Chapter 11 – Stoichiometry
11.1
11.2
11.3
11.4
What is Stoichiometry?
Stoichiometric Calculations
Limiting Reactants
Percent Yield
Section 11.1 Defining Stoichiometry
The amount of each reactant present at
the start of a chemical reaction
determines how much product can form.
• Describe the types of relationships indicated by a
balanced chemical equation.
• Explain the relationship between stoichiometry and the
principle of the conservation of mass.
• State the mole ratios from a balanced chemical
equation.
Section 11.1 Defining Stoichiometry
Key Concepts
• Balanced chemical equations can be interpreted in
terms of moles, mass, and representative particles
(atoms, molecules, formula units).
• The law of conservation of mass applies to all
chemical reactions.
• Mole ratios are derived from the coefficients of a
balanced chemical equation. Each mole ratio relates
the number of moles of one reactant or product to the
number of moles of another reactant or product in the
chemical reaction.
Stoichiometry
Study of quantitative relationships
between amounts of reactants used
and amounts of products formed by a
chemical reaction
Combo of 2 Greek words – “to measure
the elements”
Really a tool (system) for accounting for
the conservation of matter
• What goes in must come out
Stoichiometry
Rests on principle of conservation of
matter (mass)
2 Al(s) + 3 Br2(l)  Al2Br6(s)
Stoichiometry - Questions
If I have this much reactant, how much
product can I form?
If I have this much of one reactant, how
much of a second reactant do I need so
that after the reaction is completed I
won’t have any of either reactant left
over?
Conversion Map for Stoichiometry
Calculations
Mass
reactant
Moles
reactant
Requires
balanced
equation!
Stoichiometric
factor
Mass
product
Moles
product
Stoichiometry & Balanced Equation
4Fe(s) + 3O2(g)  2Fe2O3(s)
Can interpret as
• 4 atoms Fe + 3 molecules O2  2
formula units Fe2O3
• 4 mol Fe + 3 mol O2  2 mol Fe2O3
No direct information on masses
• Can use molar mass to get mass info
Stoichiometry & Balanced Equation
4Fe(s) + 3O2(g)  2Fe2O3(s)
4 mol Fe  55.85 g Fe/mol Fe = 223.4 g Fe
3 mol O2  32.00 g O2/mol O2 = 96.00 g O2
2 mol Fe2O3  159.7 g Fe2O3/mol Fe2O3 =
319.4 g Fe2O3
Total mass reactants = 319.4 g
Total mass products = 319.4 g
Stoichiometry & Balanced Equation
Example problem 11.1, p. 370
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
Interpret as molecules, moles & mass
1 mol
C3H8  C
44.09
g5
C3molecules
H8 /mol C3HO
molecule
 g
8 =
3H8 +
2 44.09
C
H8
3 3molecules
CO2 + 4 molecules H2O
5
g O2/mol
O2 = 160.0 g O2
1 mol
molOC23H32.00
+
5
mol
O

8
2
3
g CO
2/mol CO2 = 132.0 g CO2
3 mol
molCO
CO2 2 +44.01
4 mol
H2O
4 mol H2O  18.02 g H2O /mol H2O = 72.08 g H2O
Mass reactants = 44.09 + 160.0 = 204.1 g
Mass products = 132.0 + 72.08 = 204.1 g
Practice
Problems 1 (a-c), 2(a-b) page 371
Problems 44 - 45 page 392
Problems 1- 3, page 982
Mole Ratio
Ratio between the numbers of moles of
any two substances in a balanced
chemical equation
2Al(s) + 3Br2(l)  2AlBr3(s)
Possible to write 3 unique mole ratios:
Al:Br2 (2:3) Al: AlBr3 (2:2) Br2:AlBr3 (3:2)
Get 3 more ratios from inverses of
above for a total of 6 for this reaction
Mole Ratio
In general, for n species (reactant,
product) in chemical equation, have
n(n-1) mole ratios
3Fe(s) + 4H2O(l)  Fe3O4(s) + 4H2(g)
4 species, 4  3 = 12 possible mole
ratios (6 + inverse of each)
Fe:H2O
Fe:Fe3O4 Fe:H2
H2O:Fe3O4 H2O:H2 Fe3O4: H2
Mole Ratio
How many mole ratios can be written
for the following reaction?
4H2(g) + O2(g) → 2H2O(l)
A. 6
?
