Four grams of hydrocarbon (C_xH_y) on complete combustion gave 12grams of CO₂. What is the empirical formula of the hydrocarbon ? (C = 12 ; H = 1)

Weight of CxHy = 4 gm

Weight of CO₂ = 12 gm

% of carbon = $\frac{12}{44}\times \frac{{\mathrm{Wt}}_{{\mathrm{CO}}_2}}{{\mathrm{W}}_{\mathrm{O}.\mathrm{C}}}\times 100$ $=\frac{12}{44}\times \frac{12}{4}\times 100$ = 81.8 %

% of Hydrogen  =  18.2%

E.F of the compound is = C₃ H₈