Chapter 3 – Stoichiometry:
Calculations with Chemical
Formulas and Equations
Jennie L. Borders
Section 3.1 – Chemical Equations
• Stoichiometry is the area of study that
examines the quantities of substances
consumed and produced in chemical
reactions.
• Remember: Atoms are neither created nor
destroyed during any chemical reaction or
physical process.
Chemical Equations
• Below are the parts of a chemical
equation:
• The arrow is called the yield sign.
Balancing Equations
• Since atoms are neither created nor
destroyed in a chemical reaction, a
chemical equation must have an equal
number of atoms of each element on each
side of the arrow.
Balancing Equations
• When balancing equations the coefficients
must be the lowest whole-numbers
possible.
• There are some exceptions if a reaction is
particularly describing the reaction of 1
mole of a substance.
Rules/Hints for Balancing
Equations
• To balance equations, only coefficients can be
added, subscripts can never be changed or
added.
• Balance H and O last.
• If a polyatomic ion stays
together on both sides of the
arrow, then keep it together when balancing.
• If you have an odd number of a substance and
you want an even number, try doubling the
coefficient.
Sample Exercise 3.1
• The following diagram represents a
chemical reaction in which the red spheres
are oxygen atoms and the blue spheres
are nitrogen atoms.
a. Write the chemical formulas for the
reactants and products.
Sample Exercise 3.1
b. Write a balanced equation for the
reaction.
c. Is the diagram consistent with the law of
conservation of mass?
Practice Exercise
• In the following diagram, the white spheres
represent hydrogen atoms, and the blue
sphere represent nitrogen atoms. To be
consistent with the law of conservation of
mass, how many NH3 molecules should
be shown in the right box?
States of Matter
• We use the symbols (g) for gas, (l) for
liquid, (s) for solid, and (aq) for aqueous to
identify the states of matter of the
reactants and products.
• Symbols listed above the arrow indicate
catalysts. When heat is a catalyst the
Delta symbol is used, D.
Sample Exercise 3.2
• Balance this equation:
Na(s) + H2O(l)  NaOH(aq) + H2(g)
Practice Exercise
• Balance the following equations:
a. Fe(s) + O2(g)  Fe2O3(s)
b. C2H4(g) + O2(g)  CO2(g) + H2O(g)
c. Al(s) + HCl(aq)  AlCl3(aq) + H2(g)
Section 3.2 – Some Simple Patterns
of Chemical Reactivity
• In synthesis/combination reactions two or
more substances react to form one
product.
• Generic Reaction: A + B  AB
• Real Reaction: 2Mg + O2  2MgO
Decomposition Reactions
• In a decomposition reaction one
substance undergoes a reaction to
produce two or more products.
• Generic Reaction: AB  A + B
• Real Reaction: CaCO3  CaO + CO2
Writing Equations
• Remember to balance charges if you are
predicting the products of a reaction.
• Ionic compounds (a metal and a nonmetal)
tend to be solids and acids tend to be
aqueous in reaction.
Sample Exercise 3.3
• Write the balanced equations for the following
reactions:
a. The synthesis reaction that occurs when
lithium metal and fluorine gas react.
b. The decomposition reaction that occurs
when solid barium carbonate is heated.
Hint for Predicting Products
• Like in the previous exercise, carbonate
ions tend to produce CO2 when they break
down.
Practice Exercise
• Write the balance equation for the following
reactions:
a. Solid mercury (II) sulfide decomposes into
its elements when heated.
b. The surface of aluminum metal undergoes
a combination reaction with oxygen in the air.
Combustion
• Combustion reactions are rapid
reactions that produce a flame.
They require air (O2) as a reactant.
• When hydrocarbons are combusted
completely, the products are CO2 and
H2O.
• Generic Reaction: CxHy + O2  CO2 +
H2O
• Real Reaction: C3H8 + 5O2  3CO2 +
Sample Exercise 3.4
• Write the balanced equation for the
reaction that occurs when methanol,
CH3OH(l), is burned in air.
Practice Exercise
• Write the balance equation for the reaction
that occurs when ethanol, C2H5OH(l), is
burned in air.
Formula Weights
• The formula weight is a substance is the
sum of the atomic weights of each atom in
its chemical formula.
• Ex: H2SO4 =
2(1 amu) + 32 amu + 4(16 amu) = 98 amu
 If the chemical formula is that of the
molecule, then the formula weight is also
called the molecular weight.