B. 4
C. 3
D. 2
Practice
Problems 3 (a-c), 4 (a-b) page 372
Problems 47 - 53, page 392
Problems 4 - 5, page 982-3
Chapter 11 – Stoichiometry
11.1
11.2
11.3
11.4
What is Stoichiometry?
Stoichiometric Calculations
Limiting Reactants
Percent Yield
Section 11.2 Stoichiometric Calculations
The solution to every stoichiometric
problem requires a balanced chemical
equation.
• List the sequence of steps used in solving
stoichiometric problems.
• Solve stoichiometric problems (mole-to-mole, mole-tomass, mass-to-mass and variations including number
of particles) using the factor-label method.
Section 11.2 Stoichiometric Calculations
Key Concepts
• Chemists use stoichiometric calculations to predict the
amounts of reactants used and products formed in
specific reactions.
• The first step in solving stoichiometric problems is
writing the balanced chemical equation.
• Mole ratios derived from the balanced chemical
equation are used in stoichiometric calculations.
• Stoichiometric problems make use of mole ratios to
convert between mass and moles.
Stoichiometric Calculations
Start with balanced equation
Determine required mole ratios
Use mole-to-mass and mass-to-mole
relations for 3 basic conversions:
1) Mole to mole
2) Mole to mass
3) Mass to mass
Once have moles, can get # particles
Conversion Map for Stoichiometry
Calculations
Mass
reactant
Moles
reactant
Requires
balanced
equation!
Stoichiometric
factor
Mass
product
Moles
product
nA + mB  xC + yD
1
Mass
reactant
2 Moles
reactant
3 # part.
reactant
4
Mass
product
5 Moles
product
6 # part.
product
Mole to Mole Conversions
Given number of moles of a reactant
(product), how many moles of a product
(reactant) must be formed?
Use initial moles and mole ratio
mol known x [mol unknown/mol known]
= mol unknown
mole ratio from equation
Mole to Mole Conversions
2K(s) + 2H2O(l)  2KOH(aq) + H2(g)
Mol H2 if start with 4.00x10-2 mol K?
4.00x10-2 mol K  [1 mol H2/2 mol K]
= 2.00x10-2 mol H2
mole ratio from equation
Mole to Mole Conversions
How many moles of CO2 will be
produced in following reaction if initial
amount of reactants = 0.50 moles?
2NaHCO3(s) →
Na2CO3(s) + CO2(g) + H2O(g)
A.0.25
?
B. 0.3
C. 0.5
D. 1.0
Practice
Problems 11,12 page 375
Problems 61- 62, page 393
Problems 6 - 7, page 983
Mole to Mass Conversions
Step 1 – Do mole to mole conversion
using mole ratio
Step 2 – Do mole to mass conversion
using molar mass
Conversion Map for Stoichiometry
Calculations
Mass
reactant
Moles
reactant
Requires
balanced
equation!
Stoichiometric
factor
Mass
product
Moles
product
Mole to Mass Conversions
2Na(s) + Cl2(g)  2NaCl(s)
Mass NaCl produced from 1.25 mol Cl2?
1.25 mol Cl2  [2 mol NaCl/1 mol Cl2]
= 2.50 mol NaCl
2.50 mol NaCl mole
[58.44
g
NaCl/mol
NaCl]
ratio from equation
= 146 g NaCl
molar mass of NaCl
Practice
Problems 13,14 page 376
Problems 63 - 66, page 393
Problems 8 - 9, page 983
Mass to Mass Conversions
Step 1 – Do mass to mole conversion
using molar mass
Step 2 – Do mole to mole conversion
using mole ratio
Step 3 – Do mole to mass conversion
using molar mass
Conversion Map for Stoichiometry
Calculations
Mass
reactant
Moles
reactant
Requires
balanced
equation!
Stoichiometric
factor
Mass
product
Moles
product
Mass to Mass Conversions
NH4NO3(s)  N2O(g) + 2H2O(g)
Mass of H2O from 25.0 g NH4NO3(s)?
AN= NH4NO3(s)
25.0 g AN  [1 mol AN/80.04 g AN]
= 0.312 mol AN
AN AN]
0.312 mol AN 1/molar
[2 mol mass
H2O/1ofmol
= 0.624 mol H
2O ratio from equation
mole
Mass to Mass Conversions
NH4NO3(s)  N2O(g) + 2H2O(g)
Mass of H2O from 25.0 g NH4NO3(s)?
