Chapter 9
Calculations from
Chemical Equations
Accurate
measurement and
calculation of the
correct dosage are
important in
dispensing the
correct medicine
to patients
throughout the
Introduction to General, Organic, and Biochemistry 10e
world.
John Wiley & Sons, Inc
Morris Hein, Scott Pattison, and Susan Arena
Chapter Outline
9.1 A Short Review
9.5 Mass-Mass Calculations
9.2 Introduction to
Stoichiometry
9.6 Limiting Reactant and
Yield Calculations
9.3 Mole-Mole Calculations
9.4 Mole-Mass Calculations
Copyright 2012 John Wiley & Sons, Inc
9-2
Objectives for Today
 Review the “mole” concept
 Determine relationships between
moles using stoichiometry
 Develop mole and mass
relationships
9-3
REVIEWING MOLES
9-4
Molar Mass – sum of atomic masses of all atoms
in 1 mole of an element or compound; the
units are g/mol.
1 mole =
6.022x1023 molecules
6.022x1023 formula units
6.022x1023 atoms
6.022x1023 ions
9-5
What is the molar mass of Al(ClO3)3?
Al 1(26.98 g)
3Cl 3(35.45 g)
9O 9(16.00 g)
Al(ClO3)3 277.33 g/mol
atomic mass
Al
26.98
Cl
35.45
O
16.00
7-6
Calculate the mass of 2.5 moles of
aluminum chlorate.
Plan
2.5 mol Al(ClO3)3  g Al(ClO3)3
1 mol Al(ClO3)3 = 277.33 g Al(ClO3)3
Calculate
 277.33 g Al(ClO3 )3 
2.5 mol Al(ClO3 )3 
 = 690 g Al(ClO3 )3
 1 mol Al(ClO3 )3 
9-7
Calculate the moles of 3.52g of
aluminum chlorate.
Plan
3.52 g Al(ClO3)3  mol Al(ClO3)3
1 mol Al(ClO3)3 = 277.33 g Al(ClO3)3
Calculate
 1 mol Al(ClO3 )3 
2
3.52 g Al(ClO3 )3
=
1.27

10
mol Al(ClO3 )3

 277.33 g Al(ClO3 )3 
9-8
Calculate the number of formula units
contained in 12.4 g aluminum
chlorate.
Plan
12.4 g Al(ClO3)3  formula units Al(ClO3)3
1 mol Al(ClO3)3 = 277.33 g = 6.022x1023 formula units
Calculate
 6.022 1023 formula units 
22
12.4 g Al(ClO3 )3 
 = 2.69 10 formula units
 277.33 g Al(ClO3 )3 
9-9
Your Turn!
What is the mass of 3.61 moles of CaCl2?
atomic mass
a. 3.61 g
Ca 40.08
b. 272 g
Cl 35.45
c. 2.17 × 1024 g
d. 401 g
9-10
Your Turn!
How many moles of HCl are contained in 18.2 g
atomic mass
HCl?
H 1.01
a. 1.00 mol
Cl 35.45
b. 0.500 mol
c. 0.250 mol
d. 0.125 mol
9-11
Your Turn!
What is the mass of 1.60×1023 molecules of HCl?
atomic mass
a. 9.69 g
H 1.01
b. 137 g
Cl 35.45
c. 0.729 g
d. 36.5 g
9-12
USING STOICHIOMETRY
9-13
Stoichiometry deals with the quantitative
relationships between the reactants and
products in a balanced chemical equation.
9-14
1N2(g) + 3I2(s)  2NI3(s)
1 mol N2 + 3 mol I2  2 mol NI3
Mole ratios come from the coefficients in the
balanced equation:
3 mol I 2
1 mol N 2
3 mol I 2
2 mol NI3
1 mol N 2
2 mol NI3
The 3 other possibilities are the inverse of
these ratios.
9-15
Your Turn!
Which of these statements is not true about the
reaction?
1N2(g) + 3I2(s)  2NI3(s)
a. 1 mole of nitrogen is needed for every 3
moles of iodine
b. 1 gram of nitrogen is needed for every 3
grams of iodine
c. Both statements are true
9-16
Calculate the number of moles of
NI3 that can be made from 5.50
mol N2 in the reaction: 1N2(g) +
3I2(s)  2NI3(s)
Plan
5.50 mol N2  mol NI3
moles of desired substance in equation
Set-Up mole ratio =
moles of starting substance in equation
2 mol NI3
mol ratio =
1 mol N 2
Calculate
 2 mol NI3 
5.50 mol N2  
 = 11.0 mol NI3
 1 mol N 2 
9-17
Calculate the number of moles of I2
needed to react with 5.50 mol N2 in
the reaction:
1N2(g) + 3I2(s)  2NI3(s)
Plan
Set-Up
Calculate
5.50 mol N2  mol I2
 3 mol I2 
mole ratio = 

