#### Read Microsoft Word - 04 Determination of Molar Mass by Freezing Point Depression.doc text version

`Determination of Molar Mass by Freezing-Point DepressionWhen a solute is dissolved in a solvent, the freezing temperature is lowered in proportion to the number of moles of solute added. This property, known as freezing-point depression, is a colligative property; that is, it depends on the ratio of solute and solvent particles, not on the nature of the substance itself. The equation that shows this relationship is Tf = Kf × m× i where Tf is the freezing point depression, Kf is the freezing point depression constant for a particular solvent (3.9°C·kg/mol for lauric acid in this experiment1), i is the van't Hoff factor, and m is the molality of the solution (in mol solute/kg solvent). Since lauric acid is not ionic, its van't Hoff factor is essentially equal to 1.OBJECTIVESIn this experiment, you will· · · ·Determine the freezing temperature of the pure solvent, lauric acid. Determine the freezing temperature of a mixture of lauric acid and benzoic acid. Calculate the freezing point depression of the mixture. Calculate the molecular weight of benzoic acid.Figure 1MATERIALSData Collection Mechanism Temperature Probe ring stand 400 mL beaker Tissue or paper towels lauric acid, CH3(CH2)10COOH lauric acid-benzoic acid mixture hot water bath utility clamp two 18 × 150 mm test tubes (if pre-made samples are not provided by your teacher)1&quot;The Computer-Based Laboratory&quot;, Journal of Chemical Education: Software, 1988, Vol.1A, No. 2, p. 73.Adapted from Advanced Chemistry with Vernier &amp; Laboratory Experiments for Advanced Placement Chemistry by Sally Ann Vonderbrink, Ph. D. Page 1The Determination of Molar Mass by Freezing-Point DepressionPROCEDURE1. Obtain and wear goggles. 2. Set up the data collection system. a. Connect a Temperature Probe to the interface. b. Start the data collection program. c. Set up the time graph for 10 seconds per sample and 60 samples. Part I: Determine the Freezing Temperature of Pure Lauric Acid 3. Add about 300 mL of tap water with a temperature of 20-25°C to a 400 mL beaker. Place the beaker on the base of the ring stand. 4. Use a utility clamp to obtain a test tube containing hot, melted lauric acid from your instructor. Fasten the utility clamp at the top of the test tube. CAUTION: Be careful not to spill the hot lauric acid on yourself and do not touch the bottom of the test tube. 5. Insert the Temperature Probe into the hot lauric acid. Fasten the utility clamp to the ring stand so the test tube is above the water bath. 6. Begin data collection. Lower the test tube into the water bath. Make sure the water level outside the test tube is higher than the lauric acid level in the test tube, as shown in Figure 1. 7. With a very slight up-and-down motion of the Temperature Probe, continuously stir the lauric acid for the ten-minute duration of the experiment. Do not allow the temperature probe to touch the bottom of the test tube. 8. When data collection is complete, use a hot water bath to melt the lauric acid enough to safely remove the Temperature Probe. Carefully wipe any excess lauric acid liquid from the probe with a paper towel or tissue. 9. The freezing temperature can be determined by finding the mean temperature in the portion of the graph with nearly constant temperature. a. Select the data in the flat region of the graph. b. Find the mean temperature for the selected data. Record this value. c. Store the data, so that they can be used later.Adapted from Advanced Chemistry with Vernier &amp; Laboratory Experiments for Advanced Placement Chemistry by Sally Ann Vonderbrink, Ph. D.Page 2The Determination of Molar Mass by Freezing-Point Depression Part II: Determine the Freezing Point of a Solution of Benzoic Acid and Lauric Acid 10. Obtain a test tube containing a melted solution with ~1 g of benzoic acid dissolved in ~8 g of lauric acid. Record the precise masses of benzoic acid and lauric acid as indicated on the label of the test tube.Freezing PointTimeRepeat Steps 3-8. 11. The freezing point of the benzoic acid-lauric acid solution can be determined by finding the temperature at which the mixture initially started to freeze. Unlike pure lauric acid, the mixture results in a gradual linear decrease in temperature during freezing. 