Stoichiometry Notes – chapter 121
Chapter 12.1: What is stoichiometry?
Chemical reactions stop when one of the
reactants is used up.
Stoichiometry: the study of quantitative
relationships between amounts of reactants
used and products formed by a chemical
reaction.
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Based on the law of conservation of
mass
Mass of reactants must equal the mass of
the products
A balanced equation shows the
relationship between the amount of
reactant and the amount of product
We can use a balanced equation to
determine the masses involved in a
chemical reaction
Practice problem
Combustion of propane (C3H8)
____C3H8 +____O2  ____CO2 + ___H2O
Calculate the mass of each reactant and
each product. Show that the law of
conservation of mass is true.
____Fe + ____O2  ____Fe2O3
What we know from the equation:
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4 molecules of iron reacting with 3
molecules of oxygen to produce 2
molecules of iron (III) oxide
4 moles of iron reacting with 3 moles of
oxygen to produce 2 moles of iron (III)
oxide
To determine the mass of the
substances, multiply by molar mass.
4 mol x 55.9 g Fe = 223.6 g Fe
1 mol
3 mol x 32 g O2 = 96 g O2
1 mol
Total mass of reactants = 319.6 g
Mole ratio
Mole ratio is the ratio between the
numbers of moles of any two
substances in a balanced chemical
equation.
Example:
____Al + ___Br2  ____AlBr3
Write the mole ratios for each term in
the equation.
2 mol x 159.8 g Fe2O3 = 319.6 g
1 mol
2 mol Al
3 mol Br2
2 mol Al
2 mol AlBr3
Total mass of reactants equals the mass of the
Why mole ratios are important:
3 mol Br2
2 mol Al
3 mol Br2
2 mol AlBr3
Mole ratios are important in calculations
based upon a chemical equation.
2 mol AlBr3
2 mol Al
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If you know the amount of any one
substance, you can calculate the
amounts for all the others.
2 mol AlBr3
3 mol Br2
To find out how many mole ratios
there are, multiply the number of
terms by the next lower number.
Stoichiometry Notes – chapter 122
Stoichiometric mole-to-mole conversion
Example problem:
All stoichiometric calculations begin with a
balanced chemical equation.
How many moles of carbon dioxide are
produced when 10.0 moles of propane are
burned in excess oxygen in a gas grill?
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Indicates relative amounts of the
substances that react and products that
form
____C3H8 + ____O2  ____CO2 + ____H20
Requirements
1. Balanced equation (ch. 10)
2. Mole ratio (ch. 12.1)
3. Conversions for mass-mole, molerepresentative particles (ch. 11)
Example: potassium and water has a vigorous
reaction and produces hydrogen gas. How
much hydrogen gas is produced when 0.0400
moles of potassium is used?
1. Write the balanced equation
____K + ____ H2O  ____KOH + ____H2
2. Mole ratio between what is known and what
is unknown.
What we know is in moles and what we
what to know is in moles, so this is a moleto-mole conversion.
Write the unknown mole quantity in the
numerator, with the known quantity in
the denominator.
1 mole H2
2 moles K
moles of known x moles of unknown =
moles of known
0.0400 moles x 1 mol H2 = 0.0200 mol H2
2 mol K
Stoichiometric mole-to-mass conversion
If you know the number of moles or a
reactant or product, but you want to
know the mass of another reactant or
product, you do the following:
1. Write the balanced equation
2. Determine mole ratio of unknown to
known
3. Multiply given number of moles by
the mole ratio
4. Multiply moles of unknown by the
molar mass of the unknown.
Determine the mass of sodium chloride or table salt (NaCl) produced when 1.25 moles of
chlorine gas reacts vigorously with sodium.
Stoichiometry Notes – chapter 123
Mass-to-mass conversion
Steps to follow:
1.
2.
3.
4.
Write the balanced equation
Convert mass of given substance to moles
Determine the mole ratio and multiply by moles of known substance.
Convert moles of unknown to mass
Ammonium nitrate (NH4NO3) produces N2O gas and H2O when it decomposes. Determine the
mass of water produced from the decomposition of 25.0 g of solid ammonium nitrate.
Stoichiometric
calculations wrap-up
1. The specified unit
of the given
substance
determines where
you start.
2. If moles, skip step
2 and start with
step 3 (mole-tomole)
3. If mass, start with
step 2
4. The end point
depends upon the
specified unit of
the unknown. If
moles, the
calculation is
finished with step
3.
If mass, you must
continue on to step
4.
Stoichiometry Notes – chapter 124
Chapter 12.3: Limiting Reactants
In nature, rarely are reactants present in
the exact ratios specified by the balanced
equation.