 1 mol N2 
 3 mol I2 
5.50 mol N2  
 = 16.5 mol I2
 1 mol N 2 
Copyright 2012 John Wiley & Sons, Inc
9-18
How many moles of HF will be
produced by the complete reaction of
1.42 moles of H2 in the following
equation?
H2 + F2  2HF
a.
b.
c.
d.
0.710
1.42
2.00
2.84
9-19
Problem Solving Strategy for
stoichiometry problems:
1. Convert starting substance to moles.
2. Convert the moles of starting substance to
moles of desired substance.
3. Convert the moles of desired substance to
the units specified in the problem.
9-20
Copyright 2012 John Wiley & Sons, Inc
9-21
USING STOICHIOMETRY IN MOLEMOLE CALCULATIONS
9-22
How many moles of Al are needed to
make 0.0935 mol of H2?
2Al(s) + 6HCl (aq)  2AlCl3(aq) + 3H2(g)
Plan
0.0935 mol H2  mol Al
2 mol Al 

Set-Up mole ratio = 

 3 mol H2 
2 mol Al 

Calculate 0.0935 mol H2 
 =.0623 mol Al
 3 mol H2 
9-23
How many moles of HCl are needed to
make 0.0935 mol of H2?
2Al(s) + 6HCl (aq)  2AlCl3(aq) + 3H2(g)
Plan
Set-Up
Calculate
0.0935 mol H2  mol HCl
6 mol HCl
mole ratio =
3 mol H2
 6 mol HCl 
0.0935 mol H2 
 = 0.187 mol HCl
 3 mol H2 
9-24
Your Turn!
How many moles of H2 are made by the reaction
of 1.5 mol HCl with excess aluminum?
2Al(s) + 6HCl (aq)  2AlCl3(aq) + 3H2(g)
a. 0.75 mol
b. 3.0 mol
c. 6.0 mol
d. 4.5 mol
Copyright 2012 John Wiley & Sons, Inc
9-25
Your Turn!
How many moles of carbon dioxide are
produced when 3.00 moles of oxygen react
completely in the following equation?
C3H8 + 5O2  3CO2 + 9H2O
a. 5.00 mol
b. 3.00 mol
c. 1.80 mol
d. 1.50 mol
Copyright 2012 John Wiley & Sons, Inc
9-26
Your Turn!
How many moles of C3H8 are consumed when
1.81x1023 molecules of CO2 are produced in
the following equation?
C3H8 + 5O2  3CO2 + 4H2O
a. 0.100
b. 0.897
c. 6.03 × 1022
d. 5.43 × 1023
Copyright 2012 John Wiley & Sons, Inc
9-27
Objectives for Today



Review the “mole” concept
Determine relationships between
moles using stoichiometry
Develop mole and mass
relationships
9-28
MOLE-MASS CALCULATIONS
9-29
What mass of H2 (2.02 g/mol) is made
by the reaction of 3.0 mol HCl with
excess aluminum?
2Al(s) + 6HCl (aq)  2AlCl3(aq) + 3H2(g)
Plan
3.0 mol HCl mol H2  g H2
Calculate
3.0 mol HCl  3 mol H 2   1.5 mol H2
 6 mol HCl 


1.5 mol H2  2.02 g H 2   3.0 g H 2
 1 mol H 2 
9-30
How many moles of HCl are needed to
completely consume 2.00 g Al
(26.98g/mol)?
2Al(s) + 6HCl (aq)  2AlCl3(aq) + 3H2(g)
Plan
2.00 g Al  mol Al  mol HCl
Calculate
2.00 g Al  1 mol Al  = 0.0741 mol Al