12. Print a graph showing both trials. (See the graph pictured in question 6 of the Pre-Lab Questions.)DATA TABLEMass of lauric acid (g)Mass of benzoic acid (g)Freezing temperature of pure lauric acid (°C)Freezing point of the benzoic acid-lauric acid mixture (°C)PRE-LAB QUESTIONS1. What types of intermolecular forces are present in a molecular solid such as lauric acid? Describe what is happening with regard to intermolecular forces as a molecular liquid freezes. 2. If you were able to choose a solvent for this experiment from the list below, which would you choose? Justify your answer. · · · · Acetic Acid CH3COOH Benzene C6H6 tert-Butanol C4H9OH Cyclohexane C6H12 Kf Kf Kf Kf = = = = 3.90 °C·kg/mol 5.12 °C·kg/mol 9.10 °C·kg/mol 20.00 °C·kg/molPage 3Adapted from Advanced Chemistry with Vernier &amp; Laboratory Experiments for Advanced Placement Chemistry by Sally Ann Vonderbrink, Ph. D.The Determination of Molar Mass by Freezing-Point Depression 3. What is the equation for calculating molality? Why do we use molality rather than molarity as our concentration unit for this experiment? 4. Use the equation above along with the freezing-point depression equation to derive an expression for calculating the molar mass of a solute. 5. The following data were obtained in an experiment designed to determine the molar mass of a solute by freezing-point depression. The Kf of para-dichlorobenzene is 7.1 °C·kg/molMass of para-dichlorobenzene (g) Mass of unknown solute (g) Freezing temperature of pure para-dichlorobenzene (°C) Freezing point of the solution (°C)24.80 g 2.04 g 53.02 °C 50.78 °C(a) Calculate the freezing-point depression, Tf of the solution. (b) Calculate the molar mass of the unknown substance.6. Examine the graph below. What laboratory technique best prevents supercooling?Adapted from Advanced Chemistry with Vernier &amp; Laboratory Experiments for Advanced Placement Chemistry by Sally Ann Vonderbrink, Ph. D.Page 4The Determination of Molar Mass by Freezing-Point DepressionPOST-LAB QUESTIONS AND DATA ANALYSIS1. Calculate molality (m), in mol/kg, using the formula Tf = Kf × m × i. The Kf value for lauric acid is 3.9°C·kg/mol and since lauric acid is a molecular solid, i is approximately equal to 1. 2. Calculate moles of benzoic acid solute, using the molality and the mass (in kg) of lauric acid solvent. 3. Calculate the experimental molecular weight of benzoic acid, in g/mol. 4. Determine the accepted molecular weight of benzoic acid from its formula, C6H5COOH. 5. Calculate the percent error between the experimental and accepted values. 6. Explain why the pure solvent shows a level horizontal curve as solidification occurs, but the curve for the solution slopes downward slightly. 7. A student spills some of the solvent before the solute was added. What effect does this error have on the calculated molar mass of the solute? Mathematically justify your answer. 8. A different student spills some of the solution before the freezing-point was determined. What effect does this error have on the calculated molar mass of the solution? Justify your answer.Adapted from Advanced Chemistry with Vernier &amp; Laboratory Experiments for Advanced Placement Chemistry by Sally Ann Vonderbrink, Ph. D.Page 5The Determination of Molar Mass by Freezing-Point DepressionTEACHER INFORMATION1. This experiment conforms to the guidelines for the fourth laboratory experiment listed in the College Board AP Chemistry guide (the Acorn book). 