key

Unit 9: The Mole

4 Perfume Lab

3 Neutralization Challenge

10-18 Mole Worksheets

5-9 Mole Lecture

19-10 How to Ace the mole unit

Table of Contents

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.

This stamp indicates you performed this reaction successfully

I

H

H

H

O H

O H

H

H

O

O

Na H OH

C C+ C C +

baking soda acetic acidethyl acetate water

_____ g _____ g _____ g _____ g

NaHCO3 + CO2

_____ mol _____ mol _____ mol _____ mol

carbondioxide

_____ g

_____ mol

E

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

3 moles C4H10 x 2 moles C4H10

8 moles CO2 x

1 moles CO2

44 grams CO2 = 528 grams

CO2

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

4 moles C4H10 x 2 moles C4H10

13 moles O2 x

1 moles O2

32 grams O2 = 832 grams O2

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

453 g C4H10 x

58 g C4H10

1 mole C4H10 x 10 moles H2O 1 mole H2O

18 grams H2O 2 moles C4H10 x = 703 grams H2O

H-O-H

What is the % composition by mass of water?

KEY: assume one mole to make it easy.

1 g/mol

16 g/mol

1 g/mol

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

C

NO2

NO2

O2N

H H

HH

H

N

27 6.02 x 1023

54

most answers are rounded to the nearest whole number for simplicity

1.2 x 1024

227 22.7

6.02 x 1022

H (H2)

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

2

C6H15N = 72 + 15 + 14 – 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.

1. Mn + O2 MnO2 55 g 32 g 87g

key

137

54

79 103

137

7 77 70

24

175 38

116 106 110

44 36 64

264 210

Combine 12 g carbon with 4 g hydrogen.

100

18 102

46 74

For simplicity, molar masses are rounded to the nearest whole number.

103 103

4 32 36

A

You try some: 1. How many moles of O2 are required to produce 240 moles of H2O? 2. How many moles of butane (C4H10) are needed to produce 25 moles of water?

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?

How many moles of C4H10 are required to produce 100 grams of CO2?

240 moles H2O x ___________ =

Note that this can be done mostly in your head: “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. 10 moles H2O

13 moles O2

25 moles H2O x ___________ = moles H2O. 10 moles H2O

2 moles C4H10 5

17 moles O2 x ___________ x __________ = 13 moles O2

8 moles CO2 460 g CO2 44 g CO2

1 mole CO2

100 g CO2 x ___________ x __________ = 0.57 mol C4H10.

moles CO2 1 mole C4H10 4 moles CO2

312 moles H2O

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? You try one:

How many grams of butane (C4H10) are required to produce 25 grams of CO2? 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

x ___________ x __________ x __________ = ___ mol C4H10.

moles CO2 0.57

1 mole C4H10 4 moles CO2

1 g CO2 x ___________ x __________ x __________ x ____________ x __________ 150 pound body

1 pound body 1 g body

1 x 10-12

g C-14 0.18 g C 453.6 g body

1 g C 14 g C-14 mol C-14

1 mol C-14

6 x 1023 atoms C-14

My weight

Body Pounds to body grams

Body g to g C (18%)

g C to g C-14

(one in a trillion)

Grams to moles Moles to

atoms

= 5.2 x 1014 C-14 atoms in my body

x ______________ 25 g CO2 = 8.2 g C4H10. x ______________ x ______________

Start with this- it is what you are given

x ______________ 40 g O2 = x ______________ x ______________

OH moles 5 H moles 2

OH moles 2 x

H grams 2

H mole x H grams 10 2

2

2

2

22

E

x ___________ x __________ x __________ = ___ mol C4H10.

moles CO2 0.57

1 mole C4H10 4 moles CO2

1 g CO2

2

x ______________ = 84.2 mol NaCl

x ______________

x ______________ V2O5 47 g V2O5 = 0.52 mol V x ______________ V2O5

V2O5

V2O5

x ______________ V2O5 31.4 g V2O5 = 13.8 g O2 x ______________ x ______________ V2O5

V2O5

I

x ______________ 25 g P4 = 57.3 g x ______________ x ______________ 124 g P4

P4 + 5 O2 P4O10

x ______________ O2

50 g O2 = 88 g x ______________ x ______________

32 g O2

O2

Answers on next page

4 V + 5 O2 2 V2O5

Solution: First we write the balanced chemical equation:

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

x ______________ 200 g V = 357 g x ______________ x ______________ 51 g V

x ______________ O2

100 g O2 = 227 g x ______________ x ______________

32 g O2

O2

x ______________ O2

100 g O2 = 127.5 g V; x ______________ x ______________

32 g O2

O2

72.5 g V is in excess

A

x ______________ CuCl2 15 g CuCl2 = 13.0 g NaCl x ______________ x ______________ CuCl2

CuCl2

CuCl2 + 2 NaNO3 Cu(NO3)2 + 2 NaCl

134 g molar masses: 85 g 188g 58 g

x ______________ NaNO 20 g NaNO3 = 13.6 g NaCl x ______________ x ______________

NaNO

NaNO

CuCl2

x ______________ CuCl2 15 g CuCl2 = 19 g

x ______________ x ______________ CuCl2

CuCl2

______________

x 100 = 87%

23/58 x 100 = 40% Na

35/58 x 100 = 60% Na 23

35

(solution: assume a mole, and see what fraction (and therefore what percent) comes from each atom. In this case a 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) 12/44, or 27% C

The rest, or 73% O

Well, that 88 gram sample is 73 percent oxygen; that’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 ions…that 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 32/194…note that the rounding makes things a bit imperfect)

I

44 6 x 1023

18

2 H2 + O2 2 H2O

4 32

1.2 x 1024

82 18

1

72

36

mol H2

2 g H2

mol H2O

mol H2 mol H2O 18 g H2O

mol O2

1.125

0.1 1.125 9

1.2 x 1024 6 x 1023

9

32 g O2

2 mol H2O mol H2O 18 g H2O

mol O2

1.125

O2

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 – 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.