A carbon compound contains `12.8%` of carbon, `2.1%` of hydrogen and `85.1%` of bromince. The molecular weight of the compound is `187.9`. Calculatte the molecular formula of the compound. (Atomic weight of H `+1.008,` C `= 12.0` and Br = `79. 9))

Options

A. `CH_(3)Br`
B. `CH_(2)Br_(2)l`
C. `C_(2)H_(4)Br_(2)`
D. `C_(2)H_(3)Br_(3)`

Answer

C

Explanation

Hence, the empirical formula becomes `CH_(2) Br`
Atomic weight of `CH_(2) Br = 12 + 2 + 80 = 94`
As we know that
Multiple `(n)=("molecular weight")/("empirical formula weight")=187.9/92=2`
Thus, molecular formula `= ("Empirical formula")_(n)` of
moleceular formula
`= (CH_(2) Br)_(2) = C_(2) H_(4) Br_(2)`