3.0 gms of an organic compound on combustion give 8.8gm of co2 and 5.4 gm of water .the empirical of the compound is

mass of CO2 = 8.8 gm or since the molar mass of CO2 is 44gm / mole we have0.2 moles of CO2 since there is 1 mole so C in CO2we have 0.2 moles of C or 2.4 gm of Cmass of H2O = 5.4 gm or since the molar mass of water is 18 gm / mol we have0.3 moles of H2O since there are 2 moles of H in 1 moles of H2O =0.6 moles of H or 0.6gm of Hmolar ratio of C : H = 0.2 : 0.6mass of C+H = 3.0 gmor after dividing by the smallest 0.2molar ratio of C : H = 1.0 : 3.0Hence empirical formula of the compound is CH3. *this is based on combustion of alkane