# Question #343ff

Sep 27, 2016

Let the molecular formula of HC be ${C}_{x} {H}_{y}$

And the balanced equation of the combustion reaction of the HC in oxygen is

${C}_{x} {H}_{y} + \left(x + \frac{y}{4}\right) {O}_{2} \left(g\right) \to x C {O}_{2} \left(g\right) + \frac{y}{2} {H}_{2} O \left(l\right)$

So by this equation the stochiometric ratio of $C {O}_{2}$ and ${H}_{2} O$ produced on combustion is
$= \frac{x m o l}{\frac{y}{2} m o l} = \frac{44 x g}{\frac{18}{2} y g} = \frac{44 x}{9 y}$
$\text{Where " 44g/"mol"" is the molar mass of "CO_2 and 18g/"mol" " is the molar mass of } {H}_{2} O$

But by the given data this ratio is

$\frac{321 m g}{64 m g} = \frac{321}{64} \approx 5$

So equating these two we get

$\frac{44 x}{9 y} = 5$

$\implies \frac{x}{y} = \frac{45}{44} \approx \frac{1}{1}$

Since the ratio of number of atoms C and H in the HC molecule is $1 : 1$ then we can easily say the Empirical formula of HC is $\textcolor{red}{C H}$