$10$ mL of a gaseous hydrocarbon was burnt completely in $80$ mL of $O_{2}$ at NTP. The remaining gas occupied $70$ mL at NTP. This volume became $50$ mL on treatment with KOH solution. What is the formula of the hydrocarbon?

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Solution

Let formula of hydrocarbon $= C_{x} H_{y}$
Volume of hydrocarbon taken $= 10 m L$
Volume of oxygen gas added $= 80 m L$
Combustion of hydrocarbon-
$C_{x} H_{y} + (x + \frac{y}{4}) O_{2} ⟶ x C O_{2} + \frac{y}{2} H_{2} O$
From the above reaction-
$1$ mole of $C_{x} H_{y}$ reacts with $(x + \frac{y}{4})$ mole of $O_{2}$ to produce $x$ mole of $C O_{2}$ .
Therefore by avogadro's law-
$1$ mL of $C_{x} H_{y}$ reacts with $(x + \frac{y}{4})$ mL of $O_{2}$ to produce $x$ mL of $C O_{2}$ .
$10$ mL of $C_{x} H_{y}$ reacts with $10 (x + \frac{y}{4})$ mL of $O_{2}$ to produce $10 x$ mL of $C O_{2}$ .
Therefore,
Volume of residual gas $(C O_{2} + unused O_{2}) = 70 m L$
Since $K O H$ absorbs whole $C O_{2}$ and volume of gas left after $K O H$ treatment is $50$ mL
Volume of $C O_{2}$ formed $= 70 - 50 = 20 m L$
From the above reaction of combustion of $C_{x} H_{y}$ -
$10 x mL of C O_{2} = 20 mL of C O_{2}$
$x = \frac{20}{10} = 2$
Now,
Volume of unused $O_{2} = 50 m L$
$∴$ Volume of $O_{2}$ used $4 = 80 - 50 = 30 m L$
Again from the above combustion reaction-
$10 (x + \frac{y}{4}) = 30$
$2 + \frac{y}{4} = 3$
$\Rightarrow y = 4$
Hence the molecular formula of hydrocarbo will be $C_{x} H_{y} = C_{2} H_{4}$ .

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