Question

A gaseous compound of carbon and nitrogen containing 53.8% by weight of nitrogen was found to have a vapour density of 25.8. What is the molecular formula of the compound? (At. wt. C = 12, N = 14)

Answer 1

Calculation of empirical formula

% of N = 53.8%, % of C = 100 53.8 = 46·2%

Moles of N = \(\frac{53.8}{14} = 3.85\)

Moles of C = \(\frac{46.2}{12} = 3.85\)

Moles of C : Moles of N

3.85 moles : 3.85 mole

1 mole : 1 mole

1 atom : 1 atom

Empirical formula of the compound = CN

Calculation of molecular formula 

Mol. wt. of compound = 2 x V.D. = 2 x 25.8 = 51.6

Whole No. ratio, n = \(\frac{\text{Mol. wt.}}{\text{Empirical formula wt.}}\)

\(= \frac{51.6}{12 + 14}\)

\(= \frac{51.6}{26}\)

\(= 2\)

\(\therefore \) Molecular formula = (Empirical formula)_n = (CN)₂ = C₂N₂

Hence the molecular formula of the compound is C₂N₂.