A compound contain $12.8 %$ Carbon, $2.1 %$ Hydrogen, $85.1 %$ Bromine. The molecular mass of the compound is 188. Find the molecular formula of the compound.

C_{2} H_{4} B r

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C H_{2} B r_{2}

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C H B r_{2}

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C_{2} H_{4} B r_{2}

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Solution

The correct option is D $C_{2} H_{4} B r_{2}$
Divide the percentages with atomic masses-
carbon $\frac{12.8}{12}$ 1.06 1
Hydrogen $\frac{2.1}{1}$ 2.1 2
Bromine $\frac{85.1}{79.9}$ 1.06 1
Empirical formula- $C H_{2} B r$
(Empirical formula wt) $_{n} = 188$ or $(94)_{n} = 188$ or $n = 2$
So molecular formula will be - $(C H_{2} B r)_{2} = C_{2} H_{4} B r_{2}$

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