`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 mu` the molecular mass is formula will be

Options

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

Answer

B

Explanation

Percentage of C `=12/44 xx 0.44/0.30 xx 100 = 40%`
Percentage of `H=2/18 xx 0.18/0.30 = 6.67%`
Percentage of `O = 130 -(40 + 6.67) = 53.33%`
Calculation of empirical formula

Empirical formula `=CH_(2)O`
Empirical formula mass `=12 +2 xx 1+ 16 = 30`
Molecular mass = 60
`therefore n=60/30=2`
`therefore` Molecular formula mass `=(CH_(2)O)_(2) =C_(2)H_(4)O_(2)`