CHAPTER 7
Chemical Calculations 7.1-7.7
1
The Mole


A number of atoms, ions, or molecules
that is large enough to see and handle.
A mole = number of things



Avogadro’s number = 6.022 x 1023

2
Just like a dozen = 12 things
One mole = 6.022 x 1023 things
Symbol for Avogadro’s number is NA.
The Mole
3
The Mole

How do we know when we have a mole?




Molar mass - mass in grams numerically equal
to the atomic weight of the element in grams.
H has an atomic weight of 1.00794 g


1.00794 g of H atoms = 6.022 x 1023 H atoms
Mg has an atomic weight of 24.3050 g

4
count it out
weigh it out
24.3050 g of Mg atoms = 6.022 x 1023 Mg atoms
The Mole
5
The Mole

6
Example 2-1: Calculate the mass of a
single Mg atom in grams to 3 significant
figures.
? g Mg 
The Mole

7
Example 2-1: Calculate the mass of a
single Mg atom in grams to 3 significant
figures.
? g Mg  1 Mg atom
The Mole

Example 2-1: Calculate the mass of a
single Mg atom in grams to 3 significant
figures.


1 mol Mg atoms
 
? g Mg  1 Mg atom
23
 6.022 10 Mg atoms 
8
The Mole

Example 2-1: Calculate the mass of a
single Mg atom, in grams, to 3 significant
figures.

1 mol Mg atoms
? g Mg  1 Mg atom 
23
6.022

10
Mg atoms

 24.30gMg 

  4.04  10  23 g Mg
 1 mol Mg atoms 
9

 

The Mole

Example 2-2: Calculate the number of atoms in
one-millionth of a gram of Mg to 3 significant
figures.
? Mg atoms 
10
The Mole

11
Example 2-2: Calculate the number of
atoms in one-millionth of a gram of Mg to
3 significant figures.
 1 mol Mg 
6

? Mg atoms  1.00  10 g Mg
 24.30 g Mg 
The Mole

Example 2-2: Calculate the number of
atoms in one-millionth of a gram of Mg to
3 significant figures.
? Mg atoms  1.00  10
6
 6.022  1023 Mg atoms

1 mol Mg atoms

12



 1 mol Mg 

g Mg 
 24.30 g Mg 
The Mole

Example 2-2: Calculate the number of
atoms in one-millionth of a gram of Mg to
3 significant figures.
 1 mol Mg 

? Mg atoms  1.00  10 g Mg
 24.30 g Mg 
6
 6.022  1023 Mg atoms 

  2.48  1016 Mg atoms
 1 mol Mg atoms

13
The Mole

Example 2-3. How many atoms are
contained in 1.67 moles of Mg?
? Mg atoms 
14
The Mole

Example 2-3. How many atoms are
contained in 1.67 moles of Mg?
? Mg atoms  1.67 mol Mg
15
The Mole

Example 2-3. How many atoms are
contained in 1.67 moles of Mg?
 6.022 10 23 Mg atoms 

? Mg atoms  1.67 mol Mg 
1 mol Mg


16
The Mole

Example 2-3. How many atoms are
contained in 1.67 moles of Mg?
 6.022 10 23 Mg atoms 

? Mg atoms  1.67 mol Mg 
1 mol Mg


 1.00 10 24 Mg atoms
17
The Mole

Example 2-3. How many atoms are
contained in 1.67 moles of Mg?
 6.022  1023 Mg atoms 

? Mg atoms  1.67 mol Mg
1 mol Mg


 1.00  1024 Mg atoms
18
The Mole

19
Example 2-4: How many moles of Mg atoms are
present in 73.4 g of Mg?
You do it!
The Mole

Example 2-4: How many moles of Mg atoms are
present in 73.4 g of Mg?
? mol Mg  73.4 g Mg
20
The Mole

Example 2-4: How many moles of Mg atoms are
present in 73.4 g of Mg?
 1 mol Mg atoms 

? mol Mg  73.4 g Mg 
 24.30 g Mg 
21
The Mole

Example 2-4: How many moles of Mg atoms are
present in 73.4 g of Mg?
 1 mol Mg atoms 

? mol Mg  73.4 g Mg 
 24.30 g Mg 
 3.02 mol Mg
IT IS IMPERATIVE THAT YOU KNOW
HOW TO DO THESE PROBLEMS
22
Formula Weights, Molecular
Weights, and Moles

How do we calculate the molar mass of a
compound?


