LAB 4 Freezing Point PDF

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lab report of lab 4 freezing point...

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Xinyang Ye

Chem101-62

Lab Report #3

Experiment #3 Determination of Molar Mass by Freezing Point Depression

Introduction The purpose of the experiment is to determine the molar mass of an unknown subject by observing the difference in freezing point of both a pure solvent and a solvent with the unknown subject as the solute. Colligative properties are properties that depend on the amount of molecules in the solution, instead of their chemical identity. For example, the boiling point elevation, the freezing point depression and the vapor pressure lowering are such colligative properties. When there is other substance, namely solute, added into the pure solvent, the freezing point of the solution will decrease comparing to the pure solvent. That is to say, the freezing point of the pure solvent is higher than the solution with the same substance as the solvent. For example, the freezing point of water is 0 degree Celsius. When the pure water is added with sodium chloride as solute and becomes sodium chloride solution, the freezing point of such solution is lower than the pure water. Sodium chloride could be used as snow-melting agent. By using sodium chloride, the freezing point of water is lowered. Therefore, the snow on the road would not freeze until the temperature dropped below the freezing point of the sodium chloride solution. Thus, the snow on the road would start melting. Antifreeze in the car radiator serves another example of this. By using a solution instead of the pure water in the radiator, the liquid will not freeze until it reaches to the temperature below 0 degree Celsius. The larger the amount of dissolved solute in the solution, the larger the depression of the freezing point of the solvent is. That is to say, when there is more dissolved solute in the solution, the freezing point would be lower. In the lab, we have to measure the freezing point of both the pure solvent and the solution with the same substance as the solvent. We recorded the temperature at every ten seconds and plot all the data in the plot. There are three stages of the plot: 1. The temperature dropping at the beginning; 2. The temperature stays constant, resulting in a plateau on the curve, which means that the solution starts freezing; 3. The temperature starts decreasing again. The plot is about time(seconds) as the x-axis and temperature ( degree Celsius) as the y-axis. There would be two curves, one for the pure solvent and the other one for the solution. In order to measure the depression of the freezing point of the solvent when mixed with a solute, we need to first measure the freezing point of the pure solvent by measuring the stage 2 part of the plot, which is the plateau on the curve. Then, we need to measure the freezing point of the

solution with the same method. And find the difference between these two freezing points, which is the freezing point depression.

In order to find the molar mass of the solute, we need to first find the freezing point of both the pure solvent and the solution. In this experiment, we used the cyclohexanol as the solvent and solute B as our solute. Firstly, we need to find the freezing point of the pure solvent. Then, we need to measure the mass of the solute adding into the pure solvent and then measure the freezing point of the solution. Since we know the freezing point for both pure solvent and solution, we could find the freezing point depression. According to the formula: △Tf =molality, since we know the depression of the freezing point and k f is the molal kf freezing point depression constant, which is 39.4 ℃/m , we could calculate the molality of solute in the solution. Secondly, based on the moles of solute = (molality)(kg of the solvent used), we could find the moles of solute used in the solution. Then, since we know the mass of the solute added into the solution, we could calculate the molar mass according to the formula: molar mass = grams of solute added/ moles of solute.

Calculation and discussion

Temperature of pure solvent (°C)

Time(Seconds)

temperature with solute (°C)

0

31.719243

26.1221844

10.2

30.5680981

24.4750222

19.8

29.0959968

23.848183

30

27.5144564

22.5451737

40.2

27.0501511

21.0965337

49.8

26.4468308

20.2747042

60

25.542284

19.9686768

70.2

25.0552112

19.6385451

79.8

24.6607266

18.5963106

90

24.3590546

18.0725631

100.2

24.0572607

17.8338192

109.8

23.6622487

17.426895

120

23.4762455

16.9224803

130.2

23.1971936

16.5609531

139.8

23.1274495

16.4400579

150

23.1506989

15.9798042

160.2

23.1274495

15.9312842

169.8

22.9179526

15.5908147

180

22.6150289

15.0533924

190.2

22.3586287

14.6359822

199.8

22.1020165

14.5127789

210

21.7984868

14.4388993

220.2

21.4710439

14.3403005

229.8

21.1667642

14.216699

239.4

20.8386407

14.0931303

Temperature vs Time(sec) of pure solvent and solution with solute 35 30

Temperature(℃)

25 20 Temperature of pure solvent temperature with solute

15 10 5 0

Time (seconds)

The freezing points of the pure solvent is around 23℃ because the plateau of the curve locates at 23 ℃ . The freezing point of the solution with the same substance as the solvent is around 14.5℃ , because the plateau part appears to be at 14.5℃. Therefore, we could calculate the △ T f : △ T f =T pure solvent −T with solute △ T f =23 ℃−14.5 ℃ △ T f =8.5℃ .

Therefore, the freezing point depression is 8.5

℃ .

In this experiment, we used cyclohexanol as the solvent and solute B as solute. According to the formula: △ T f =k f × m where kf is the molal freezing point depression constant, also known as cryoscopic constant, we could find the m, which is the moles of the solute added to ℃ the solution. The cryoscopic constant for cyclohexanol is 39.4 . m

△ Tf ℃ =8.5 ℃ ÷ 39.4 =0.22 mol / kg kf m Therefore, the molality of the solute is 0.22 mol/kg. m=

Since we know the molality of the solute, we could calculate the moles of solute by the mass of the solvent used. The solvent we used in the experiment is cyclohexanol and the mass of the cyclohexanol is 9.34 grams, which is 0.00934 kg. Thus, we could use the formula: Moles of solute= (molality)(kg of solvent used) Moles of solute=(0.22 mol/kg)(0.00934 kg) =0.0021 mol

Since the mass of the solute we added into the solution is 0.75 grams and we already know the moles of solute, which is 0.0021 mol, we could calculate the molar mass by the formula: Molar Mass=

grams of solute added moles of solute Molar Mass= 0.75 grams/0.0021 mol =357.1 g/mol

Therefore, the molar mass of the solute B is 357.1 grams/mol. According to the given molar mass of both Hexadecanol (245.5 grams/mol) and Octadecanol (270.5 grams/mol), the calculated molar mass of solute B (357.1 grams/mol) is closest to the molar mass of Octadecanol. However, the difference of molar mass between solute B and Octadecanol is a little bit too large. Therefore, there can be an error during the experiment.

Conclusions The purpose of this experiment is to find the molar mass of the unknown by measuring the freezing point depression. We found when the solute added into the pure solvent, the freezing point of the solution did go down. Thus, based on the formulas and the data we collected through the lab, we could calculate the molar mass of the unknown solute B, which is 357.1 grams/mol. However, it differs a lot from the molar mass of the given solutes (Hexandecanol and Octandecanol). Thus, there should be some errors during the lab. So, the results did not 100% match the expected results. There could be two reasons for the deviated result: 1. There might be some water contamination during the lab. The test tube is not hundred percent dried. With other substances existing in the solution, the freezing point would be affected largely. 2. There was not a very clear plateau section on the curve for the solution. So, the freezing point of the solution could be somewhere even further beyond the time we recorded, or we made a wrong estimation of freezing point for the solution. During this lab, theoretically, I have learned how to calculate the molality of a solute by given freezing point depression, the moles of the solute by given both mass of the solvent and molality and molar mass by given both moles of the solute and the mass of the solute. Also, the molal freezing

point depression constant, kf, differs by solvent. That is to say, with different substance as solvent, the kf comes with different values. Additionally, I have learned how to read thermometer and how to plot the data into the graph I desire....