Chapter 12 - Stoichiometry
“SUPER
DIMENSIONAL
ANALYSIS”
Review of moles…..
• 1 mole = 6.022 x 1023 particles
• Molar mass =
X grams
1 mole
(
X = molar mass
of substance
Use atomic mass
)
Calculate molar mass of CaBr2
Ca: 40.1 x 1 = 40.1 g/mol
Br: 79.9 x 2 = 159.8 g/mol
= 199.9 g/mol
What is mass of 0.250 mol CO2?
C:
O:
12
16
x 1 = 12 g/mol
x 2 = 32 g/mol
= 44 g/mol
0.250 mol CO2 44 g CO2
1 mol CO2
= 11 g CO2
Stoichiometry
• Defn – study of mass relationships
between reactants and products in a
chemical reaction
• What does this mean?
– How much of something (products) can be
made?
– How much of starting materials were there?
Stoichiometry
balanced chemical reaction
You must have a _____________________
to do stoichiometry calculations.
To make a basic cheeseburger:
2 buns + 1 meat patty + 1 cheese
1 cheeseburger
1 c-burger
1 c-burger
2 buns
1 meat patty
1 meat patty
2 buns
1 c-burger
1 cheese
1 cheese
2 buns
1 meat patty
1 cheese
Mole Ratio
• Defn – ratio btwn # of moles of any two
substances in a balanced chem rxn
Ex reaction
2 Al + 3 Br2  2 AlBr3
• give all the mole ratios
2 mol Al
3 mol Br2
2 mol AlBr3
3 mol Br2
2 mol Al
2 mol Al
2 mol Al
3 mol Br2
2 mol AlBr3
2 mol AlBr3
2 mol AlBr3
3 mol Br2
4 steps to a basic stoichiometry
problem
1) Balance the equation
2) Identify the given, convert to moles
3) Identify unknown, do a mole to mole
ratio between given and unknown (KEY
STEP)
4) Convert unknown to unit specified in
problem
Stoichiometry Flow Chart
Molar
Mass
A
grams A
Mole B
Ratio A
moles A
moles B
Molar
Mass
B
grams B
examples
B
• How many moles of H2 are formed when
65 g HCl is used?
A
A
• If 3.7 mol KBr reacts with calcium, how
many moles of CaBr2 are formed?
B
Mole A to Mole B Conversions
Molar
Mass
A
grams A
Mole B
Ratio A
moles A
moles B
Molar
Mass
B
grams B
Mole A to Mole B Conversions
1 C3H8 + __
5 O2  __
3 CO2 + __
4 H2O
__
10.0 mol
? mol
B
A
• How many moles of CO2 are produced when
10.0 moles of O2 is used?
10.0 mol O2
3 mol CO2
5 mol O2
Mole B
Ratio A
= 6 mol CO2
Mole A to Mass B Conversion
Molar
Mass
A
grams A
Mole B
Ratio A
moles A
moles B
Molar
Mass
B
grams B
Mole A to Mass B Conversion
A
2 Na + ___
1 Cl  ___
2 NaCl
___
2
B
?g
1.25 mol
• How many grams of sodium chloride is
formed when 1.25 moles of sodium react
w/ chlorine gas?
1.25 mol Na
2 mol NaCl
2 mol Na
Mole B
Ratio A
58.5 g NaCl
= 73.1 g NaCl
1 mol NaCl
Molar
Mass B
Mass A to Mole B Conversion
Molar
Mass
A
grams A
Mole B
Ratio A
moles A
moles B
Molar
Mass
B
grams B
Mass A to Mole B Conversion
B
A
2 Na + ___
1 Cl  ___
2 NaCl
___
2
? mol
50 g
• How many moles of chlorine gas is
needed to make 50 g NaCl?
50 g NaCl
1 mol NaCl
58.5 g NaCl
Molar
Mass A
1 mol Cl2
= 0.43 mol
2 mol NaCl
Cl2
Mole B
Ratio A
Mass A to Mass B Conversion
Molar
Mass
A
grams A
Mole B
Ratio A
moles A
moles B
Molar
Mass
B
grams B
Mass A to Mass B Conversion
A
B
1
1 N O + ___
2 HO
___NH
NO

___
4
3
2
2
25 g
?g
• Determine the mass of water formed from
decomposition of 25.0 g NH4NO3
25 g NH4NO3 1 mol NH4NO3
2 mol H2O
18 g H2O
80 g NH4NO3 1 mol NH4NO3 1 mol H2O
= 11.25 g H2O
Limiting Reactant
• In basic stoichiometry problems, you are
provided with one given quantity. In LR
problems, you are given both reactants.
