Calculate the empirical and molecular formula of an organic compound whose percentage composition is C=70.54%, H=5.87% and O=23.52%

Given:

Molecular weight of compound=136 g.

Percentage composition is C=70.54%, H=5.87% and O=23.52%

Step-1: Calculate the empirical formula

• Empirical formula: Empirical formula is the simplest formula of the compound which gives the simplest whole number ratio of the atoms of the various elements present in one molecule of the compound.

Element	% Of Element	Atomic mass (g)	Atomic ratio	Simplest ratio
Carbon (C)	70.54	12	70.5412=5.87	5.871.47=3.99≈4
Hydrogen (H)	5.87	1	5.871=5.87	5.871.47=3.99≈4
Oxygen (O)	23.52	16	23.5216=1.47	1.471.47=1

As the simplest ratio of Carbon is 4, Hydrogen is 4 and Oxygen is 1, therefore the empirical formula comes out to be C₄H₄O_.

Step-2: Calculate the empirical and molecular mass

Empirical formula weight: Empirical formula weight is obtained by the addition of the atomic weight of the various atoms present in the empirical formula.

Atomic weight of Carbon= 12 g

Atomic weight of Hydrogen= 1 g

Atomic weight of Oxygen= 16 g

Empirical formula weight= 4(Atomic weight of Carbon) + 4(Atomic weight of Hydrogen) + Atomic weight of Oxygen

Empirical formula weight= 4(12) + 4(1) + 16= 68 g

Molecular formula weight: Molecular formula weight is obtained by the addition of the atomic weight of the various atoms present in the molecular formula.

Molecular weight =136 g (given)

Step-3: Calculate the molecular formula

• Molecular formula: Molecular formula is the actual formula of the compound which gives the actual number of the atoms present in the molecule of the compound.
• Relationship between empirical formula and molecular formula is given as follows: Molecular formula = n × Empirical formula, where n is an integer such as 1,2,3, etc.
• n=MolecularweightofthecompoundEmpiricalformulamass
• n=13668=2

As stated above, Molecular formula = n× Empirical formula, and empirical formula is C₄H₄O.

Therefore, the molecular formula is (C₄H₄O)₂ or C₈H₈O₂.