Question

An alcohol contains 64.81% C, 13.60% H, and 21.59% O. The molecular

mass of this compound is 74 g/mole. Determine the Empirical formula

and Molecular formula.

Answer 1

Molecular Mass of H = 1 g/mole

Molecular Mass of O = 16 g/mole

Molecular Mass of C = 12 g/mole

Molecular Mass of Compound = 74 g/mole

% of C present in Compound = 64.81%

Mass of C in the compound = (64.81/100) * 74 = 48.02g

So no of C atoms = 48.02/12 = 4

% of H present in Compound = 13.60%

Mass of H in the compound = (13.60/100) * 74 = 10g

So no of H atoms = 10/1 = 10

% of O present in Compound = 21.59%

Mass of O in the compound = (21.59/100) * 74 = 15.99g

So no of O atoms = 15.99/16 = 1

So Empirical Formula of Compound = (C₄H₁₀O)_n

and Molecular Formula = C₄H₁₀O