Stoichiometry
Newport High School
Academic Chemistry
Mrs. Teates
Lesson 1 – Introduction to
Stoichiometry
 Lesson
Essential Questions:
• What is stoichiometry and how is it
used to describe reactions?
Vocabulary: stoichiometry, composition
stoichiometry, reaction stoichiometry,
mole, Avogadro’s number, molar
mass
Stoichiometry Definition
 Stoichiometry
• mass relationships between
substances in a chemical reaction
 Composition stoichiometry deals with
the mass relationships of elements in
compounds.
 Reaction stoichiometry involves the
mass relationships between reactants
and products in a chemical reaction.
What is the Mole?

A counting number (like a dozen)
• Can be used to measure anything (DVDs,
cars, eggs, molecules)
1
mol = 6.02  1023 items
A
large
amount!!!!
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!
Molar Mass
Mass
of 1 mole of an element or
compound.
Atomic
mass tells the...
• atomic mass units per atom
(amu)
• grams per mole (g/mol)
Round
to 2 decimal places
Molar Mass Examples
carbon
aluminum
12.01 g/mol
26.98 g/mol
65.39 g/mol
zinc
Molar Mass Examples
water
 H2O
 2(1.01) + 16.00 = 18.02 g/mol
sodium
chloride
 NaCl
 22.99 + 35.45 = 58.44 g/mol
Molar Mass Examples
sodium
bicarbonate
 NaHCO3
 22.99 + 1.01 + 12.01 + 3(16.00)
= 84.01 g/mol
sucrose
 C12H22O11
 12(12.01) + 22(1.01) + 11(16.00)
= 342.34 g/mol
Molar Conversions
molar
mass
6.02  1023
MASS
NUMBER
MOLES
IN
GRAMS
OF
PARTICLES
(g/mol)
(particles/mol)
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
Molar Conversion Examples
How
many molecules are in
2.50 moles of C12H22O11?
2.50
mol
6.02  1023
molecules
1 mol
= 1.51  1024
molecules
C12H22O11
Molar Conversion Examples
the mass of 2.1  1024
molecules of NaHCO3.
Find
2.1  1024
1 mol
84.01 g
molecule
s
6.02  1023 1 mol
molecules
= 290 g NaHCO3
Percentage Composition
Calculate
because it is useful to
know the percentage by mass of
an element in a compound
m ass_ of _ elem ent_ in _ sam ple_ of _ com pound
*100  percentage_ com position
m ass_ of _ sam ple_ of _ com pound
or
m ass_ of _ elem ent_ in _ 1m ol_ of _ com pound
*100  percentage_ com position
m olar_ m ass_ of _ com pound
Sample Problems
Sample
Problem J on page 243
Practice problem #1
Mole Ratio
A
mole ratio is a conversion
factor that relates the amounts in
moles of any two substances
involved in a chemical reaction
Example:
2Al2O3(l) → 4Al(s) +
,
,
3O2(g)
Mole Ratios: 2 mol Al2O3 2 mol Al2O3
4 mol Al
4 mol Al
3 mol O2
3 mol O2
Converting Between Amounts
in Moles
Stoichiometry Calculations
Lesson 2 – Ideal
Stoichiometric Calculations
Lesson
Essential Questions:
• How are calculations used to
describe chemical reactions?
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
Stoichiometry Steps
1. Write a balanced equation.
2. Identify known & unknown.
3. Line up conversion factors.
• Mole ratio - moles
molesmoles
moles
• Molar mass - moles  grams
• Molarity moles  liters soln
• Molar volume - moles  liters gas
Core step in all stoichiometry problems!!
4. Check answer.
Stoichiometry Calculations
Conversion of Quantities in Moles
Conversions of Quantities in Moles
Sample Problem A
In a spacecraft, the carbon dioxide exhaled by astronauts
can be removed by its reaction with lithium hydroxide,
LiOH, according to the following chemical equation.
CO2(g) + 2LiOH(s) → Li2CO3(s) + H2O(l)
How many moles of lithium hydroxide are required to react
with 20 mol CO2, the average amount exhaled by a person
each day?
Conversions of Quantities in Moles
Sample Problem A Solution
CO2(g) + 2LiOH(s) →
Li2CO3(s) + H2O(l)
Given: amount of CO2 = 20 mol
Unknown: amount of LiOH (mol)
mol LiOH
Solution: mol CO2 
 mol LiOH
mol CO2
2 mol LiOH
20 mol CO2 
 40 mol LiOH
1 mol CO2
Conversions of Amounts in Moles
to Mass
Conversions of Amounts in Moles to Mass
Sample Problem B
In photosynthesis, plants use energy
from the sun to produce glucose,
C6H12O6, and oxygen from the reaction
of carbon dioxide and water.
What mass, in grams, of glucose is
produced when 3.00 mol of water react
with carbon dioxide?
Conversions of Amounts in Moles to Mass
Sample Problem B Solution
Given: amount of H2O = 3.00 mol
Unknown: mass of C6H12O6 produced (g)
Solution:
Balanced Equation: 6CO2(g) + 6H2O(l) → C6H12O6(s) + 6O2(g)
mol ratio
mol H2O 
3.00 mol H2O 
mol C6H12O6
mol H2O
molar mass factor