0.624 mol H2O  [18.02 g H2O/mol H2O]
= 11.2 g H2O
molar mass of H2O
Practice
Problems 15,16 page 377
Problems 60, 67- 72, pages 393-4
Problems 10 - 11, page 983
Chapter 11 – Stoichiometry
11.1
11.2
11.3
11.4
What is Stoichiometry?
Stoichiometric Calculations
Limiting Reactants
Percent Yield
Section 11.3 Limiting Reactants
A chemical reaction stops when one of
the reactants is used up.
• Identify the limiting reactant in a chemical equation.
• Identify the excess reactant, and calculate the
amount remaining after the reaction is complete.
• Calculate the mass of a product when the amounts of
more than one reactant are given.
Section 11.3 Limiting Reactants
Key Concepts
• The limiting reactant is the reactant that is completely
consumed during a chemical reaction. Reactants that
remain after the reaction stops are called excess
reactants.
• To determine the limiting reactant, the actual mole
ratio of the available reactants must be compared with
the ratio of the reactants obtained from the coefficients
in the balanced chemical equation.
• Stoichiometric calculations must be based on the
limiting reactant.
Limiting Reactants
Typical case: reactants not present in
exact stoichiometric ratios
• Mole ratio given by coefficients of
reactants in balanced chemical equation
N2 + 3H2  2NH3
Stoichiometric ratio of reactants for
ammonia synthesis is 1 N2: 3 H2
If ratio  1:3 then one reactant is limiting
Limiting and Excess Reactants
Ammonia Synthesis
N2(g) + 3H2(g)  2NH3(g)
Case of 1:1 N2:H2 mole ratio
Before Reaction
3N2
3H2
After Reaction
2NH3
N22 is
is limiting
excess reactant
reactant
H
2N2
Limiting and Excess Reactants
Limiting reactant
• Limits the extent of reaction
• Determines the amount of product
Excess reactant
• Always have “left over”
• Often a reactant is deliberately used in
excess to:


Drive reaction to completion (equilibrium)
Speed up reaction (kinetics)
Limiting and Excess Reactants
Limiting and excess air (oxygen) in
combustion reaction
Air limiting –
soot (carbon)
Air excess –
CO & H O
Limiting Reactant - Identifying
Step 1 – If masses given, convert to
moles
Step 2 – Compare actual mole ratio of
reactants to mole ratio determined from
balanced equation
• Ratios same - no limiting reactant
• Ratios different - compare to identify
limiting reactant
Limiting Reactant
S8(l) + 4Cl2(g)  4S2Cl2(l)
200.0 g sulfur reacts with 100.0 g Cl2
Part a: g of product produced?
Step 1 – Convert masses to moles
100.0 g Cl2  [1 mol Cl2 /70.91 g Cl2]
= 1.410 mol Cl2 1/molar mass of S8
1/molar
mass of
Cl82]
200.0 g S8  [1 mol
S8 /256.5
gS
= 0.7797 mol S8
Limiting Reactant
S8(l) + 4Cl2(g)  4S2Cl2(l)
200.0 g sulfur reacts with 100.0 g Cl2
Part a: g product ?
Step 2 – Compare actual mole ratio to
mole ratio from equation
Equation: Cl2:S8 = 4:1
Actual: 1.410 mol Cl2:0.7797 mol S8
= 1.808:1
Compare ratios  Cl2 is limiting
Limiting Reactant
S8(l) + 4Cl2(g)  4S2Cl2(l)
200.0 g sulfur reacts with 100.0 g Cl2
Part a: g product ?
Cl2 limiting reactant, have 1.410 mol
Calculate mol product using mole ratio
1.410 mol Cl2  [4 mol S2Cl2 /4 mol Cl2]
molar
mass
of
S
Cl
2
2
= 1.410 mol S2Cl2
Mole ratio
1.410 mol S2Cl2  [135.0 g S2Cl2/mol S2Cl2]
= 190.4 g S2Cl2
Limiting Reactant
S8(l) + 4Cl2(g)  4S2Cl2(l)
200.0 g sulfur reacts with 100.0 g Cl ?
Part b: g of sulfur actually reacted?
Cl2 limiting reactant, have 1.410 mol
Calculate mol sulfur using mole ratio
1.410 mol Cl2  [1 mol S8 /4 mol Cl2]
molar
mass
of
S
8
= 0.3525 mol S8
Mole
0.3525 mol S8  [256.5
g Sratio
8 /mol S8]
= 90.42 g S8 reacts
109.6 g excess
Limiting Reactant
S8(l) + 4Cl2(g)  4S2Cl2(l)
200.0 g sulfur reacts with 100.0 g Cl ?
