The ratio of mass percent of `C` and `H` of an organic compound `C_(X)H_(Y)O_(Z)` is `6:1`. If one molecule of the above compound `C_(X)H_(Y)O_(2)` contains half as much oxygen as required to burn one molecule of compound `C_(X)H_(Y)` completely to `CO_(2)` and `H_(2)O` .The empirical formula of the compound `C_(X)H_(Y)O_(Z)` is:
`C_(3)H_(6)O_(3)`
`C_(2)H_(4)O`
`C_(3)H_(4)O_(2)`
`C_(2)H_(4)O_(3)`