The ratio of mass percent of C and H of an organic compound (CxHyOz) is 6:1. If one molecule of the above compound (CxHyOz) contains half as much oxygen as required to burn one molecule of compound CxHy completely to CO2 and H20. The empirical formula of compound CxHyOz is : A. O C3H402 B. C2H403 YOU C. C3H603 D.O C2H40

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:

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:

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 -