Stoichiometry – Ch. 8
Stoichiometric
Calculations
Proportional Relationships
2 1/4 c. flour
1 tsp. baking soda
1 tsp. salt
1 c. butter
3/4 c. sugar
3/4 c. brown sugar
1 tsp vanilla extract
2 eggs
2 c. chocolate chips
Makes 5 dozen cookies.
I
have 5 eggs. How many cookies
can I make?
Ratio of eggs to cookies
5 eggs 5 doz.
2 eggs
= 12.5 dozen cookies
Proportional Relationships
 Stoichiometry
• mass relationships between
substances in a chemical reaction
• based on the mole ratio
 Mole
Ratio
• indicated by coefficients in a
balanced equation
2 Mg + O2  2 MgO
Stoichiometry Steps
1. Write a balanced equation.
2. Identify known & unknown.
3. Line up conversion factors.
• Mole ratio - moles
molesmoles
moles
• Molar mass - moles  grams
Core step in all stoichiometry problems!!
4. Check answer.
Stoichiometry Problems
If you have 45 kg of isoamyl alcohol
and enough acetic acid to react with
all of the isoamyl alcohol, what is the
maximum number of kg of isoamyl
acetate that can be made?
C5H11OH + CH3COOH 
CH3COOC5H11 + H20
Stoichiometry Problems
Equation balanced?
C5H11OH + CH3COOH 
CH3COOC5H11 + H20
YEA!!!
Stoichiometry Problems
What do you know?
45 kg of C5H11OH
? kg of isoamyl acetate
Molar mass of C5H11OH = 88.17g/mol
Molar mass of CH3COOC5H11= 130.21
g/mol
1 mol C5H11OH : 1 mol CH3COOC5H11
1000 g = 1 kg
Stoichiometry Problems
45 kg  1mol C5H11OH  1000 g
88.17 g C5H11OH 1kg
= 510.mol C5H11OH
510.mol C5H11OH 1 mol CH3COOC5H11
1 mol C5H11OH
510. mol CH3COOC5H11
=
Stoichiometry Problems
510. mol CH3COOC5H11  130.21 g
1kg
1 mol CH3COOC5H11 1000g
=66.4 kg isoamyl acetate
66 kg isoamyl acetate is the maximum
amount that can be produced
Stoichiometry Problems
Magnesium burns in oxygen to
produce magnesium oxide. How
much magnesium will burn in the
presence of 189 ml of oxygen?
The density of oxygen is 1.429
g/L.
2Mg + O2 → 2MgO
Stoichiometry Problems
189 ml O2
D = 1.429 g/L
1 mol O2 = 2 mol Mg
MM O2 = 32.00 g/ mol
MM Mg = 24.30 g/ mol
Stoichiometry Problems
189 ml O2 • 1.429 g • 1 L • 1 mol O2
1L
1000 ml 32.00 g O2
= 8.44 X 10-3 mol O2
8.44 X 10-3 mol O2 •
2 mol Mg
1 mol O2
= 16.88 X 10-3 mol Mg
16.88 X 10-3 mol Mg • 24.30 g Mg =
1 mol Mg
0.410 g Mg
Limiting Reactants
Available
Ingredients
• 4 slices of bread
• 1 jar of peanut butter
• 1/2 jar of jelly
Limiting
Reactant
• bread
Excess
Reactants
• peanut butter and jelly
Limiting Reactants
Limiting
Reactant
• used up in a reaction
• determines the amount of
product
Excess
Reactant
• added to ensure that the other
reactant is completely used up
• cheaper & easier to recycle
Limiting Reactants- Steps
1. Write a balanced equation.
2. Determine # of moles present for
each reactant.
3. Mole ratios but this time not for
what you are looking for, but for
reactants.
Limiting Reactants- Steps
4. Use the limiting reactant to solve
the problem as usual .
Limiting Reactants
Example problem page 286
CO + H2 → CH3OH
Not balanced!
CO+ 2H2 → CH3OH
1C
1C
1O
1O
4H
4H
BALANCED!!!!!!!!!!!
Limiting Reactants
Looking for kg of CH3OH produced
152.5 kg CO
MM CO = 28.01 g/mol
24.5 kg H2
MM H2 = 2.02 g/mol
1mol CO = 2mol H2
MM CH3OH = 32.05g/mol
1mol CO = 1mol CH3OH
2mol H2 = 1mol CH3OH
Limiting Reactants
Determine # of moles present for each
reactant.
152.5 kg CO x 1mol CO x 1000g =
28.01 g CO
1 kg
5.444 x 103 mol CO
24.5 kg H2 x
1mol
x 1000 g
2.02 g H2
1 kg
1.213 x 104 mol H2
=
Limiting Reactants
Mole ratios but this time not for product for
reactants
1 mol CO = 2 mol H2
1.213 x 104 mol H2 x 1mol CO =
2 mol H2
6.065 x 103 mol CO
Need 6.065 X 103 mol CO
Have 5.444 X 103 mol CO
Limiting Reactants
CO is the limiting reactant!!!!
Not enough to react with all of the H2
present.
Use the limiting reactant to solve the
problem
Limiting Reactants
5.444 X 103 mol CO  1 mol CH3OH 
1 mol CO
32.05 g CH3OH  1 kg
1 mol CH3OH
1000 g
= 174.5 kg CH3OH
Percent Yield
measured in lab
actual yield
% yield 
 100
theoretical yield
calculated on paper
Percent Yield
When
45.8 g of K2CO3 react
with excess HCl, 46.3 g of KCl
are formed. Calculate the
theoretical and % yields of KCl.
K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
?g
actual: 46.3 g
Percent Yield
K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
?g
actual: 46.3 g
Theoretical Yield:
45.8 g 1 mol 2 mol 74.55
K2CO3 K2CO3
KCl g KCl
= 49.4
138.21 g 1 mol 1 mol g KCl
K2CO3 K2CO3 KCl
Percent Yield
K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
49.4 g
actual: 46.3 g
Theoretical Yield = 49.4 g KCl
% Yield =
46.3 g
49.4 g
 100 = 93.7%