Chemical Equations
Balancing equations and
applications
Objectives
 Write and balance chemical equations
 Perform calculations using moles and molar
masses
 Determine percent composition from formula
 Determine empirical formula from percent
composition
 Determine molecular formula
The chemical equation
aA + bB = cC + dD
coefficient
Reactant
side
ELEMENT or
COMPOUND
Product
side
The Law of Conservation of Matter states that matter
is neither created nor destroyed
All the atoms on the left must be the same as those
on the right
Chemical book-keeping
 The key to writing correct equations is to ask
the question, “Have I gained or lost any
atoms?”
 Another thing is to put down the correct
formula for each reactant or product
 Formulas cannot be changed in order to
balance the equation
 In the reaction of hydrogen with oxygen to produce
water, the reactants are the elements H2 and O2,
and the product is H2O
The big number
multiplies every
atom after it
 Count the atoms: 4 H and 2 O
The subscript
only multiplies
the atom before
it
4 H and 2 O
Balance the equations
 CH4 + O2 = CO2 + H2O
 CH4 + 2O2 = CO2 + 2H2O
 C3H8 + O2 = CO2 + H2O
 C3H8 + 5O2 = 3CO2 + 4H2O
 N2 + H2 = NH3
 N2 + 3H2 = 2NH3
 Do balancing equation exercises
Molecules or moles
 The numbers (coefficients) in chemical
equation can refer to molecules
 But for practical applications, we need a
more useful number: we cannot count
molecules
The Mole
 The mole is a unit of quantity used in
chemistry to measure the number of atoms
or molecules
 DEFINITION:
 The number of atoms in exactly 12 g of 12C
 A mole of anything always has the same
number of particles: atoms, molecules or
potatoes – 6.02 x 1023 – Avogadro’s number
Mole conversions
Atomic and molecular weights
 Two scales:
 Atomic mass unit scale
 The mass of an individual atom or molecule in
atomic mass units (amu)
 Molar mass scale
 The mass of a mole of atoms or molecules in
grams
 Confusing?...
The Good News
 The weight of an atom in amu has the same
numerical value as its molar mass in grams




The atomic mass of carbon is 12 amu
The molar mass of carbon is 12 g
The formula mass of H2O is 18 amu
The molar mass of H2O is 18 g
Examples
 How many atoms are in 6.94 g of lithium if
the atomic mass of Li is 6.94 amu?
 6.02 x 1023
 What is the molar mass of H2O if the atomic
mass of H = 1 amu and O = 16 amu
 18 g

How many H atoms in 1 mol of CH4?
1 mol = 6.02 x 1023 particles
How many moles of O2 in 64 g of oxygen?

Atomic mass of O = 16 AMU
Calculate the formula mass of Na2SO4 in AMU.
Use atomic mass Na = 23 AMU, O = 16 AMU, S = 32 AMU
Significance of formula unit
 Ionic compounds do not contain molecules.
Simplest formula is the formula unit
 Covalent compounds, the molecular formula is
the formula unit
Percent composition and empirical
formula
 Chemical analysis gives the mass % of each
element in the compound
 Molar masses give the number of moles
 Obtain mole ratios
 Determine empirical formula
Determining percent composition
 Percent composition is obtained from the actual
masses.
Example:
Sample contained 0.4205 g of C and 0.0795 g of H.
Total mass = 0.5000 g (0.4205 + 0.0795)
Therefore: in 100 g there are:
100 g
x0.4205 = 84.10 g C (84.10 %)
.5000 g
100 g
x0.0795 = 15.90 g H (15.90 %)
.5000 g
Percent composition: 84.10 % C, 15.90 % H
Percent composition from formula
 What is percent composition of C5H10O2?
 1 mol C5H10O2 contains 5 mol C, 10 mol H and 2 mol O
atoms
 Mass of each element
12.01 g C
= 60.05 g C
1 mol C
1.008 g H
10 mol H
= 10.08 g H
1 mol H
16.00 g O
2 mol O
= 32.00 g O
1 mol O
5 mol C

Total mass = 102.13 g
Convert masses into percents
%C=
60.05 g C
x100 = 58.80 % C
102.13 g C5 H10 O2
%H=
10.08 g H
x100 = 9.870 % H
102.13 g C5 H10 O2
%O=
32.00 g O
x100 = 31.33 % O
102.13 g C5 H10 O2
 Percent composition:
58.80 % C + 9.870 % H + 31.33 % O = 100.00%
Empirical formula from percent
composition: 84.1 % C, 15.9 % H
1.
Convert percents into moles
2.
Determine


84.10 g C
84.10 g of C ≡ 7.00 mol C 12.00 g/mol
15.9 g H
15.9 g of H ≡ 15.8 mol H
1.008 g/mol
mole ratio
15.8 mol H
 2.26 :1
Mole ratio H:C =
7.00 mol C
Simplest formula (decimal form): C1H2.26
Make smallest integers by multiplying
C4H9
May require rounding. Errors in real data cause problems

Do percent composition and empirical formula exercises
What is percent C content of C2H6?

