A carbon compound contains 12.8% carbon, 1% hydrogen, 85.1% bromine by mass. The molecular weight of the compound is 187.9. Calculate its molecular formula.

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

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C H_{4} B r_{2}

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C H B r_{2}

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C H_{2} B r

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Solution

The correct option is A $C_{2} H_{4} B r_{2}$

Element % Atomic mass Moles Simplified
mole ratio

C 12.8 12 $\frac{12.8}{12} = 1.06$ 1

H 2.1 1 $\frac{2.1}{} = 2.1$ 2

Br 85.1 80 $\frac{85.1}{80} = 1.06$ 1

Empirical formula of the compound = $C H_{2} B r$
Empirical formula mass = 94
$n = \frac{molecular mass}{empirical mass} = 2$
Hence the molecular formula = $C_{2} H_{4} B r_{2}$

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12.8

2.1

85.1

187.9

H = 1.008, C = 12.0, B r = 79.9

12.8 %

2.1 %

85.1 %

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