Question

An organic compound ($C_x H_{2y} O_y$) was burnt with twice the amount of oxygen needed for complete combustion to $CO_2$ and $H_2O$. The hot gases when cooled to $0^{\circ}C$ and $1$ atm pressure, measured $2.24\, L$. The water collected during cooling weight $0.9 \,g$. The vapour pressure of pure water at $20^{\circ}C$ is $17.5\, mm \,Hg$ and is lowered by $0.104 \,mm$ when $50\, g$ of the organic compound are dissolved in $1000 \,g$ of water.
Give the molecular formula of the organic compound.

Answer 1

From lowering of vapour pressure information :
$ \frac{0.104}{17.5} =\chi_2 =\frac{n_2}{n_1+n_2} $
$\Rightarrow \frac{n_1}{n-2}+1 =168.27$
$\Rightarrow \frac{n_1}{n-2}=167.27$
$\Rightarrow \frac{100}{18} \times \frac{M}{50} =167.27$
$\Rightarrow M=150\, g/mol$
Also, the combustion reaction is :
$ C_xH_{2y}O_y + xO_2 \rightarrow \, xCO_2 +yH_2O$
$\because \, 18 \,y \,g$ of $ H_2O$ is produced from $1.0$ mole of compound.
$\therefore 0.9\, g \,$ of $\, H_2O$ will be produced from $\frac{0.9}{18y}=\frac{1}{20y} mol $
$\Rightarrow $ At the end, moles of $O_2$left =$\frac{x}{20y}$
moles of $CO_2 \,$ formed $=\frac{x}{20y}$
$\Rightarrow $ Total moles of gases at STP =$\frac{2x}{20y} =\frac{2.24}{22.4}$
$\Rightarrow x=y$
$\Rightarrow $ Molar mass; $150 = 12x + 2x + 16x = 30x$
$\Rightarrow x=\frac{150}{30}=5$
$\Rightarrow $ Formula $=C_5H_{10}O_5$