Stoichiometry
Applying equations to calculating
quantities
Learning objectives
 Use chemical equations to predict quantity
of one substance from given quantity of
another
 Determine percentage yield
 Identify limiting reactant
The chemical equation
aA + bB = cC + dD
Reactant
side
coefficient
ELEMENT or
COMPOUND
Product
side
The Law of Conservation of Matter states that matter
is neither created nor destroyed
All atoms on left must be same as those on right
Working with equations:
STOICHIOMETRY
 Predict how much product is obtained from
given amount of reactant
 Predict how much reactant is needed to give
required amount of product
 Predict how much of one reactant is
required to give optimum result with given
amount of another reactant
Stoichiometry with equations:
The roadmap



Equations are in moles, but we measure in grams
Three conversions required:
A is given substance (reactant or product); B is target
substance (reactant or product)
1. Must convert grams A to moles A using molar mass
2. Use coefficients in equation to get moles B from moles A
3. Convert moles B to grams B using molar mass
Mass A
Mass/molar mass
Moles A
Mole:mole ratio
Moles B
Moles x molar
mass
Mass B
Remember the mole triangle
Mole:mole ratio: the central step
 Tells us molar ratio of substances in
balanced chemical equation
xX + yY = cC + dD
 Mole:mole ratio (target:given) B:A = b/a
Mass A
Mass/molar mass
Moles A
Mole:mole ratio
Moles B
Moles x molar
mass
Mass B
Mole:mole ratio examples
Types of problems:
Moles A → moles B
(A is given; B is target)
xX + yY = cC + dD




Single step
Require mole:mole ratio:
Always target/given (B/A)
In balanced equation a mol A ≡ b mol B
Mass A
Mass/molar mass
Moles A
Mole:mole ratio
Moles B
moles B
=
moles A
Moles x molar
mass
b mol B
a mol A
Mass B
Mole:mole ratio problems
 CH4 + 2O2 = CO2 + 2H2O
 How many moles of H2O are produced from 5 moles of CH4?
 Identify given substance: A = CH4
 Identify target substance: B = H2O
 Mole:mole ratio is

moles B
moles A
=
b mol B
a mol A
 Identify coefficients: CH4 a = 1; H2O b = 2
moles H 2O 2 mol H 2O

moles CH 4 1 mol CH 4
Moles
A
→
mass
B
xX + yY = cC + dD
Two steps on road map
1. Convert moles A → moles B:
Mole:mole ratio (target/given):
moles B
moles A
2.
=
b mol B
a mol A
Convert moles B → mass B using molar mass B
mass B (g) = moles B (mol) x molar mass B (g/mol)
Mass A
Mass/molar mass
Moles A
Mole:mole ratio
Moles B
Moles x molar
mass
Mass B
Mole:mass/mass:mole problems
CH4 + 2O2 = CO2 + 2H2O
 What mass of H2O is produced from burning 5 moles CH4?
 How many moles of CO2 is produced when 64 g O2 is consumed?
Mass A → mass B
xX + yY = cC + dD
All three steps
1.
Mass A → moles A using molar mass A
moles A (mol) =
2.
3.
mass A (g)
molar mass A (g/mol)
Moles A → moles B using mole:mole ratio moles B
Moles B → mass B using molar mass B
moles A
=
b mol B
a mol A
mass B (g) = moles B (mol) x molar mass B (g/mol)
Mass A
Mass/molar mass
Moles A
Mole:mole ratio
Moles B
Moles x molar
mass
Mass B
2 NaOH + Cl2 = NaOCl + NaCl + H2O
 What mass NaOH required to react with 142 g Cl2?
 Molar mass Cl2 = 35.5 x 2 = 71.0 g/mol
 Molar mass NaOH = 23.00 + 16.00 + 1.01 = 40.01 g/mol
Summary of stoichiometry problems

Maximum of three conversions required
1. Must convert grams A to moles A using molar mass
2. Use coefficients in equation to get moles B from
moles A
3. Convert moles B to grams B using molar mass

Maximum of three pieces of information
required
1. Molar mass of given substance (maybe)
2. Molar mass of target substance (maybe)
3. Balanced chemical equation (always)
Work this example
 CH4 + 2O2 = CO2 + 2H2O
 What mass of CO2 is produced by the
complete combustion of 16 g of CH4
 Atomic weight H = 1, C = 12, O = 16
44 g
 Do stoichiometry exercises
Reaction Yield
 The actual yield from a chemical reaction is
normally less than predicted from the
stoichiometry.
 Incomplete reaction
 Product lost in recovery
 Competing side reactions
 Percent yield is:
actual yield
% yield=
theoretical yield
Worked example
 Actual yield of product is 32.8 g after reaction of 26.3 g of C4H8 with
excess CH3OH to give C5H12O.
 What is theoretical yield? Use stoichiometry to get mass of product:
convert mass  moles  moles  mass
 Theoretical yield = 41.4 g
 Percent yield = 32.8/41.4 x 100 %
Mass A
Mass/molar mass
Moles A
Mole:mole ratio
Moles B
Moles x molar
mass
Mass B
CH4 + 2O2 = CO2 + 2H2O
 What is percent yield if 40 g of CO2 is
produced by the complete combustion of 16
g of CH4?
 First calculate theoretical yield as previously
 Theoretical yield CO2 = 44 g
 Actual yield CO2 = 40 g
actual yield
% yield=
theoretical yield
 Percent yield = 40 g/44 g x 100% = 91%
Limiting reactant
 Balanced equations involve precisely the
right amounts of each reactant. None is left
over
 Real-life situations usually involve an excess
of one reactant
 Burning CH4 in excess O2
 Reacting Zn with excess HCl
Identifying the limiting reactant
 In balanced equations
each reactant will give
the same amount of
products
 When one reactant is
limiting (deficient) the
amount of product will
be determined by its
amount
Identifying the limiting reactant
 Which reactant gives the least amount of product
CO(g) + 2 H2(g) = CH3OH(g)
 Given 3 mol CO and 5 mol H2
 3 mol CO → 3 mol CH3OH
 (mol:mol ratio CH3OH:CO = 1)
 5 mol H2 → 2.5 mol CH3OH
 (mol:mol ratio CH3OH:H2 = 0.5)
 H2 is limiting
 Maximum product = 2.5 mol CH3OH