Chapter 9
Calculations from
Chemical Equations
Accurate
measurement
and calculation
of the correct
dosage are
important in
dispensing the
correct medicine
to patients
Introduction to General, Organic, and Biochemistry 10e
throughout the
John Wiley & Sons, Inc
world.
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
Molar Mass
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
Copyright 2012 John Wiley & Sons, Inc
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Molar Mass
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
Copyright 2012 John Wiley & Sons, Inc
atomic mass
Al
26.98
Cl
35.45
O
16.00
7-4
Molar Mass
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 
Copyright 2012 John Wiley & Sons, Inc
9-5
Molar Mass
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 
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Molar Mass
Calculate the number of formula units contained in
12.4 g aluminum chlorate.
12.4 g Al(ClO3)3  formula units Al(ClO3)3
Plan
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 
Copyright 2012 John Wiley & Sons, Inc
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Your Turn!
What is the mass of 3.61 moles of CaCl2?
a. 3.61 g
Ca
b. 272 g
Cl
24
c. 2.17 × 10 g
d. 401 g
Copyright 2012 John Wiley & Sons, Inc
atomic mass
40.08
35.45
9-8
You Turn!
How many moles of HCl are contained in 18.2 g HCl?
a. 1.00 mol
atomic mass
H 1.01
b. 0.500 mol
Cl 35.45
c. 0.250 mol
d. 0.125 mol
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Your Turn!
What is the mass of 1.60×1023 molecules of HCl?
a. 9.69 g
atomic mass
H 1.01
b. 137 g
Cl 35.45
c. 0.729 g
d. 36.5 g
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Stoichiometry
Stoichiometry deals with the quantitative relationships
between the reactants and products in a balanced
chemical equation.
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
3 mol I 2
1 mol N 2
2
1 mol N 2
2 mol NI3
2 mol NI3
The 3 other possibilities are the inverse of these ratios.
Copyright 2012 John Wiley & Sons, Inc
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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
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9-12
Using the mole ratio
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 
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9-13
Using the mole ratio
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
5.50 mol N2  mol I2
 3 mol I2 
Set-Up
mole ratio = 

 1 mol N2 
Calculate
 3 mol I2 
5.50 mol N2  
 = 16.5 mol I2
 1 mol N 2 
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Your Turn!
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. 0.710
b. 1.42
c. 2.00
d. 2.84
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Stoichiometry
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.
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Stoichiometry
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Mole-Mole Calculations
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 
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Mole-Mole Calculations
How many moles of HCl are needed to make 0.0935
mol of H2?
2Al(s) + 6HCl (aq)  2AlCl3(aq) + 3H2(g)
0.0935 mol H2  mol HCl
6 mol HCl
mole ratio =
Set-Up
3 mol H2
Plan
Calculate
 6 mol HCl 
0.0935 mol H2 
 = 0.187 mol HCl
 3 mol H2 
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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
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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
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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
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Mole-Mass Calculations
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 
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Mole-Mass Calculations
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 
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Mole-Mass Calculations
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 
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Mole-Mass Calculations
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 
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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
a. 0.248
O 16.00
b. 0.559
c. 1.50
d. 3.00
Copyright 2012 John Wiley & Sons, Inc
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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
a. 54.7g
Cl 35.45
b. 72.9g
c. 109g
d. 219g
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Mass-Mass Calculations
Now we will put it all together.
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)
7.00 g Al mol Al  mol Br2  g Br2
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Mass-Mass Calculations
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 
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Mass-Mass Calculations
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 
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Your Turn!
What mass of oxygen is consumed when 54.0g of water
is produced in the following equation?
2H2 + O2  2H2O
atomic mass
H 1.01
a. 0.167 g
O 16.00
b. 0.667 g
c. 1.50 g
d. 47.9 g
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Your Turn!
What mass of H2O is produced when 12.0g of HCl react
completely in the following equation?
6HCl + Fe2O3  2FeCl3 + 3H2O
a. 2.97 g
atomic mass
b. 39.4 g
H 1.01
O 16.00
c. 27.4 g
Cl 35.45
d. 110. g
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Limiting Reactant
Determine the number of
that can be made
given these quantities of reactants and the reaction
equation:
+
+


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Limiting Reactant
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.
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Excess Reactant
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.
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Limiting reactant
Figure 9.2
The number of
bicycles that can
be built from these
parts is determined
by the “limiting
reactant” (the
pedal assemblies).
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Limiting Reactant Calculations
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.
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Limiting Reactant Calculation
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.
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Limiting Reactant Calculation continued
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?
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Limiting Reactant Calculation continued
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
Copyright 2012 John Wiley & Sons, Inc
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Limiting Reactant Calculation
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
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Limiting Reactant Calculation continued
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 
15.0 g Al x 
  2 mol Al 

 26.98 g Al  
 63.55 g Cu 

=
 1 mol Cu 
53.0 g 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 
  3 mol CuSO  1 mol Cu  = 9.96 g Cu

4 
 159.58 g CuSO4  
3. Compare answers.
CuSO4 is the limiting reagent. The theoretical yield of
Cu is 9.96 g.
Copyright 2012 John Wiley & Sons, Inc
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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
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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
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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
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Percent Yield
To determine the efficiency of a process for making a
compound, chemists compute the percent yield of the
reaction.
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.
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Percent Yield
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 
  2 mol P   1 mol PCl3  = 22.2g PCl3

 
 30.97 g P  
Compute the % Yield.
Actual Yield
17.2 g
% Yield =
 100%=
100%=77.5%
Theoretical Yield
22.2 g
Copyright 2012 John Wiley & Sons, Inc
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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.%
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