Question

0.90 g

of a solid organic compound (molecular weight

= 90 gm

) containing

C, H

and

O

was heated with oxygen corresponding to a volume of

224 mL

at STP. After the combustion, the total volume of the gases was

560 mL

at STP. On treatment with

KOH

, the volume decreased to

112 mL

. Determine the molecular formula of the compound.

Accepted Answer

Let the given organic compound $= C_{x} H_{y} O_{z}$

Thus, molecular weight $= [12 (x) + 1 (y) + 16 (z)] = 90 gm .$

The chemical equation for combustion is,

$C_{x} H_{y} O_{z} + (x + \frac{y}{4} - \frac{z}{2}) O_{2} \to {xCO}_{2} + \frac{y}{2} H_{2} O 1 (x + \frac{y}{4} - \frac{z}{2}) x$

Number of moles of hydrocarbon $= \frac{0.9}{90} = 0.01$

Volume of oxygen required is equal to $(0.01 \times 22400 mL (x + \frac{y}{4} - \frac{z}{2}))$

Volume of oxygen left over $= [224 mL - (0.01 \times 22400 mL (x + \frac{y}{4} - \frac{z}{2}))]$

Volume of ${CO}_{2}$ released $= 0.01 \times 22400 mL \times' x' = 224 x mL$ .

So after the combustion,

Total volume of gas $=$ Volume of oxygen remaining $+$ Volume of ${CO}_{2} = 560 mL$

$[224 mL - (22400 mL \times 0.01 \times (x + \frac{y}{4} - \frac{z}{2})) + 224 x mL] = 560 mL [224 mL - (224 mL \times (x + \frac{y}{4} - \frac{z}{2})) + 224 x mL] = 560 mL [224 (- x - \frac{y}{4} + \frac{z}{2} + x)] = 336 mL \frac{z}{2} - \frac{y}{4} = 1.5 2 z - y = 6 \dots (2)$

${CO}_{2}$ is absorbed in $KOH$ . Hence, volume of ${CO}_{2}$ absorbed $= 224 x = 560 - 112 = 448 \Rightarrow x = 2$

Putting known values and solving equation $(1)$ and $(2)$ Simultaneously, we get

$y = 2 and z = 4$ .

Hence, the carbon compound is $C_{2} H_{2} O_{4}$ .

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