The weight percentage of $\mathrm{C}$ and $\mathrm{H}$ in a hydrocarbon is in the ratio of $4: 1$. What is…
The weight percentage of $\mathrm{C}$ and $\mathrm{H}$ in a hydrocarbon is in the ratio of $4: 1$. What is its empirical formula?
$\mathrm{CH}$
$\mathrm{CH}_2$
$\mathrm{CH}_3$
$\mathrm{CH}_4$
Solution
$\mathrm{M}(\mathrm{c})=12.01, \mathrm{M}(\mathrm{H})=1.0$ $\mathrm{m}(\mathrm{c})=4 \times \mathrm{m}(\mathrm{H})$ (Given) If $\mathrm{n}(\mathrm{c})=1$ and $\mathrm{n}(\mathrm{H})=3$ $\Rightarrow \mathrm{m}(\mathrm{c})=12.01$ and $\mathrm{m}(\mathrm{H})=3$ and $\mathrm{C}: \mathrm{H}$ mass ratio $=4: 1$ $\Rightarrow$ Empirical formula $=\mathrm{CH}_3$.