This document introduces the concept of stoichiometry in chemistry, which is the calculation of quantities in chemical reactions from a balanced chemical equation. It discusses how stoichiometry allows chemists to determine the amounts of reactants and products using concepts like moles, molar masses, and mole ratios derived from balanced equations. Specific calculations covered include mole-mole, mass-mole, mole-mass, and mass-mass problems. It also introduces related concepts like theoretical yield and percent yield.

1. STOICHIOMETRY
Or How I Learned to Love Counting
Atoms

2. Equations are the recipes that tell chemists what amounts
of reactants to mix and what amounts of products to
expect. You can determine the quantities of reactants and
products in a reaction from the balanced equation.
When you know the quantity of one substance in a
reaction, you can calculate the quantity of any other
substance consumed or created in the reaction. (Quantity
usually means the amount of a substance expressed in
grams or moles. But quantity could just as well be in
liters, tons, or molecules.)
The calculation of quantities in chemical reactions is a
subject of chemistry called stoichiometry.

3. Calculations using balanced equations are called
stoichiometric calculations. For chemists, stoichiometry is
a form of bookkeeping.

4. INTERPRETING CHEMICAL
EQUATIONS
As you may recall, ammonia is widely used as a fertilizer.
Ammonia is produced industrially by the reaction of
nitrogen with hydrogen.
N2(g) + 3H2(g) 2NH3(g)
What kinds of information can be derived from this
equation?

5. Do you see that mass and atoms are conserved in this
chemical reaction? Mass and atoms are conserved in
every chemical reaction.
The mass of the reactants equals the mass of the products.
The number of atoms of each reactant equals the number
of atoms for that reactant in the product(s).
Unlike mass and atoms, however, molecules, formula
units, moles and volumes of gases will not necessarily be
conserved - although they may be.
Only mass and atoms are conserved in every chemical
reaction.

6. Interpreting Chemical Reactions
Again, the formation of ammonia from hydrogen and
nitrogen is
3H2 + N2 2 NH3

7. Mole-Mole Calculations
• The most important interpretation of this
equation is that 1 mol of nitrogen reacts
with 3 mol of hydrogen to form 2 mol of
ammonia.
• With this interpretation, you can relate
moles of reactants to moles of product.
• The coefficients from the balanced equation
are used to write conversion factors called
mole ratios.

8. The mole ratios are used to calculate the
number of moles of product from a given
number of moles of reactant or to calculate the
number of moles of reactant from a given
number of moles of product.
Three of the mole ratios for this equation are
1 mol N2 2 mol NH3 3 mol H2
3 mol H2 1 mol N2 2 mol NH3

9. In the mole ratio below, W is the unknown
quantity. The value of a and b are the
coefficients from the balanced equation. Thus a
general solution for a mole-mole problem is
given by
Given Mole ratio Calculated
x mol G x = mol W
b mol W xb
a mol G a
From balanced
equation

10. Using the ammonia reaction, answer the following
question.
How many moles of ammonia are produced when 0.60
mol of nitrogen reacts with hydrogen?
0.60 mol N2 x
2 mol NH3
1 mol N2
= 1.2 mol NH3
Given Mole Ratio

11. MASS-MOLES
CALCULATIONS
Balances don’t tell you numbers in moles but in grams.
As such, there are two related stoichiometry calculations:
Moles - Mass
&
Mass - Moles

12. In a mole-mass problem you are asked to calculate the
mass (usually in grams) of a substance that will react with
or be produced from a given number of moles of a second
substance.
moles A moles B mass B
moles A x mole ratio of x molar mass of B
B
A
(If, in an example, you are told something in is excess,
just ignore that substance and solve the problem with the
needed substances.)

13. Example:
Plants use carbon dioxide and water to form
glucose (C6H12O6) and oxygen. What mass, in grams, of
glucose is produced when 3.00 mol of water react with
carbon dioxide?
Answer:
1. Write the balanced equation
6CO2(g) + 6H2O(l) -> C6H12O6(s) + 6O2(g)

14. 2. Determine what you need to find/know.
Unknown: mass of C6H12O6 produced
Given: amount of H2O = 3.00 mol
3. Determine conversion factors
moles H2O x x
moles C6H12O6
moles H2O
grams C6H12O6
1 mole C6H12O6
= grams C6H12O6

15. 4. Solve
3.00 moles H2O x x
1 mol C6H12O6
6 moles H2O
180 g C6H12O6
1 mole C6H12O6
= 90.0 g C6H12O6

16. In a mass-mole problem you are asked to calculate the
moles of a substance that will react with or be produced
from a given number of grams of a second substance.
mass A moles A moles B
1 mole A
molar mass A
mass A x x mole ratio
B
A

17. Worksheet questions
Mol  mass
mol A  mol B  mass B
75.0 mol C7H6O3 x
1 mol C9H8O4
1 mol C7H6O3
180 g C9H8O4
1 mol C9H8O4
x
= 13500 g C9H8O4 x
1 kg
1000 g
= 13.5 kg C9H8O4

