Chapter 12
Stoichiometry
Mr. Mole
Molar Mass of Compounds
• Molar mass (MM) of a compound - determined by
adding up the atomic masses
20
17
of each element
Cl
Ca
40.08
– Ex. Molar mass of CaCl2
35.45
– MM of Calcium = 40.08 g/mol
– MM of Chlorine = 35.45g g/mol x 2 = 70.9 g/mol
– Molar Mass of CaCl2=
40.08 g/mol Ca + 70.9 g/mol Cl
= 110.98 g/mol CaCl2
Atoms or
Molecules
Flowchart
Divide by 6.02 X 1023
Multiply by 6.02 X 1023
Moles
Divide by atomic/molar
mass from periodic
table
Multiply by
atomic/molar mass
from periodic table
Mass
(grams)
Practice
• Calculate the Molar Mass of calcium phosphide
– Formula = Ca3P2
– Masses elements:
Ca = 40.08 g/mol * 3
P = 30.97 g/mol * 2
– Molar Mass = 182.18 g/mol
Calculations
molar mass
Grams
Avogadro’s number
Moles
particles
22.4 L
Volume
Everything must go through Moles!!!
Chocolate Chip Cookies!!
1 cup butter
1 cup packed brown sugar
2 eggs
2 1/2 cups all-purpose flour
2 cups semisweet chocolate
chips
Makes 3 dozen
How many eggs are needed to make 3 dozen cookies? 2
How many eggs to make 9 dozen cookies?
6
How much brown sugar would I need if I had 4 eggs? 2
Cookies and Chemistry…Huh!?!?
• Just like chocolate chip
cookies have recipes,
chemists have recipes called
equations
• Instead of using cups and
teaspoons, we use
moles
• Lastly, instead of eggs,
butter, sugar, etc. we use
compounds
Chemistry Recipes
• Balanced reactions tell us how much
reactant will react to get a product
– like the cookie recipe
– Be sure you have a balanced equation before you start!
• Ex: 2 Na + Cl2  2 NaCl
• Reaction tells us by mixing 2 moles of sodium with 1
mole of chlorine we will get 2 moles of sodium
chloride
• What if we wanted 4 moles of NaCl? 10 moles?
50 moles?
Practice
• Write the balanced reaction for hydrogen gas reacting
with oxygen gas.
2 H2 + O 2  2 H2 O
– How many moles of reactants are needed? 2 H2, 1 O2
– What if we wanted 4 moles of water? 4 H2, 2 O2
– What if we had 3 moles of oxygen, how much
hydrogen would we need to react, and how much
water would we get? 6 H2, 6 H2O
– What if we had 50 moles of hydrogen, how much
oxygen would we need, and how much water
25 O2, 50 H2O
produced?
Stoichiometry is…
• Greek for “measuring elements”
Pronounced “stoy kee ah muh tree”
• Defined as: calculations of the
quantities in chemical reactions,
based on a balanced equation.
• There are 4 ways to interpret a
balanced chemical equation
#1. In terms of Particles
• An Element is made of atoms
• A Molecular compound (made of
only nonmetals) is made up of
molecules
• Ionic Compounds (made of a metal
and nonmetal parts) are made of
formula units
Example: 2H2 + O2 → 2H2O
• Two molecules of hydrogen and one
molecule of oxygen form two molecules of
water.
• Another example: 2Al2O3 Al + 3O2
2 formula units
Al2O3 form 4 atoms Al
and 3 molecules O2
Now read this: 2Na + 2H2O  2NaOH + H2
#2. In terms of Moles
• Coefficients tell us
how many moles of each substance
2Al2O3 Al + 3O2
–2 mol Al2O3, 4 mol Al, 3 mol O2
2Na + 2H2O  2NaOH + H2
• Remember: A balanced equation is a
Molar Ratio
#3. In terms of Mass
• The Law of Conservation of Mass applies
• We can check mass by using moles.
Be + 2F  BeF2
1 mole Be
9.01 g Be
1 mole Be
2 mole F
19.00 g F
1 mole F
= 9.01 g Be
+
= 38.00 g F
36.04 gg Be
H2 + 2F
O2
47.01
reactants
In terms of Mass (for products)
Be + 2F  BeF2
1 moles BeF2 47.01 g BeF2
= 47.01 g BeF2
1 mole BeF2
47.01 g Be + 2F = 47.01 g BeF2
reactant
=
product
Mass of reactants and products.
must be equal
#4. In terms of Volume
• At STP, 1 mol of any gas = 22.4 L
2H2 + O2  2H2O
(2 x 22.4 L H2) + (1 x 22.4 L O2)  (2 x 22.4 L H2O)
67.2 Liters of reactant ≠ 44.8 Liters of product!
**Mass and atoms
are ALWAYS conserved
however, molecules, formula units, moles, and
volumes will not necessarily be conserved!