add atomic weights of each atom
The molar mass of propane, C3H8, is:
3  C  3  12.01 amu  36.03 amu
8  H  8  1.01 amu  8.08 amu
Molar mass
23
 44.11 amu
Formula Weights, Molecular
Weights, and Moles

24
The molar mass of calcium nitrate,
Ca(NO3)2 , is:
You do it!
Formula Weights, Molecular
Weights, and Moles
1 Ca  1 40.08 amu  40.08 amu
2  N  2 14.01 amu  28.02 amu
6  O  6 16.00 amu  96.00 amu
Molar mass
25
 164.10 amu
Formula Weights, Molecular
Weights, and Moles

26
One Mole of

Cl2 or 70.90g

C3H8 You do it!
Contains
6.022 x 1023 Cl2 molecules
2(6.022 x 1023 ) Cl atoms
Formula Weights, Molecular
Weights, and Moles

27
One Mole of

Cl2 or 70.90g

C3H8 or 44.11 g
Contains
6.022 x 1023 Cl2 molecules
2(6.022 x 1023 ) Cl atoms
6.022 x 1023 C3H8 molecules
3 (6.022 x 1023 ) C atoms
8 (6.022 x 1023 ) H atoms
Formula Weights, Molecular
Weights, and Moles

Example 2-5: Calculate the number of
C3H8 molecules in 74.6 g of propane.
? C3H8 molecules  74.6 g C3H8 
28
Formula Weights, Molecular
Weights, and Moles

Example 2-5: Calculate the number of
C3H8 molecules in 74.6 g of propane.
? C3 H 8 molecules  74.6 g C3 H 8 
 1 mole C3 H 8 


 44.11 g C3 H 8 
29
Formula Weights, Molecular
Weights, and Moles

Example 2-5: Calculate the number of
C3H8 molecules in 74.6 g of propane.
? C3H8 molecules  74.6 g C3H8 
 1 mole C3H8  6.022 1023 C3H8 molecules


44.11 g C3H8
 44.11 g C3H8 
30

 

Formula Weights, Molecular
Weights, and Moles

Example 2-5: Calculate the number of
C3H8 molecules in 74.6 g of propane.
? C3 H 8 molecules  74.6 g C3 H 8 
 1 mole C3 H 8  6.022 10 23 C3 H 8 molecules


44.11 g C3 H 8
 44.11 g C3H 8 
24
1.02  10 molecules
31

 

Formula Weights, Molecular
Weights, and Moles

32
Example 2-8. Calculate the number of O
atoms in 26.5 g of Li2CO3.
You do it!
Formula Weights, Molecular
Weights, and Moles

Example 2-8. Calculate the number of O
atoms in 26.5 g of Li2CO3.
1 mol Li 2 CO3
? O atoms  26.5 g Li 2 CO3 

73.8 g Li 2 CO3
6.022 10 23 form. units Li 2 CO3
3 O atoms


1 mol Li 2 CO3
1 formula unit Li 2 CO3
6.49 10 23 O atoms
33
Percent Composition


34
% composition = mass of an individual
element in a compound divided by the
total mass of the compound x 100%
Determine the percent composition of C
in C3H8.
mass C
%C
100%
mass C3 H 8
3 12.01 g

100%
44.11 g
 81.68%
Percent Composition and
Formulas of Compounds

35
What is the percent composition of H in C3H8?
You do it!
Percent Composition

What is the percent composition of H in C3H8?
mass H
%H
 100%
mass C3 H 8
8 H

 100%
C3H 8
8  1.01 g

 100%  18.32%
44.11 g
or
36
18.32%  100%  81.68%
Percent Composition

Example 2-10: Calculate the percent
composition of Fe2(SO4)3 to 3 significant
figures.
You do it!
37
Percent Composition

Example 2-10: Calculate the percent
composition of Fe2(SO4)3 to 3 sig. fig.
2  Fe
2  55.8 g
% Fe 
100% 
100%  27.9% Fe
Fe 2 (SO 4 ) 3
399.9 g
3 S
3  32.1 g
%S 
100% 
100%  24.1% S
Fe 2 (SO 4 ) 3
399.9 g
12  O
12 16.0 g
% O 
100% 
100%  48.0% O
Fe 2 (SO 4 ) 3
399.9 g
Total
38
 100%
Percent Composition
39
Empirical and Molecular Formulas



