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The ration of mass per cent of C and H o...

The ration of mass per cent 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)` compleltely to `CO_(2) and H_(2)O`. The empirial formula of compound `C_(x)H_(y)O_(z)` is:

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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 -

The ratio of mass percent of C and H of an organic compound ( C_xH_yO_z ) is 6 : 1. If one molecule of the above compound ( C_xH_yO_z ) contains half as much oxygen as required to burn one molecule of compound C_xH_y completely to CO_(2) and H_2O . The empirical formula of compound C_xH_yO_z is: