An organic compound has an empirical formula `CH_(2)`O. Its vapour density is 45. The molecular formula of compound is

Options

A. `CH_(2)`O
B. `C_(2)H_(5)O`
C. `C_(2)H_(4)O_(2)`
D. `C_(3)H_(6)O_(3)`

Answer

D

Explanation

Empirical formula of the compound is - `CH_(2)`O
Vapour density - 45
therefore Molecular weight = 2 `xx `Vapour Density
= 2` xx` 45 = 90
Empirical formula weight `(CH_(2)O)` = 12 +2 + 16 =30
`rarr n=("Molecular weight")/("Empirical formula "weight)=90/30=3`
therefore Molecular formula - n` xx `empirical formula
3 x `CH_(2)`O
`C_(3)H_(6)O_(3)`