40
Empirical Formula - smallest whole-number ratio
of atoms present in a compound
Molecular Formula - actual numbers of atoms of
each element present in a molecule of the
compound
We determine the empirical and molecular
formulas of a compound from the percent
composition of the compound.
Empirical And Molecular
Formulas
41
Empirical Formulas



Example 2-11: A compound contains 24.74% K,
34.76% Mn, and 40.50% O by mass. What is its
empirical formula?
Make the simplifying assumption that we have
100.0 g of compound.
In 100.0 g of compound there are:



42
24.74 g of K
34.76 g of Mn
40.50 g of O
Empirical Formulas
1 mol K
? mol K  24.74g K 
 0.6327mol K
39.10g K
43
Empirical Formulas
1 mol K
? mol K  24.74g K 
 0.6327mol K
39.10g K
1 mol Mn
? mol Mn  34.76g Mn 
 0.6327mol Mn
54.94g Mn
44
Empirical Formulas
1 mol K
? mol K  24.74g K 
 0.6327mol K
39.10g K
1 mol Mn
? mol Mn  34.76g Mn 
 0.6327mol Mn
54.94g Mn
1mol O
? mol O  40.50g O 
 2.531mol O
16.00g O
obtain smallest whole number ratio
45
Empirical Formulas
1 mol K
? mol K  24.74g K 
 0.6327mol K
39.10g K
1 mol Mn
? mol Mn  34.76g Mn 
 0.6327mol Mn
54.94g Mn
1mol O
? mol O  40.50g O 
 2.531mol O
16.00g O
obtain smallest whole number ratio
for K 
46
0.6327
1K
0.6327
for Mn 
0.6327
 1 Mn
0.6327
Empirical Formulas
? mol K
 24.74 g K 
1 mol K
 0.6327 mol K
39.10 g K
1 mol Mn
? mol Mn  34.76 g Mn 
 0.6327 mol Mn
54.94 g Mn
1mol O
? mol O  40.50 g O 
 2.531 mol O
16.00 g O
obtain smallest whole number ratio
0.6327
for K 
1K
0.6327
47
0.6327
for Mn 
 1 Mn
0.6327
2.531
for O 
4O
0.6327
thus the chemical formula is KMnO 4
Empirical Formulas

48
Example 2-12: A sample of a compound
contains 6.541g of Co and 2.368g of O.
What is empirical formula for this
compound?
You do it!
Empirical Formulas

Example 2-12: A sample of a compound
contains 6.541g of Co and 2.368g of O.
What is empirical formula for this
compound?
1 mol Co
? mol Co  6.541 g Co 
 0.1110 mol Co
58.93 gCo
1mol O
? mol O  2.368 g O 
 0.1480 mol O
16.00 g O
find smallest whole number ratio
49
Empirical Formulas

Example 2-12: A sample of a compound
contains 6.541g of Co and 2.368g of O.
What is empirical formula for this
compound?
0.1110
0.1480
for Co 
 1 Co for O 
 1.333O
0.1110
0.1110
multipy both by 3 to turn fraction to whole number
1 Co  3  3 Co 1.333 O  3  4 O
Thus the compound' s formula is :
Co3O 4
50
Molecular Formulas

Example 2-13: A compound is found to
contain 85.63% C and 14.37% H by mass. In
another experiment its molar mass is found to
be 56.1 g/mol. What is its molecular formula?

short cut method
1 mol contains 56.1 g
85.63% is C and 14.37% is H
56.1 g  0.8563  48.0 g of C
56.1 g  0.1437  8.10 g of H
51
Determination of Molecular
Formulas
convert masses to moles
52
1 mol C
48.0 g of C 
 4 mol C
12.0 g C
1 mol H
8.10 g of H 
 8 mol H
1.01 g H
Thus the formula is :
C4 H8
Calculations Based on Chemical
Equations
53
Calculations Based on Chemical
Equations

Fe2O3 + 3 CO  2 Fe + 3 CO2

Example 3-1: How many CO molecules
are required to react with 25 formula
units of Fe2O3?
3 CO molecules
? CO molecules= 25 formulaunits Fe2O3 
1 Fe2O3 formulaunit
 75 moleculesof CO
54
Calculations Based on Chemical
Equations

Example 3-2: How many iron atoms can
be produced by the reaction of 2.50 x 105
formula units of iron (III) oxide with excess
carbon monoxide?
? Fe atoms= 2.5010 formulaunits Fe2O3
5
55
Calculations Based on Chemical
Equations

