Empirical and Molecular Formula

Empirical formula is the simplest formula which shows its composition by mass. It is sometimes known as simplest formula.

Empirical formula is the formula which shows the simplest whole number ratio of atoms of each element in a compound.

Empirical formula is usually determined by considering experimental data. That is why it is called “empirical” which means “based on experimentation”.

Molecular formula is the formula which shows the actual number of atoms of each element in a compound.

Compound Empirical formula Molecular formula
Benzene CH C6H6
Butane CH2 C4H8
Ethene CH2 C2H4
Hydrogen peroxide HO H2O2
Water H2O H2O

CALCULATIONS OF EMPIRICAL FORMULA

Example 1:

Solution

A compound was analyzed and found to contain 13g of calcium, 10.8g of oxygen and 0.675g of hydrogen. Calculate its empirical formula.
Elements Ca O H
Composition by mass(g) 13 0.8 0.675
Relative atomic mass40 16 1
Divide mass by relative atomic mass 13/40 =0.325 10.8/16 = 0.675 0.675/1 = 0.675
Divide by smallest number 0.325/0.325 = 1 0.675/0.325 = 2 0.675/0.325 = 2
Whole number ratio 1 2 2
The empirical formula is Ca(OH)2

NOTE

In some calculations the stage of dividing by the smallest number does not result to a simple whole number ratio. In this case, the values obtained should not be rounded up or rounded down. But one must look at the numbers and see if there is some factor that could be multiplied by to get each one a whole number.

Example 2:


Calculate the empirical formula of a compound with the following percentage composition: C = 39.13%, O = 52.17% and H= 8.70%.

Soln.

Element C H O
% composition 39.13 8.70 52.17
Relative atomic mass 12 1 16
% composition/R.A.M 39.13/12 =3.26 8.70/1 = 8.70 52.17/16 = 3.26
Divide by smallest number 3.26/3.26 =1 8.70/3.26 = 2.66 3.26/3.26 = 1
Multiply by 3 to get whole number 1×3 =3 2.66 ×3 =8 1×3=3
Whole number ratio 3 8 3
The empirical formula is C3H8O3.

The Molecular Formula Calculations

Molecular formula relies on empirical formula and relative molecular masses of a compound.

Molecular formula = n(empirical formula).
Where n= a whole number
and
n can be obtained by relation
Whole number (n) = Relative Molecular Mass of the compound
                   Relative Molecular Mass of empirical formula

If you have been given a vapour density instead of relative molecular mass then the relation between them is:

Relative molecular mass =Molar mass = 2 × Vapour density

Example 3:

A certain compound contains 73.47% carbon and 10.20% hydrogen by mass, the remaining percentage is of oxygen. If this compound has a relative molecular mass of 98. Calculate its molecular formula. Soln.
Elements C H O
Percentage composition 73.47 10.20 16.33
Relative atomic mass 12 1 16
% composition divide by R.A.M 73.47/12 = 6.1225 10.20/1 = 10.20 16.33/16 = 1.02
Divide by smallest number 6.225/1.02 = 6 10.20/1.02 =10 1.02/1.02 =1
Whole number ratios 6 10 1
The empirical formula is C6H10O Thus
Molecula formula = n(empirical formula)
 n = Mr of the compound
     Mr of the empirical formula
Mr of the empirical formula is = (12×6) +(10×1) + 16
= 98
Therefore, n=98/98 =1
The molecular formula = 1(C6H10O)
The molecular formula is C6H10O.

A compound contains 40% carbon, 6.67% hydrogen, and 53.3% oxygen by mass. What is its empirical formula?(CH2O)

Empirical formula and water of crystallization

Water of crystallization refers to the water molecules that are tightly bound within the crystal lattice of certain compounds. This phenomenon is commonly observed in inorganic salts and plays a crucial role in their physical properties.

Read here.

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