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

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Solution

Element % mass atomic mass no of atoms simplest ratio rounding of ratio

C 82.76 12 $\frac{82.76}{12} = 6.89$ $\frac{6.89}{6.89} = 1$ $1 \times 2 = 2$

H 17.24 1 $\frac{17.24}{1} = 17.24$ $\frac{17.24}{6.89} = 2.5$ $2.5 \times 2 = 5$

Empirical formula is $C_{2} H_{5}$
Molecular weight = 2 $\times$ Vapour density
Molecular weight = n $\times$ (Empirical formula weight)
58 = n $\times$ (12 $\times$ 2 + 1 $\times$ 5)
n = 2
So, molecular formula = $C_{4} H_{10}$

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Element	% mass	atomic mass	no of atoms	simplest ratio	rounding of ratio
C	82.76	12	$\frac{82.76}{12} = 6.89$	$\frac{6.89}{6.89} = 1$	$1 \times 2 = 2$
H	17.24	1	$\frac{17.24}{1} = 17.24$	$\frac{17.24}{6.89} = 2.5$	$2.5 \times 2 = 5$