A gaseous hydrocarbon contains $82.76$ % of carbon. Given that its vapour density is $29$ , find its molecular formula. $[C = 12, H = 1]$ .

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Solution

Element %composition Mol. mass No. of C atoms Simplest Rounding
ratio off
C 82.76 12 6.89 1 $\times$ 2 2
H 17.24 1 17.24 2.5 $\times$ 2 5
Number of carbon atoms present = $\frac{82.76}{12} = 6.89$
Number of hydrogen atoms present = $\frac{17.24}{1} = 17.24$
simplest ratio = $\frac{6.89}{6.89} = 1$
simplest ratio = $\frac{17.24}{6.89} = 2.5$

$∴$ Empirical fromula is $C_{2} H_{5}$
Empirical formula mass= $2 \times 12 + 5$ = 29
$∴$ V.D. = 29
Mol. mass = 2 $\times$ V.D.
=58
$n = \frac{M o l e c u l a r m a s s}{E m p i r i c a l f o r m u l a m a s s}$
= $\frac{58}{29} = 2$
$∴$ Molecular formula = $C_{4} H_{10}$ .

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A gaseous hydrocarbon contains 82.76% of carbon. Given that its vapour density is 29, find its molecular formula. [C=12,H=1].

A gaseous hydrocarbon contains $82.76$ % of carbon. Given that its vapour density is $29$ , find its molecular formula. $[C = 12, H = 1]$ .