1. Chapter 13 Properties of Solutions Preliminary Lesson For PRACTICE FRQ’S Worksheet 1998 B

2. A chemist is asked to determine the molecular formula of an organic compound. His lab has determined through mass spectrometry that the compound contains just three elements (carbon, hydrogen, and oxygen) and the chemical composition by mass is: 52.13% carbon 13.15% hydrogen 34.72% oxygen

3. Solution Knowing the % composition of the compound allows the chemist to determine the empirical formula: 1. Assume a 100 g sample and replace the percentages with gram units: 52.13 g carbon (C) 13.15 g hydrogen (H) 34.72 g oxygen (O)

4. 2. Divide each element mass by its molar mass to determine the mole ratio: 52.13 g C ÷ 12.01 g/mol = 4.34 mol 13.15 g H ÷ 1.01 g/mol = 13.02 mol 34.72 g O ÷ 16.0 g/mol = 2.17 mol 3. Then divide each mole value by the lowest mole value: carbon 4.34/2.17 = 2.0 hydrogen 13.02/2.17 = 6.0 oxygen 2.17/2.17 = 1.0

5. 4. Use these values for the subscripts in the empirical formula: C 2 H 6 O To determine the molecular formula of the compound, the chemist decides to add a specific mass of the compound to the polar solvent, water, and measure the freezing point depression. He knows the molal freezing point depression constant, K d or K f is 1.86 kg·K·mol -1 for water (also 1.86 0 C/m, m = molality).

6. Using ΔT f = K f m he can determine the mole value of the mass of the compound and then solve for the molar mass of the compound (g/mol). He then divides the empirical mass, determined by the empirical formula, into the molar mass of the compound and finds the molecular mass to be twice that of the empirical mass. He then knows the subscripts in the molecular formula are twice those in the empirical formula (C 2 H 6 O)…molecular formula is C 4 H 12 O 2. Note: when using ΔT f, 0 C = K

7. The chemist does not recognize this organic formula so decides to run another type of analysis. He decides to vaporize the same mass of the compound that he placed in solution and use the ideal gas law to determine the molar mass by measuring the volume of the vaporized compound at a specific temperature and pressure: PV = nRT n = moles moles = mass/molar mass So: PV = (mass/molar mass)RT

8. Or: molar mass = (mass)RT/PV He finds that the molar mass in this analysis is exactly ½ of the molar mass in the aqueous solution analysis. This means the molecular formula would equal the empirical formula (C 2 H 6 O). How is this possible? …dimerization! Question: What is a dimer?