Sample Exercise 3.5
• Calculate the formula weight of
a. sucrose (C12H22O11)
b. calcium nitrate
Practice Exercise
• Calculate the formula weight of
a. aluminum hydroxide
b. methanol (CH3OH)
Percent Composition
• Percent Composition is the percentage by
mass contributed by each element in a
substance.
Percent Composition = mass of element x
100
mass of compound
Sample Exercise 3.6
• Calculate the percent of carbon, hydrogen,
and oxygen (by mass) in C12H22O11.
Practice Exercise
• Calculate the percentage of nitrogen, by
mass, in calcium nitrate.
Section 3.4 – Avogadro’s Number
and the Mole
• In Chemistry the unit for dealing with the
number of atoms, ions, or molecules in a
common-sized sample is the mole,
abbreviated mol.
• 1 mol = 6.02 x 1023 representative
particles
Sample Exercise 3.7
• Arrange the following samples in order of
increasing numbers of carbon atoms:
12g C12, 1 mol C2H2, 9 x 1023 molecules of
CO2.
Practice Exercise
• Arrange the following samples in order of
increasing number of O atoms: 1 mol H2O,
1 mol CO2, 3 x 1023 molecules O3.
Sample Exercise 3.8
• Calculate the number of H atoms in 0.350
mol of C6H12O6.
Practice Exercise
• How many oxygen atoms are in
a. 0.25 mol of calcium nitrate?
b. 1.50 mol of sodium carbonate?
Molar Mass
• A mole is always the same number (6.02 x
1023), but 1 mole sample of different
substances will have different masses.
• The mass of a single atom of an element
(in amu) is numerically equal to the mass
(in grams) of 1 mole of that element.
• The mass in grams of 1 mole
of a substance is called the
molar mass of the substance.
Sample Exercise 3.9
• What is the mass in grams of 1.000 mol of
glucose, C6H12O6?
Practice Exercise
• Calculate the molar mass of calcium
nitrate.
Sample Exercise 3.10
• Calculate the number of moles of glucose
in 5.380g of glucose.
Practice Exercise
• How many moles of sodium bicarbonate
are in 508g of sodium bicarbonate (also
known as sodium hydrogen carbonate)?
Sample Exercise 3.11
• Calculate the mass, in grams. Of 0.433
mol of calcium nitrate.
Practice Exercise
• What is the mass, in grams, of
a. 6.33 mol sodium bicarbonate?
b. 3.0 x 10-5 mol of sulfuric acid?
Sample Exercise 3.12
a. How many molecules are in 5.23g of
glucose?
b. How many oxygen atoms are in this
sample?
Practice Exercise
a. How many molecules are in 4.20g of
nitric acid?
b. How many O atoms are in this sample?
Section 3.5 – Empirical Formulas
from Analyses
• An empirical formula is the lowest whole
number ratio of elements in chemical
formula.
• The ratio of the number of moles of each
element in a compound gives the
subscripts in the compound’s empirical
formula.
Empirical Formula
1. When given percentages, assume 100g
so that the percentages can be grams.
2. Divide the mass of each element by its
molar mass.
3. Divide all answers by the lowest number
to get subscripts.
4. If you get a .5, then multiply all numbers
by 2. If you get a .33 or .66, then multiply
all numbers by 3.
Sample Exercise 3.13
 Ascorbic acid (vitamin C) contains 40.92%
C, 4.58% H, and 54.5% O by mass. What
is the empirical formula of ascorbic acid?
Practice Exercise
• A 5.325g sample of methyl benzoate, a
compound used in perfumes, contains
3.758g C, 0.316g H, and 1.251g of O.
What is the empirical formula of this
substance?
Molecular Formula
• The molecular formula is the actual
formula of a molecule. It can be the same
as the empirical formula.
• When the molecular formula is not the
same, it is a whole number multiple of the
empirical formula.
Molecular Formula
• If you know the empirical formula and the
mass of the molecular formula, then you
can calculate the molecular formula.
Whole # = molecular weight
empirical weight
Sample Exercise 3.14
• Mesitylene, a hydrocarbon in crude oil,
has an empirical formula of C3H4. The
molecular weight of the substance is 121
amu. What is the molecular formula of
mesitylene?
Practice Exercise
• Ethylene glycol, the substance used in
antifreeze, is composed of 38.7%C, 9.7%
H, and 51.6% O by mass. Its molar mass
is 62.1 g/mol.
a. What is the empirical formula of
ethylene glycol?
