Question

The molecular formula of a gas with vapour density 15 and empirical formula \(\mathrm{CH}_{3}\) is

Short answer

Answer: The molecular formula of the gas is C2H6.

Step-by-step solution

Determine the empirical formula weight

The empirical formula is \(\mathrm{CH}_3\). We need to compute its molecular weight using the atomic weights of Carbon (C) and Hydrogen (H): Weight of 1 Carbon atom = 12.01 atomic units (a.u.) Weight of 1 Hydrogen atom = 1.008 a.u. Empirical formula weight = 1 Carbon atom + 3 Hydrogen atoms Empirical formula weight = 12.01 a.u. + 3(1.008 a.u.) = 12.01 + 3.024 = 15.034 a.u.

Calculate molecular weight using vapor density

Vapor density is half the molecular weight of the gas. So, the molecular weight of the gas can be found by multiplying the vapor density by 2: Molecular weight (MW) = 2 x vapor density Molecular weight (MW) = 2 x 15 Molecular weight (MW) = 30

Find the ratio of Molecular weight to empirical formula weight

We need to divide the molecular weight of the gas by the empirical formula weight to find the multiple n that corresponds to the molecular formula: n = Molecular weight / Empirical formula weight n = 30 / 15.034 n ≈ 1.998 ≈ 2

Determine the molecular formula

Now that we have the multiple n, we can multiply the empirical formula by that value to get the molecular formula: Molecular formula = (Empirical formula) x n Molecular formula = (\(\mathrm{CH}_3\)) x 2 Therefore, the molecular formula of the gas is \(\mathrm{C}_2\mathrm{H}_6\).