2. Lauric acid, CH3(CH2)10COOH, has an accepted melting point of 44.0°C and a molecular mass of 122.0 g/mol. It is also called dodecanoic acid. 3. The Tf value limits the number of significant figures of the final answer for molecular mass to two or more depending on your temperature probe or thermometer. Thus, the final molecular mass in the sample data is expressed as 120 g/mol not 119 g/mol. 4. Test tube sizes 18 × 150 mm, 20 × 150 mm, or 25 × 150 mm work well. 5. These can be prepared well in advance. For Part I, put about 8 g of lauric acid in each tube (once you've done the first one, you can guess for the rest since the amount doesn't matter). Number each tube 1-2, 1-2, etc.. For Part II, mix about 8 grams of lauric acid with about 1 gram of benzoic acid per test tube. Label each test tube with a number 2-1, 2-2, etc. and the precise mass of lauric acid and benzoic acid and make sure that the label will be above the water level of the water bath. These filled test tubes may be reused, as long as your students avoid cross-contaminating the test tubes with the Temperature Probe. Stopper the test tubes and store them for future use. 6. It is a good idea to have hot plates with water baths warming up, before students arrive, to save time.7. Prepare a separate hot water bath in a central location for the students to use to free the probes that have been frozen in test tubes.HAZARD ALERTSLauric acid: Slightly toxic by ingestion; body tissue irritant; combustible. Hazard Code: C--Somewhat hazardous. Benzoic acid: Slightly toxic by ingestion; body tissue irritant; combustible. Hazard Code: C--Somewhat hazardous. The hazard information reference is: Flinn Scientific, Inc., Chemical and Biological Catalog Reference Manual, P.O. Box 219, Batavia, IL 60510, (800) 452-1261, www.flinnsci.comAdapted from Advanced Chemistry with Vernier &amp; Laboratory Experiments for Advanced Placement Chemistry by Sally Ann Vonderbrink, Ph. D.Page 6The Determination of Molar Mass by Freezing-Point DepressionSAMPLE DATA TABLEMass of lauric acid (g) 8.01 gMass of benzoic acid (g)1.00 gFreezing temperature of pure lauric acid (°C)44.0°CFreezing point of the benzoic acid-lauric acid mixture (°C)39.9°CAnswers to PRE-LAB QUESTIONS1. What types of intermolecular forces are present in a molecular solid such as lauric acid? Describe what it happening with regard to both energy and intermolecular forces as a molecular liquid freezes. Since lauric acid is molecular, the most prevalent IMFs are London dispersion forces (induced dipoleinduced dipole). As the sample cools, the temperature decreases hence the average kinetic energy of the molecules decreases and the molecules slow down. At some point, enough heat is removed so that the attractive forces of the molecules bring them closer together, and their positions become fixed and the substance freezes. 2. If you were able to choose a solvent for this experiment from the list below, which would you choose? Justify your answer. · · · · Acetic Acid CH3COOH Benzene C6H6 tert-Butanol C4H9OH Cyclohexane C6H12 Kf Kf Kf Kf = = = = 3.90 °C·kg/mol 5.12 °C·kg/mol 9.10 °C·kg/mol 20.00 °C·kg/molCyclohexane. The larger the value of the freezing-point depression constant, the more precise the molar mass can be determined considering the small values of temperature change for this type of experiment.Adapted from Advanced Chemistry with Vernier &amp; Laboratory Experiments for Advanced Placement Chemistry by Sally Ann Vonderbrink, Ph. D.Page 7The Determination of Molar Mass by Freezing-Point Depression 3. What is the equation for calculating molality? Why do we use molality rather than molarity as our concentration unit for this experiment? mass solute (g) moles solute MM Students should receive full credit for either version m= = kg solvent kg solvent We use molality for colligative property experiments since it is temperature independent and is essentially a ratio of masses (solute : solvent)which does not change with temperature changes. Molarity incorporates the volume of the solution which can expand or contract with changes in temperature. 4. Use the equation above along with the freezing-point depression equation to derive an expression for calculating the molar mass of a solute. mass solute (g) moles solute MM and T = K × m × i m= = f f kg solvent kg solvent g T f = K f × MM × i ; where i  1 kg g T f (kg ) = K f T f (kg ) MM = K f g MM K f g ( solute)  MM = T f (kg solvent ) 5. The following data were obtained in an experiment designed to determine the molar mass of a solute by freezing-point depression. The Kf of para-dichlorobenzene is 7.1 °C·kg/molMass of para-dichlorobenzene (g) Mass of unknown solute (g) Freezing temperature of pure para-dichlorobenzene (°C) Freezing point of the solution (°C) 24.80 g 2.04 g 53.02 °C 50.78 °C(a) Calculate the freezing-point