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Q.

A carbon compound contains 12.8% of carbon, 2.1% of hydrogen and 85.1% of bromine. The molecular weight of compound is 187.9. Calculate the molecular formula of the compound.

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a

${CH}_{3} Br$

b

$C_{2} H_{3} {Br}_{3}$

c

$C_{2} H_{4} {Br}_{2}$

d

${CH}_{2} {Br}_{2}$

answer is C.

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Detailed Solution

$\begin{array}{l} C = \frac{12.8}{12} = 1.06, \frac{1.06}{1.06} = 1 \\ H = \frac{2.1}{1} = 2.1, \frac{2.1}{1.06} = 2 \\ Br = \frac{85.1}{80} = 1.06, \frac{1.06}{1.06} = 1 \\ ∴ EF = {CH}_{2} Br n = \frac{187.9}{94} = 2 ∴ mF = C_{2} H_{4} {Br}_{2} \end{array}$

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