Question

$0.92$ g of an organic compound containing carbon, hydrogen and oxygen was analysed by combustion method. The increase in the mass of the U-tube and the potash bulbs at the end of the operation was found to be $1.08$ g and $1.76$ g, respectively. Determine the percentage composition of the compound.

• A. $C=32.17\%;H=33.04\%$ and $O=34.79\%$
• B. $C=42.17\%;H=13.04\%$ and $O=44.79\%$
• C. $C=52.17\%;H=13.04\%$ and $O=34.79\%$
• D. $C=52.17\%;H=23.04\%$ and $O=24.79\%$

Answer 1

The correct option is C $C=52.17\%;H=13.04\%$ and $O=34.79\%$

Increase in the mass of U-tube $=1.08$ g
Mass of water formed $=1.08$ g
Since, $18$ g of water contains $2$ g of hydrogen.
$1.08$ g of water contains $\frac{2}{18}\times 1.08$ g of hydrogen.
Mass of the compound taken for analysis $=0.92$ g
Percentage of hydrogen $=\frac{2}{18}\times \frac{1.08}{0.92}\times \times 100=13.04$
Increase in the mass of potash bulbs $=1.76$ g
Mass of carbon dioxide formed $=1.76$ g
Since, $44$ g of carbon dioxide contains $12$ g carbon.
$1.76$ g of carbon dioxide contains $=\frac{12}{44}\times 1.76$ g carbon.
Percentage of carbon $=\frac{12}{44}\times \frac{1.76}{0.92}\times 100=52.17$
Percentage of oxygen $=100-$(percentage of C + percentage of H)
$=100-(52.17+13.04)$
$=100-65.21=34.79$
Composition of the compound: $C=52.17\%;H=13.04\%$ and $O=34.79\%$