0.30g of an organic compound contaiining C, H and Oxygen an combustion yields 0.44g `CO_2` and 0.18g `H_2O`. If one mol of compound weight 60, then molecular formula of the compound is

Options

A. `C_2H_4O_2`
B. `C_4H_6O`
C. `C_3H_8O`
D. `CH_2O`

Answer

A

Explanation

Percentage of `C = 12/44 xx 0.44/0.30 xx 100 = 40%`
Percentage of `H= 12/18 xx 0.18/0.30 xx 100 = 6.6%`
Percentage of `O = 100-(40 + 6.6) = 53.4%`

Hence empirical formula `= CH_2O`
`n = ("Molecular mass")/("Emperical formula mass") =60/30 = 2`
`rArr ` Molecular formula of the compound ` = (CH_2O)_2 = C_2H_4O_2`