Example 3-2: How many iron atoms can
be produced by the reaction of 2.50 x 105
formula units of iron (III) oxide with excess
carbon monoxide?
? Fe atoms= 2.5010 formulaunits Fe2O3
5
2 Fe atoms


1 formulaunits Fe2O3
56
Calculations Based on Chemical
Equations

Example 3-2: How many iron atoms can
be produced by the reaction of 2.50 x 105
formula units of iron (III) oxide with excess
carbon monoxide?
? Fe atoms= 2.5010 formulaunits Fe2O3
5
2 Fe atoms
5

 5.0010 Fe atoms
1 formulaunits Fe2O3
57
Calculations Based on Chemical
Equations

Example 3-3: What mass of CO is
required to react with 146 g of iron (III)
oxide?
1 molFe2O3
? g CO = 146g Fe2O3 
159.7 g Fe2O3
58
Calculations Based on Chemical
Equations

Example 3-3: What mass of CO is
required to react with 146 g of iron (III)
oxide?
1 molFe2O3
3 molCO
? g CO = 146g Fe2O3 

159.7 g Fe2O3 1 molFe2O3
59
Calculations Based on Chemical
Equations

Example 3-3: What mass of CO is
required to react with 146 g of iron (III)
oxide?
1 mol Fe2O3
3 molCO
? g CO = 146g Fe2O3 

159.7 g Fe2O3 1 mol Fe2O3
28.0 g CO

 76.8 g CO
1 molCO
60
Calculations Based on Chemical
Equations

Example 3-4: What mass of carbon dioxide can
be produced by the reaction of 0.540 mole of
iron (III) oxide with carbon monoxide?
3 molCO 2
? g CO 2  0.540 molFe2O3 
1 mol Fe2O3
61
Calculations Based on Chemical
Equations

Example 3-4: What mass of carbon dioxide can
be produced by the reaction of 0.540 mole of
iron (III) oxide with carbon monoxide?
3 molCO 2
44.0 g CO 2
? g CO 2  0.540 mol Fe2O3 

1 mol Fe2O3 1 molCO 2
62
Calculations Based on Chemical
Equations

Example 3-4: What mass of carbon
dioxide can be produced by the reaction of
0.540 mole of iron (III) oxide with carbon
monoxide?
3 mol CO 2
44.0 g CO 2
? g CO 2  0.540 mol Fe 2 O 3 

1 mol Fe 2 O 3 1 mol CO 2
= 71.3 g CO 2
63
Calculations Based on Chemical
Equations

64
Example 3-5: What mass of iron (III) oxide
reacted with carbon monoxide if the
carbon dioxide produced by the reaction
had a mass of 8.65 grams?
You do it!
Calculations Based on Chemical
Equations

Example 3-5: What mass of iron (III) oxide
reacted with carbon monoxide if the
carbon dioxide produced by the reaction
had a mass of 8.65 grams?
1 molCO2 1mol Fe2O3
? g Fe2O3  8.65g CO2 


44.0g CO2 3 molCO2
159.7 g Fe2O3
 10.5 g Fe2O3
1 mol Fe2O3
65
Calculations Based on Chemical
Equations

66
Example 3-6: How many pounds of carbon
monoxide would react with 125 pounds of
iron (III) oxide?
You do it!
Calculations Based on Chemical
Equations
454 g Fe2 O3
? lb CO = 125lb Fe2 O3 
1 lb Fe 2 O 3
1 mol Fe 2 O3
3 molCO



159.7 g Fe 2 O3 1 mol Fe2 O 3
28 g CO
1 lb CO

 65.7 lb CO
1 mol CO 454 g CO
67
YOU MUST BE PROFICIENT WITH THESE
TYPES OF PROBLEMS!!!
Limiting Reactant Concept

Kitchen example of limiting reactant concept.
1 packet of muffin mix + 2 eggs + 1 cup of milk
 12 muffins

68
How many muffins can we make with the
following amounts of mix, eggs, and milk?
Limiting Reactant Concept

69
Mix Packets Eggs
Milk
1
1 dozen
1 gallon
limiting reactant is the muffin mix
2
1 dozen
1 gallon
3
1 dozen
1 gallon
4
1 dozen
1 gallon
5
1 dozen
1 gallon
6
1 dozen
1 gallon
7
1 dozen
1 gallon
limiting reactant is the dozen eggs
Limiting Reactant Concept

