04 determining molar mass by fp depression

Adapted from Advanced Chemistry with Vernier & Laboratory Experiments for Advanced Placement Chemistry by Sally Ann Vonderbrink, Ph. D.

Determination of Molar Mass by Freezing-Point Depression

When 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 (3.9Ckg/mol for lauric acid in this experiment1), i is the vant Hoff factor, and m is the molality of the solution (in mol solute/kg solvent). Since lauric acid is not ionic, its vant Hoff factor is essentially equal to 1.

OBJECTIVES In 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 1

MATERIALS Data Collection Mechanism lauric acid, CH3(CH2)10COOH Temperature Probe lauric acid-benzoic acid mixture ring stand hot water bath400 mL beaker utility clampTissue or paper towels two 18 150 mm test tubes (if pre-made samples are not provided by your teacher)

1The Computer-Based Laboratory, Journal of Chemical Education: Software, 1988, Vol.1A, No. 2, p. 73.

The Determination of Molar Mass by Freezing-Point Depression


PROCEDURE

1. 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 20-25C 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.



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.

Repeat 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 TABLE

Mass of lauric acid in the benzoic acid-lauric acid mixture (g)

Mass of benzoic acid in the benzoic acid-lauric acid mixture (g)

Freezing temperature of pure lauric acid (C)

Freezing point of the benzoic acid-lauric acid mixture (C)

PRE-LAB QUESTIONS

1. 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 Kf = 3.90 Ckg/mol Benzene C6H6 Kf = 5.12 Ckg/mol tert-Butanol C4H9OH Kf = 9.10 Ckg/mol Cyclohexane C6H12 Kf = 20.00 Ckg/mol

Time

Freezing Point



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 Ckg/mol

Mass of para-dichlorobenzene (g) 24.80 g

Mass of unknown solute (g) 2.04 g

Freezing temperature of pure para-dichlorobenzene (C) 53.02 C

Freezing point of the solution (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?



POST-LAB QUESTIONS AND DATA ANALYSIS 1. Calculate molality (m), in mol/kg, using the formula Tf = Kf m i. The Kf value for lauric acid is

3.9Ckg/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.



TEACHER INFORMATION

1. 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 44.0C 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 youve done the first one, you can guess for the rest since the amount doesnt 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. You may also make one large batch of the 8g:1g ratio, but be sure to melt the mixture to ensure complete mixing and pour it into the tubes. Each tube will need to be labeled with the precise masses of each of the two substances.

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 ALERTS Lauric acid: Slightly toxic by ingestion; body tissue irritant; combustible. Hazard Code: CSomewhat hazardous. Benzoic acid: Slightly toxic by ingestion; body tissue irritant; combustible. Hazard Code: CSomewhat 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.com



SAMPLE DATA TABLE

Mass of lauric acid in the lauric acid-benzoic acid mixture (g) 8.01 g

Mass of benzoic acid in the lauric acid-benzoic acid mixture (g) 1.00 g

Freezing temperature of pure lauric acid (C) 44.0C

Freezing point of the benzoic acid-lauric acid mixture (C) 39.9C

Answers to PRE-LAB QUESTIONS

1. 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 dipole-induced 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 Kf = 3.90 Ckg/mol Benzene C6H6 Kf = 5.12 Ckg/mol tert-Butanol C4H9OH Kf = 9.10 Ckg/mol Cyclohexane C6H12 Kf = 20.00 Ckg/mol

Cyclohexane. 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.



3. What is the equation for calculating molality? Why do we use molality rather than molarity as our concentration unit for this experiment?

masssolute (g)moles solutekg solvent kg solvent

MMm = = Students should receive full credit for either version

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.

masssolute (g)moles solute and kg solvent kg solvent

; where 1

( ) ( )

( )( )

f f

f f

f f f f

f

f

MMm T K m i

gMMT K i ikg

gT kg K T kg MM K gMM

K g soluteMM

T 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 Ckg/mol

Mass of para-dichlorobenzene (g) 24.80 g

Mass of unknown solute (g) 2.04 g

Freezing temperature of pure para-dichlorobenzene (C) 53.02 C

Freezing point of the solution (C) 50.78 C

(a) Calculate the freezing-point depression, Tf of the solution. ( ) 53.02 50.78 CfT = (b) Calculate the molar mass of the unknown substance.

