An organic compound contains `C, H`, and `O`. Its elemental analysis gave `C(38.71%)` and `H (9.67%)`. The empirical formula of the compound will be

Options

A. `CH_(4)O`
B. `CH_(3)O`
C. `CHO`
D. `CH_(2)O`

Answer

B

Explanation

`% O = 100% - (% C + % H)`
`= 100% - 48.38% = 51.62%`
Considering `100 g` of compound, we have, `m_(C) = 38.71 g`, `m_(H) = 9.67 g, m_(O) = 51.62g`
Thus,
`n_(C) : n_(H) : n_(O) = (38.71 g)/(12 g mol^(-1)) : (9.67 g)/(1 g mol^(-1)) : (51.62 g)/(16 g mol^(-1))`
`= 3.226 : 9.67 : 3.226`
`= (3.226)/(3.226):(9.67)/(3.226):(3.226)/(3.226)`
`= 1:3:1`
Empirical formula is `CH_(3)O`.