Question

On complete combustion, $0.246g$ of an organic compound gave $0.198g$ of $C{O}_2$ and $0.1014g$ of $H_{2}O$ . The ratio of carbon and hydrogen atoms in the compound is

• A. 1 : 3
• B. 1 : 2
• C. 2 : 5
• D. 2 : 7

Answer 1

Answer: (C)

$\%$ of $C=\frac{12}{44}\times \frac{\text{Weight of CO2 × 100}}{\text{weight of organic compound}}$
$=\frac{12}{44}\times \frac{0.198\times 100}{0.246}=21.95\%$
$\%$ of $H=\frac{2}{18}\times \frac{\text{weight of H2O×100}}{\text{weight of organic compound}}$
$=\frac{2}{18}\times \frac{0.1014\times 100}{0.246}=4.57$
$\%$ of $O=\left(100-21.95-4.57\right)=73.47$
Moles of $C=\frac{21.95}{12}=1.83$
Atoms of $C=1.83\times N_A$
Moles of $H=\frac{4.57}{1}=4.57$
Atoms of $C=4.57\times N_A$
Moles of $O=\frac{73.47}{16}=4.59$
Atoms of $O=4.59\times N_A$
Simplest ratio $:C:H:O=1.83:4.57:4.59$
$=1:2.5:2.5=2:5:5$
Thus, the ratio of atoms of carbon and hydrogen in the compound is $2:5$