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Question

# 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):

A
CH3Br
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B
CH2Br2
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C
C2H4Br2
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D
C2H3Br3
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Solution

## The correct option is C C2H4Br2Moles of C =12.812=1.067 Moles of H =\dfrac{2.1}{1.008}=2.08\$ Moles of Br =85.179.9=1.065 C=1.0671.065=1, H=2.081.065=2, Br=1.0651.065=1Empirical formula =CH2Br Molecular formula =(CH2Br)nEmpirical mass =93.9Therefore n=187.993.9=2Molecular formula =C2H4Br2

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