A closed mixture of gases contains Helium ($\mathrm{He}$) and Methane ($\mathrm{CH}_4$) gases in a m

Question

A closed mixture of gases contains Helium ($\mathrm{He}$) and Methane ($\mathrm{CH}_4$) gases in a mass ratio of 1:2 (w/w). What is the exact molar ratio of the two gases in the mixture?

Professor PC Thomas Classes · Answer

(b) : Let the mass of $\mathrm{He}$ be $w$ and the mass of $\mathrm{CH}_4$ be $2w$. The molar mass of $\mathrm{He}$ is $4 \mathrm{~g/mol}$ and that of $\mathrm{CH}_4$ is $16 \mathrm{~g/mol}$. The molar ratio is $n_{\mathrm{He}} : n_{\mathrm{CH}_4} = (w/4) : (2w/16) = (w/4) : (w/8)$. Simplifying this gives $2 : 1$.