Stoichiometry
Hein and Arena
Version 1.1
Eugene Passer
Chemistry Department
1 College
Bronx Community
© John Wiley and Sons, Inc.
Chapter Outline
9.1 A Short Review
9.4 Mole-Mass Calculations
9.2 Introduction to
Stoichiometry:
The Mole-Ratio Method
9.5 Mass-Mass Calculations
9.6 Limiting-Reactant and Yield
Calculations
9.3 Mole-Mole Calculations
2
A Short Review
3
• The molar mass of an element is its
atomic mass in grams.
• It contains 6.022 x 1023 atoms
(Avogadro’s number) of the element.
4
The molar mass of an element or compound is
the sum of the atomic masses of all its atoms.
5
of a substance
of molecules
gramsgrams
of anumber
monoatomic
element
molar
number
mass
of=moles
=
molar
mass
=
23
number
6.022
of
x 10 ofmolecules/mole
the element
substance
number
ofmoles
moles
the
6
Avogadro’s
Number of
Particles
23
10
6.022 x
Particles
1 MOLE
Molar Mass
7
23
23atoms
1 mole
11mole
mole
= 6.022
==6.022
6.022
x 10
xx10
1023
molecules
ions
8
• For calculations of mole-mass-volume
relationships.
– The chemical equation must be balanced.
– The coefficient in front of a formula
represents the number of moles of the
reactant or product.
The equation is balanced.

2 Al + Fe2O3  2 Fe + Al2O3
2 mol
1 mol
2 mol
1 mol
9
Introduction to Stoichiometry:
The Mole-Ratio Method
10
• Stoichiometry: The area of chemistry
that deals with the quantitative
relationships between reactants and
products.
• Mole Ratio: a ratio between the moles
of any two substances involved in a
chemical reaction.
– The coefficients used in mole ratio
expressions are derived from the
coefficients used in the balanced
equation.
11
Examples
12
N2 + 3H2  2NH3
1 mol
3 mol
2 mol
1 mol N 2
3 mol H 2
13
N2 + 3H2  2NH3
1 mol
3 mol
2 mol
3 mol H 2
2 mol NH 3
14
• The mole ratio is used to convert the
number of moles of one substance to
the corresponding number of moles of
another substance in a stoichiometry
problem.
• The mole ratio is used in the solution of
every type of stoichiometry problem.
15
The Mole Ratio Method
1. Convert the quantity of starting substance
to moles (if it is not already 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.
16
Step 1 Determine the number of moles of
starting substance.
Identify the starting substance from the
data given in the problem statement.
Convert the quantity of the starting
substance to moles, if it is not already in
moles.
 1 mole 
moles = grams 

 molar mass 
17
How many moles of NaCl are present in 292.215
grams of NaCl? The molar mass of NaCl =58.443 g.
 1 mole 
moles = grams 

 molar mass 
 1 mole NaCl 
moles NaCl = 292.15 grams NaCl 
= 5.000 moles NaCl

 58.443 g NaCl 
18
How many moles of NaCl are present in 292.215
grams of NaCl? The molar mass of NaCl =58.443 g.
 1 mole 
moles = grams 

 molar mass 
 1 mole NaCl 
moles NaCl = 292.15 grams NaCl 
 = 5.0000 moles NaCl
 58.443 g NaCl 
19
Step 2 Determine the mole ratio of the
desired substance to the starting
substance.
The number of moles of each substance
in the balanced equation is indicated by
the coefficient in front of each substance.
Use these coefficients to set up the mole
ratio.
moles of desired substance in the equation
mole ratio =
moles of starting substance in the equation
20
Step 2 Determine the mole ratio of the
desired substance to the starting
substance.
Multiply the number of moles of starting
substance (from Step 1) by the mole ratio
to obtain the number of moles of desired
substance.
 moles of desired substance 


in
the
equation

moles of desired substance = moles of starting substance 
 moles of starting substance 


in
the
equation


21
In the following reaction how many moles of PbCl2
are formed if 5.000 moles of NaCl react?
2NaCl(aq) + Pb(NO3)2(aq)  PbCl2(s) + 2NaNO3(aq)
 moles of desired substance in the equation 
moles of desired substance = moles of starting substance 

 moles of starting substance in the equation 
 1 mol PbCl 2 
moles of PbCl 2 = 5.000 moles NaCl 
  2.500 mol PbCl 2
 2 mol NaCl 
22
Step 3. Calculate the desired substance in
the units specified in the problem.
• If the answer is to be in moles, the
calculation is complete
• If units other than moles are wanted,
multiply the moles of the desired
substance (from Step 2) by the
appropriate factor to convert moles to
the units required.
23
Step 3. Calculate the desired substance in
the units specified in the problem.
 molar mass 
1. To calculate grams: grams = moles x 

