ______________

= 8.2 g C4H10.

Note that the mole is an amount, so we can determine the number of molecules, or even atoms or electrons for any of these problems. Challenge question: How many Carbon-14 atoms are in your body? Carbon-14 has a natural abundance of about one in a trillion (one trillion = 1012) carbon atoms. A human body is on average 18 percent carbon by mass. Be sure to show your work with cancelled units

453.6 g body x __________ 0.18 g C 150 pound body x ___________ 1 pound body 1 g body My weight

Body Pounds to body grams

Body g to g C (18%)

6 x 1023 1 x 10-12 atoms C-14 g C-14 mol C-14 x __________ x 1____________ x __________ 1gC g C to g C-14 (one in a trillion)

14 g C-14 Grams to moles

mol C-14

Moles to atoms

= 5.2 x 1014 C-14 atoms in my body

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10 grams H 2 x

mole H 2 2 moles H 2 O x  5 moles H 2 O 2 grams H 2 2 moles H 2



2 x ______________

= 84.2 mol NaCl

x ______________

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V2O5 x ______________ V2O5

V2O5 31.4 g V2O5x ______________ V2O5

x ______________ V2O5

47 g V2O5

= 0.52 mol V

x ______________

= 13.8 g O2

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

P4 + 5 O2  P4O10 25 g P4

50 g O2

x ______________ 124 g P4 O2 x ______________ 32 g O2

x ______________

x ______________ O2

x ______________

= 57.3 g

= 88 g

Answers on next page

Solution: First we write the balanced chemical equation: 4 V + 5 O2 ď&#x192; 2 V2O5

200 g V

x ______________ 51 g V

O2 100 g O2 x ______________ 32 g O2

x ______________

= 357 g

x ______________ O2

x ______________

= 227 g

Finally, we see how much vanadium is needed to react with 100 g of oxygen- the rest of the vanadium is excess. O2 100 g O2 x ______________ 32 g O2

x ______________ O2

x ______________

= 127.5 g V;

72.5 g V is in excess

CuCl2 + 2 NaNO3  Cu(NO3)2 + 2 NaCl molar masses: 134 g

85 g

188g

58 g

15 g CuCl2

CuCl2 x ______________ CuCl2

x ______________ CuCl2

x ______________

= 13.0 g NaCl

20 g NaNO3

NaNO x ______________ NaNO

x ______________ NaNO

x ______________

= 13.6 g NaCl

CuCl2

15 g CuCl2

CuCl2 x ______________ CuCl2

______________

x ______________ CuCl2

x 100

= 87%

x ______________

= 19 g

12/44, or 27% C The rest, or 73% O

(solution: assume a mole, and see what 23/58 x 100 = 40% Na fraction (and therefore what percent) comes from each atom. In this case a 35/58 x 100 = 60% Na mole has a mass of 58 g (23 for sodium and 35 for chlorine, rounding to whole numbers), of which 23 come from Na, and 35 from Cl)

Well, that 88 gram sample is 73 percent oxygen; thatâ&#x20AC;&#x2122;s 64 grams (.73 x 88).

Molar mass = 144 + 22 + 176 = 342 g/mole, so

C: 144/342 or 42% H: 22/342 or 6% O: the rest, or 52%

OK, an aluminum sulfate molecule has 2 aluminum ions, and 3 sulfate ionsâ&#x20AC;Śthat means Al2S3O12 for a molar mass of 54 + 96 + 192 = 342 g/mol. Al: 42% (144/342)

S: 28% (96/342) O: 30% (the rest) 56 + 64 = 120 g/mol, 56/120 iron or 47% 112 + 48 = 160 g/mol, 112/160 iron or 70% 56 + 12 + 48 = 116, 56/116 iron or 49%

96 + 10 + 56 + 32 = 194 g/mol

C: 49% (96/194, rounding to the nearest whole number) H: 5% (10/194) N: 29% (56/194) O: 17% (the rest, or 32/194â&#x20AC;Śnote that the rounding makes things a bit imperfect)

6 x 1023

2 H2 + O2  2 H2O

1.2 x 1024 82

1.2 x 1024

6 x 1023

72 36

mol H2 2 g H2 mol O2 32 g O2 O2

mol H2O mol H2 2 mol H2O mol O2

18 g H2O 9 mol H2O 18 g H2O 1.125 mol H2O 1.125

0.1

1.125

The mole: supplementary material

Homework Assignment Due Thursday March 10 at the beginning of class Bring to class the full text of an article that best answers the question:

how do we detect odors? This being a chemistry class, be aware that you are looking for a molecular answer to this question. The best articles would show a molecular interaction â&#x20AC;&#x201C; presumably a lock and key sort of relationship- between the molecule being smelled, and whatever part of us (is it in the nose?) that smells it. This article may continue to explain how this signal (or is it a collection of signals?) is processed (presumably in the brain). This assignment is worth up to five points, depending how thoroughly it answers the question.