Chem 113 Abstract PDF

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CHEM 113

17 January 2018 Lab #2: Molecular Mass by Freezing Point Depression

The purpose of this lab was to measure freezing-point depression and determine the molecular weight of a given compound. Freezing-point depression refers to the decrease of a solvent’s freezing point when a solute is added. Freezing-point depression is a colligative property, meaning that it is dependent upon the solute to solvent particle ratio. This relationship can be explained using the following equation: ∆Tf = i · Kf · m. To determine the molecular mass of the samples, a balance was used. A thermometer was used to measure the freezing points of separate solutions, each containing an unknown compound. After using measuring the mass of the test tube containing water and an unknown compound, the test tube was placed in a constant temperature bath. Once the substance began to freeze, the freezing point was recorded. This same procedure was repeated for the second compound sample. The freezing-point equation was used to determine the molecular weight of the second sample. The equation considers molality and freezing point depression as variables, m and K f, respectively. The freezing-point of each solution was found by determining the temperature at which the solution begins to freeze. This temperature was recorded based on observation of when the liquid started to freeze. We found that the freezing points of the samples were slightly lower than that of water’s freezing point, 0oC. Mass of water Freezing point temperature of water Freezing point temperature of sample 1 Freezing point temperature of sample 2

10g 0oC -2.1oC -3.8oC

Data Analysis 1. Calculate the change in temperature of FP sample 1. ΔTf = T(pure solvent) – T(solution) = (0oC) - (-2.1oC) = 2.1oC 2. Calculate the molar mass of FP sample 2. Kf of water = 1.86oC/m ΔTf = Kf · m m = ΔTf / Kf = 3.8oC / (1.86oC/m) = 2.043m m = (# of moles) / (solvent mass in kg) # of moles = m · mass of solvent in kg = 2.043m · 0.01kg = 0.02043 moles molecular weight of FP sample 2 = 2.000g / (0.02043) = 97.895 g/mol 3. What is the van’t Hoff factor (i) of the unknown substance, given the following table data? Kf of water 1.86 °C/m freezing point of pure water 0.0 °C molality of solution 0.512 mol/kg

ΔT = iKfm = i · (1.86oC/m) · (0.512 mol/kg) = 2.9oC i = (2.9oC) / (1.86 · 0.512) i = 2.9 / 0.95232 i = 3.045

CHEM 113 freezing point of solution –2.9 °C

17 January 2018...