`0.30g` of an organic compound containing `C, H`, and `O` an combustion yields `0.44g` of `CO_(2)` and `0.18g` of `H_(2) O`. If its molecular mass is `60 g` then molecular formula will be

Options

A. `C_(4) H_(8) O_(4)`
B. `C_(3) H_(6) O_(3)`
C. `C_(2) H_(4) O_(2)`
D. `CH_(2)O`

Answer

C

Explanation

Since 1 mol of `C` always yields 1 mol of `CO_(2)`,
`n_(C)` in `CO_(2) = n_(CO_(2)) = (0.44g)/(44g mol^(-1)) = 0.01 mol`
`:. Mass_(C) = (n_(C)) (MM_(C)) = (0.01 mol) (12g mol^(-1)) = 0.12g`
Since 1 mol of `H` always yield `1//2` mol of `H_(2) O`, we have
`m_(H)` in `H_(2) O = 2xx n_(H_(2)O) = 2 xx ((0.18g)/(18g mol^(-1))) = 0.02 mol`
`:. m_(H) = ((n_(H)) (MM_(H)) = (0.02 mol) (1g mol^(-1)) = 0.02g`
Hence `m_(O) = (m_(OC)) - (m_(C) + m_(H))` (`OC` is organic compound)
`= (0.30g) - (0.12g + 0.02g)`
`= 0.16g`
`:. n_(O) = (0.16g)/(16g mol^(-1)) = 0.01 mol`
Mole ratious of elements `C, H` and `O = 0.01 : 0.02 : 0.01`
`= 1 : 2 : 1`
`:. EF = CH_(2) O` and `EFM = 30 u`
Now, `n = (MM)/(EFM) = (60u)/(30u) = 2`
`:. MF = 2(EF) = 2(CH_(2)O) = C_(2)H_(4) O_(2)`