Molar mass C = 12 g/mol; molar mass H = 1 g/mol
Empirical formula with more than two
elements

Percent composition of vitamin C is:

40.9 % C, 4.58 % H, 54.5 % O
1. Convert into moles
2. Determine mole ratios
3. Find lowest whole numbers
Inaccuracy can lead to ambiguous or
incorrect formulas
 What if H:C is 2.20 rather than 2.26? An
error of only 3 %
 Formula becomes C5H11 rather than C4H9
 What if H:C is 2.30 rather than 2.26? An
error of only 2 %
 Formula becomes C3H7
 Sometimes chemical intuition is required: we
know there is FeO, Fe3O4 and Fe2O3; so a
formula FeO3 would indicate an error
Rounding or not: the role of chemical
intuition
 Formulae are always written with integers
 Experimental ratios are always fractions
 Two choices:
 Round to nearest whole number
 Multiply top and bottom to find ratio of whole
numbers with same value
 Choice depends on the type of substance
Hydrocarbons: A case for not
rounding
 There are millions of different hydrocarbons
 What if H:C is 2.20 rather than 2.26? An error of
only 3 %
 Formula becomes C5H11 rather than C4H9
 What if H:C is 2.30 rather than 2.26? An error of
only 2 %
 Formula becomes C3H7
 All formulae are reasonable
 So how do I know what the composition is?
 Additional knowledge about the substance is
helpful: melting point, boiling point, molar mass
Inorganic compounds: Rounding
makes sense
 Inorganic compounds tend to have few
compositions
 Iron forms three oxides: FeO, Fe3O4 and
Fe2O3
 Experimental formula FeO1.75 would indicate
Fe2O3 not FeO2
Practice empirical formula problem
 A compound contains 62.1 % C, 5.21 % H, 12.1 % N and
20.7 % O. What is the empirical formula?
Empirical and molecular formula
 Percent composition gives the empirical
(simplest) formula. It says nothing about the
molecular formula.
 Molecular formula describes number of
atoms in the molecule
 May be much larger than the empirical formula
in the case of molecular covalent compounds
 For ionic compounds empirical formula =
“molecular” formula
Elements and compounds can have molecular
formula different from simplest formula
Substance Empirical
formula
Molecular
formula
Substance Empirical
formula
Molecular
formula
Sulphur S
S8
Phosphorous
P4
Benzene CH
C6H6
Acetylene CH
C2H2
Ethylene CH2
C2H4
Cyclohexane
C6H12
P
CH2
Determination of molecular formula
 Require:
1. Empirical formula from percent composition
analysis
2. Molar mass from some other source
 Number of empirical formula units in molecule:
Molar mass
n
Empirical formula mass


There are n (AaBbCc) in molecule:
Molecular formula is AnaBnbCnc
Molecular formula of vitamin C
 Empirical formula of vitamin C is C3H4O3
 Molar mass vitamin C is 176.12 g/mol
 Mass of empirical formula = 88.06 g/mol
 (3 x 12.01 + 4 x 1.008 + 3 x 16.00)
 Number of formula units per molecule =
Molar mass vitamin C
176.12
n

2
Empirical formula mass vitamin C 88.06
 Molecular formula = 2(C3H4O3) = C6H8O6
Ionic solids do not have molecular
formulas
 Infinite lattices like ionic solids and
covalently bonded lattices are not molecular.
 The formula used for an ionic compound is
the same as the empirical formula – with
one or two exceptions
 Hg2Cl2 rather than HgCl
Structural formula provides more
information
 The molecular formula indicates the number of
atoms in the molecule
 The structural formula indicates how those atoms
are arranged
 C2H6O is the molecular formula for ethanol and
dimethyl ether
 Structural formula for ethanol is CH3CH2OH
 Structural formula for ether is CH3OCH3
 In a recipe we would need to use the structural
formula to identify the correct reagent
In some cases the mole contents of the
compound are obtained indirectly
 Analysis of the
hydrocarbon is performed
by combustion.
 Mole ratios of the elements
are derived from the mole
ratios of the combustion
products CO2 + H2O
(1 mol H2O ≡ 2 mol H)
(1 mol CO2 ≡ 1 mol C)
The molecular or empirical formula can
be used to determine the percent
composition





Formula of aspirin is C9H8O4
Molar mass is 180 g
Mass of C = 108 g
Mass of H = 8 g
Mass of O = 64 g
 % C = 108/180 x 100 %
 % H = 8/180 x 100 %
 % O = 64/180 x 100 %