Question

During an analysis of $255\text{mg}$ of a molecule that contains only $C,H$ and $O$, $561\text{mg}C{O}_2$ is produced along with $306\text{mg}{H}_{2}O$ on combustion. If the molar mass of the compound is $180\text{g/mol}$, what is the molecular formula of the compound ?

• A. $C_{16}{H}_4{O}_9$
• B. $C_{10}{H}_{21}{O}_9$
• C. $C_5{H}_{21}{O}_3$
• D. $C_9{H}_{24}{O}_3$

Answer 1

The correct option is D $C_9{H}_{24}{O}_3$

$C_x{H}_y{O}_z$
on combustion we have : (according general combustion reaction)
$C_x{H}_y{O}_z+(x+\frac{y}{4}-\frac{z}{2})O_2\to xC{O}_2+\frac{y}{2}{H}_{2}O$

moles of $C$ are equal to moles of $C{O}_2$ so :
$n_{C{O}_2}=\frac{561}{44}=12.75\times {10}^{-3}$
$n_C=n_{C{O}_2}=12.75\times {10}^{-3}\therefore w_C=12\times 12.75=153\text{mg}$

moles of $H$ are equal to twice of moles of $H_{2}O$ so :
$n_H=2\times n_{H_{2}O}=2\times \frac{306}{18}\times {10}^{-3}=34\times {10}^{-3}\therefore w_H=34\text{mg}$

$w_C+w_H=153+34=187\text{mg}$

$w_O=w_{\text{compound}}-187=255-187=68\text{mg}$

so moles of $O$ :
$n_O=\frac{68}{16}=4.25\times {10}^{-3}=4.25\text{mmol}$
therefore the ratio $C:H:O=12.75:34:4.25$
$=3:8:1$

therefore Empirical formula : $C_3{H}_{8}O$
Empirical mass $=60$
Empirical mass $=\frac{\text{molecular mass}}{n}$

(molar mass of compound $=180\text{g/mol}$ and n =integer)

$n=\frac{\text{molecular mass}}{\text{empirical mass}}=\frac{180}{60}=3$

so we get $n=3$

molecular formula $n=n\times \text{empirical formula}=3\times C_3{H}_{8}O$

so, Molecular formula $=C_9{H}_{24}{O}_3$