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_(X)H_(Y

1.	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.
Answer» `C_(3)H_(6)O_(3)` `C_(2)H_(4)O` `C_(3)H_(4)O_(2)` `C_(2)H_(4)O_(3)` SOLUTION :`C_(X)H_(Y) + (X + (Y)/(4)) O_(2) RARR XCO_(2) + (Y)/(2)H_(2)O` according to given information `Z= (1)/(2) (2X + (Y)/(2))` `:. X=2, Y= 4, Z=3`