Stoichiometry Notes and Flow Diagram
The word stoichiometry comes from the Greek stoikheion
(meaning element) + metry (measurement of), and thus describes a
method of measuring the elements of a chemical reaction. The
method allows you to determine how much of all the other
chemicals you will need or can produce when you know anything
about the amount of at least one chemical in a reaction.
The Key to all stoichiometry problems is a balanced chemical
equation, because the balanced chemical equation gives the correct
ratio of species (parts) of the chemical equation. (think about the
PB&J sandwiches). This ratio known as the mole ratio is the one
conversion factor that is used in every single stoichiometry
calculation.
Example.
2 C2H6 (g) + 7 O2 (g) 4 CO2 (g) + 6 H2O (g)
The mole ratios are
2 C2H6
7 O2
2 C2H6
4 CO2
2 C2H6
6 H2O
7 O2
4 CO2
7 O2
6 H2O
4 CO2
6 H2O
If you wanted to know how many moles of oxygen you would
require to burn 5.5 moles of ethane you would do the following…..
5.5 moles C2H6 (g)
7 O2
2 C2H6
= 19.25 moles O2
Stoichiometry Notes and Flow Diagram
The word stoichiometry comes from the Greek stoikheion
(meaning element) + metry (measurement of), and thus describes a
method of measuring the elements of a chemical reaction. The
method allows you to determine how much of all the other
chemicals you will need or can produce when you know anything
about the amount of at least one chemical in a reaction.
The Key to all stoichiometry problems is a balanced chemical
equation, because the balanced chemical equation gives the correct
ratio of species (parts) of the chemical equation. (think about the
PB&J sandwiches). This ratio known as the mole ratio is the one
conversion factor that is used in every single stoichiometry
calculation.
Example. ___ C2H6 (g) + ___ O2 (g) ___ CO2 (g) + ___ H2O (g)
The mole ratios are
C2H6
O2
C2H6
CO2
C2H6
H2O
O2
CO2
O2
H2O
CO2
H2O
If you wanted to know how many moles of oxygen you would
require to burn 5.5 moles of ethane you would do the following…..
5.5 moles C2H6 (g)
=
moles O2
Flowchart
The flowchart below shows you how to do nearly any stoichiometry problem by finding your starting
and ending point and following the instructions along the arrows connecting them. For all situations
consider species A to be whatever material you know about and species B to be whatever material you
are trying to find out about.
A-Side (what you know)
B-side (what you want to know)
Mass A
Mass B
÷ Molar Mass A
#Particles A
÷ Avagadro’s #
Moles A
x Molar Mass B
x Mole Ratio (B/A)
Moles B
x (P/RT)
Volume Gas A
If temperature and pressure stay constant x Mole Ratio (B/A)
x Avagadro’s #
# Particles B
x (RT/P)
Volume Gas B
Flowchart
The flowchart below shows you how to do nearly any stoichiometry problem by finding your starting
and ending point and following the instructions along the arrows connecting them. For all situations
consider species A to be whatever material you know about and species B to be whatever material you
are trying to find out about.
A-Side (what you know)
B-side (what you want to know)
Mass A
Mass B
÷ ____________
#Particles A
÷ ___________
Moles A
x ____________
x _____________
Moles B
x _________
x _______
Volume Gas A
If temperature and pressure stay constant x ___________ (B/A)
# Particles B
x _______
Volume Gas B
Limiting Reactants
Any situation or problem where you have data about one species in a reaction and need to know something about one of the other species
is a stoichiometry problem. The most common type of stoichiometry problem in real chemistry situations is a limiting reactant problem.
Limiting reactants are just what the name implies: reactants that limit the extent of the chemical reaction. They do this by running out
first.
New Deinitions:
Limiting Reactant: reactant that runs out first during a chemical reaction thus limiting the product.
Excess Reactant: any reactant that exists in greater amounts than the limiting reactant requires so it will be l
left over when the reaction is complete.
Theoretical Yield: the amount of product that should be produced stoichiometrically.
You can recognize a limiting reactant situation because you have quantitative data about more than one reactant.
There are many ways to approach limiting reactant problems. Below is one method that works.
1. Do a separate stoichiometry calculation starting with each known reactant.
2. The reactant that produces the least product is the limiting reactant and the amount produced is the theoretical yield.
If you also need to know how much is left over of the excess reactant, do the following:
3. Use stoichiometry starting with your limiting reactant to determine how much of each excess reactant is used up.
4.
Subtract the used amount of the excess reactant from the initial amount and you have what is left over.
Example: If you have 10.0 g of aluminum foil and 20.0 g of copper (II) chloride, how much copper will you produce? What reactant is
left over and how much of that reactant is left over?
First recognize that this is a limiting reactant situation because you have data for both aluminum and copper (II) chloride.
2 Al + 3 CuCl2  3 Cu + 2AlCl3
Step 1. 10.0 g Al  1 mole  3 Cu  63.55 g
= 35.3 g Cu
Step 2: Since copper (II) chloride produces less copper
26.98 g  2 Al  1mole
copper (II) chloride is the limiting reactant
20.0 g CuCl2  1 mole  3 Cu
 63.55 g = 9.45 g Cu
aluminum is the excess reactant
134.45 g  3 CuCl2  1mole
and the theoretical yield is 9.45 g Cu
Step 3. 20.0 g CuCl2  1 mole  2 Al
 26.98 g = 2.67 g Al
Step 4. 10.0 g Al - 2.67 g Al = 7.33g Al
134.45 g  3 CuCl2  1mole
7.33 g aluminum remains.