0.3g of an organic compound on combustion liberated 0.18g of water vapour and 0.44 g of carbonioxide. Calculate the percentage composition of the compound

Explanation

Weight of the compound =W = 0.3g
Weight of `CO_(2)= W_(1) = 0.44g`
Weight of `H_(2)O = W_(2) = 0.18g`
% of carbon `=(W_(1) xx 12 xx 100)/(W xx 44) = (0.44 xx 12 xx 100)/(0.3 xx 44) = 40.0%`
% of hydrogen `= (W_(2) xx 2xx 100)/(W xx 18) = (0.18 xx 2 xx 100)/(0.3 xx 18) = 6.67%`
After calculating the percentage composition of the given elements in the compound, if their sum is not 100, then the difference is assumed as oxygen
`:.` % of oxygen `=[100-(40+6.67)]= 53.33%`