1
mo
l


 18.02 g H 2O 
90.10 grams H 2O = 5.000 mol H 2O 

1
mol
H
O
2


24
Step 3. Calculate the desired substance in
the units specified in the problem.
 6.022 x 1023atoms 
2. To calculate atoms: atoms = moles 

1
mo
l


23

6.022
x
10
Na atoms 
3.011 x 1024 Na atoms = 5.000 moles Na atoms 

1
mol
Na
atoms


25
Step 3. Calculate the desired substance in
the units specified in the problem.
 6.022 x 1023molecules 
3. To calculate molecules: molecules = moles x 

1
mol


23

6.022
x
10
H 2O molecules 
24
3.011 x 10 molecules H 2O = 5.000 moles H 2O 

1
mol
H
O
2


26
Mole-Mole
Calculations
27
Phosphoric Acid
• Phosphoric acid (H3PO4) is one of the
most widely produced industrial
chemicals in the world.
• Most of the world’s phosphoric acid is
produced by the wet process which
involves the reaction of phosphate
rock, Ca5(PO4)3F, with sulfuric acid
(H2SO4).
Ca5(PO4)3F(s) + 5H2SO4  3H3PO4 + HF + 5CaSO
4
28
Calculate the number of moles of phosphoric acid
(H3PO4) formed by the reaction of 10 moles of sulfuric
acid (H2SO4).
Ca5(PO4)3F + 5H2SO4  3H3PO4 + HF + 5CaSO4
1 mol
5 mol
3 mol
1 mol
5 mol
Step 1 Moles starting substance: 10.0 mol H2SO4
Step 2 The conversion needed is
moles H2SO4  moles H3PO4
Mole Ratio
3 mol H 3PO4
10 mol H 2SO4 x
= 6 mol H 3PO4
5 mol H 2SO4
29
Calculate the number of moles of sulfuric acid
(H2SO4) that react when 10 moles of Ca5(PO4)3 react.
Ca5(PO4)3F + 5H2SO4  3H3PO4 + HF + 5CaSO4
1 mol
5 mol
3 mol
1 mol
5 mol
Step 1 The starting substance is 10.0 mol Ca5(PO4)3F
Step 2 The conversion needed is
moles Ca5(PO4)3F  moles H2SO4
Mole Ratio
5 mol H 2SO4
10 mol Ca 5 (PO 4 )3F x
= 50 mol H 2SO4
1 mol Ca 5 (PO4 )3F
30
Mole-Mass
Calculations
31
1. The object of this type of problem is
to calculate the mass of one
substance that reacts with or is
produced from a given number of
moles of another substance in a
chemical reaction.
2. If the mass of the starting substance
is given, we need to convert it to
moles.
32
3. We use the mole ratio to convert
moles of starting substance to moles
of desired substance.
4. We can then change moles of desired
substance to mass of desired substance
if called for by the problem.
33
Examples
34
Calculate the number of moles of H2SO4 necessary to
yield 784 g of H3PO4.
Ca5(PO4)3F+ 5H2SO4  3H3PO4 + HF + 5CaSO4
Method 1 Step by Step
Step 1 The starting substance is 784 grams of H3PO4.
Step 2 Convert grams of H3PO4 to moles of H3PO4.
 1 mol H 3PO4 
784 g H 3PO4 
 = 8.00 mol H 3PO4
 98.0 g H 3PO4 
Step 3 Convert moles of H3PO4 to moles of H2SO4 by the
mole-ratio method.
 5 mol H 2SO4 
8.00 mol H 3PO4 
 = 13.3 mol H 2SO4
 3 mol H 3PO4 
35
Mole Ratio
Calculate the number of moles of H2SO4 necessary to
yield 784 g of H3PO4
Ca5(PO4)3F+ 5H2SO4  3H3PO4 + HF + 5CaSO4
Method 2 Continuous
The conversion needed is
grams H3PO4  moles H3PO4  moles H2SO4
Mole Ratio
 1 mol H 3PO4 
784 g H 3PO4 

