The ratio 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 contains half as much oxygen as required to burn one molecule of the compound C_(x)H_(y) completely to CO_(2) and H_(2)OThe empirical formula of the compound C_(x)H_(y)O_(z) is

The ratio of mass per cent of C and H of an organic compound (C_(x)H_(

1.	The ratio 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 contains half as much oxygen as required to burn one molecule of the compound C_(x)H_(y) completely to CO_(2) and H_(2)OThe empirical formula of the compound C_(x)H_(y)O_(z) is
Answer» `C_(2)H_(4)O_(3)` `C_(3)H_(6)O_(3)` `C_(2)H_(4)O` `C_(3)H_(4)O_(2)` Solution :Mole RATIO of C & H in 1 mole of `C_(x)H_(y)O_(z)= (6//12)/(1//1)= 1:2= x: y` . As 2Z mole of O (or z mole `O_(2)`) is used to combust one mole of `C_(x)H_(y)` `overset("1 mole")(C_(x)H_(y)) + overset("z mole")(O_(2)) rarr CO_(2) + H_(2)O` , applying POAC for C, H and O, we GET, `z=x + (y)/(4)`, if x=1 and y= 2` therefore z=1.5` i.e., `C_(2)H_(4)O_(3)` should be the EMPIRICAL formula