$7.5 m L$ of a hydrocarbon gas was exploded with an excess of oxygen. On cooling, it was found to have undergone a contraction of 1 $5 m L$ . If the vapour density of the hydrocarbon is $14$ , determine its molecular formula.(C=12, H=1)

C H_{4}

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C_{2} H_{4}

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C_{3} H_{8}

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C_{4} H_{10}

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Solution

The correct option is C $C_{2} H_{4}$
Let the molecular formula of hydrocarbon be $C_{x} H_{y}$
Vapour Density= $\frac{M o l . w t}{2}$
$∴$ Molecular weight= $14 \times 2$ = $28 g$

Now, $12 x + y = 28 g (i)$
Combustion of Hydrocarbon follows following reaction:
$C_{x} H_{y} + (x + \frac{y}{4}) O_{2} ⟶ x {C O}_{2} + (y / 2) H_{2} O$
This implies, $1$ mole of $C_{x} H_{y}$ $\equiv$ $x$ moles of ${C O}_{2}$

But $\frac{v o l u m e o f {C O}_{2} g e n e r a t e d}{v o l u m e o f o r i g i n a l h y d r o c a r b o n} = \frac{15}{7.5} = 2 : 1$

Therefore, $x = 2$

Therefore, from ${e q}^{n} (i)$ , $12 \times 2 + y = 28 ∴ y = 4$
$∴$ Hydrocarbon is $C_{2} H_{4}$

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7.5mL of a hydrocarbon gas was exploded with an excess of oxygen. On cooling, it was found to have undergone a contraction of 15mL. If the vapour density of the hydrocarbon is 14, determine its molecular formula.(C=12, H=1)

$7.5 m L$ of a hydrocarbon gas was exploded with an excess of oxygen. On cooling, it was found to have undergone a contraction of 1 $5 m L$ . If the vapour density of the hydrocarbon is $14$ , determine its molecular formula.(C=12, H=1)