Combustion of 0.202 g of a carbon compound gave 0.361 g of carbon dioxide and 0.147g of water. Determine the empirical formula of the compound. 

Explanation

`%C = (0.361xx12xx100)/(0.202xx44) = 48.76% , %H = (0.147xx2xx100)/(0.202xx18) = 8.07% , %O = 100-56.83 = 43.17`
C : H : O = `(48.76)/(12) : (8.07)/(1) : (43.17)/(16) = 4.06 : 8.07 : 2.7 = 3:6:2:`
The empirical formula of the compound is `C_(3)H_(6)O_(2)`.