Stoichiometry
Using the Balanced Equation
What is STOICHIOMETRY?
Stoichiometry is the study of
quantitative relationships
between amounts of reactants
used and products formed by a
chemical reaction.
What does the balanced equation
really mean?
2H2 + O2  2H2O
Particles: 2 molecules H2 + 1 molecule O2  2 molecules H2O
Moles:
2 moles H2 + 1 mole O2  2 moles H2O
Which is more useful?
moles
What does the balanced equation
really mean?
2H2 + O2  2H2O
Mass: ____gH
4.0 2 + ____gO
32.0 2  ____gH
36.0 2O
What scientific law does this illustrate?
Law of Conservation of Mass
2H2 + O2  2H2O
Like a recipe, the balanced equation tells you
in what ratios the ingredients must be mixed.
Therefore, for every 2 moles of H2 you will
need 1 mole of O2.
Mathematically, this can be written
2 moles H2 = 1 mole O2
What can you do with equalities?
Dimensional Analysis
2H2 + O2  2H2O
2 moles H2 = 1 mole O2
This type of equality derived from a balanced
equation can be expressed as a mole ratio.
The mole ratio is a ratio between the numbers
of moles of any two substances in a
balanced chemical equation.
2H2 + O2  2H2O
How many conversion factors (called
mole ratios) can be written from this 1
equation?
2 moles H2 = 1 mole O2
2 moles H 2
1 mole O2
or
1 mole O2
2 moles H 2
2 moles H2 = 2 mole H2O
2 mole H 2O
2 moles H 2
or
2 moles H 2O
2 moles H 2
1 mole O2 = 2 moles H2O
2 moles H 2O
1 moles O2
or
2 moles H 2O
1 mole O2
2H2 + O2  2H2O
How many moles of O2 are needed to react
with 6.4 mol H2?
6.4 mol H2
1
2
mol O2
mol H2
=
3.2
mol O2
2H2 + O2  2H2O
How many moles of H2O are produced from
6.4 mol H2?
6.4 mol H2
2
2
mol H2O
mol H2
=
6.4
mol H2O
Multi-step Stoichiometry
EVERY stoichiometry problem involves moving
from 1 substance to another!
How many molecules of H2 does it take to make 647 g of H2O?
2H2 + O2  2H2O
? molecules
647 g
The only way to move from 1 substance to another is the mole ratio!
Multi-step Stoichiometry
You have three tools for stoichiometry problems!
(so far!)
Tools of Stoichiometry
1. Mole Ratio
2. Molar Mass
3. Avogadro's #
balanced equation coefficients
periodic table mass = 1 mol
6.02  1023 particles = 1 mol particles
moles A  moles B
mass A (g)  1 mole A
particles A  1 mole A
Every stoichiometry problem uses
the mole ratio!!!
Multi-step Stoichiometry
To answer the original question, you need to
use all 3 tools.
How many molecules of H2 does it take to make 647 g of H2O?
2H2 + O2  2H2O
? molecules
647 g
This problem not only involves changing substances
(mole ratio), but also mass (molar mass) and molecules
(Avogadro’s number).
2H2 + O2  2H2O
? molecules
647 g
At worst, stoichiometry problems will be 3
steps, so start be drawing all 3 steps!
Then put in your given and target.
647 g H2O
= ?
molecules
H2
2H2 + O2  2H2O
? molecules
647 g
Next, EVERY problem must have a mole ratio
step, so put it in the middle!
647 g H2O
2 mol H2
= ?
2 mol H2O
molecules
H2
• You don’t want H2O, so put moles H2O in the bottom of
the middle step (remember you have to be in moles to
move from 1 substance to another).
• You want to change to H2, so put moles of H2 on top (mole
ratio is moles to moles).
• The numbers come from the balanced equation!
2H2 + O2  2H2O
? molecules
647 g
Next, decide if you need a first and last step.
647 g H2O 1 mol H2O 2 mol H2
18.0 g H2O 2 mol H2O
= ?
molecules
H2
Since we need to be in moles of H2O for the
middle step, we need to change g H2O to
moles of H2O using the molar mass.
2H2 + O2  2H2O
? molecules
647 g
Since, we need molecules and the middle step
gets us to moles H2, we need to use
Avogadro’s number.
647 g H2O 1 mol H2O 2 mol H2
18.0 g H2O 2 mol H2O
6.021023 molecule H2
1 mol H2
Answer = 2.161025 molecules H2
= ?
molecules
H2
Multi-step Stoichiometry
Every problem is started the same way.
1. Draw 3 steps and put in your given and
target.
2. Put the mole ratio step in the middle.
3. Decide if your need the first step (you
need it if you don’t start in moles).
