A carbon compound contains 12.8% of carbon; 2.1% of hydrogen and 85.1% of bromine. The molecular mass of compound is 187.9. Calculate the molecular formula of the compound (Atomic mass H = 1.008, C = 12, Br = 79.9):

C H_{3} B r

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C H_{2} B r_{2}

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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 C $C_{2} H_{4} B r_{2}$
Moles of C $= \frac{12.8}{12} = 1.067$

Moles of H $=$ \dfrac{2.1}{1.008}=2.08$

Moles of Br $= \frac{85.1}{79.9} = 1.065$

$C = \frac{1.067}{1.065} = 1$ , $H = \frac{2.08}{1.065} = 2$ , $B r = \frac{1.065}{1.065} = 1$

Empirical formula $= C H_{2} B r$

Molecular formula $= (C H_{2} B r) n$

Empirical mass $= 93.9$
Therefore $n = \frac{187.9}{93.9} = 2$

Molecular formula $= C_{2} H_{4} B r_{2}$

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