 98.0 g H 3PO4 
 5 mol H 2SO4 

 = 13.3 mol H 2SO4
 3mol H 3PO4 
36
Calculate the number of grams of H2 required to form
12.0 moles of NH3.
N2 + 3H2  2NH3
Method 1 Step by Step
Step 1 The starting substance is 12.0 moles of NH3
Step 2 Calculate moles of H2 by the mole-ratio method.
 3 mol H 2 
12.0 mol NH 3 
 = 18.0 mol H 2
 2 mol NH 3 
Mole Ratio
Step 3 Convert moles of H2 to grams of H2.
 2.02 g H 2 
18.0 mol H2 
 = 36.0 g H2
 1 mol H 2 
37
Calculate the number of grams of H2 required to form
12.0 moles of NH3.
N2 + 3H2  2NH3
Method 2 Continuous
The conversion needed is
moles NH3  moles H2  grams H2
Mole Ratio
 3 mol H 2   2.02 g H 2 
12.0 mol NH 3 
  1mol H  = 36.0 g H 2
2 
 2 mol NH 3  
38
Mass-Mass
Calculations
39
Solving mass-mass stoichiometry problems
requires all the steps of the mole-ratio
method.
1. The mass of starting substance is
converted to moles.
2. The mole ratio is then used to determine
moles of desired substance.
3. The moles of desired substance are
converted to mass of desired substance.
4. Baby Aufort is Boy!
40
Calculate the number of grams of NH3 formed by the
reaction of 112 grams of H2.
N2 + 3H2  2NH3
Method 1 Step by Step
Step 1 The starting substance is 112 grams of H2. Convert
112 g of H2 to moles.
grams  moles
 1 mol H 2 
112 g H 2 
  55.4 moles H 2
 2.02 g H 2 
Step 2 Calculate the moles of NH3 by the mole ratio
method.
 2 mol NH 3 
55.4 moles H 2 
 = 36.9 moles NH 3
 3 mol H 2 
41
Calculate the number of grams of NH3 formed by the
reaction of 112 grams of H2.
N2 + 3H2  2NH3
Method 1 Step by Step
Step 3 Convert moles NH3 to grams NH3.
moles  grams
 17.0 g NH 3 
36.9 moles NH 3 
 = 629 g NH3
 1 mol NH 3 
42
Calculate the number of grams of NH3 formed by the
reaction of 112 grams of H2.
N2 + 3H2  2NH3
Method 2 Continuous
grams H2  moles H2  moles NH3  grams NH3
 1 mol H 2   2 mol NH 3   17.0 g NH 3 
112 g H 2 
 = 629 g NH3


 2.02 g H 2   3 mol H 2   1 mol NH 3 
43
Limiting-Reactant and
Yield Calculations
44
Limiting Reactant
45
• The limiting reactant is one of the
reactants in a chemical reaction.
• It is called the
limiting reactant
because the amount of it present is
insufficient to react with the amounts
of other reactants that are present.
• The limiting reactant limits the amount
of product that can be formed.
46
How manyFrom
bicycles
From
eight three
wheels
From
pedal
four
fourassemblies
frames four
can be assembled
bikes three
can be
bikes
constructed.
bikes
cancan
be be
constructed.
constructed.
from the parts
shown?
The limiting part is the
number of pedal
assemblies.
47
9.2
H2 + Cl2  2HCl
+
4 molecules Cl2 can
form 8 molecules HCl
9.3
7 molecules H2 can
form 14 molecules HCl

Cl
limiting
3 molecules
of H
2 remain
2 is the
H2 is in excess
reactant
48
Steps Used to Determine the
Limiting Reactant
49
1. Calculate the amount of product (moles or
grams, as needed) formed from each reactant.
2. Determine which reactant is limiting. (The
reactant that gives the least amount of
product is the limiting reactant; the other
reactant is in excess.
3. Calculate the amount of the other reactant
required to react with the limiting reactant,
then subtract this amount from the starting
quantity of the reactant. This gives the
amount of the substance that remains
unreacted.
50
Examples
51
How many moles of HCl can be produced by reacting
4.0 mol H2 and 3.5 mol Cl2? Which compound is the
limiting reactant?
H2 + Cl2 → 2HCl
Step 1 Calculate the moles of HCl that can form
from each reactant.


