Empirical and Molecular Formulas Notes
(Source: http://www.ausetute.com.au/empirical.html)
Key Concepts
Empirical Formula of a compound shows the ratio of elements present in a
compound.
Molecular Formula of a compound shows how many atoms of each element
are present in a molecule of the compound.
The empirical formula mass of a compound refers to the sum of the atomic
masses of the elements present in the empirical formula.
The Molecular Mass (formula mass, formula weight or molecular weight) of a
compound is a multiple of the empirical formula mass.
MM = n x empirical formula mass
Empirical Formula can be calculated from the percentage (or percent)
composition of a compound.
Examples of Empirical and Molecular Formula
If carbon and hydrogen are present in a compound in a ratio of 1:2, the empirical formula for
the compound is CH2.
The empirical formula mass of this compound is: 12.0 + (2 x 1.0) = 14.0 g/mol
If we know the molecular mass of the compound is 28.0 g/mol then we can find the molecular
formula for the compound.
MM = n x empirical formula mass
28.0 = n x 14.0
n=2
So the molecular formula for the compound is 2 x empirical formula, ie, 2 x (CH 2) which is
C2H4
There are many compounds that can have the empirical formula CH2.
These include:
C2H4
C3H6
C3H6
C4H8
C4H8
(ethene or ethylene) molecular mass=28.0g/mol and n=2
(propene or propylene) molecular mass=42.0g/mol and n=3
(cyclopropane) molecular mass=42.0g/mol and n=3
(butene or butylene) molecular mass=56.0g/mol and n=4
(cyclobutane) molecular mass=56.0g/mol and n=4
Empirical and Molecular Formulas Notes
Calculating Empirical Formula from Percentage Composition
1.
2.
3.
Table
Assume 100g of sample
Convert all percentages to a mass in grams, eg, 21% = 21g, 9% = 9g
Find the relative atomic mass (r.a.m) of each element present using the Periodic
4.
Calculate the moles of each element present: n = mass ÷ r.a.m
5.
Divide the moles of each element by the smallest of these to get a mole ratio
6.
If the numbers in the mole ratio are all whole numbers (integers) convert this to an
empirical formula
7.
If the numbers in the mole ratio are NOT whole numbers, you will need to further
manipulate these until the mole ratio is a ratio of whole numbers (integers)
Recognizing the decimal equivalent of common fractions is very helpful!
Common Decimal
Equivalent Fraction
Mole Ratio Example
0.125
1/8
1 : 1.125 converts to 1 : 9/8
multiply throughout by 8 to give 8 : 9
0.25
1/4
1 : 0.25 converts to 1 : 1/4
multiply throughout by 4 to give 4 : 1
0.33
1/3
1 : 1.33 converts to 1 : 4/3
multiple throughout by 3 to give 3 : 4
0.375
3/8
1 : 1.375 converts to 1 : 11/8
multiply throughout by 8 to give 8 : 11
0.5
1/2
2 : 1.5 converts to 2 : 3/2
multiply throughout by 2 to give 4 : 3
0.625
5/8
1 : 1.625 converts to 1 : 13/8
multiply throughout by 8 to give 8 : 13
0.66
2/3
2 : 1.66 converts to 2 : 5/3
multiply throughout by 3 to give 6 : 5
0.875
7/8
1 : 0.875 converts to 1 : 7/8
multiply throughout by 8 to give 8 : 7
Empirical and Molecular Formulas Notes
Example 1
A compound is found to contain 47.25% copper and 52.75% chlorine.
Find the empirical formula for this compound.
element
mass in grams
r.a.m
moles = mass ÷ r.a.m
divide throughout by lowest number
Cu
Cl
47.25
52.75
63.6
35.5
47.25 ÷ 63.6 = 0.74 52.75 ÷ 35.5 = 1.49
0.74 ÷ 0.74 = 1
1.49 ÷ 0.74 = 2.01 = 2
Empirical formula for this compound is CuCl 2
Example 2
A compound with a molecular mass of 34.0g/mol is known to contain 5.88% hydrogen and
94.12% oxygen.
Find the molecular formula for this compound.
First, find the empirical formula of the compound.
element
mass in grams
r.a.m
moles = mass ÷ r.a.m
divide throughout by the smallest number
H
5.88
94.12
1.0
16.0
5.88 ÷ 1.0 = 5.88 94.12 ÷ 16.0 = 5.88
5.88 ÷ 5.88 = 1
Empirical formula is HO
Calculate the empirical formula mass: 1.0 + 16.0 = 17.0 g/mol
Molecular Mass = n x empirical formula mass
34.0 = n x 17.0
n = 34.0 ÷ 17.0 = 2
Molecular Formula is 2 x (HO) which is H2O2
O
5.88 ÷ 5.88 = 1
Empirical and Molecular Formulas Notes