0.9g of a solid organic compound (mol. wt. 90) containing carbon, hydrogen, and oxygen was heated with oxygen corresponding to a volume of 224mL at STP

Question

$0.9 g$ of a solid organic compound (mol. wt. 90) containing carbon, hydrogen, and oxygen was heated with oxygen corresponding to a volume of $224 \cdot 0 mL$ at S.T.P.

After combustion the total volume of the gases was $560 mL$ at S. T.P. On treatment with $KOH$ solution the volume decreased to $112 mL$ at S.T.P. Determine the molecular formula of the compound.

Answer 1

SOLUTION : Wt. of compound $=0.9 g =\frac{0 \cdot 9}{90}$ mole $=0.01$ mole

Vol. of $CO _{2}$ formed $=(560-112) mL =448 mL$ at S.T.P.

Mole of $CO _{2}$ formed $=\frac{448}{22400}$ mole $=0.02$ mole

Volume of $O _{2}$ taken $=224 mL$,

The volume of gas not absorbed by $KOH$ must be $O _{2}$ which remained unreacted, i.e., $112 mL$.

$\therefore$ Vol. of $O_{2}$ reacted $=(224-112) mL .=112 mL$ at S.T.P.

Mole of $O _{2}$ reacted $=\frac{112}{22400}$ mole $=0.005$ mole

Let the formula of the compound be $C_{x} H_{y} O_{z}$

Reaction :

$C _{x} H _{y} O _{z}+\left(x+\frac{y}{4}-\frac{z}{2}\right) O _{2}=x CO _{2}^{-}+\frac{y}{2} H _{2} O$

1 mole $\left(x+\frac{y}{4}-\frac{z}{2}\right)$ mole $x$ mole $\quad \frac{y}{2}$ mole

$\therefore 0.01$ mole $0.01\left(x+\frac{y}{4}-\frac{z}{2}\right)$ mole $0.01 x$ mole $.01 \frac{y}{2}$ mole

Mole of $CO _{2}: 0 \cdot 01 x=0 \cdot 02: \quad \therefore \quad x=2$

Mole of $O _{2}: 0 \cdot 01\left(x+\frac{y}{4}-\frac{z}{2}\right)=0.005 \quad$ or, $2+\frac{y}{4}-\frac{z}{2}=\frac{0 \cdot(1) 05}{0 \cdot() i}=\frac{1}{2}$

$y-2 z=-6$

$\therefore$ or, Mol. wt. of the compound $=90 ; \quad$ Mol. wt. of $C_{x} H_{y} O_{z}=12 x+y+16 z$

$\therefore 12 x+y+16 z=90$ or, $y+16 z=90-12 \times 2=66($ here $x=2$ )

or, $y+16 z=66$

Subtraction (I) from (II), $18 z=72$ or $z=4$

From (I), $y=2 z-6=2 \times 4-6=2$

Therefore, formula of the compound $= C _{2} H _{2} O _{4}$.