Empricial formula of a compound is `CH_(2)O` and its molecular mass is 90, the molecular formula of the compound is

Options

A. `C_(3)H_(6)O_(3)`
B. `C_(2)H_(4)O_(2)`
C. `C_(6)H_(12)O_(6)`
D. `CH_(2)O`

Answer

A

Explanation

Empirical formula mass, `CH_(2)O = 30`
`n = ("Mol. Mass")/("Empirical formula mass") = (90)/(30)` = 3
Molecular formula = `("Empirical formula")_(n)`
`= (CH_(2)O)_(3) = C_(3)H_(6)O_(3)`