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 `

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_(Z))`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 compound `C_(X)H_(Y)O_(Z)` is
A. `C_(2)H_(4)O`
B. `C_(3)H_(4)O_(2)`
C. `C_(2)H_(4)O_(3)`
D. `C_(3)H_(6)O_(3)`

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_(Z))`cont

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