Practice:
• Show that the following equation
follows the Law of Conservation of
Mass (show the atoms balance,
and the mass on both sides is
equal)
2Al2O3 Al + 3O2
Atoms: 4 Al and 6 O
=
4 Al and 6 O
Mass: Reactant: 312 g/mol
=
Products: 108 g/mol + 204 g/mol
Practice
1). Balance the equation and interpret it in terms of
atoms, moles and mass. Show that the law of
conservation is observed.
- N2 + 3 H2--> 2 NH3
Atoms: 2 N and 6 H
=
2 N and 6 H
Moles: 1 N2 and 3 H2
≠
2 NH3
Mass: Reactants: 28.02 g/mol + 6.06 g/mol =
Products: 34.08 g/mol
Section 12.2
Chemical Calculations
Mole to Mole conversions
2Al2O3 Al + 3O2
– each time we use 2 moles of Al2O3 we will also
make 3 moles of O2
2 moles Al2O3
3 mole O2
or
3 mole O2
2 moles Al2O3
This is how we can convert from mols to mols.
This is why we need balanced equations.
Mole to Mole conversions
• How many moles of O2 are
produced when 3.34 moles of
Al2O3 decompose?
2Al2O3 Al + 3O2
3.34 mol Al2O3
3 mol O2
2 mol Al2O3
= 5.01 mol O2
Conversion factor from balanced equation
If you know the amount of ANY chemical in the reaction,
you can find the amount of ALL the other chemicals!
Practice:
2C2H2 + 5O2  4CO2 + 2H2O
• If 3.84 moles of C2H2 are burned, how many moles of O2 are
needed?
3.84 moles C2H2 5 moles O2
9.6 mol O2
2 moles C2H2
•How many moles of C2H2 are needed to produce 8.95
mole of H2O?
8.95 moles H2O 2 moles C2H2
2 moles H2O
8.95 mol C2H2
Steps to Calculate
Stoichiometric Problems
1.
2.
3.
4.
Correctly balance the equation.
Convert the given amount into moles.
Set up mole ratios.
Use mole ratios to calculate moles of
desired chemical.
5. Convert moles back into final unit.
Mole-Mass Conversions
• Most of the time in chemistry, the amounts are given in
grams instead of moles
• We still use the mole ratio, but now we also use
molar mass to get to grams
• Example: How many grams of chlorine are required to react
with 5.00 moles of sodium to produce sodium chloride?
2 Na + Cl2  2 NaCl
5.00 moles Na 1 mol Cl2 70.90g Cl2
2 mol Na 1 mol Cl2
= 177g Cl2
Practice
• Calculate the mass in grams of Iodine required
to react completely with 0.50 moles of
aluminum.
Al + 3I  AlI3
0.50 mol Al
3 mol I
126.90 g I
1 mol Al
1 mol I
= 190.35 g I
Mass-Mole
• We can also start with mass and convert to moles
• We use molar mass and the mole ratio to get to
moles of the compound of interest
– Calculate the number of moles of ethane (C2H6) needed
to produce 10.0 g of water
– 2 C2H6 + 7 O2  4 CO2 + 6 H20
10.0 g H2O 1 mol H2O 2 mol C2H6 = 0.185 mol
18.0 g H2O 6 mol H20
C2H6
Practice
• Calculate how many moles of oxygen are
required to make 10.0 g of aluminum oxide
• 2Al + 3O  Al2O3
10.0 g Al2O3
1 mol Al2O3
101.96 g Al2O3
0.29 g O2
3 mol O2
1 mol Al2O3
Mass-Mass Problem:
6.50 grams of aluminum reacts with oxygen. How
many grams of aluminum oxide are formed?
4Al + 3O2
6.50 g Al
2Al2O3
1 mol Al
2 mol Al2O3 101.96 g Al2O3
26.98 g Al
4 mol Al
1 mol Al2O3
(6.50 x 1 x 2 x 101.96) ÷ (26.98 x 4 x 1) =
= ? g Al2O3
12.3 g Al2O3
are formed
Another example:
• If 10.1 g of Fe are added to a solution
of Copper (II) Sulfate, how many
grams of solid copper would form?
2Fe + 3CuSO4  Fe2(SO4)3 + 3Cu
10.1 g Fe
1 mol Fe
3 mol Cu
63.55 g Cu
55.85 g Fe
2 mol Fe
1 mol Cu
Answer = 17.2 g Cu
Volume-Volume Calculations:
• How many liters of CH4 at STP are required
to completely react with 17.5 L of O2 ?