26.98
g
Al


 6 mol HCl 
0.0741 mol Al 
  0.0222 mol HCl
 2 mol Al 
9-31
What mass of Al(NO3)3 (213g/mol) is
needed to react with .093 mol
Na2CO3?
3Na2CO3(aq) + 2Al(NO3)3(aq)  Al2(CO3)3(s) + 6NaNO3(aq)
Plan
0.093 mol Na2CO3  mol Al(NO3)3  g Al(NO3)3
 2 mol Al(NO3 )3 
Calculate .093 mol Na 2CO3 
  .062 mol Al(NO3 )3
 3 mol Na 2CO3 
 213.00g Al(NO3 )3 
0.062 mol Al(NO3 )3 
  13 g Al(NO3 )3
 1 mol Al(NO3 )3 
9-32
How many moles of Al2(CO3)3 are
made by the reaction of 3.45g Na2CO3
(105.99 g/mol) with excess Al(NO3)3?
3Na2CO3(aq) + 2Al(NO3)3(aq)  Al2(CO3)3(s) + 6NaNO3(aq)
Plan
3.45g Na2CO3  mol Na2CO3  g Al2(CO3)3
Calculate
 1 mol Na 2CO3 
3.45g Na 2CO3 
  0.0326 mol Na 2CO3
 105.99g Na 2CO3 
 1 mol Al2 (CO3 )3 
0.0326 mol Na 2CO3 
  0.0109 mol Na 2CO3
 3 mol Na 2CO3 
9-33
Your Turn!
How many moles of oxygen are consumed when
38.0g of aluminum oxide are produced in the
following equation?
atomic mass
4Al(s) + 3O2(g)  2Al2O3(s) Al 26.98
O 16.00
a. 0.248
b. 0.559
c. 1.50
d. 3.00
Copyright 2012 John Wiley & Sons, Inc
9-34
Your Turn!
What mass of HCl is produced when 1.81x1024
molecules of H2 react completely in the
following equation?
atomic mass
H2(g) + Cl2(g)  2HCl(g) H 1.01
Cl 35.45
a. 54.7g
b. 72.9g
c. 109g
d. 219g
Copyright 2012 John Wiley & Sons, Inc
9-35
Objectives for Today
 Further analyze moles and mass
using stoichiometry
 Use stoichiometry to examine
limiting reagents and yield
9-36
MASS-MASS CALCULATIONS
9-37
Mass-Mass Calculations
Now we will put it all together.
9-38
What mass of Br2 (159.80 g/mol) is
needed to completely consume 7.00 g
Al (26.98 g/mol)?
2Al(s) + 3Br2(l)  2AlBr3(s)
Plan
7.00 g Al mol Al  mol Br2  g Br2
Calculate


7.00 g Al  1 mol Al 
 26.98g Al 
 0.259 mol Al
 3 mol Br2   0.389 mol Br
0.259 mol Al 
2

 2 mol Al 
 159.80g Br2 
0.389 mol Br2 
  62.2 g Br2
 1 mol Br2 
9-39
What mass of Fe2S3 (207.91g/mol) can
be made from the reaction of 9.34 g
FeCl3 (162.20 g/mol) with excess Na2S?
2FeCl3(aq) + 3Na2S(aq)  Fe2S3(s) + 6NaCl(aq)
Plan
9.34 g FeCl3  mol FeCl3  mol Fe2S3  g Fe2S3
Calculate
 1 mol FeCl3 
9.34 g FeCl3
  0.0576 mol FeCl3
 162.20g FeCl3 
 1 mol Fe2S3 
0.0576 mol FeCl3
  0.0288 mol Fe2S3
 2 mol FeCl3 
 207.91g Fe2S3 
0.0288 mol Fe2S3 
  5.99 g Fe2S3
 1 mol Fe2S3 
9-40
Your Turn!
What mass of oxygen is consumed when 54.0g
of water is produced in the following
atomic mass
equation?
H 1.01
2H2 + O2  2H2O O 16.00
a. 0.167 g
b. 0.667 g
c. 1.50 g
d. 47.9 g
Copyright 2012 John Wiley & Sons, Inc
9-41
Your Turn!
What mass of H2O is produced when 12.0g of
HCl react completely in the following
equation?
6HCl + Fe2O3  2FeCl3 + 3H2Oatomic mass
H 1.01
a. 2.97 g
O 16.00
b. 39.4 g
Cl 35.45
c. 27.4 g
d. 110. g
Copyright 2012 John Wiley & Sons, Inc
9-42
LIMITING REACTANT
9-43
Determine the number of
that can be
made given these quantities of reactants and
the reaction equation:
+
+


9-44
The limiting reactant is the reactant that limits
the amount of product that can be made.
The reaction stops when the limiting reactant is
used up.
What was the limiting reactant
in the reaction:

+
The small blue balls.
Copyright 2012 John Wiley & Sons, Inc
9-45
The excess reactant is the reactant that remains
when the reaction stops.
There is always left over excess reactant.
What was the excess reactant in the reaction:
+

