A carbon compound contains $12.8$ % of carbon, $2.1$ % of hydrogen and $85.1$ % of bromine. The molecular weight of the compound is $187.9$ . Calculate the molecular formula of the compound. (Atomic wts: $H = 1.008, C = 12.0, B r = 79.9$ )

C H_{3} B r

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C H_{2} {B r}_{2} I

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C_{2} H_{4} {B r}_{2}

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C_{2} H_{3} {B r}_{3}

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Solution

The correct option is A $C_{2} H_{4} {B r}_{2}$
Element $%$ At. wt. relative number simple ratio
C 12.8 12 $\frac{12.8}{12} = 1.06$ $\frac{106}{1.06} = 1$
H 2.1 1 $\frac{2.1}{1} = 2.1$ $\frac{2.1}{1.06} = 2$
Br 85.1 80 $\frac{85.1}{80} = 1.06$ $\frac{1.06}{1.06} = 1$
$E F = C H_{2} B r$
Wt. of $E F = 12 + 2 + 80 = 94$
$n = \frac{M . W}{w t . o f E F} = \frac{187.9}{94} = 2$
$M F = (E F)_{n} = (C H_{2} B r)_{2} = C_{2} H_{4} B r_{2}$

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12.8 %

2.1 %

85.1 %

[C = 12, H = 1, C l = 35.5]

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