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Stoichiometry
stoi·chi·om·e·try noun
Literally – measuring the components
Calculations of the quantitative
relationships between reactants and
products in a chemical reaction.
The Mole and Chemical Reactions:
The Macro-Nano Connection
2 H2 + O 2 
2 H2 molecules
2 moles H2 molecules
4.0 g H2
1 O2 molecule
1 mole O2 molecules
32.0 g O2
2 H2 O
2 H2O molecules
2 moles H2O molecules
36.0 g H2O
Stoichiometric Relationships
EXAMPLE How much iron(III) oxide
and aluminum powder are required to field
weld the ends of two rails together?
Assume that the rail is
132 lb Fe/yard and
that the weld is 1/10
inch wide.
weld
Photo by Mike Condren
The mass of iron in 1/10 inch of this rail is:
1 in  1yd  132lb  454g  166g
10
36in 1yd
1lb
Fe2O3 + 2 Al  Al2O3 +2 Fe
1mol Fe 1mol Fe2O3 159.68g Fe2O3
166g Fe 


 237g Fe2O3
55.84g Fe
2mol Fe
1mol Fe2O3
1mol Fe 2mol Al 26.98g Al
166g Fe 


 80.g Al
55.84g Fe 2mol Fe 1mol Al
1.
2.
3.
4.
5.
C
Ge
Si
Sn
Ti
Limiting Reactant
reactant that limits the amount of product
that can be produced
Limiting Reactant
EXAMPLE What is the number of moles of CaSO4 (S)
that can be produced by allowing 1.0 mol SO2, 2.0 mol
CaCO3, and 3.0 mol O2 to react?
2SO2(g) + 2CaCO3(s) + O2(g)  2CaSO4(S) + 2CO2(g)
balanced equation relates:
2SO2(g)  2CaCO3(s)  O2(g)
have only:
1SO2(g)  2CaCO3(s)  3O2(g)
not enough SO2 to use all of the CaCO3 or the O2
not enough CaCO3 to use all of the O2
SO2 is the limiting reactant
Compare Amounts of Product
2mol CaSO 4
2mol CaCO3 
 2mol CaSO4
2mol CaCO3
2mol CaSO 4
1mol SO 2 
 1mol CaSO 4
2mol SO 2
2mol CaSO 4
3mol O 2 
 6mol CaSO 4
1mol O 2
SO2 limits the amount of product formed
Mass of Product
• 95.0 g of chlorine and 27.0 g of phosphorus
react to form PCl3. What mass of PCl3 is
formed?
P4 (s) + 6 Cl2 (g)  4 PCl3 (l)
4mol PCl3 137.32g PCl3
1mol Cl2
95.0g Cl2 


 122.7g PCl3
70.90g Cl2 6mol Cl2
1mol PCl3
4mol PCl3 137.32g PCl3
1mol P4
27.0g P4 


 119.7g PCl3
123.88g P4
1mol P4
1mol PCl3
27.0g phosphorus gives the smaller yield.
Phosphorus is the limiting reactant
it limits the amount of product formed
Excess Reactant
• We can also calculate the amount of the nonlimiting reactant that is used up, and thus how
much is left over
1mol P4
6mol Cl2 70.90g Cl2
27.0g P4 


 92.72g Cl2
123.88g P4 1mol P4 1mol PCl3
92.72 g Cl2 are required to react with all the P4
Thus 95.0-92.72=2.3 g Cl2 are left over as excess
Theoretical Yield
The amount of product produced by a
reaction based on the amount of the limiting
reactant
Actual Yield
The amount of product actually produced in
a reaction
Percent Yield
actual yield
% yield =
 100%
theoretical yield
For example, if only 103.5 g of PCl3
were actually produced
103.5g PCl3
% yield 
100%  86.5% yield
119.7g PCl3
EXAMPLE
A rocket fuel, hydrazine, is
produced by a reaction of Cl2 with excess
NaOH and NH3. (a) What theoretical yield
can be produced from 1.00 kg of Cl2?
2NaOH + Cl2 + 2NH3  N2H4 + 2NaCl + 2H2O
(a) to calculate the theoretical yield, use the net equation for the overall process
(1.00 kg Cl2) (1 kmol Cl2)(1 kmol N2H4)(32.0 kg N2H4)
(70.9 kg Cl2) (1 kmol Cl2) (1 kmol N2H4)
molar mass
balanced
equation
molar mass
= 0.451 kg N2H4
EXAMPLE
(b) What is the actual yield
if 0.299 kg of 98.0% N2H4 is produced for
every 1.00 kg of Cl2?
2NaOH + Cl2 + 2NH3  N2H4 + 2NaCl + 2H2O
(a) theoretical yield
#kg N2H4 = 0.451 kg N2H4
(b) actual yield
(0.299 kg product)
(98.0 kg N2H4)
(100 kg product)
purity factor
= 0.293 kg N2H4
EXAMPLE (c) What is the percent yield
of pure N2H4?
2NaOH + Cl2 + 2NH3  N2H4 + 2NaCl + 2H2O
(a) theoretical yield = 0.451 kg N2H4
(b) actual yield
= 0.293 kg N2H4
(c) percent yield
0.293 kg
% yield =
 100% = 65.0 % yield
0.451kg
1.
2.
3.
4.
16 %
32 %
50 %
65 %
Combustion Analysis
Example Benzoic acid is known to contain only
C, H, and O. A 6.49-mg sample of benzoic acid was
burned completely in a C-H analyzer. The increase
in the mass of each absorption tube showed that
16.4-mg of CO2 and 2.85-mg of H2O formed. What
is the empirical formula of benzoic acid?
# mg C =
(16.4-mg of CO2 )(12.01-mg C)
= 4.48-mg C
(44.01-mg CO2)
# mg H =
(2.85-mg of H2O )(2.01-mg H)
(18.01-mg H2O)
= 0.318-mg H
# mg O = 6.49 - 4.48 - 0.318 = 1.69 mg O
Empirical Formula
4.48mg C 
1mmol C
 0.373mmol C / 0.106  3.53mmol C  2  7
12.01mg C
0.318mg H 
1.69mg O 
1mmol H
 0.315mmol H / .106  2.99mmol H  2  6
1.008mg H
1mmol O
 0.106mmol O / .106  1.00mmol O  2  2
16.00mg O
C7H6O2