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Empirical and Molecular Formula

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 (CH₂) which is C₂H₄

There are many compounds that can have the empirical formula CH₂.
These include:

C₂H₄ (ethene or ethylene) molecular mass=28.0g/mol and n=2
C₃H₆ (propene or propylene) molecular mass=42.0g/mol and n=3
C₃H₆ (cyclopropane) molecular mass=42.0g/mol and n=3
C₄H₈ (butene or butylene) molecular mass=56.0g/mol and n=4
C₄H₈ (cyclobutane) molecular mass=56.0g/mol and n=4

Calculating Empirical Formula from Percentage Composition

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 Table
Calculate the moles of each element present: n = mass ÷ r.a.m
Divide the moles of each element by the smallest of these to get a mole ratio
If the numbers in the mole ratio are all whole numbers (integers) convert this to an empirical formula

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)

Recognising 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

Example 1

A compound is found to contain 47.25% copper and 52.75% chlorine.
Find the empirical formula for this compound.

element Cu Cl

mass in grams 47.25 52.75

r.a.m 63.6 35.5

moles = mass ÷ r.a.m 47.25 ÷ 63.6 = 0.74 52.75 ÷ 35.5 = 1.49

divide throughout by lowest number 0.74 ÷ 0.74 = 1 1.49 ÷ 0.74 = 2.01 = 2

Empirical formula for this compound is CuCl₂

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 H O

mass in grams 5.88 94.12

r.a.m 1.0 16.0

moles = mass ÷ r.a.m 5.88 ÷ 1.0 = 5.88 94.12 ÷ 16.0 = 5.88

divide throughout by the smallest number 5.88 ÷ 5.88 = 1 5.88 ÷ 5.88 = 1