Question

$0.3g$ of an organic compound containing $C,H$ and $O$ on combustion yields $0.44g$ of ${CO}_2$ and $0.18g$ of $H_{2}O$, with two $O$ atoms per molecule. Molecular formula of the compound is

• A. $C_2{H}_4{O}_2$
• B. ${CH}_2{O}_2$
• C. $C_3{H}_6{O}_2$
• D. $C_4{H}_8{O}_2$

Answer 1

The correct option is A $C_2{H}_4{O}_2$

We know that, in 44g of carbon di oxide, 12g of carbon is there.

Therefore in 0.44g of carbon di oxide, 0.12g of carbon will be there.

Hence, percent of carbon in the organic compound$=\frac{0.12}{0.3}\times 100$$=40\%$

Also, since in 18g of water, 2g of hydrogen is there. Hence, in 0.18g of water, 0.2g of hydrogen will be there.

Hence the percent of hydrogen in the oorganic compound$=\frac{0.2}{0.18}\times 100$$=6.6\%$

Hence, per cent of oxygen in the organic compound$=100-(40+6.6)$$=53.34\%$

Now, C:H:O$=\frac{40}{12}:\frac{6.66}{1.0}:\frac{53.34}{16}$$=3.33:6.66:3.33$$=1:2:1$

Hence the empirical formula of the compound$=C{H}_{2}O$

Now, since, in the question, it is mentioned that for each molecule, there are 2 atoms of oxygen, hence, we multiply the empirical formula by 2 to get its molecular formula.

Hence the molecular formula of the organic compound$=C_2{H}_4{O}_2$.

Hence, (a) is the correct answer.