Review
 Chemical formulas are very important; they:
 state which elements are in the molecule
 give the exact number of atoms of each element that are
in the molecule
 may give an indication as to how the elements are
bonded together
The law of definite proportions
 When elements combine to form compounds, they do
so in a definite proportion.
 it is also sometimes called the law of constant
composition.
 The percentage composition of an element can be
expressed as:
Examples
 Water has the chemical formula H2O. In terms of mass,
its molecule is always made up of 11% hydrogen and
thus the remaining mass (89%) must be oxygen.
 Notice how much more massive the oxygen atom is in
this image of the water molecule.
 Similarly, CO2 is always made up of 27.3% carbon.
Percentage composition of compounds
 The percentage composition is the percentage of an
element in a compound, in relation to its total mass.
Example 1
 Calculate the percentage composition (by mass) of
Al2O3.
MM(Al2O3) = 2(27 g/mol) + 3(16 g/mol) = 102 g/mol
% Al = 2(27 g/mol)/102 g/mol x 100% = 52.9%
% O = 3(16 g/mol)/102 g/mol = 47.1%
So Al2O3 is 52.9% aluminum and 47.1% oxygen.
Example 2
 Calculate the percentage composition by mass of sodium
sulphate, Na2SO4.
MM(Na2SO4) = 2(23 g/mol) + 1(32.1 g/mol) +4(16
g/mol) = 142.1 g/mol
% Na = 2(23 g/mol)/142.1 g/mol = 32.4%
% S = 1(32.1 g/mol)/142.1 g/mol = 22.6%
% O = 4(16 g/mol)/142.1 g/mol = 45.0%
So, sodium sulphate is 32.4% sodium, 22.6% sulphur
and 45.0% oxygen.
Check Your Understanding
1. 13.0 g of tin is burned in the presence of oxygen to form 16.5 g of
tin (IV) oxide. Which of the following represents the percentage
composition of tin by mass in this compound?
a) 33.5%
b) 44.1%
c) 78.8%
d) 85.1%
2. 5.3 g of mercury (II) oxide is decomposed to form 4.9 g of
mercury. What is the percentage composition (mass) of mercury
in the compound?
a) 7.5%
b) 16.0%
c) 59.6%
d) 92.5%
Empirical Formula
 A chemical's empirical formula indicates the
simplest ratio in which the different atoms are
combined. It is also called the simplest formula.
 For example the empirical formula for acetylene
(C2H2) is CH, since there is 1 C for every H.
Empirical Formula (con’t)
 However, benzene also has the empirical formula CH;
but its chemical formula is C6H6.
 Benzene and acetylene are not the same chemical;
each compound is unique in terms of physical and
chemical properties.
Steps for finding the empirical
formula:
1. Consider the amounts you are given as being in
units of grams.
2. If percentages are given instead of grams, assume
them to be grams (since percentages add up to
100, then your sample's mass will be 100 g).
3. Convert the grams to moles for each element.
4. Find the smallest whole number ratio of moles
for each element.
Example
3
A sample of calcium chloride contains 1.82 g of calcium
and 3.23 g of chlorine. What is the empirical formula
for the compound?
Ca: 1.82 g/40.1 g/mol = 0.0454 mol
Cl: 3.23 g/35.5 g/mol = 0.0910 mol
mole ratio = Cl:Ca = 0.0910 mol:0.0454 mol = 2:1
So there are 2 mol of Cl for every 1 mol of Ca
The empirical formula is CaCl2.
Example 4
The elemental analysis of an unknown organic compound
returns the following data:
%C = 64.6% %H = 10.8% %O = 24.6%
What is the empirical formula for this compound?
Assuming a 100 g sample, divide the percentages by the molar masses
of the elements to determine the number of moles of each element
present in the formula:
C: 64.6 g / 12.001 g/mol = 5.38 mol C
H: 10.8 g / 1.008 g/mol = 10.7 mol H
O: 24.6 g / 16.00 g/mol = 1.54 mol O
Divide the number of moles by the smallest number of moles to
determine the relative ratios of the elements.
C:H:O = 5.38 / 1.54 : 10.7 / 1.54 : 1.54 / 1.54 = 3.49:6.95:1.00
This gives the formula C3.49H6.95O1.00
Example 4 (con’t)
 Remember that elements only combine in whole
number ratios.
 If the ratios are close to whole numbers then we know
the empirical formula.
 If they are not, then you must multiply the ratios by a
whole number until all are relatively close to an integer
value.
 Here, we multiply the above formula by 2.
 Therefore, the empirical formula is C6.98H13.9O2.00
You can round this off to C7H14O2
Check Your Understanding
3. Suppose a compound is analyzed and reveals that is
composed of 85.7% carbon and 14.3% hydrogen. Find the
empirical formula for this compound.
4. Determine the empirical formula for a compound which is
broken down into 6.16 g of sulphur and 8.84 g sodium.
Molecular Formula
 The molecular formula indicates the number of
atoms of each element in a molecule. It is determined
using the empirical formula and the molar mass.
Steps for finding the molecular
formula:
1. Determine the compound's empirical formula.
2. Divide the compound's molar mass by the mass
of the empirical formula.
3. Multiply each subscript in the empirical formula
by this number.
Example 5
 Hydrogen peroxide is made of 5.03% hydrogen and 79.87% oxygen. Its
molar mass is 34.01 g/mol. Find its molecular formula.
First, you need to find the empirical formula:
Percentages are given, so you can assume that you are working with a 100
g sample that is 5.03 g of hydrogen and 79.87 g of oxygen.
H: 5.03 g/1.0079 g/mol = 4.99 mol
O: 79.87 g/15.9994 g/mol = 4.99 mol
mole ratio = H:O = 4.99 mol:4.99 mol = 1:1
So there is 1 mol of H for each mole of O.
The empirical formula is HO.
Now, you can find the molecular formula:
MM(HO) = 17.0073 g / mol
MM(hydrogen peroxide) = 34.01 g/mol
MM(hydrogen peroxide)/MM(HO)=34.01 g/mol /17.0073 g/mol
=1.9997 =2
Therefore, each subscript in the empirical formula is multiplied by 2. That
gives H2O2.
So, the molecular formula of hydrogen peroxide is H2O2.
Check Your Understanding
5. If 4.04 g of nitrogen combine with 11.46 g of oxygen to produce a
compound with a molar mass of 108.0 g/mol, what is the molecular
formula of this compound?
Check Your Understanding
6. Determine the molecular formula of a compound with an empirical
formula of NH2 and a molar mass of 32.06 g/mol.
Here is a comparison between three molecular formulas that have
the same empirical formula of CH2O. Notice how they have vastly
different properties.
Substance
Formaldehyde
Acetic acid
Glucose
Empirical formula
CH2O
CH2O
CH2O
Molecular formula
CH2O
C2H4O2
C6H12O6
biological
preservative
vinegar
sweetener
State
gas
liquid
solid
Molar mass
(g/mol)
30
60
180
Melting point
(°C)
-92
16
140
Boiling point
(°C)
-19.5
118
decomposes
Use
Physical
properties