Practice Exercise
b. What is the molecular formula for the
previous question?
Combustion Analysis
• For combustion reactions, we can
calculate the number of moles (subscripts)
of C and H from the amounts of CO2 and
H2O.
• If any other elements are present, then we
can subtract that mass to get the mass of
only C and H.
Sample Exercise 3.15
• Isopropyl alcohol, a component of rubbing
alcohol, is composed of C, H, and O.
Combustion of 0.255g of isopropyl alcohol
produces 0.561g of CO2 and 0.306g H2O.
Determine the empirical formula.
Practice Exercise
a. Caproic acid, the smell in dirty socks, is
composed of C, H, and O atoms.
Combustion of a 0.225g sample of this
compound produces 0.512g CO2, 0.209g
H2O. What is the empirical formula of
caproic acid?
Practice Exercise
b. Caproic acid has a molar mass of 116
g/mol. What is the molecular formula?
Section 3.6 – Quantitative Information from
Balanced Equations
• The coefficients in a balanced equation
indicate both the relative numbers of
moles in the reaction.
• A mole ratio is the ratio of the coefficients
of 2 substances in a chemical equation.
• You must use the mole ratio to convert
from one substance to another.
Sample Exercise 3.16
• How many grams of water are produced in
the oxidation of 1.00g of glucose?
C6H12O6(s) + 6O2(g)  6CO2(g) + 6O2(g)
Practice Exercise
• How many grams of O2 can be prepared
from 4.50g KClO3?
2KClO3(s)  2KCl(s) + 3O2(g)
Sample Exercise 3.17
• Lithium hydroxide reacts with carbon
dioxide to form lithium carbonate and
water. How many grams of carbon dioxide
can be absorbed by 1.00g of lithium
hydroxide?
Practice Exercise
• What mass of O2 is consumed in the
combustion of 1.00g of propane (C3H8)?
Section 3.7 – Limiting Reagents
• The reactant that is completely consumed
in a reaction is called the limiting reactant
or limiting reagent.
• The other reactants are called the excess
reactants or reagents.
• The amount of product is always based on
the limiting reagent.
Sample Exercise 3.18
• How many moles of NH3 can be formed
from 3.0 mol of N2 and 6.0 mol of H2?
N2(g) + 3H2(g)  2NH3(g)
Practice Exercise
• A mixture of 1.50 mol of Al and 3.00 mol of
Cl2 is allowed to react.
2Al(s) + 3Cl2(g)  2AlCl3(s)
a. Which is the limiting reagent?
Practice Exercise
• How many moles of AlCl3 are formed?
• How many moles of the excess reactant
remain at the end of the reaction?
Sample Exercise 3.19
• Suppose a fuel cell is set up with 150g of
hydrogen gas and 1500g of oxygen gas.
How many grams of water can be formed?
2H2(g) + O2(g)  2H2O(g)
Practice Exercise
• A strip of zinc metal with a mass of 2.00g
is placed in an aqueous solution
containing 2.50g of silver nitrate, causing
the following reaction to occur:
Zn(s) + 2AgNO3(aq)  2Ag(s) + Zn(NO3)2(aq)
a. Which reactant is limiting?
Practice Exercise
b. How many grams of Ag will form?
c. How many grams of Zn(NO3)2 will form?
d. How many grams of the excess reagent
will be left?
Percent Yield
• The percent yield of a reaction relates the
actual yield (done in the lab) to the
theoretical yield (calculated using
stoichiometry).
% Yield =
actual yield
x 100
theoretical yield
Percent Yield
• Since the theoretical yield is the maximum
amount of product that can be produced
under ideal conditions, then the actual
yield should never be larger than the
theoretical yield.
• Percent yield should never
exceed 100%.
Sample Exercise 3.20
• Adipic acid, H2C6H8O4, is used to produce
nylon. Adipic acid is made by the following
reaction:
2C6H12(l) + 5O2(g)  2H2C6H8O4(l) + 2H2O(g)
a. Assume that you carry out this reaction
starting with 25.0g of cyclohexane and that
cyclohexane is the limiting reagent. What
is the theoretical yield of the adipic acid?
Sample Exercise 3.20
b. If you obtain 33.5g of adipic acid from
your reaction, what is the percent yield of
adipic acid?
Practice Exercise
a. If you start with 150g of Fe2O3 as the
limiting reagent, what is the theoretical yield
of Fe?
Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g)
b. If the actual yield of Fe in your test was
87.9g, what is the percent yield?