Part b: g of sulfur actually reacted?
Know that all Cl2(g) reacted (100.0 g)
Previously calculated 190.4 g product
S2Cl2(l) produced
Cons. mass: 300.0 g = starting mass =
final mass = m[S8(l)] + m[4S2Cl2(l)]
m[S8(l)]= 300.0 g -m[4S2Cl2(l)] =109.6 g
Limiting Reactant
Excess reactant in following reaction if
start with 50.0g of each reactant?
P4(s) + 5 O2(g) → P4O10(s)
A. O2
B. P4
C. None – quantities are stoichiometric
50.0 g P4  [1 mol P4 /123.90 g P4]= 0.404 mol P4
?
50.0 g O2  [1 mol O2 /32.00 g O2] = 1.56 mol O2
Need 5  0.404 mol = 2.02 mol O2  O2 limiting
Practice
Problems 23, 24 page 383
Problems 77 - 82, page 394
Problems 12 - 13, page 983
Chapter 11 – Stoichiometry
11.1
11.2
11.3
11.4
What is Stoichiometry?
Stoichiometric Calculations
Limiting Reactants
Percent Yield
Section 11.4 Percent Yield
Percent yield is a measure of the
efficiency of a chemical reaction.
• Calculate the theoretical yield of a chemical reaction
from data.
• Explain the difference between a theoretical yield of a
chemical reaction and an actual yield.
• Provide reasons why the theoretical yield and actual
yield can be different from each other.
• Determine the percent yield for a chemical reaction.
• Explain why the percent yield can exeed 100%
Section 11.4 Percent Yield
Key Concepts
• The theoretical yield of a chemical reaction is the
maximum amount of product that can be produced
from a given amount of reactant. Theoretical yield is
calculated from the balanced chemical equation.
• The actual yield is the amount of product produced.
Actual yield must be obtained through
experimentation.
• Percent yield is the ratio of actual yield to theoretical
yield expressed as a percent. High percent yield is
important in reducing the cost of every product
produced through chemical processes.
Percent Yield
Actual (real world) reactions don’t
produce amount of product expected
from masses of reactant and the
balanced chemical equation
(stoichiometry)
Types of Yields
Theoretical Yield
• Maximum amount of product that can be
produced from given amounts of
reactants
• Calculated using stoichiometry
Actual Yield
• Amount of product actually obtained
when the reaction is carried out
Types of Yields
Theoretical Yield and Actual Yield can
differ because
• Reaction stops before limiting reactant
all used up
• Product is not all collected – sticks to
reaction vessel, filter paper, etc
• Reactions other than the intended
reaction occur, producing other products
• Impurities in reactants
Alternate Reaction Paths
Combustion of natural gas (methane)
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
Know this reaction takes place if have
enough O2 – get nice blue flame
Know if limit O2, get yellow flame and
soot (carbon) production
Same reactants yield different products
Percent Yield
Percent yield is a way of expressing the
efficiency of a reaction
Percent yield = 100  actual yield
theoretical yield
Theoretical yield computed using same
procedures used in previous sections
(stoichiometry)
% Percent Yield
Theoretical yield of chemical reaction =
25 g. % yield if actual yield = 22 g ?
A. 88% B. 44% C. 50% D. 97%
?
% yield = 100  22 g / 25 g = 88%
Percent Yield
In commercial production of chemicals,
attempt to produce as close to 100%
yield as possible
• Optimum use of materials
• Minimizes or eliminates need to
separate product from unused reactants
and from unwanted products
Percent Yield for Multi-Stage
Attempt to produce as close to 100%
yield as possible
• Critical for multi-stage reaction
processes where product of one
reaction used as one of the reactants of
a subsequent reaction
• A three-stage reaction with 90% yield
from each stage gives only 73% overall
yield (0.9 x 0.9 x 0.9 = 0.73)
>100% Percent Yield
Actual Yield: Amount of product
actually obtained when from reaction
Obtained from a laboratory
measurement – most common is
measuring mass on a balance
Assumption made is that the only “stuff”
you’re measuring is the product of
interest
>100% Percent Yield
Possible to get more “stuff” you think is
product than predicted if:
Haven’t completely separated
intended product from other products
in reaction or from leftover reactants
and/or solvent (insufficient drying)
Product reacts with air or water vapor
before it is weighed
Practice
Problems 28 - 30 page 387
Problem 35, page 388
Problems 90 - 97, page 395
Problems 14 -15, page 983