Question

On complete combustion, $ 0.246\, g $ of an organic compound gave $ 0.198\, g $ of $ CO_{2} $ and $ 0.1014\, g $ 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 $CO_{2}$ $\times$ 100}}{\text{weight of organic compound}}$
$=\frac{12}{44}\times\frac{0.198\times100}{0.246}=21.95\%$
$\%$ of $H=\frac{2}{18}\times\frac{\text{weight of $H_{2}O\times100$}}{\text{weight of organic compound}}$
$=\frac{2}{18}\times\frac{0.1014\times100}{0.246}=4.57$
$\%$ 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 $2:5$