Reaction continues until one of the
reactants is used up.
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Amount of the product produces
depends upon the reactant that is
limited.
Limiting reactant: reactant that
limits the amount of product
formed from the reaction
o Reactant that runs out first
Excess reactant: reactant “leftover” after the reaction occurs
Example:
Disulfur dichloride (S2Cl2) is used to vulcanize
rubber. If 200.0 g of sulfur reacts with
100.0 g of chlorine, what mass of disulfur
dichloride is produced?
____S8 + ____Cl2  ____S2Cl2
1. 200 g
100 g
?g
S8
+ Cl2
 S2Cl2
2. 200g S8 x 1 mol S8 = 0.779 mol S8
256.56 g
100g Cl2 x 1 mol Cl2 = 1.41 mol Cl2
71 g
3. 1.41 mol Cl2 available = 1.81 mol Cl2
0.779 mol S8 available 1 mol S8
4. From the balanced formula: 4 mol Cl2
1 mol S8
Steps to determine the Limiting Reactant
1. Write the balanced equation (with
given amounts)
2. Determine the number of moles of
each reactant
3. Find the mole ratio of the reactants
4. Find the mole ratio from the
balanced equation and determine
the limiting reactant
5. Use the moles of limiting reactant to
find the mass of the product
Only 1.81 mol of Cl2 is available for every
1 mol of S8, instead of the 4 mol of Cl2
required. Chlorine is the limiting reactant.
5. Determine the amount of the product
that can be produced.
Multiply the given number of moles of
the limiting reactant (1.41 mol Cl2) by
the mole ratio that relates disulfur
dichloride and chlorine.
1.41 mol Cl2 x 4 mol S2Cl2 = 1.41 mol S2Cl2
4 mol Cl2
1.41 mol S2Cl2 x 135 g S2Cl2 = 190.4 g S2Cl2
1 mol S2Cl2
Stoichiometry Notes – chapter 125
Determining how much of the other reactant
is left over
1. Multiply moles of limiting reactant by
mole ratio (from balanced equation) to
find the number of moles [of excess
reactant] used.
2. Multiply moles of excess used by the
molar mass to find the mass used.
3. Subtract the amount used from the
amount given.
Continuing on with S2Cl2
How much of the sulfur is left over?
1. Calculate the mass of sulfur needed
to react completely with 1.41 mol of
Cl2 using a mole-to-mass calculation.
1.41 mol Cl2 x 1 mol S8 = 0.353 mol S8
4 mol Cl2
Practice problem
2. 0.353 mol S8 x 256.56 g S8 = 90.6 g S8
1 mol S8
3. 200g S8 given ― 90.6g S8 used =
109.4g S8 in excess
The reaction between solid white phosphorus
and oxygen produces solid tetraphosphorus
decoxide (P4O10).
Extra work space, if needed.
a. Determine the mass of tetraphosphorus
decoxide formed if 25.0 g of
phosphorus (P4) and 50.0 g of oxygen
(O2) are combined.
b. How much of the excess reactant
remains after the reaction stops?
Why have an excess of 1 reactant instead of
just calculating and using exact amounts?
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Some reactants will not continue until
all of the reactants are used up.
o Inefficient and wasteful
By using an excess of (the least
expensive) reactant, the reaction is
driven to continue until all of the
limiting reactant is used up.
May also speed up the reaction
Stoichiometry Notes – chapter 126
Chapter 12.4: Percent Yield
Most reactions never succeed in producing the
predicted amount of product
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Many reaction stop before all of the
reactants are used up
Actual amount of product is less than
expected
Some liquid reactants or products may
adhere to the surfaces of the containers
or evaporate
Solid product can be left on filter paper
or lost while purifying
Products other than the intended ones
may be formed by competing reactions
Theoretical yield: maximum amount of
product that can be produced from a given
amount of reactant.
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Rarely happens!
Actual yield: amount of product actually
produced with the chemical reaction is carried
out in an experiment
Percent yield: ratio of the actual yield to the
theoretical yield, expressed as a percent.
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Measures efficiency of the reaction
Percent yield =
actual yield
x 100
theoretical yield from calc.
Calculating Percent Yield
1. Write the balanced equation
2. Find the theoretical yield using a mass-tomass conversion
3. Divide actual yield by theoretical yield and
multiply by 100
Extra space, if needed.
Practice Problem
When potassium chromate (K2CrO4) is
added to a solution containing .500 g
silver nitrate (AgNO3), solid silver
chromate (Ag2CrO4) is formed.
1. Determine the theoretical yield of
the silver chromate precipitate.
2. If 0.455 g of silver chromate is
obtained, calculate the percent
yield.