depression, Tf of the solution. T f = ( 53.02 - 50.78 ) °C (b) Calculate the molar mass of the unknown substance. Note that i  1 °C · kg × 2.04 g K f g ( solute) g mol  MM = = = 260 T f (kg solvent ) 0.02480 kg × 2.24°C mol Note that the precision of your thermometer will dictate the number of sig. figs reported on the molar mass. 7.1Adapted from Advanced Chemistry with Vernier &amp; Laboratory Experiments for Advanced Placement Chemistry by Sally Ann Vonderbrink, Ph. D.Page 8The Determination of Molar Mass by Freezing-Point Depression 6. Examine the graph below. What laboratory technique best prevents supercooling?Stirring constantly during the cooling process.Adapted from Advanced Chemistry with Vernier &amp; Laboratory Experiments for Advanced Placement Chemistry by Sally Ann Vonderbrink, Ph. D.Page 9The Determination of Molar Mass by Freezing-Point DepressionAnswers to POST-LAB QUESTIONS AND DATA ANALYSIS1. Calculate molality (m), in mol/kg, using the formula Tf = Kf × m × i. The Kf value for lauric acid is 3.9°C·kg/mol and since lauric acid is a molecular solid, i is approximately equal to 1.T f = K f × m × i; where i  1 m = T f Kf = 4.1°C mol = 1.05 °C · kg kg 3.9 mol2. Calculate moles of benzoic acid solute, using the molality and the mass (in kg) of lauric acid solvent. Answers will vary. For the sample data, mol 1.05 × 0.00801 kg solvent = 0.00841 mol benzoic acid kg 3. Calculate the experimental molecular weight of benzoic acid, in g/mol.°C · kg    3.9  (1.00 g ) K f g ( solute) g g mol    MM = = = 118.8 = 120 mol mol T f (kg solvent ) ( 4.1°C )( 0.00801 kg ) Since the change in temperature is reported as 2 SF. 4. Determine the accepted molecular weight of benzoic acid from its formula, C6H5COOH. The accepted molar mass is the molar mass calculated from the chemical formula given and is equal to 122 g/mol. 5. Calculate the percent error between the experimental and accepted values.% error = 122 - 120 122 × 100 = 0.02 % error6. Explain why the pure solvent shows a level horizontal curve as solidification occurs, but the curve for the solution slopes downward slightly. A pure substance maintains a constant as it freezes. When a solution freezes, the pure solvent freezes first. As it solidifies, the remaining solution is more concentrated so its freezing point is lower. 7. A student spills some of the solvent before the solute was added. What effect does this error have on the calculated molar mass of the solute? Mathematically justify your answer., therefore a smaller number for kg T f (kg solvent ) of solvent in the denominator will result in a larger calculated molar mass.Adapted from Advanced Chemistry with Vernier &amp; Laboratory Experiments for Advanced Placement Chemistry by Sally Ann Vonderbrink, Ph. D.The calculated molar mass will be too high. MM =K f g ( solute)Page 10The Determination of Molar Mass by Freezing-Point Depression 8. A different student spills some of the solution before the freezing-point was determined. What effect does this error have on the calculated molar mass of the solution? Justify your answer.The calculated molar mass will unaffected. The spill will not alter the ratio of solute molecules to solvent molecules; therefore the molality of the solution remains the same before and after the spill.Adapted from Advanced Chemistry with Vernier &amp; Laboratory Experiments for Advanced Placement Chemistry by Sally Ann Vonderbrink, Ph. D.Page 112000 AP® CHEMISTRY FREE-RESPONSE QUESTIONSYour responses to the rest of the questions in this part of the examination will be graded on the basis of the accuracy and relevance of the information cited. Explanations should be clear and well organized. Examples and equations may be included in your responses where appropriate. Specific answers are preferable to broad, diffuse responses. Answer BOTH Question 5 below AND Question 6 printed on page 11. Both of these questions will be graded. The Section II score weighting for these questions is 30 percent (15 percent each). 5. The molar mass of an unknown solid, which is nonvolatile and a nonelectrolyte, is to be determined by the freezing-point depression method. The pure solvent used in the experiment freezes at 10°C and has a known molal freezing-point depression constant, Kf . Assume that the following materials are also available.  