This question has me befuddled. Thiophene is a sulfur-containing hydrocarbon som

Question

This question has me befuddled.

Thiophene is a sulfur-containing hydrocarbon sometimes used as asolvent in place of benzene. Combustion of 2.348g sample ofthiophene produces 4.913g CO2 1.005g H2O and1.788 g SO2. When a .867g sample of thiophene isdissolved in 44.56g of benzene, the freezing point is lowered by1.183oC. What is the molecular formula of thiphene?

I know how to get the answer, I just don't understand why.Apparently you find the molality of benzene, amount of solute,molar mass of thiophene, and the empirical formula from the massesof the "combustion products."

Can anyone clarify this series of calculations for me so Iunderstand the conceptual basis for a question like this?

Explanation / Answer

Here in the problem we need molecular formula forThiophene. Firstly we have the data concerned to the depression infreezing point. Formula :                  Tf = Kf * m Where Tf is the depression in freezing point           Kf is the molal depression in freezing constant           m isthe molality Data :          Tf = 1.183oC           Kf  = 5.12 oC/m for Benzene        m   = 1.183oC /  5.12oC/m                 = 0.231 m Mass of solute = 0.867 g Mass of solvent = 44.56 g                         = 0.04456 kg No.of mols        = 0.231 m * 0.04456 kg                         = 0.0103 mols Molar mass       = 0.867 g /0.0103 mols                         = 84.17 g /mol Now we know the molar mass . By knowing the emperical Formulawe can have emperical weight. From that molecular formula canbe obtained. Mass of C = 4.913 g / 44 g / mol * 12.0 g /mol                  = 1.33 g Mols of C = 1.33 g / 12 g / mol                  = 0.11 mols Mass of H = 1.005 g * 1.08 g *2/ mol /18 g /mol                  =0.1206 g Mols of H = 0.1206 g / 1.08 g / mol                =   0.111 Mass of S= 1.788 g * 32 g / mol / 64 g / mol                = 0.897 g Mols of S = 0.027 mols Upon simplification we get C = 4 , H = 4 , S= 1 Emperical formula =C4H4S It's molecular weigth is almost similar to the empericalweight. Molecular formula=  C4H4S Hope it will help u to understand the solution .

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