A gaseous hydrocarbon has 85% carbon and vapour density of 28. The possible molecular formula of the hydrocarbon will be :

C_{3} H_{6}

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C_{2} H_{4}

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C_{5} H_{12}

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C_{4} H_{8}

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Solution

The correct option is C $C_{4} H_{8}$
$V . D . = \frac{M o l . M a s s}{2}$

$M o l . M a s s = 28 \times 2 = 56 g m / m o l$

Therefore, the gaseous hydrocarbon is $C_{4} H_{8}$

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Similar questions

A gaseous hydrocarbon contains 82.76 % of carbon. Given that its vapour density is 29, find its molecular formula

82.8 %

29

[H = 1, C = 12]

30

8.8 g C O_{2}

{5.4}_{g} H_{2} O

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