4. Decide if you need the last step (you need
it if you don’t want to end in moles).
Another Example
How many grams of O2 are needed to use up
13.6 moles of H2?
2H2 + O2  2H2O
13.6 mol
13.6 mol H2
?g
1 mol O2
2 mol H2
Don’t need this step!
32.0 g O2
1 mol O2
= 218 g O2
Molar Volume
(4th stoichiometry tool)
• Molar Volume
– The volume occupied by 1 mole of any gas at
STP (standard temperature (0°C) and pressure
(1 atm)) is 22.4 L!
• 22.4 L = 1 mole
Tools of Stoichiometry
1.
2.
3.
4.
5.
Mole Ratio
Molar Mass
Avogadro's #
Molar Volume (at STP)
Formula
balanced equation coefficients
periodic table mass = 1 mol
6.02  1023 particles = 1 mol particles
22.4 L = 1 mol
from formula
moles A  moles B
mass A (g)  1 mole A
particles A  1 mole A
mol A  liters A
atoms A  molecule/fu
How many moles of water can be
made from 3.4 L of O2 at STP?
2H2(g) + O2(g)  2H2O(g)
3.4 L O2
1 mol O2
22.4 L O2
2 mol H2O
1 mol O2
= .30 mol H2O
Limiting Reactants
Cake Recipe
2 cups flour
1 cup butter
1tsp. salt
1/2 cup sugar
If I have 6 cups of flour, 4 cups of butter, 112
tsp. of salt, and 429 cups of sugar, how many
cakes can I make?
Limiting Reactants
Cake Recipe
You Have in the Cupboard
2 cups flour
1 cup butter
1tsp. Salt
1/2 cup sugar
6 cups flour
6 cups flour
4 cup butter
112 tsp. salt
429 cups sugar
1 cakes
= 3 cakes
2 cups flour
4 cups butter
1 cakes
= 4 cakes
1 cups butter
Limiting Reactants
Cake Recipe
You Have in the Cupboard
2 cups flour
1 cup butter
1tsp. Salt
1/2 cup sugar
112 tsp. salt
6 cups flour
4 cup butter
112 tsp. salt
429 cups sugar
1 cakes
= 112 cakes
1 tsp. salt
429 cups sugar
1 cakes
= 858 cakes
.5 cups sugar
Limiting Reactants
Flour controls or limits how many cakes we
can make because after we make 3 cakes we
run out of flour.
It also controls how much of the other
ingredients we need!
Limiting Reactants
How much butter is left over?
First find how much was used and then subtract
how much is left.
FLOUR controls how much we use.
6 cups flour
1 cup butter
= 3 cups butter
2 cups flour
4 cups butter – 3 cups butter = 1 cup butter left
Limiting Reactants
Limiting reactant - the reactant that makes
the least. It controls everything about the
reaction because when it is used up the
reaction stops.
Chemical reactions work the same way!!!!
Limiting Reactants
2H2 + O2  2H2O
2.50 g
3.40 g
How much water can be made?
2.50 g H2
3.40 g O2
1 mol H2
2 mol H2O
18.0 g H2O
1 mol O2
2 mol H2O
18.0 g H2O
32.0 g O2
1 mol O2
1 mol H2O
= 23
You must find which reactant controls
2.0 g H2
2 mol H2
1 mol H2O
(limiting reactant) the reaction.
g H2O
= 3.83 g H2O
O2 is therefore the limiting reactant and 3.83 g of water is produced!
Limiting Reactants
2H2 + O2  2H2O
2.50 g
3.40 g
How much H2 is left over?
1. Find out how much H2 was used up by
the 3.40 g O2
2. Then subtract that amount from 2.50 g
to find out how much H2 is left
Limiting Reactants
2H2 + O2  2H2O
2.50 g
3.40 g
How much H2 is left over?
3.40 g O2
1 mol O2
2 mol H2
2.0
g H2
32.0 g O2
1 mol O2
1 mol H2
2.50 g
.43 g
2.07 g H2 left over
= .43 g H2
Percent Yield
When we calculated that 3.83 g of H2O could be made in
the previous sample problem, we didn’t actually
perform the reaction and measure the resulting water.
We made theoretical prediction of how much could be
made.
Actual Yield – do the reaction to find the yield
Theoretical Yield – do the math to predict the yield
Percent Yield
When doing stoichiometric calculations, you
are calculating what is known as a
theoretical yield - the maximum amount that
could be produced during the reaction.
Under ideal conditions, this is exactly what
will happen.
In reality, however, the actual yield will be
less than what is predicted.
Percent Yield
actual yield
percent yield 
100
theoretical yield