# Question 67dbe

Oct 5, 2015

Empirical Formula is ${C}_{4} {H}_{5}$ and the molecular formula is ${C}_{8} {H}_{10}$.

#### Explanation:

$\text{Number of moles of organic compound} = \frac{0.2612}{106}$
$\implies n = 2.4642 \times {10}^{-} 3 m o l$
$\text{Number of moles of } C {O}_{2} = \frac{0.8661}{44}$
$\implies n = 0.0197 m o l$
$\text{Number of moles of } {H}_{2} O = \frac{0.2250}{18}$
$\implies n = 0.0125 m o l$
The equation for the reaction is:
${C}_{a} {H}_{b} + a \cdot b \cdot {O}_{2} \to a \cdot C {O}_{2} + \frac{b}{2} \cdot {H}_{2} O$

Mole ratio of CO_2:"organic compund"= 0.0197/(2.4642xx10^-3
$= 8 : 1$
This means $8 m o l$ of $C {O}_{2}$ is produced for burning $1 m o l$ of the organic compound. The value of $a$ is therefore 8.
Mole ratio of H_2O: "organic compound" = 0.0125/(2.4642xx10^-3#
$\approx 5 : 1$
This shows $5 m o l$ of ${H}_{2} O$ is produced. So, $\frac{b}{2} = 5$, therefore $b$ is equal to 10.
Hence, the organic compound has molecular formula $C 8 H 10$. It's empirical formula would hence be ${C}_{4} {H}_{5}$. The name of this compound is Xylene, or dimethyl benzene.

**n.b. this is not an orthodox method. If this question is asked in an exam, you may not be awarded any marks for using this method.