A 0.10 g sample of an organic compound containing carbon, hydrogen, and sulphur is subjected to complete combustion, yielding 0.22 g of CO ₂ and 0.072 g of H ₂ O . A separate 0.10 g sample of the same compound is subjected to Carius estimation, yielding 0.233 g of a white precipitate of barium sulphate. The empirical formula of the organic compound is: (Given molar masses in g mol ⁻¹ : C = 12 , H = 1 , O = 16 , S = 3

Step-by-step solution

Mass of carbon in the compound = 12 44 0.22 g = 0.06 g Mass of hydrogen in the compound = 2 18 0.072 g = 0.008 g Mass of sulphur in the compound = 32 233 0.233 g = 0.032 g Total mass of C, H, and S = 0.06 + 0.008 + 0.032 = 0.10 g . Since the sum of the masses of C, H, and S is equal to the mass of the sample taken ( 0.10 g ), there is no oxygen present in the compound. Moles of C = 0.06 12 = 0.005 mol Moles of H = 0.008 1 = 0.008 mol Moles of S = 0.032 32 = 0.001 mol The molar ratio of C : H : S is 0.005 : 0.008 :