Chapter 11
Stoichiometry
Chocolate Chip Cookie Recipe
Makes 42 cookies
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½ c butter
½ c brown sugar
½ c granulated sugar
1 egg
½ tsp vanilla
1 cup flour
½ tsp salt
½ tsp baking soda
½ c chopped walnuts
½ c chocolate chips
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What is the product
of this recipe?
 How would you
adjust the
ingredients in the
recipe to make 7
dozen cookies?
 What would you do
if you only had ¼ c
butter?
Stoichiometry
Mass-mass relationships
 Quantitative study of chemical reactions

Balanced Equations give us:
the relationship between the reactants
and the products
 the ratios in which they combine
 The coefficients represent the number
of moles
 2 H2 + 1 O2  2 H20
 2 moles of hydrogen react w/ 1 mole of
oxygen to form 2 moles of water

Law of Conservation of Mass

Basis behind stoichiometry
 Mass of reactants= mass of product
 Ex. 4Fe + 3O2  2 Fe2O3
– 4 moles of iron react w/ 3 moles of oxygen to
form 2 moles of rust
– Convert moles to grams
4 moles Fe
55.8 g Fe
= 223.4
g Fe
1mole Fe
3 moles O2
Total mass
of reactants:
319.4 g
32.0 g O2
1 mole O2
= 96.0
g O2
2 moles Fe2O3
158.7 g Fe2O3
1 mole Fe2O3
= 319.4 g Fe2O3
(mass of products)
Mass of reactants= mass of products
Trail Mix
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In order to make trail mix, you need raisins, peanuts,
and M&M's. What makes it good trail mix is having
the right ratios! In the bags there are 1 dozen raisins
(R), 2 dozen peanuts (P), and 3 dozen M&M's (MM).
This is a perfect combination to make 1 package of
trail mix (TM).
If we were to write an equation to represent this
recipe it would look like this:
1 dozen raisins + 2 dozen peanuts + 3 dozen M&M's
= 1 trail mix
or 1 R + 2 P + 3 MM  1 TM
1 R + 2 P + 3 MM1 TM
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With this in mind, what if you only had ½ the contents of your
bags…
How many dozen peanuts would you need to mix them with 1/2
dozen raisins?
By looking at the equation, you can come up with a conversion
of 2 dozen peanuts for every 1 dozen raisins.
SO…
½ dozen raisins x 2 dozen peanuts = 1 dozen peanuts needed
1 dozen raisins
(or ½ your bag!)
(Just like halving a recipe!)
1 R + 2 P + 3 MM1 TM
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How many dozen M&M's would you
need to combine them with 6 dozen
peanuts?
6 dozen peanut
3 dozen MM
2 dozen
peanuts
9 dozen m &m’s
1 R + 2 P + 3 MM1 TM
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How many dozen raisins would you
need to mix with 4 dozen peanuts?
4 dozen peanut
1 dozen raisin
2 dozen peanut
= 2 dozen raisins
1 R + 2 P + 3 MM1 TM
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How many dozen peanuts would you
need to mix them with 18
M&M's?(careful!)
18 mm’s
1 dozen mm
12 mm’s
2 dozen
peanuts
3 dozen mm’s
1 dozen peanuts
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Chemists use ratios such as these ALL
the time in the laboratory. Instead of
dozens, they use moles – something
we've already looked at. Just as in the
equation above, the coefficients in front
of the compounds provide the "mole
ratios" the chemists need.
Mole Ratios
Ratio between the numbers of moles of
any 2 substances in a balanced
equation
 You can have many different ratio
scenarios for each equation
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2 Al + 3 Cl2  2 AlCl3
Ratio of Al to Cl
2 moles Al
3 moles Cl
Ratio of Cl to Al
3 moles Cl
2 moles Al
Ratio of Cl to product
3 moles Cl
2 moles AlCl3
Ratio of product to Cl
2 moles AlCl3
3 moles Cl
Ratio of Al to product
2 moles Al
2 moles AlCl3
Ratio of product to Al
2 moles AlCl3
2 moles Al
Stoichiometry Answers these ?’s:
How much of 1 reactant is needed to
combine w/ a given amount of another
reactant
 How much product will be produced w/
a given amount of reactant
 How much reactant is needed to
produce a given amount of product
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Steps in Stoichiometry
1.
2.
3.
Write the balanced equation
Write the given information under the
equation (molar mass, # of moles,
grams of known)
Convert grams of known substance to
moles using molar mass
4.
Determine the mole ratio using the
coefficients from balanced equation
moles of unknown
moles of known
5. Convert the moles of unknown to
grams using molar mass
Known mass g
(from problem)
1 mole known
g known
unknown moles
known moles
(molar mass known) ( coef. from eqn.)
(molar mass
unknown)
g unknown
1 mole unknown
How many grams of AlCl3 are produced if
3.0g of Cl2 react with excess aluminum?
1.) Balanced Equation:
2 Al
2.) Write in
known info
+
3 Cl2 
2 AlCl3
3.0 g
?g
3 mole Cl2
2 moles AlCl3
71.0 g/mole
133.5 g/mole
3. Convert known grams to moles, put in
mole ratio (from coefficients) and
convert unknown moles to grams.
3.0g Cl2
(Mass
known)
1 mole Cl2
2 moles AlCl3 133.5 g AlCl3
71.0 g Cl2
3 moles Cl2
(Molar mass
known)
(Mole
Ratio)
1 mole AlCl3
(molar mass
unknown)
3. Multiply the top line and divide by the
bottom (making sure to use
parentheses correctly)
(3.0 x 2 x 133.5)
(71.0 x 3)
= 3.8 g of AlCl3 formed
Stoichiometry of Shuttle Launch
NASA TV (Discovery Last Launch)
 You Tube (STS 119)
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Percent Yield
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Quantities found in stoichiometry are
theoretical or predicted amounts (the amount
you are supposed to get if all the conditions
are perfect)
 During actual experimentation, you will come
up w/ less than expected
 % Yield = actual amt (experiment) x 100
Theoretical amt (calculated)
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You calculated that you should get
82.2g of NaCl in a reaction. When you
performed the experiment you produced
only 30.7 g NaCl. What is your % yield?
% Yield = 30.7 x 100
82.2
37.3%
Limiting Reactant (reagent)
Reactant that is totally consumed during
a reaction
 Limits the extent of a reaction
 Determines the amount of product that
can be formed
 Once it is gone, the reaction stops
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Excess Reactant (reagent)
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Reactant that remains after the reaction
stops
Limiting Reactant Problems

