The general formula of a carbohydrate is `C_(x)(H_(2)O)_(y)`. 3.1 g of the carbohydrate on heating in the absence of oxygen yields 1.24 g of carbon. The molecular mass of carbohydrate is 180u. Calculate its molecular formula.

Explanation

The carbohydrate may be regarded as the hydrate of carbon
Percentage of carbon (C) in carbohydrate `= ((1.24g))/((3.10g))xx100=40`
Percentage of water `(H_(2)O)` in carbohydrate `= 100 - 40 = 60`
Moles of `C=((40 g))/(("12 g mol"^(-1)))=3.33` mol
Moles of `H_(2)O=((60g))/(("18 g mol"^(-1)))=3.33` mol
`:.` Mole ratio of `C :H_(2)O : : 1 : 1`
Empirical formula of carbohydrate `= C (H_(2)O)`
Empirical formula mass `= 12 + 18 = 30` u
Molecular mass = 180u (given) , `n = ("Molecular mass")/("Empirical formula mass")=(("180 u"))/(("30 u"))=6`
`:.` Molecular formula of carbohydrate `= 6 xx (CH_(2)O)=C_(6)H_(12)O_(6)`.