Question

An organic compound $C _{x} H _{2 y} O _{y}$ was burnt with twice the amount of oxygen needed for complete combustion to $CO _{2}$ and $H _{2} O$. The hot gases when cooled to $0^{\circ} C$ and 1 atm pressure measured $2 \cdot 24$ litres. The water vapour collected during cooling weighed $0.9 g$. If the molecular weight of the compound be 150, calculate the molecular formula of the compound.

Answer 1

Reaction :

Let us suppose that the mole of the organic compound taken $=z$ mole

and mole of $O _{2}$ required for combustion $=z x$ mole

mole of $O_{2}$ actually is (from the problem)

$=2 \times$ mole of $O_{2}$ required for combustion $=2 z x$

Hence. mole of $O _{2}$ unreacted $=2 z x-z x=z x$.

Careful analysis of the problem then indicates that $2 \cdot 24$ litres of gas obtained after combustion and cooling at S.T.P. is a mixture of $CO _{2}$ formed and the amount of unreacted $O _{2}$.

Vol. of gas $=2 \cdot 24$ Litres at S.T.P.

Therefore Mole of the gas $=\frac{2 \cdot 24}{22 \cdot 4} \text { mole }=0 \cdot 1 \text mole $

Mole of $CO _{2}$ formed + mole of unreacted $O _{2}=z x+z x=2 z x$

$2 z x=0 \cdot 1$

Mole of water formed $=\frac{0 \cdot 9}{18}$ mole $=0 \cdot 05$ mole, $z y=0 \cdot 05, \therefore 2 z y=0 \cdot 1 \quad \ldots$ (II) From (I) and (II), $\quad 2 z y=2 z x ; y=x$

Given that mol. wt. of the compound $=150 ; \quad \therefore \quad C_{x} H_{2 y} O_{y}=150$

or, $12 x+2 y+16 y=150$ or, $12 x+18 y=150$ or, $30 x=150 \quad(\because x=y)$

$\therefore x=5 \therefore$ Formula of the compound $= C _{5} H _{10} O _{5}$.