6.7 & 6.9 Empirical & Molecular
Formulae
pp. 289 – 293, 296 - 300
Empirical & Molecular Formulae
• The chemical word equation for the reaction
between aluminum and bromine liquid is as follows:
aluminum + bromine liquid → aluminum bromide
• Its balanced chemical equation is:
2Al(s) + 3Br2(l) → Al2Br6(s)
Empirical & Molecular Formulae
• At first glance, the formula for aluminum bromide in
the equation, Al2Br6 appears to be a strange one.
• Aluminum’s valence is 3 and bromine’s valence is 1
and using the cross-over rule, the formula would end
up as being AlBr3.
• However, through chemical analysis, the molecular
formula is Al2Br6.
• The molecular formula is the exact formula of the
compound.
• The empirical formula is the simplest whole
number ratio of the elements’ atoms in a
compound.
– The empirical formula of aluminum bromide is AlBr3.
• In many cases, the empirical formula is the
molecular formula of a compound.
– Water, H2O, is such a case.
• The elements’ respective subscripts in the formula
show a mole ratio relationship.
• The formula for water, H2O, shows us that on a
macroscopic level, there are 2 moles of hydrogen
for every 1 mole of oxygen atoms.
• To calculate an empirical or a molecular formula for
a compound, the given masses for each element
must be converted into their respective whole
number mole values.
Sample Calculation
• What is the empirical formula of a compound which
is found by analysis to contain 2.2% hydrogen,
26.7% carbon and 71.1% oxygen?
• If we assume a 100g sample,
G: mH = 2.2g
mC = 26.7g
R: Empirical Formula
mO = 71.1g
CHO = ?
A: n = m ÷ M
S: nH = 2.2 g ÷ 1.01 g/mol = 2.18 mol
nC = 26.7 g ÷ 12.01 g/mol = 2.22 mol
nO = 71.1 g ÷ 16.00 g/mol = 4.44 mol
Mole Ratio:
H
2.18 mol
:
C
: 2.22 mol
:
O
: 4.44 mol
Dividing each by the lowest mole value, 2.18
1 mol
: 1.02 mol
: 2.04 mol
If the values are within 0.1 of a whole number,
the numbers can be rounded off to the nearest
whole number
1 mol
: 1 mol
: 2 mol
Therefore, the empirical formula is HCO2
Example # 2
• A hydrocarbon, upon analysis, shows the following
mole to mole ratio relationship between carbon
and hydrogen
Mole Ratio:
C
0.166 mol
:
H
: 0.444 mol
Dividing each by the lowest mole value, 0.166
1 mol
: 2.67 mol
2.67 cannot be rounded, so try multiplying
each number by 2
2 mol
: 5.3 mol
5.3 cannot be rounded, so try multiplying
each number from the previous page by 3
3 mol
: 8.02 mol
Round each to the nearest whole number
3 mol
:
8 mol
Therefore, the empirical formula is C3H8
Molecular Formulae
• The molecular formula of a compound gives you the
actual composition of a molecule.
• Its calculation is not much different from that of the
empirical formula.
• Once you calculate the empirical formula you need
to do one further calculation:
x = Mcompound ÷ Mempirical formula
• You will then multiply each subscript in the
empirical formula by x
Sample Calculation
• The empirical formula of a compound is determined
to be CH. Its molecular molar mass is 104.16 g/mol.
Determine its molecular formula.
G: MCH = 13.02 g/mol
Mcompound = 104.16 g/mol
R: Molecular formula of compound
A: x = Mcompound ÷ Mempirical formula
S: x = 104.16 g/mol ÷ 13.02 g/mol
=8
P: Therefore the molecular formula is C8H8
Homework
• Read pp. 289 – 292
– Answer # 1 – 11 p. 293
• Read pp. 304 – 305
– Answer # 1 – 9 p. 300