Note that 1C kg7.1 2.04 g( ) gmol 260

( ) 0.02480 kg 2.24 C molf

f

i

K g soluteMM

T kg solvent

= = =

Note that the precision of your thermometer will dictate the number of sig. figs reported on the molar mass.



6. Examine the graph below. What laboratory technique best prevents supercooling?

Stirring constantly during the cooling process.



Answers to POST-LAB QUESTIONS AND DATA ANALYSIS 1. Calculate molality (m), in mol/kg, using the formula Tf = Kf m i. The Kf value for lauric acid is

3.9Ckg/mol and since lauric acid is a molecular solid, i is approximately equal to 1.

; where 1

4.1 C mol 1.05C kg kg3.9mol

f f

f

f

T K m i iT

mK

= = = =

2. 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,

mol1.05 0.00801 kg solvent = 0.00841 mol benzoic acidkg

3. Calculate the experimental molecular weight of benzoic acid, in g/mol.

( )( )( )

C kg3.9 1.00 g( ) g gmol 118.8 120( ) 4.1 C 0.00801 kg mol molf

f

K g soluteMM

T kg solvent

= = = = 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.

122 120% error = 100 0.02 % error

122 =

6. 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.

The calculated molar mass will be too high. ( )

( )f

f

K g soluteMM

T kg solvent= , therefore a smaller number for kg

of solvent in the denominator will result in a larger calculated molar mass.



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.

2000 AP CHEMISTRY FREE-RESPONSE QUESTIONS

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.-10-

Your responses to the rest of the questions in this part of the examination will be graded on the basis of the accuracyand relevance of the information cited. Explanations should be clear and well organized. Examples and equationsmay 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. TheSection 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 thefreezing-point depression method. The pure solvent used in the experiment freezes at 10C and has a knownmolal freezing-point depression constant, Kf . Assume that the following materials are also available.

test tubes stirrer pipet thermometer balance beaker stopwatch graph paper hot-water bath ice

(a) Using the two sets of axes provided below, sketch cooling curves for (i) the pure solvent and for (ii) thesolution as each is cooled from 20C to 0.0C.

(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 testtube. What effect would this have on the calculated value of the molar mass of the solid (i.e., too large, toosmall, or no effect)? Justify your answer.

(d) Show the setup for the calculation of the percentage error in a students result if the student obtains a valueof 126 g mol1 for the molar mass of the solid when the actual value is 120. g mol1.

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AP Chemistry 2000 Scoring Standards


Question 5(10 points)

(a)2 pts.

Notes: One point is earned for each correct graph. The first graph should show aline that drops to 10C, holds steady at 10C, and then falls steadily to 0C. Theremust be a discernable plateau at 10C to earn this point. The second graph shouldshow a line that drops to below 10C, levels off (or slants down a bit), and then fallsmore sharply to 0C.

(b) (i) Measure mass of solute, mass of solvent, mass of solution 1 pt.(two of three must be shown)

Measure the Tfp (or the freezing point of the solution) 1 pt.

Volume of solution (without density), molality, or number of moles do not earn points

(ii) Given: T = iKf m (or T = Kf m) 2 pts.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 earnedfor all three equations. Solute and solvent must be clearly identified inthe equations.

(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.

Time

Tem

pera

ture

(oC

)

0

5

10

15

20

Pure Solvent

Time

Tem

pera

ture

(oC

)

0

5

10

15

20

Solution

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AP Chemistry 2000 Scoring Standards


Question 5(continued)

(c) The molar mass is too small. 1 pt.

If some of the solvent evaporates, then the (kg solvent) term used in the 1 pt.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 betoo small.

or

If some of the solvent evaporates, then the concentration (molality) of thesolute will be greater than we think it is. More moles of solute results in asmaller molar mass (or since T = iKf m, then the Tobs would be greaterthan it should be). Since the molar mass of the unknown solute is inverselyproportional to T, an erroneously high value for T implies an erroneouslylow value for the molar mass (calculated molar mass would be too small).

(d) % error = 1

11

molg120)molg120molg126(

100% 1 pt.

or

% error = 1

1

molg120molg6

100%

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04 determining molar mass by fp depression

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