Example 3-7: Suppose a box contains 87 bolts,
110 washers, and 99 nuts. How many sets,
each consisting of one bolt, two washers, and
one nut, can you construct from the contents of
one box?
 1 bolt   87 sets
  55sets
110 washers 1 set
2 washers
  99 sets
99 nuts1 set
1 nut
87 bolts1 set
70
the maximumnumber we can makeis 55 sets
determinedby thesmallest number
Limiting Reactant Concept

Example 3-8: What is the maximum mass of
sulfur dioxide that can be produced by the
reaction of 95.6 g of carbon disulfide with 110. g
of oxygen?
CS2  3 O2  CO2  2 SO 2
71
Limiting Reactant Concept

Example 3-8: What is the maximum mass of
sulfur dioxide that can be produced by the
reaction of 95.6 g of carbon disulfide with 110. g
of oxygen?
CS2  3 O2  CO2  2 SO 2
1 mol
72
3 mol
1 mol
2 mol
Limiting Reactant Concept

Example 3-8: What is the maximum mass of
sulfur dioxide that can be produced by the
reaction of 95.6 g of carbon disulfide with 110. g
of oxygen?
CS2  3 O 2  CO 2  2 SO 2
1 mol 3 mol
1 mol 2 mol
76.2 g 3(32.0 g) 44.0 g 2(64.1 g)
73
Limiting Reactant Concept

Example 3-8: What is the maximum mass of
sulfur dioxide that can be produced by the
reaction of 95.6 g of carbon disulfide with 110. g
of oxygen?

Determine which mass makes the most product
CS2  3 O2  CO2  2 SO 2
1 molCS2
? molSO 2  95.6 g CS2 
76.2g
74
Limiting Reactant Concept
CS2  3 O 2  CO 2  2 SO 2
1 molCS2
? molSO 2  95.6 g CS2 
76.2g
2 molSO 2
64.1 g SO 2


 161g SO 2
1 molCS2
1 molSO 2
75
Limiting Reactant Concept
CS2  3 O 2  CO 2  2 SO 2
1 molCS2 2 molSO 2 64.1 g SO 2
? molSO 2  95.6 g CS2 


 161g SO 2
76.2g
1 molCS2 1 molSO 2
1 molO 2 2 molSO 2 64.1 g SO 2
? molSO 2  110 g O 2 


 147 g SO 2
32.0g O 2 3 molO 2 1 molSO 2




76
Which is limiting reactant?
Limiting reactant is O2.
What is maximum mass of sulfur dioxide?
Maximum mass is 147 g.
Percent Yields from Reactions

Theoretical yield is calculated by assuming
that the reaction goes to completion.


Actual yield is the amount of a specified pure
product made in a given reaction.


77
Determined from the limiting reactant calculation.
In the laboratory, this is the amount of product that
is formed in your beaker, after it is purified and
dried.
Percent yield indicates how much of the
product is obtained from a reaction.
actual yield
% yield =
 100%
theoretical yield
Percent Yields from Reactions

Example 3-9: A 10.0 g sample of ethanol,
C2H5OH, was boiled with excess acetic acid,
CH3COOH, to produce 14.8 g of ethyl acetate,
CH3COOC2H5. What is the percent yield?
CH 3COOH + C2 H 5OH  CH 3COOC2 H 5  H 2O
1. Calculate the theoretical yield
78
Percent Yields from Reactions
CH 3COOH + C 2 H 5OH  CH 3COOC2 H 5  H 2 O
1. Calculate the theoretical yield
88.0g CH 3COOC2 H 5
? g CH 3COOC2 H 5 = 10.0g C 2 H 5OH 
46.0 g C 2 H 5OH
 19.1 g CH 3COOC2 H 5
79
Percent Yields from Reactions
CH 3COOH + C 2 H 5OH  CH 3COOC2 H 5  H 2 O
1. Calculat e t he t heoret ical yield
88.0g CH 3COOC2 H 5
? g CH 3COOC2 H 5 = 10.0g C 2 H 5OH 
46.0 g C 2 H 5OH
 19.1 g CH 3COOC2 H 5
2. Calculat e t hepercentyield.
80
Percent Yields from Reactions
CH 3COOH + C 2 H 5OH  CH 3COOC2 H 5  H 2 O
1. Calculate the theoretical yield
88.0g CH 3COOC2 H 5
? g CH 3COOC2 H 5 = 10.0g C 2 H 5OH 
46.0 g C 2 H 5OH
 19.1 g CH 3COOC2 H 5
2. Calculate thepercentyield.
14.8g CH 3COOC2 H 5
% yield =
100%  77.5%
19.1g CH 3COOC2 H 5
81