The excess reactant was the larger blue ball.
Copyright 2012 John Wiley & Sons, Inc
9-46
Technique for solving limiting reactant
problems:
1. Convert reactant 1 to moles or mass of
product
2. Convert reactant 2 to moles or mass of
product
3. Compare answers. The smaller answer is the
maximum theoretical yield.
Copyright 2012 John Wiley & Sons, Inc
9-47
Calculate the number of moles of water that
can be made by the reaction of 1.51 mol H2
with 0.932 mol O2
2H2(g) + O2(g)  2H2O(g)
1. Calculate the theoretical yield of H2O assuming H2
is the limiting reactant and that O2 is the excess
reactant.
2. Calculate the theoretical yield of H2O assuming that
O2 is the limiting reactant and that H2 is the excess
reactant.
9-48
Assuming that H2 is limiting and O2 is excess:
 2 mol H 2O 
1.51 mol H 2  
 =1.51 mol H 2O
 2 mol H 2 
Assuming that O2 is limiting and H2 is excess:
 2 mol H 2O 
0.932 mol O2  
 =1.86 mol H 2O
 1 mol 02 
So what is the maximum yield of H2O?
9-49
How much H2 and O2 remain when the reaction
stops?
H2: Limiting Reactant – None remains. It was
used up in the reaction.
O2: Excess Reactant – Calculate the amount of
O2 used in the reaction with H2. Then subtract
that from the original amount.
 1 mol O2 
1.51 mol H 2 x 
 =0.755 mol O2
 2 mol H 2 
0.932 mol O2 to start - 0.755 mol O2 = 0.177 mol of excess O2
9-50
Calculate the mass of copper that can
be made from the combination of 15.0
g aluminum with 25.0 g copper(II)
sulfate.
2Al(s) + 3CuSO4(aq)  Al2(SO4)3(aq) + 3Cu(s)
Plan
15 g Al  mol Al  mol Cu  g Cu
25 g CuSO4  mol CuSO4  mol Cu  g Cu
Compare answers. The smaller number is the
right answer.
Copyright 2012 John Wiley & Sons, Inc
9-51
2Al(s) + 3CuSO4(aq)  Al2(SO4)3(aq) + 3Cu(s)
1. Assume Al is limiting and CuSO4 is in excess.
 1 mol Al   3 mol Cu   63.55 g Cu 
15.0 g Al x 
= 53.0 g Cu
 
 

26.98
g
Al
2
mol
Al
  1 mol Cu 

 
2. Assume CuSO4 is limiting and Al is in excess.
 1 mol CuSO4  3 mol Cu   63.55 g Cu 
25.0 g CuSO4 
= 9.96 g Cu



 159.58 g CuSO4  3 mol CuSO4   1 mol Cu 
3. Compare answers.
CuSO4 is the limiting reagent. The theoretical yield of Cu
is 9.96 g.
9-52
Your Turn!
True/False:
You can compare the quantities of reactant
when you work a limiting reactant problem.
The reactant you have the least of is the
limiting reactant.
a. True
b. False
Copyright 2012 John Wiley & Sons, Inc
9-53
Your Turn!
Which is the limiting reactant when 3.00 moles
of copper are reacted with 3.00 moles of silver
nitrate in the following equation?
Cu + 2AgNO3  Cu(NO3)2 + 2Ag
a. Cu
b. AgNO3
c. Cu(NO3)2
d. Ag
Copyright 2012 John Wiley & Sons, Inc
9-54
Your Turn!
What is the mass of silver (107.87 g/mol)
produced by the reaction of 3.00 moles of
copper with 3.00 moles of silver nitrate?
Cu + 2AgNO3  Cu(NO3)2 + 2Ag
a. 162g
b. 216g
c. 324g
d. 647g
Copyright 2012 John Wiley & Sons, Inc
9-55
PERCENT YIELD
9-56
Actual Yield
% Yield =
 100
Theoretical Yield
The theoretical yield is the result calculated using
stoichiometry.
The actual yield of a chemical reaction is the experimental
result, which is often less than the theoretical yield due to
experimental losses and errors along the way.
9-57
Calculate the % yield of PCl3 that results from
reacting 5.00 g P with excess Cl2 if only 17.2 g
of PCl3 were recovered.
2P + 3Cl2 2PCl3
Compute the expected yield of PCl3 from 5.00 g P with excess Cl2.
 1 mol P   2 mol PCl3   137.33 g PCl3 
5.00 g P x 
=
 



 30.97 g P   2 mol P   1 mol PCl3 
22.2g PCl3
Compute the % Yield.
Actual Yield
17.2 g
% Yield =
 100%=
100%=77.5%
Theoretical Yield
22.2 g
Copyright 2012 John Wiley & Sons, Inc
9-58
Your Turn!
In a reaction to produce ammonia, the
theoretical yield is 420. g. What is the percent
yield if the actual yield is 350. g?
A. 83.3%
B. 20.0%
C. 16.7%
D. 120.%
Copyright 2012 John Wiley & Sons, Inc
9-59
Objectives for Today

Further analyze moles and mass
using stoichiometry
 Use stoichiometry to examine
limiting reagents and yield
9-60