`0.30g` of an organic compound containing `C`, `H` and Oxygen on combustion yields `0.44g CO_(2)` and `0.18g H_(2)O`. If one mole of compound weighs `60`, then molecular formula of the compound is

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)xx100=40%`
Percentage of `=(2)/(18)xx(0.18)/(0.30)xx100=6.6%`
percentate of `O=100-(40+6.6)=53.4%`

Hence, empirical formula `=CH_(2)O`
`n=("Molecular mass")/("Empirical formula mass")=(60)/(30)=2`
`rArr` Molecular formula of compound `=(CH_(2)O)_(2)`
`=C_(2)H_(4)O_(2)`