test tubes  beaker  stirrer  stopwatch  pipet  graph paper  thermometer  hot-water bath  balance  ice(a) Using the two sets of axes provided below, sketch cooling curves for (i) the pure solvent and for (ii) the solution as each is cooled from 20°C to 0.0°C.(b) Information from these graphs may be used to determine the molar mass of the unknown solid. (i) Describe the measurements that must be made to determine the molar mass of the unknown solid by this method. (ii) Show the setup(s) for the calculation(s) that must be performed to determine the molar mass of the unknown solid from the experimental data. (iii) Explain how the difference(s) between the two graphs in part (a) can be used to obtain information needed to calculate the molar mass of the unknown solid. (c) Suppose that during the experiment a significant but unknown amount of solvent evaporates from the test tube. What effect would this have on the calculated value of the molar mass of the solid (i.e., too large, too small, or no effect)? Justify your answer. (d) Show the setup for the calculation of the percentage error in a student's result if the student obtains a value of 126 g mol-1 for the molar mass of the solid when the actual value is 120. g mol-1. Copyright © 2000 College Entrance Examination Board and Educational Testing Service. All rights reserved. AP is a registered trademark of the College Entrance Examination Board. GO ON TO THE NEXT PAGE. -10Page 12AP® Chemistry 2000  Scoring StandardsQuestion 5 (10 points)Pure Solvent Solution(a)20 202 pts.Temperature (oC)10Temperature (oC) Time1515105500TimeNotes: One point is earned for each correct graph. The first graph should show a line that drops to 10°C, holds steady at 10°C, and then falls steadily to 0°C. There must be a discernable plateau at 10°C to earn this point. The second graph should show a line that drops to below 10°C, levels off (or slants down a bit), and then falls more sharply to 0°C.(b)(i) Measure mass of solute, mass of solvent, mass of solution (two of three must be shown) Measure the Tfp (or the freezing point of the solution) · Volume of solution (without density), molality, or number of moles do not earn points (ii) Given: T = iKf m (or T = Kf m) m = (mol solute)/(kg solvent) moles = g/(molar mass) Combine to get: molar mass = (i)(Kf)(g solute)/(T)(kg solvent) Notes: One point is earned for any two equations, and two points are earned for all three equations. &quot;Solute&quot; and &quot;solvent&quot; must be clearly identified in the equations.1 pt.1 pt.2 pts.(iii) the difference in the vertical position of the horizontal portions of the graphs 1 pt. is equal to Tfp , the change in freezing point due to the addition of the solute.Copyright © 2000 College Entrance Examination Board and Educational Testing Service. All rights reserved. AP is a registered trademark of the College Entrance Examination Board.Page 13AP® Chemistry 2000  Scoring StandardsQuestion 5 (continued)(c)The molar mass is too small. If some of the solvent evaporates, then the (kg solvent) term used in the equation in (b) (ii) is larger than the actual value. If the (kg solvent) term used is too large, then the value calculated for the molar mass will be too small. or If some of the solvent evaporates, then the concentration (molality) of the solute will be greater than we think it is. More moles of solute results in a smaller molar mass (or since T = iKf m, then the Tobs would be greater than it should be). Since the molar mass of the unknown solute is inversely proportional to T, an erroneously high value for T implies an erroneously low value for the molar mass (calculated molar mass would be too small).(126 g mol -1 - 120 g mol -1 ) 120 g mol -11 pt. 1 pt.(d)% error =× 100%1 pt.or % error =6 g mol -1 120 g mol -1× 100%Copyright © 2000 College Entrance Examination Board and Educational Testing Service. All rights reserved. AP is a registered trademark of the College Entrance Examination Board.Page 14Page 15Page 16`

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Microsoft Word - 04 Determination of Molar Mass by Freezing Point Depression.doc