An organic compound contains `C,H` and `O`. If `C (%):H^(%) = 6:1`, what is the simplest formula of the compound, given that one mole of the compound contains half as much oxygen as would be required to burn all the `C` and `H` atoms in it to `CO_(2)` and `H_(2)O`?

Options

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

Answer

B

Explanation

Molar ratio of `C:H = (6)/(12) : (1)/(1) = 2:1`
Formula of compound, `C_(x) H_(2x) O_(y)`
`C_(x) H_(2x) O_(y) + O_(2) rarr xCO_(2) + xH_(2)O`
From the equation `(3x)/(2) = y`
`x : y = 2 : 3`
Formula : `C_(2)H_(4)O_(3)`