2
mol
HCl
4.0 mol H 2 
  8.0 mol HCl
 1 mol H 2 
3.5 mol Cl2  2 mol HCl   7.0 mol HCl
 1 mol Cl 2 
Step 2 Determine the limiting reactant.
The limiting reactant is Cl2 because it
52
produces less HCl than H2.
How many moles of silver bromide (AgBr) can be
formed when solutions containing 50.0 g of MgBr2
and 100.0 g of AgNO3 are mixed together? How
many grams of the excess reactant remain unreacted?
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Step 1 Calculate the grams of AgBr that can form
from each reactant.
The conversion needed is
g reactant → mol reactant → mol AgBr → g AgBr
 1 mol MgBr2   2 mol AgBr   187.8 g AgBr 
50.0 g MgBr2  
 
  1 mol AgBr   102 g AgBr

 184.1 g MgBr2   1 mol MgBr2  
 1 mol AgNO3   2 mol AgBr  187.8 g AgBr 
110.5 g AgBr
100.0 g AgNO3 



53
 169.9 g AgNO3   2 mol AgNO3  1 mol AgBr 
How many moles of silver bromide (AgBr) can be
formed when solutions containing 50.0 g of MgBr2
and 100.0 g of AgNO3 are mixed together? How
many grams of the excess reactant remain unreacted?
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Step 2 Determine the limiting reactant.
The limiting reactant is MgBr2 because it
forms less Ag Br.
 1 mol MgBr2   2 mol AgBr   187.8 g AgBr 
50.0 g MgBr2  
 
  1 mol AgBr   102 g AgBr

 184.1 g MgBr2   1 mol MgBr2  
 1 mol AgNO3   2 mol AgBr  187.8 g AgBr 
110.5 g AgBr
100.0 g AgNO3 



54
 169.9 g AgNO3   2 mol AgNO3  1 mol AgBr 
How many grams of the excess reactant (AgNO3)
remain unreacted?
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Step 3 Calculate the grams of unreacted AgNO3.
First calculate the number of grams of
AgNO3 that will react with 50 g of MgBr2.
The conversion needed is
g MgBr2 → mol MgBr2 → mol AgNO3 → g AgNO3
 1 mol MgBr2  2 mol AgNO3   169.9 g AgNO3 
50.0 g MgBr2  
  92.3 g AgNO3


 184.1 g MgBr2  1 mol MgBr2   1 mol AgNO3 
The amount of AgNO3 that remains is
100.0 g AgNO3 - 92.3 g AgNO3 = 7.7 g AgNO553
Reaction Yield
56
The quantities of products calculated
from equations represent the maximum
yield (100%) of product according to the
reaction represented by the equation.
57
Many reactions fail to give
a 100% yield of product.
This occurs because of side reactions
and the fact that many reactions are
reversible.
58
• The theoretical yield of a reaction is
the calculated amount of product that
can be obtained from a given amount
of reactant.
• The actual yield is the amount of
product finally obtained from a given
amount of reactant.
59
• The percent yield of a reaction is the
ratio of the actual yield to the theoretical
yield multiplied by 100.
actual yield
x 100 = percent yield
theoretical yield
60
Silver bromide was prepared by reacting 200.0 g of
magnesium bromide and an adequate amount of silver
nitrate. Calculate the percent yield if 375.0 g of silver
bromide was obtained from the reaction:
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Step 1 Determine the theoretical yield by
calculating the grams of AgBr that can be
formed.
The conversion needed is
g MgBr2 → mol MgBr2 → mol AgBr → g AgBr
 1 mol MgBr2 
200.0
g
MgBr


2 
184.1
g
MgBr
2 

 2 mol AgBr   187.8 g AgBr 
408.0 g AgBr



 1 mol MgBr2   1 mol AgBr 
61
Silver bromide was prepared by reacting 200.0 g of
magnesium bromide and an adequate amount of silver
nitrate. Calculate the percent yield if 375.0 g of silver
bromide was obtained from the reaction:
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Step 2 Calculate the percent yield.
must have same units
actual yield
percent yield =
x 100
theoretical yield
must have same units
375.0 g AgBr
x 100 = 91.9%
percent yield =
408.0 g AgBr
62
63