CH4 + 2O2  CO2 + 2H2O
1 mol O2 1 mol CH4 22.4 L CH4
17.5 L O2
22.4 L O2 2 mol O2 1 mol CH4
= 8.75 L CH4
Stoichiometry Song - Mark Rosengarten
Section 12.3
Limiting Reagent & Percent Yield
• OBJECTIVES:
–Identify the limiting reagent in a
reaction.
“Limiting” Reagent
• If you are given one dozen loaves of bread, a
gallon of mustard, and three pieces of salami,
how many salami sandwiches can you make?
• The limiting reagent is the
reactant you run out of first.
• The excess reagent is the
one you have left over.
• The limiting reagent determines
how much product you can make
How do you find out which is limited?
• The chemical that makes the
least amount of product
is the “limiting reagent”.
• Limiting reagent problems will give
you 2 amounts of chemicals
• You must do two stoichiometry
problems; one for each reagent given.
• If 2 products are given, pick one and use it for both
calculations
• If 10.6 g of copper reacts with 3.83 g sulfur,
is the(copper (I)
how many grams of theCu
product
sulfide) will be formed? Limiting
Reagent,
2Cu + S  Cu
2S
since it
1 mol Cu2S 159.16 g Cu2S
1
mol
Cu
produced less
10.6 g Cu
Cu
63.55g Cu 2 mol
1 mol Cu2S
product.
= 13.3 g Cu2S
1
mol
S
3.83 g S
32.06g S
1 mol Cu2S 159.16 g Cu2S
1 mol S
1 mol Cu2S
= 19.0 g Cu2S
Finding the Amount of Excess
• By calculating the amount of the reactant
needed to completely react with the limiting
reactant, we can subtract that amount from
the given amount to find the amount of
excess.
• Can we find the amount of excess potassium
in the next problem?
Finding Excess Practice
• 15.0 g of potassium reacts with 15.0 g of iodine.
2 K + I2  2 KI
• We found that Iodine is the limiting reactant.
15.0 g I2
1 mol I2
2 mol K
39.1 g K
254 g I2
1 mol I2
1 mol K
= 4.62 g K
USED!
15.0 g K – 4.62 g K = 10.38 g K EXCESS
Given
amount
of excess
reactant
Amount of
excess reactant
actually used
Note that we started with the
limiting reactant! Once you
determine the LR, you should
only start with it!
Another example:
• If 10.3 g of aluminum are reacted with 51.7 g of CuSO4, how
much copper (grams) will be produced?
2Al + 3CuSO4 → 3Cu + Al2(SO4)3
10.3 g Al
1 mol Al
3 mol Cu 63.55 g Cu
= 36.39 g Cu
26.98 g Al 2 mol Al 1 mol Cu
51.7 g CuSO4 1 mol CuSO4
63.55 g Cu
3 mol Cu
159.62 g CuSO4 3 mol CuSO4 1 mol Cu
= 20.58 g Cu
CuSO4 limiting reactant
How much excess reactant do we
have from the last problem?
• CuSO4 limiting reactant
51.7 g CuSO4 1 mol CuSO4
26.98 g Al
2 mol Al
159.62 g CuSO4 3 mol CuSO4 1 mol Al
= 5.83 g Al
actually used
• Excess Al = 10.3 g – 5.83 g =
4.47 g Al excess
Limiting Reactant: Recap
1. You can recognize a limiting reactant problem
because there is MORE THAN ONE GIVEN AMOUNT.
2. Convert ALL of the reactants to the SAME product
(pick any product you choose.)
3. The lowest answer is the limiting reactant
4. The other reactant(s) are in EXCESS.
5. To find the amount of excess, subtract the amount
used from the given amount.
6. If you have to find more than one product, be sure to
start with the limiting reactant. You don’t have to
determine which is the LR over and over again!
The Concept of:
A little different type of yield than
you had in Driver’s Education
class.
What is Yield?
• Yield is the amount of product made in a
chemical reaction.
• There are three types:
1. Theoretical yield- what the balanced equation
tells should be made
2. Actual yield- what you actually get in the lab
when the chemicals are mixed
3. Percent yield = Actual
x 100%
Theoretical
Example:
• 6.78 g of copper is produced when 3.92 g of Al are
reacted with excess copper (II) sulfate.
2Al + 3 CuSO4  Al2(SO4)3 + 3Cu
• What is the actual yield?
= 6.78 g Cu
• What is the theoretical yield?
3.92 g Al 1 mol Al
3 mol Cu 63.55 mol Cu
= 13.8 g Cu
26.98 g Al 2 mol Al 1 mol Cu
What is the percent yield?
6.78 g Cu
13.8 g Cu
X 100
= 49.1 %
Details on Yield
• Percent yield tells us how “efficient” a
reaction is.
• Percent yield can not be bigger than
100 %.
• Theoretical yield will always be larger
than actual yield!
– Why? Due to impure reactants; competing side
reactions; loss of product in filtering or
transferring between containers; measuring