18. Mass-Mass Calculations
• No laboratory balance can measure substance
directly in moles
• Instead, the amount of a substance is usually
determined by measuring its mass in grams
• From the mass of a reactant or product, the
mass of any other reactant or product in a
given chemical equation can be calculated
• The mole interpretation of a balanced equation
is the basis for this conversion

19. If the given sample is measured in grams, the mass
can be converted to moles by using the molar mass
Then the mole ratio from the balanced equation can
be used to calculate the number of moles of the
unknown
If it is the mass of the unknown that needs to be
determined, the number of moles of the unknown
can be multiplied by the molar mass.
As in mole-mole calculations, the unknown can be
either a reactant or a product

20. Mass-mass problems can be solved in basically the same way as
mole-mole problems.
1. The mass G is changed to moles of G (mass G mol G) by
using the molar mass of G.
Mass G X = mol G
2. The moles of G are changed to moles of W (mol G mol W) by
using the mole ratio from the balanced equation.
Mol G X = mol W
1 mol G
molar mass G
b mol W
a mol G
3. The moles of W are changed to grams of W (mol W mass W)

21. mass A x x x
1 mole A
grams A
moles B
moles A
grams B
1 mole B
mole ratiomolar mass of A molar mass of B
The route for solving mass-mass problems is:
mass A moles A moles B mass B
given

22. Example:
Calculate the number of grams of NH3 produced
by the reaction of 5.40 g of hydrogen with an excess of
nitrogen.
Solution:
1. Write the balanced equation
N2 + 3H2 2NH3
2. Write what you know:
Unknown: g NH3; g H2 -> g NH3
Given: 5.40 g H2

23. 3. Determine conversion factors
g H2 mol H2 mol NH3 g NH3
4. Solve
5.40 g H2 x x x
1 mol H2
2.0 g H2
2 mol NH3
3 mol H2
17.0 g NH3
1 mol NH3
given
= 30.6 g NH3
changes given
to moles
mole ratio change moles of
wanted to grams

24. Practice Problems
5.00 g CaC2 x
1 mole CaC2
64.1 g CaC2
x
1 mole C2H2
1 mole CaC2
x
26.0 g C2H2
1 mole C2H2
= 2.03 g C2H2

25. Worksheet questions
a. 384 g O2 x
1 mol O2
32g O2
x
2 mol NO2
1 mol O2
x
46g NO2
1 mol NO2
= 1104 g NO2 = 1.10x103
g NO2
given Molar mass A Mole ratio Molar mass B

26. OTHER STOICHIOMETRIC
CALCULATIONS
As you already know, a balanced equation indicates the
relative number of moles of reactants and products.
From this foundation, stoichiometric calculations can be
expanded to include any unit of measurement that is
related to the mole.
The given quantity can be expressed in number of
representative particles, units of mass, or volumes of
gases at STP.

27. The following equation summarizes these steps for a
typical stoichiometric problem
aG bW
(given quantity) (wanted quantity)

29. Using the ammonia reaction equation, determine the
number of liters of ammonia that can be produced from 5
grams of nitrogen at STP.
N2 + 3H2 2NH3
5g N2
x
1 mole N2
28g N2
x
2 mole NH3
1 mole N2
x
22.4 L NH3
1 mole NH3
=

30. PERCENT YIELD
When an equation is used to calculate the amount of
product that will form during a reaction, a value
representing the theoretical yield is obtained.
The theoretical yield is the maximum amount of
product that could be formed from given amounts of
reactants.
In contrast, the amount of product that actually forms
when the reaction is carried out in the laboratory is called
the actual yield. The actual yield is often less than the
theoretical yield.

31. The percent yield is the ratio of the actual yield to
the theoretical yield expressed as a percent. The percent
yield measures the efficiency of the reaction.
Percent yield = x 100
theoretical yield
A percent yield should not normally be larger than 100%.
Many factors can cause percent yields to be less than
100%.

32. Example:
Calcium carbonate is decomposed by heating, as
shown in the following equation.
CaCO3(s) CaO(s) + CO2(g)
a. what is the theoretical yield of CaO if 24.8 g of
CaCO3 is heated?
b. What is the percent yield if 13.1 g CaO is
produced?

33. Solution:
1. List the knowns and unknowns in a.
known: mass of CaCO3 = 24.8 g
1 mol CaCO3 = 1 mol CaO (from
balanced equation)
1 mol CaO = 56.1 g (molar mass)
unknown: theoretical yield of CaO = ? g CaO

34. 2. Solve for the unknown.
24.8 g CaCO3 x x x
1 mol CaCO3
100.1 g CaCO3
1 mol CaO
1 mol CaCO3
56.1 g CaO
1 mol CaO
given amount molar mass mole ratio molar mass
= 13.9 g CaO
Again, this is the theoretical yield, the amount you would
make if the reaction were 100% accurate.

35. 3. Determine % yield for b.
actual yield = 13.1 g CaO
theoretical yield = 13.9 g CaO
Percent yield = x 100
actual yield
theoretical yield
Percent yield = x 100 = 94.2%
13.1 g CaO
13.9 g CaO