Question

An organic compound contains $C,H$ and $O$. $0.3$ g of this compound on combustion yielded $0.44$ g of $C{O}_2$ and $0.18$ g of $H_{2}O.$ If the weight of $1$ mole of compound is $60$, then the molecular formula is :

• A. $C_2{H}_4{O}_2$
• B. $C_4{H}_6{O}_2$
• C. $C_2{H}_{6}O$
• D. $C_2{H}_{4}O$

Answer 1

Answer: (A)

$C_2{H}_4{O}_2$

A given organic compound contains $C,H$ and $O$.

$0.3$ g of the compound on combustion yielded $0.44$ g $C{O}_2$ and $0.18$ g of $H_{2}O$.

$100$ g of this compound on combustion will yield $146.7$ g ($3.33$ moles) $C{O}_2$ and $60$ g ($3.33$ moles) of $H_{2}O$.

Thus, the number of moles of $C$ and $H$ present in $100$ g of sample are $3.33$ and $6.67$ respectively.

The amount of oxygen present in the sample is $100-(40+6.67)=53.3$ g or $3.33$ moles.

Thus, the number of moles of $C,H$ and $O$ present in $100$ g of sample are $3.33,6.67$ and $3.33$ respectively.

Thus, the whole number ratio of $C,H$ and $O$ is $1:2:1$.

Thus the empirical formula of the substance is $C{H}_{2}O$.

$n=\frac{\text{molar mass of the substance}}{\text{empirical formula mass}}=\frac{60}{30}=2$

$\text{Molecular formula}=n\times empiricalformula=2\times C{H}_{2}O=C_2{H}_4{O}_2$