Question

$0.92gm$ 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 $1.08gm$ and $1.76gm$, respectively. Determine the percentage composition of the compound.

Answer 1

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