STOICHIOMETRY
Ch. 12
Page 352
WHAT IS “STOICHIOMETRY”?
A way of figuring out how much of a product can
be made from a given amount of reactant
 Based on the Law of Conservation of Mass – in a
chemical reaction, the mass of the reactants
equals the mass of the products.

WHAT DO THE COEFFICIENTS IN A
CHEMICAL REACTION REPRESENT?
1.
2.
The # of particles (atoms, formula units, molecules,
ions)
The # of moles of each particle
EX: 4Fe + 3O2  2Fe2O3
4 mol Fe
3 mol O2
2 mol Fe2O3
4 atoms Fe 3 molec. O2 2 formula units
Fe2O3
4FE + 3O2  2FE2O3
A balanced chemical equation ALSO tells us
indirectly about the masses of the reactants and
products (using the # moles of each substance and the
molar masses).
 How many grams of Fe are used in this reaction?
 How many grams of O2?
 How many grams of Fe2O3?
(Let’s do these on the board.)

4FE + 3O2  2FE2O3
Do our answers for the masses of the reactants
and products in the equation above uphold the
Law of Conservation of Mass?
 YES!

4Fe + 3O2  2Fe2O3
(223.4 g + 96.0 g) 
319.4 g
=
319.4 g
319.4 g
WHAT IS A “MOLE RATIO”?


A balanced equation not only tells us the
number of particles and number of moles. It
also indicates the relationships among all
substances involved. We use a mole ratio to
show these relationships.
A mole ratio is a ratio between the # of moles
of any two substances in a balanced chemical
equation.
MOLE RATIO:
THE
“BRIDGE”
MOLE RATIOS ARE THE KEY TO
CALCULATIONS BASED UPON A CHEMICAL
EQUATION!
 From a balanced chemical equation, if you know the
amount of one reactant, you can calculate the amount
of any other reactant in the equation and the
maximum amount of product you can obtain.
 The “mole ratios” are the bridges we use to get from
one kind of substance to another.

4FE + 3O2  2FE2O3
How do you know how many mole ratios are available
in an equation??? Just multiply the total number of
substances (n) by the next lower number (n-1).
 From this equation, there are 3 substances. Multiply 3
x 2 and you will get 6 available mole ratios.
 Using the equation above, let’s write all the available
mole ratios (on the board).

SEC. 2: USING STOICHIOMETRIC
CALCULATIONS

1.
2.
3.
What are the tools needed for stoichiometric
calculations?
A balanced chemical equation
Mole ratios
Mass-to-mole conversions
TYPES OF STOICHIOMETRIC
CALCULATIONS

1.
2.
3.
There are 3 types of stoichiometric calculations
we will discuss:
mole-to-mole conversions
mole-to-mass conversions
mass-to-mass conversions
For all of these conversions, we will use the
following flow-chart:
Grams of
known
(mass)
Divide by
Molar mass
Moles of
known
Use “Mole
Ratio”
Moles of
unknown
Multiply by
Molar mass
Grams of
unknown
(mass)
STEPS IN STOICHIOMETRIC
CALCULATIONS
1.
2.
3.
4.
Write a balanced chemical equation.
Determine the moles of the known using the
mass-mole conversion
Determine the moles of the unknown
substance by using a mole ratio
Determine the mass of the unknown using a
mole-mass conversion
COMPLETE THE FOLLOWING
PRACTICE PROBLEMS
Page 877: #5 – 10
 Page 379: #67 – 68
 Page 380: #72 -73

ANSWERS
Pg. 877
5.
0.600 mol HCl
6.
2.29 mol Cr2O3
7.
924 g CO2
8.
314 g H2SO4
9.
1.31 g CO2
10. 195.7 g aspirin
Pg. 379
67. (a.) K2CrO4 + Pb(NO3) 
PbCrO4 + 2KNO3
(b.) 80.8 g PbCrO4
68.
(a.) N2H2 + H2O2  N2 + 2H2O
(b.) 3.00 x 102 g N2H2
Pg. 380
72. (a.) Pb + PbO2 + 2H2SO4 
2PbSO4 + 2H2O
(b.) 73.2 g PbSO4
73. (a.) CO + 2H2  CH3OH
(b.) 5.67 g H2
12.3: LIMITING REACTANTS, PG. 364
Rarely in nature are reactants in a chemical
reaction present in the exact ratios specified by
the balanced equation.
 Generally, one or more reactants are in excess
and the reaction proceeds until all of one reactant
is used up.

