Question

A 3.1 sample of an unknown organic gas molecules composed of carbon, hydrogen and oxygen undergoes complete combustion to produce 4.4g CO2 and 2.7g H20. Calculate the empirical and molecular formular of the gas molecule if the molar mass is 62g/mol

Answer 1

mass of C in CO₂ = 12/44 × 4.4 = 1.2g

mass of H in H₂O = 2/18 × 2.7 = 0.3g

mass of O in the organic compound = 3.1g - (1.2 + 0.3)g = 1.6g

C = 1.2/12 = 0.1/0.1 = 1

H = 0.3/1 = 0.3/0.1 = 3

O = 1.6/16 = 0.1/0.1 = 1

$\therefore$ the empirical formula of the compound is CH₃O

(CH₃O)_x = 62

(12 + (3×1) + 16)x = 62

x(31) = 62

x = 2

Molecular formula = (CH₃O)₂ = C₂H₆O₂

$\therefore$ the molecular formula of the compound is C₂H₆O₂