Molar Mass by Freezing Point Depression
AP Chemistry
Dec. 14,2009
Purpose: The purpose of this experiment is to determine the molar mass of an unknown
substance by measuring the freezing point depression of a solution of the unknown substance and
BHT.
Uncalculated Data Table 1:
Mass of empty test tube #1
Mass of test tube #1 plus BHT
Mass of weighing paper
Mass of weighing paper plus cetyl alcohol
Mass of empty test tube #2
Mass of test tube #2 plus BHT
Mass of unknown
19.05g
27.06g
2.419g
3.512g
18.964g
27.355g
1.129g
Uncalculated Data Table 2:
Time, in seconds
20
40
60
80
100
120
140
160
180
200
220
240
260
280
Crystals begin to form.
Pure BHT
77.3
74.3
71.7*
69.8
69.7
70.4
70.2
70.1
70
-
Temperature in °C
BHT + Cetyl Alcohol
83
80.1
77.3
74.8
72.5
70.5
68.3
66.6*
65
64.8
64.8
64.6
64.4
64.2
BHT + Unknown
75.6
72.1
67.9
67.2*
66.4
66.6
66.4
66.2
66
65.2
-
Calculated Data Table 1:
BHT
70 °C
8.903 °C/m,
-
Freezing Point, °C
ATfp, °C
Kfp, °C/m/
Molar mass, g/mole
BHT + Cetyl Alcohol
65 °C
5°C
-
Calculated Data Table 2:
Mass of BHT
Mass of cetyl alcohol
Mass of BHT
BHT + Unknown
66°C
4°C
299.472 g/mole
8.037g
1.093g
8.391g
Graph:
Temperature vs Cooling Time for Solvent and
Solution
85
80
O
O
Pure BHT
75
BHT + Cetyl Alcohol
H 70
BHT + Unknown
65
60
50
100
150
200
250
300
Time of Cooling
Calculations:
Mass of BHT:
(Mass of test tube #1 plus BHT) - (Mass of empty test tube #1)
27.06g-19.05g = 8.037g
Mass of cetyl alcohol:
(Mass of weighing paper plus cetyl alcohol) - (Mass of weighing paper)
3.512g-2.419g=1.093g
Mass of BHT (second time):
(Mass of test tube #2 plus BHT) - (Mass of empty test tube #2)
27.355g-18.964g = 8.391g
Percent Error:
299.472-484.49x100
284.49
= 5.266% en-or
1. A: Look at graph.
B:
m=
1.093g
(.00803 7kg)(242.16g)
= .5616°C/m
ATfp= 70°-65° = 5°
5 = kfp(.5616°C/m)
kfp= 8.903°C/m
C: molar mass: (8.9030C/m)(1.129g) = 299.472g/mol
(.008391 kg)(4°)
Conclusion: The purpose of this experiment is to determine the molar mass of an unknown
substance by measuring the freezing point depression of a solution of the unknown substance and
BHT. First, in measuring the cetyl alcohol plus the BHT, we took the masses of both substances
along with the molar mass of cetyl alcohol to determine the freezing point constant. This was
calculated to be 8.903 °C/m/. We then used the calculated constant, the mass of BHT and the
unknown, and the change in temperature to calculate the molar mass of the unknown substance.
This was calculated to be stearic acid with a 299.472 g/mol molar mass. The actual value for the
molar mass of stearic acid is 284.49 g/mol, which gave us a percent error of 5.266%.
Error Analysis: Several errors occurred within the lab. For example, parcels of the BHT were
lost while trying to transfer the BHT into the test tube both times. This affected our mass of the
test tubes plus BHT, which in turn affected the mass of the BHT. By losing BHT, our molar
mass appeared smaller than it actually should be. This same error occurred as well when we
transferred the cetyl alcohol into the test tube.
Post-Lab Questions:
2. The least precise measurement in the experiment was the recording of temperature every 20
seconds. This limits our significant digits because the thermometers can only read to one
decimal, and since they are being read in the region from 60-85, there are effectively three
significant digits.
3. When measuring the temperature of the pure BHT and the BHT plus unknown, we found
evidence of supercooling. In our measurements our data showed an original decrease in
temperature, then it rose slightly, and then proceeded to drop very gradually again.
4. It is advantageous to choose a solvent that has a large value for kf p because the larger the kfp
the larger the drop in freezing point will be. This will give us more significant digits in drop in
freezing point and will allow for better precision and accuracy.
5. The pure BHT shows a constant horizontal curve as solidification occurs which indicates that
the pure BHT will freeze at a constant temperature. The decrease in the solution's slope
indicates the solute is decreasing the freezing point. Then, the concentration of the dissolved
solute steadily increases as the solvent freezes, causing the freezing point to continually
decrease.