When a chemical reaction is carried out in the
laboratory, the same principle applies.
 Usually, one or more reactants are in excess,
while one is limited.
 The amount of product depends upon the
reactant that is limited.

The limiting reactant limits the extent of the
reaction and, thereby, determines the amount of
product.
 A portion of the other reactants remains after the
reaction stops.
 These left-over reactants are called excess
reactants.

In section 2 of this chapter, we did calculations
based on having the reactants present in the
ratio described by the balanced chemical
equation.
 How can you calculate the amount of product
formed when one reactant limits the amount of
product and the other is in excess??

STEPS TO DETERMINE THE LIMITING
REACTANT
create a balanced equation for the
reaction
2. use stoichiometry to calculate how
much product is produced by each
reactant
1.
NOTE: It does not matter which product is
chosen, but the same product must be used for
both reactants so that the amounts can be
compared.
3. The reactant that produces the lesser
amount of product is the "limiting
reactant."
EXAMPLE LIMITING REACTANT CALCULATION

A 2.00 g sample of ammonia is mixed with 4.00 g of
oxygen. Which is the limiting reactant and how much
excess reactant remains after the reaction has stopped?
TO DETERMINE THE MASS OF THE
EXCESS REACTANT
After determining the limiting reactant, do a
mass of limiting reactantmass of excess
reactant calculation (grams of limiting
reactant  grams excess reactant)
(THIS IS HOW MUCH OF THE EXCESS
REACTANT ACTUALLY REACTS.)
2.
Subtract that answer from the amount
available in the reaction and you will have the
mass of excess reactant left over.
1.
FIRST, WE NEED TO CREATE A
BALANCED EQUATION FOR THE
REACTION:

4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)
NEXT WE CAN USE STOICHIOMETRY TO CALCULATE HOW
MUCH PRODUCT IS PRODUCED BY EACH
REACTANT.
NOTE: IT DOES NOT MATTER WHICH PRODUCT
IS CHOSEN, BUT THE SAME PRODUCT MUST BE USED FOR
BOTH REACTANTS SO THAT THE AMOUNTS CAN BE COMPARED.
The reactant that produces the lesser amount of product ,in this case
is the oxygen, which is thus the "limiting reactant."
NEXT, TO FIND THE AMOUNT OF EXCESS REACTANT, WE
MUST CALCULATE HOW MUCH OF THE NON-LIMITING
REACTANT (AMMONIA) ACTUALLY DID REACT WITH THE
LIMITING REACTANT (OXYGEN).
We're not finished
yet though.
1.70 g is the amount of ammonia that reacted,
not what is left over. To find the amount of
excess reactant remaining, subtract the amount
that reacted from the amount in the original
sample.
Determining the Limiting Reactant
• In the reaction below, 40.0 g of sodium
hydroxide (NaOH) reacts with 60.0 g of
sulfuric acid (H2SO4).
Additional Assessment Questions
Topic
16
Question 1
Balance the following equation. How many
moles of KClO3 are needed to produce 50
moles of O2?
Answer
12.4: PERCENT YIELD
In many calculations we have been practicing,
we have been asked to calculate the amount of
product that can be produced from a given
amount of reactant.
 The answer we obtained is called the
theoretical yield: the max. amount of product
that can be produced from a given amount of
reactant.

A chemical reaction rarely produces the
theoretical yield of product.
 When we conduct experiments, we determine
the actual yield: the amount of product
actually produced when the chemical reaction is
carefully carried out in an experiment.
 We measure efficiency by calculating percent
yield with the following formula:

% yield = __actual yield (from experiment)_ x 100
theoretical yield (from calculations)