g C6H12O6
mol C6H12O6
 g C6H12O6
1 mol C6H12O6
180.18 g C6H12O6

=
6 mol H2O
1 mol C6H12O6
90.1 g C6H12O6
Conversions of Mass to
Amounts in Moles
Conversions of Mass to Amounts in Moles
Sample Problem D
The first step in the industrial manufacture of nitric
acid is the catalytic oxidation of ammonia.
NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced)
The reaction is run using 824 g NH3 and excess
oxygen.
a. How many moles of NO are formed?
b. How many moles of H2O are formed?
Conversions of Mass to Amounts in Moles
Sample Problem D Solution
Given: mass of NH3 = 824 g
Unknown: a. amount of NO produced (mol)
b. amount of H2O produced (mol)
Solution:
Balanced Equation: 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
molar mass factor
a.
b.
mol NH3
g NH3 
g NH3
g NH3 
mol ratio
mol NO

mol NH3
 mol NO
mol NH3
mol H2O

g NH3
mol NH3
 mol H2O
Conversions of Mass to Amounts in Moles
Sample Problem D Solution
molar mass factor
mol ratio
a. 824 g NH3 
1 mol NH3
17.04 g NH3

4 mol NO
 48.4 mol NO
4 mol NH3
b. 824 g NH3 
1 mol NH3
17.04 g NH3

6 mol H2O
4 mol NH3
 72.5 mol H2O
Mass-Mass to Calculations
Mass-Mass to Calculations
Sample Problem E
Tin(II) fluoride, SnF2, is used in some
toothpastes. It is made by the reaction of tin
with hydrogen fluoride according to the
following equation.
Sn(s) + 2HF(g) → SnF2(s) + H2(g)
How many grams of SnF2 are produced from
the reaction of 30.00 g HF with Sn?
Mass-Mass to Calculations
Sample Problem E Solution
Given: amount of HF = 30.00 g
Unknown: mass of SnF2 produced (g)
Solution:
molar mass factor
mol ratio
molar mass factor
mol SnF2
g SnF2
mol HF
g HF 


 g SnF2
g HF
mol HF
mol SnF2
1 mol SnF2
156.71 g SnF2
1 mol HF
g HF 


20.01 g HF
2 mol HF
1 mol SnF2
= 117.5 g SnF2
More Practice for Homework
Page
320 #4-16
Lesson 3 – Limiting Reactants
and Percent Yield

Lesson Essential Questions:
• How does the limiting reactant affect the
percentage yield and actual yield in a
chemical reaction?
Vocabulary: limiting reactant, excess
reactant, theoretical yield, actual yield,
percent yield
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
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
Limiting Reactants
1. Write a balanced equation.
2. Write your known and unknown.
3. For each reactant, calculate the amount of
product formed.
4. Smaller answer indicates:
• limiting reactant
• amount of product
Limiting Reactant Practice
Sample
Problem F on page 313.
• Practice problem #1 on page
313.
Sample Problem G on pages
314-315.
• Practice problems #1 & 2 on
page 315.
Homework pg. 321 #22-25
Percent Yield
measured in lab
actual yield
% yield 
 100
theoretical yield
calculated on paper
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
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
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%
Percent Yield Practice
Sample
problem H on page 317.
• Practice problem #1 & 2 on
page 318.
Homework
Page
321 #22-30, 31-36, 38
for the test – Worksheet
titled Stoichiometry Review
Problems
Review
Works Cited
Modern
Chemistry Textbook
www.nclark.net
http://mrsj.exofire.net/chem/