Chapter 9
Stoichiometry
9.1 Intro. To Stoichiometry
• What is Stoichiometry?
– The study of the quantitative relationships that
exist in chemical formulas and reactions.
• Stoicheion- element
• Metron- measure
2 Types of Stoichiometry:
• Composition stoichiometry- mass relationships of
elements in a single compound. (ch3)
• Reaction stoichiometry- mass relationships
between reactants and products in a chemical rxn.
• Solved by using ratios in Balanced chemical equations.
Reaction-Stoichiometry Problems
• There are 3 types of Stoichiometry problems:
– 1. Mass-Mass problems
– 2. Mass-Volume problems
– 3. Volume-Volume problems
In general, every problem will be solved in 3 steps:
Quantity of given →mols of given →mols of unknown → quantity of unknown
Mole Ratios
• Mole Ratio- a conversion factor that relates the
amounts in moles of any 2 substances involved in
a chemical reaction.
• The coefficient in balanced chem. Equns.
represent MOLE ratios for the substances in the
rxn.
– Ex.- N2H4 +2H2O2 → N2 + 4H2O
Means: 1mol N2H4 +2mol H2O2 → 1mol N2 + 4mol
H2O
Molar Mass (review)
• From Ch 7:
• “The mass in grams of 1 mole of a substance..” (M).
• Molar Mass of…
–
–
–
–
–
–
Ca 40 amu –> 40 g/mol
C 12 amu -> 12 g/ mol
H2O2 34 amu -> 34 g/mol
NaCl
CaCl2
C6H12O6
9.2 Idea Stoichiometric
Calculations
• Knowing that the coefficients show MOLE
RATIOS… we can now solve problems relating
moles of one substance to moles of another.
• Ex
– N2H4 +2H2O2 → N2 + 4H2O
How many mols of N2H4 are needed to react with
2 mols of H2O2 ?
More mol- mol practice
• 2S + 3O2 → 2SO3
How many moles of S are needed to react
with 2.5 mols of O2?
• Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2O
How many moles of Cu(NO3)2 are produced from
12.3 moles of HNO3?
Mass-Mass Problems
• In these problems, you will be given the mass
of one substance and asked to solve for the
mass of another substance.
• Your mantra is:
– Mass to moles, moles to moles, moles to mass.
• You will need a balanced chem. equation, the
molar mass of the known and the unknown
substance to solve.
Mass-mass practice
• 2H2 + O2 → 2H2O
– How many grams of water will be produced from
23.5 g of oxygen?
– *Mass to moles, moles to moles, moles to mass*
• Convert g to moles of O2, then mol O2 to mol
H2O, and finally, mols of H2O to g of H2O.
More mass-mass practice
• Zn + H2SO4 → ZnSO4 + H2
– How many g of H2SO4 will react with 9.5 g of Zn?
• HCl + NaOH → NaCl + H2O
– How many g of NaCl will be produced from 51.2 g of NaOH?
• 2Mg + O2 → 2 MgO
– How many g of Mg are needed to produce 12.3g MgO?
Mass-Volume Problems
In these problems you will be given the MASS of
one substance and asked to find the VOLUME of
a gas.
You will be going from grams to liters (g → L).
Your Mantra is:
Mass to moles, moles to moles, moles to volume.
To convert from mols to volume you will use the
molar volume (22.4L/ mol).
Mass-Volume practice
• 2H2 + O2 → 2H2O
– How many liters of hydrogen will be required to
produce 23.5 g of water?
– Mass to moles, moles to moles, moles to volume.
• Convert g to moles of H2O, then mol H2O to
mol H2, and finally, mols of H2 to L of H2.
– Remember, use the molar volume: 22.4 L/mol
More mass-volume practice
• Zn + H2SO4 → ZnSO4 + H2
– How many L of H2 will be produced from 9.5 g of Zn?
• Sn + 2HF → SnF2 + H2
– How many L of HF will be required to react with 30.0g of
Sn?
• NH4NO3 → N2O + 2H2O
– How many L of N2O will be produced from 52.6 g NH4NO3?
• Mg + O2 → MgO2
– How many L of O2 are needed to produce 12.3g MgO2?
Volume-Volume Problems
In these problems you will be given the VOLUME
of one gas and asked to find the VOLUME of a
different gas.
Your Mantra is:
VOLUME to moles, moles to moles, moles to
volume.
To convert from mols to volume you will use the molar
volume (22.4L/ mol).
Volume - Volume practice
• 2H2 + O2 → 2H2O
– How many liters of hydrogen will be required to
react with 23.5 L of O2?
– volume to moles, moles to moles, moles to
volume.
• Convert L to moles of O2, then mol O2 to mol
H2, and finally, mols of H2 to L of H2.
– Remember, use the molar volume: 22.4 L/mol
9.3 Limiting Reactants & Percent Yields
• The amount of product made depends on how
much reactants are available. Sometimes, one
of the reactants limits the number of products
made…
– How many bikes can be made from 11 bike frames
but only 7 tires?
– Did the # of bike frames or the # of tires limit how
many bikes were made?
• The reactant that limits the amount of product
formed is called the limiting reactant.
• The reactant that is not completely used up is
called the excess reactant.
Identifying Limiting Reactants
• There are 3 general steps to identifying the
limiting reactant.
– 1. Begin with a balanced chemical equation.
– 2. Calculate the amount of product formed by
EACH of the reactants. (pick only one product)
– 3. The reactant that produces the least amount of
product is the limiting reactant.
Limiting Reactant Problems
• Cu + 2AgNO3 → Cu(NO3)2 + 2Ag
• Which is the limiting reactant if you have 6.0g
Cu and 12.5g AgNO3 ? (hint: solve for Ag)
– You will perform 2 mass-mass problems.
• 1. mass of Cu to mass of Ag
• 2. mass of AgNO3 to mass of Ag.
Percent Yield
The amount we calculate and what we actually
make as the product are not necessarily the
same amount.
Theoretical yield – amount of product that
should be produced based on calculation.
Actual yield – the amount ACTUALLY obtained
from the reaction (in lab).
• We need to know how much of the expected
was made during the reaction. …did we only
make 5% or 55%?
• Percent Yield – what % of the predicted
amount was actually made.
% yield = (actual yield ÷ expected yield) X 100%
Percent Yield practice problems
• You burn Mg in O2 and produce 6.5g MgO.
You expected to make 8.2g of MgO. What is
your percent yield?
• Determine the %yield for 3.74g Na and excess
O2 if 5.34g of Na2O2 is recovered?
We don’t know the theoretical yield so
you will have to calculate it first, then solve
%yield.