Question

When 12.4 g of an organic compound known to be 70.58% C, 5.9% H, and 23.50% O by mass is dissolved in 630.3 g of cyclohexane, the freezing point is 3.58 °C. The normal freezing point of cyclohexane is 6.59 °C. What is the molecular formula for the organic compound? Assume that the organic compound is a molecular solid and does not ionize in water. K values for various solvents are given in the colligative constants table. molecular formula: [C,H,O

Answer 1

We know that; ATP = Krim According to the question; TF = Freezing point depresion = 66-694218) - (3:58+243) = 3.01K Kf = Molar freezing point depression constant = 20.2 K.kg.mol- m = molality - W x 1000 W₂ = mass of solute = 12.4 g M2 = Molecular weight of solute = ? wi= mass of solvent = 630.38 i ATF = kfim =) 3:01 = 20.2 x 12.4 X1000 M2 x 630.3 => Mo - 20:2X12.4 x1000 3.01 x 630.3 = 132.026 gm/mor Emperical formula determination: Imol - 5.881 mox C:70.58 gm x 12gm - - 5.841 mox H: 5.9 gm x. Toogm

imol 10:23.50 gm x = 1.468 16 gm - mol . 50 Less Emperical value is 1.968 Empenical Number for C, H, O is - c. 5.881 - 4 1.468 H = = 3.978 ~ - 41 5.841 1.468 1.468 1.468 = 1 O = i Emperical formula = C4 Hq o Emperical = 48+4.04+ 16 = 68.04 Molewian weight calculation : formula =1.972 Molecular weight 130.026 Emperical weight 68004 . Molecular formula is- (C4H90) x2 - Cg Hg O2