Question

A sample of an unknown compound, composed only of carbon and hydrogen, produced 2.56g of CO2 and 0.786g of H2O in a combustion analysis. What is the empirical formula of the unknown compound?

Answer 1

Find moles of C:

$$n(C) = n(CO_2)= \frac{m}{M} = \frac{2.56}{44.01} = 0.0582 mol$$

Find moles of H

$$n(H) = 2\times n(H_2O) = 2\times \frac {m(H_2O)}{M(H2O)} = 2\times \frac{0.786}{18.02} = 0.0872 mol$$

Find ratio n(C):n(H)

$$n(C):n(H) = 0.0582:0.0872 = \frac{0.0582}{0.0582}:\frac{0.0872}{0.0582} = 1: 1.5 = 2:3$$

Empirical formula is

$$C_2H_3$$