Unit 08 – Moles and
Stoichiometry
I. Molar Conversions
A. What is the Mole?
A counting
number (like a dozen)
Avogadro’s
number (6.02  1023
particles) (SI unit)
1
mol = molar mass
A
large amount!!!!
A. What is the Mole?
1
mole of hockey pucks would
equal the mass of the moon!
1
mole of basketballs would
fill a bag the size of the earth!
1
mole of pennies would
cover the Earth 1/4 mile deep!
B. Molar Mass
o
Molar Mass- the mass of a mole of
any element or compound (in grams)
oRound to 2 decimal places
o Also called:
oGram Formula mass – sum of
the atomic masses of all the
atoms in a formula of a
compound
oFormula weight
B. Molar Mass Examples
water
• H2O
• (1.01g x 2) + 16.00g = 18.02 g
sodium
chloride
• NaCl
• 22.99g + 35.45g = 58.44 g
B. Molar Mass Examples
sodium
hydrogen carbonate
• NaHCO3
• 22.99g + 1.01g + 12.01g +
(16.00g x 3) = 84.01 g
sucrose
• C12H22O11
• (12.01g x12) + (1.01g x 22) +
(16.00g x11)= 342.34 g
C. Number of Particles in a Mole
1 mole = 6.02 × 10 23 representative particles
(also called Avogadro’s Number)
What is a representative particle?
How the substance normally exists:
1. Atom- rep. particle for most elements
2. Ions – if atom is charged
3. Molecule- rep. particle for covalent
compounds and diatomic molecules
“BrINCl HOF”
4. Formula unit- rep. particle for ionic
compounds
D. Volume of a Mole of Gas
The
Volume of a gas varies with a
change in temperature or pressure.
Measured at standard temperature
and pressure (STP)
0°C at 1 atmosphere (atm)
1 mole of any gas occupies a volume
of 22.4L
The Mole
Road
Map
E. Molar Conversion Examples
How
many moles of carbon
are in 26 g of carbon?
26 g
C
1 mol C
12.01 g C
= 2.2 mol C
E. Molar Conversion Examples
How
many molecules are in
2.50 moles of C12H22O11?
2.50
mol
6.02 
1023
= 1.51  1024
molecules molecules
1 mol
C12H22O11
E. Molar Conversion Examples
Find
the number of molecules
of 12.00 L of O2 gas at STP.
12.00 L
1 mol
22.4 L
6.02 x 1023
molecules
1 mol
= 3.225 x 1023 molecules
II. Stoichiometric
Calculations
A. 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
A. 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
B. Stoichiometry Steps
1. Write a balanced equation.
2. Identify known & unknown.
3. Convert known to moles (IF
NECESSARY) Line up conversion
factors.
4. Use Mole ratio – from equation
5. Convert moles to unknown unit
(IF NECESSARY)
6. Calculate and write units.
C. Stoichiometry Problems
Calculate
the number of grams of
NH3 produced by the reaction of
5.40 g of hydrogen with an
excess of nitrogen.
N2 + 3H2 → 2NH3
III. Stoichiometry in the
Real World
A. 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
A. 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
A. Limiting Reactants
1. Write a balanced equation.
2. For each reactant, calculate the
amount of product formed.
3. Smaller answer indicates:
• limiting reactant
• amount of product
A. Limiting Reactants
Using the following equation identify the
limiting reagent.
 How many moles of ammonia (NH3)
can be produced from the reaction of
28.2 L of nitrogen and 25.3 L of
hydrogen?
N2 + 3H2
28.2 L
25.3 L

2NH3
? mol
A. Limiting Reactants
N2 + 3H2
28.2 L
25.3 L
28.2 1 mol 2 mol
L N2 N2
NH3
22.4
L N2
1 mol
N2

2NH3
? mol
= 2.5 mol NH3
A. Limiting Reactants
N2 + 3H2
28.2 L

25.3 L
25.3 1 mol
L H2
H2
22.4 L
H2
2 mol
NH3
2NH3
? mol
= 0.753 mol
3 mol
NH3
H2
A. Limiting Reactants
N2: 2.5 mol NH3
H2: 0.753 mol NH3
Limiting reactant: H2
Excess reactant: N2
Product Formed: 0.753 mol NH3
Limiting Reactants
Mg + 2HCl → MgCl2 + H2
How many grams of
magnesium chloride are
produced from the reaction
of 2.08 mol of Mg and 2.08
mol of HCl?
B. Percent Yield
measured in lab
actual yield
% yield 
 100
theoretical yield
calculated on paper
B. 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
B. 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
B. 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%