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Oxidation of Alkanes

An organic co...

Question

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

C_{5} H_{10} O_{5}

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C_{10} H_{20} O_{10}

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C_{4} H_{8} O_{4}

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None of these

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Solution

The correct option is D None of these
The molecular formula of the compound is $C_{x} H_{2 y} O_{y}$
$C_{x} H_{2 y} O_{y} + x O_{2} \to x C O_{2} + y H_{2} O$ .
The amount of oxygen is twice the required amount i.e 2x.
The hot gases when cooled to $0^{0} C$ and 1 atm pressure measured 2.24L.
This corresponds to 0.1 mole as 22.4 L of any gas at NTP corresponds to 1 mole.
$x + x + y = 0.1$
$2 x + y = 0.1$ .....(1)
The water collected during cooling weighs 0.9 g. This correspnds to $\frac{0.9}{18} = 0.05$ moles.
Thus $y = 0.05$ .....(2)
Substitute equation (2) in equation (1)
$2 x + 0.05 = 0.1$
$2 x = 0.05$
$x = 0.025$ ......(3)
The ration $x : y$ is $0.025 : 0.05$ or $1 : 2$ .
Hence, the empirical formula of the organic compound is $C H_{4} O_{2}$ .
The empirical formula mass is $12 + 4 + 32 = 48 g / m o l$ .
$\frac{p^{0} - p}{p^{0}} = \frac{W_{2}}{M_{2}} \times \frac{M_{1}}{W_{1}}$

$\frac{0.104}{17.5} = \frac{50}{M_{2}} \times \frac{0.018}{1000}$

$\frac{0.104}{17.5} = \frac{50}{M_{2}} \times \frac{0.018}{1000}$

$M_{2} = 0.151 k g = 151 g$
$r = \frac{Molecular mass}{empirical mass} = \frac{151}{48} = 3.1 ≃ 3$
Therefore, the molecular formula of the organic gas is $3 \times C H_{4} O_{2} = C_{3} H_{12} O_{6}$ .

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