Question

Empirical formula of a compound is \(\mathrm{CH}_{3} \mathrm{~N}\). If the empirical formula weight is equal to onefourth of its vapour density, find out the molecular formula of the compound.

Short answer

Answer: The molecular formula of the compound is C8H24N8.

Step-by-step solution

Calculate the empirical formula weight

First, let's calculate the weight of the empirical formula \(\mathrm{CH}_{3} \mathrm{N}\): Weight of carbon (C) = 12 Weight of hydrogen (H) = 1 Weight of nitrogen (N) = 14 Empirical formula weight (E) = 12 + (3 × 1) + 14 = 12 + 3 + 14 = 29

Calculate the vapor density

The empirical formula weight (E) is equal to one fourth of its vapor density (D). So, Vapor density (D) = 4 × Empirical formula weight (E) = 4 × 29 = 116

Determine the molecular weight

Since the vapor density is equal to half the molecular weight, we have: Molecular weight (M) = 2 × Vapor density (D) M = 2 × 116 = 232

Determine the molecular formula

Now, let's find the ratio of molecular weight (M) to empirical formula weight (E): Ratio = \(\frac{M}{E}\) = \(\frac{232}{29}\)= 8 The molecular formula of the compound is obtained by multiplying the empirical formula by the ratio (8). So, Molecular formula = 8 × \(\mathrm{CH}_{3} \mathrm{N}\) = \(\mathrm{C}_{8} \mathrm{H}_{24} \mathrm{N}_{8}\). Therefore, the molecular formula of the compound is \(\mathrm{C}_{8} \mathrm{H}_{24} \mathrm{N}_{8}\).