+
8 car
bodies
+
48 tires

8 cars
+
16 excess
tires
Steps:
1. Balance equation
2. Using known amounts of each
reactant, solve for mass of product (do
2 separate stoichiometry problems)
3. The one that produces the least
amount of product is the limiting
reactant.
75 g of CaO react with 30.0 g HCl to
produce calcium chloride and water.
What is the limiting reactant? How much
water is formed?
 Balance equation and write in known info
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CaO
75 g
1 mole
56.1 g/mole
+
2HCl 
CaCl2 +
H20
30.0 g
?g
2 mole
1 mole
36.5 g/mole
18.0 g/mole
Do
2 stoichiometry problems, w/ each
reactant as a known and the unknown being
water
75 g CaO
1 mole
CaO
1 mole H2O
18.0 g H2O
56.1 g CaO
1 mole CaO
1 mole H2O
=24 g H2O
Do
2 stoichiometry problems, w/ each
reactant as a known and the unknown being
water
30.0 g HCl
1 mole HCl
1 mole H2O
18.0 g H2O
36.5 g HCl
2 mole HCl
1 mole H2O
7.40 g H2O
HCl is the limiting reactant, producing
7.40 g of water
 CaO is in excess (meaning some will be
left over after the reaction)
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Finding mass of excess reactant that
remains:
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One can find the mass of excess
reactants that will remain by doing
another stoichiometry problem using
your limiting reactant as your known
and the excess reactant as your
unknown
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Mass excess starting- mass excess used
(stoich. Calc)= mass of left over excess react.
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Using HCl as the known and CaO as
the unknown, calc. the amt. Of CaO
used.
30.0 g HCl
1 mole HCl
36.5 g HCl
1 mole CaO
2 mole HCl
56.1 g CaO
1 mole CaO
=23.1 CaO used
75 g CaO – 23.1g used = 51.9 g leftover
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N2 + 3H2  2 NH3
If you have 100.0 g of N2 and 25.0 g of H2,
which is your limiting reactant in the reaction?
 How many grams of NH3 can be produced
from that limiting reactant?
 Calculate the mass of excess reactant that
remains after the reaction is complete.
Ch. 11 Test Review (stoich song)
Vocabulary
– stoichiometry
– Limiting reactant
– Excess reactant
– Mole ratio
– Actual yield (experimental)
– Theoretical yield (calculated)
– % yield
Calculations
– mole ratios
– Law of conservation of mass
– Stoichiometry
– Limiting reactant problem
– % yield
Essay
Practical application of stoichiometry
4HCl + O2  2 H2O + 2 Cl2
Interpret it in terms of moles
 What is the mole ratio between water
and oxygen?
 Show that mass is conserved
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_ Cu + _ AgNO3  _ Cu(NO3)2 + _ Ag
Balance the equation
 How many grams of copper are needed
to react with 12.0 g of AgNO3?
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_ Al + _ Cl2  __AlCl3
Balance
 If you begin with 3.2 g of aluminum and
5.4g of chlorine what is the limiting
reactant?
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__Mg+ _ HCl  _MgCl2 + _ H2
Balance
 If you reacted 10.0g of magnesium with
excess hydrochloric acid. How many
grams of MgCl2 will be formed?
 After actually reacting this you formed
29.5 g of MgCl2, what is the percent
yield?
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