-
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.
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The Determination of Molar Mass by Freezing-Point Depression
Adapted from Advanced Chemistry with Vernier & Laboratory
Experiments for Advanced Placement Chemistry by Sally Ann
Vonderbrink, Ph. D.
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.
-
The Determination of Molar Mass by Freezing-Point Depression
Adapted from Advanced Chemistry with Vernier & Laboratory
Experiments for Advanced Placement Chemistry by Sally Ann
Vonderbrink, Ph. D.
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
-
The Determination of Molar Mass by Freezing-Point Depression
Adapted from Advanced Chemistry with Vernier & Laboratory
Experiments for Advanced Placement Chemistry by Sally Ann
Vonderbrink, Ph. D.
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?
-
The Determination of Molar Mass by Freezing-Point Depression
Adapted from Advanced Chemistry with Vernier & Laboratory
Experiments for Advanced Placement Chemistry by Sally Ann
Vonderbrink, Ph. D.
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.
-
The Determination of Molar Mass by Freezing-Point Depression
Adapted from Advanced Chemistry with Vernier & Laboratory
Experiments for Advanced Placement Chemistry by Sally Ann
Vonderbrink, Ph. D.
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
-
The Determination of Molar Mass by Freezing-Point Depression
Adapted from Advanced Chemistry with Vernier & Laboratory
Experiments for Advanced Placement Chemistry by Sally Ann
Vonderbrink, Ph. D.
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.
-
The Determination of Molar Mass by Freezing-Point Depression
Adapted from Advanced Chemistry with Vernier & Laboratory
Experiments for Advanced Placement Chemistry by Sally Ann
Vonderbrink, Ph. D.
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.
-
The Determination of Molar Mass by Freezing-Point Depression
Adapted from Advanced Chemistry with Vernier & Laboratory
Experiments for Advanced Placement Chemistry by Sally Ann
Vonderbrink, Ph. D.
6. Examine the graph below. What laboratory technique best
prevents supercooling?
Stirring constantly during the cooling process.
-
The Determination of Molar Mass by Freezing-Point Depression
Adapted from Advanced Chemistry with Vernier & Laboratory
Experiments for Advanced Placement Chemistry by Sally Ann
Vonderbrink, Ph. D.
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.
-
The Determination of Molar Mass by Freezing-Point Depression
Adapted from Advanced Chemistry with Vernier & Laboratory
Experiments for Advanced Placement Chemistry by Sally Ann
Vonderbrink, Ph. D.
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.
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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.
Page 12
aarthurRectangle
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AP Chemistry 2000 Scoring Standards
Copyright 2000 College Entrance Examination Board and
Educational Testing Service. All rights reserved.AP is a registered
trademark of the College Entrance Examination Board.
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
Page 13
aarthurRectangle
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AP Chemistry 2000 Scoring Standards
Copyright 2000 College Entrance Examination Board and
Educational Testing Service. All rights reserved.AP is a registered
trademark of the College Entrance Examination Board.
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%
Page 14
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Page 15
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Page 16
04 Determining Molar Mass by FP Depression.pdfAPC FR 00
Q5.pdfap_frq00_copyright.pdf2000 Advanced Placement Program
sg_chemistry_00.pdfquestion1.pdfQuestion 1Question 1
question4.pdfQuestion 4Question 4
question5.pdfQuestion 5Question 5
question6.pdfQuestion 6Question 6(continued)
question7.pdfQuestion 7Question 7
question8.pdfQuestion 8Question 8
sample_chemistry_00_q5.pdf2000 AP Chemistry Student
SamplesQuestion 5Sample A - 10 PointsSample B - 8 PointsSample C -
6 Points