Question

$180\text{g}$ of a compound containing $C,H$ and $O$ reacts fully with excess oxygen and produces $264\text{g}$ of $C{O}_2$ and $108\text{g}$ of $H_{2}O$. If the ratio of molar mass of the compound to the empirical formula mass is $6$ then find the molecular formula of given compound.

Answer 1

Mass of carbon produced $=\frac{12}{44}\times 264=72\text{g}$
Number of moles of carbon $=\frac{72}{12}=6\text{mol}$
Mass of hydrogen produced: $=\frac{2}{18}\times 108=12\text{g}$
Number of moles of hydrogen $\left(H\right)$$=\frac{12}{1}=12\text{mol}$
Mass of oxygen in $180\text{g}$ of sample $=180-72-12=96\text{g}$
Number of moles of oxygen atoms $=\frac{96}{16}=6\text{mol}$
So, ratio of moles in $180\text{g}$ of sample:
$n_C:n_H:n_O=1:2:1$
Emprical formula of the given compound is $C{H}_{2}O$.
Molecular formula of the given compound is $=\text{n}\times \text{emprical formula}$
Where, $n$ is ratio of the molar mass to the empirical formula mass which is $6$.
Molecular formula of the given compound is $C_6{H}_{12}{O}_6$.