Before you can solve the problem, you
have to determine which of the two given
quantities to use as your given.
Limiting Reactant
You need to choose the one that will run out
limiting reactant
first, known as the __________________.
It controls how much product you can
make. The other reactant is known as the
excess reactant
_________________
because there will
be some left over.
1 cheese + 2 bread slices  1 cheese
sandwich
Given: 10 pieces of cheese, 50 bread slices
How many cheese sandwiches can you make?
10
What is the limiting reactant? cheese
What is the excess reactant? bread
How many pieces of excess reactant used? 20
How many pieces of excess reactant left over?
30
How to find limiting reactant
1) Convert both amounts of reactants to
moles
2) Divide the mole amount of each reactant
by its coefficient in the balanced equation
3) Compare two numbers. The one that is
smaller is the limiting reactant. Other one
is excess reactant.
2 Al
____
3 CuCl2  ___
3 Cu + ___
2 AlCl3
+ ___
Find the limiting reactant if 6.9 g Al and 0.35 mol CuCl2
are available.
1) Convert both amounts of reactants to moles.
Reactant #1:
6.9 g Al
Reactant #2:
1 mol Al
27 g Al
= 0.256 mol Al
= 0.35 mol CuCl2
2 Al
____
3 CuCl2  ___
3 Cu + ___
2 AlCl3
+ ___
2) Divide each mole amount by its coefficient in
the balanced equation
Reactant #1:
0.256 mol Al
2
= 0.128
Reactant #2:
0.35 mol CuCl2
3
= 0.117
CuCl2 is the limiting reactant
2 Al
____
3 CuCl2  ___
3 Cu + ___
2 AlCl3
+ ___
Find the limiting reactant if 6.2 g Al and 48.5 g CuCl2
are available.
1) Convert both amounts of reactants to moles.
Reactant #1:
6.2 g Al
1 mol Al
27 g Al
= 0.230 mol Al
Reactant #2:
48.5 g CuCl2 1 mol CuCl2
134.5 g CuCl2
= 0.360 mol CuCl2
2 Al
____
3 CuCl2  ___
3 Cu + ___
2 AlCl3
+ ___
2) Divide each mole amount by its coefficient in
the balanced equation
Reactant #1:
Reactant #2:
0.230 mol Al
0.360 mol CuCl2
2
3
= 0.115
= 0.120
Al is the LR
CuCl2 is excess reactant
Based on the LR in #4, how many grams of
copper will be produced?
2 Al
____
3 CuCl2  ___
3 Cu + ___
2 AlCl3
+ ___
6.2 g Al
1 mol Al
3 mol Cu
63.5 g Cu
27 g Al
2 mol Al
1 mol Cu
= 21.9 g Cu
Percent Yield
• You buy a 500 g ketchup bottle, do you
ever use all 500 g?
No. Some is still left on the sides you
cannot retrieve. Not all 100% is used.
Percent Yield
• Theoretical Yield – max amount of product
that can be produced (what you expect to
get)
• Actual Yield – amt of product you actually
produced (always less than theoretical)
• Percent Yield =
actual
x 100
theoretical
Ex problem #1
• Joe does experiment to form carbon
dioxide. The maximum he can obtain is
34.5 grams. He performs the experiment
and only obtains 18.6 grams. What is his
percent yield?
A
T
18.6 g
=
34.5 g
X 100 = 53.9%
Ex problem #2
__K
1 2CrO4 + __
2 AgNO3  __KNO
2
1
3 + __Ag
2CrO4
?g
0.500 g
• What is the theoretical yield of Ag2CrO4 if
0.500 g AgNO3 reacts with excess K2CrO4?
0.5 g K2CrO4 1 mol K2CrO4 1 mol Ag2CrO4
331.7 g
170 g K2CrO4 1 mol K2CrO4 1 mol Ag2CrO4
= 0.49 g Ag2CrO4
Ex problem #2
• If 0.455 g of Ag2CrO4 is produced, what is
the percent yield?
A
T
0.455 g
=
X 100 = 93.2%
0.49 g