Archer G11
Partner: Jack
17 November 2011
Determination of the Molar Mass of Volatile Liquids
Purpose: The purpose of this experiment is to determine the molar mass of three volatile liquids. The
liquid is vaporized in a Beral-type pipet with capillary head then rapidly cool to condense the vapor. The
mass of the condensed vapor is determined then used in a form of ideal gas equation to find the molar
mass of each volatile liquid. The significance of this lab is that forensic scientist could use the molar mass
of the volatile liquid together with the empirical formula of a compound to determine the molecular
formula of that compound.
Hypothesis: The hypothesis is that the molar mass of the volatile liquid can be found by vaporizing the
volatile liquid in a Beral-type pipet. The mass of the condensed vapor can be used in a form of the ideal
gas equation to find the molar mass of the volatile liquid. The temperature of the pipet was heated so it
was high, so the molecules behave like an ideal gas because they have a lot of energy to overcome
intermolecular forces.
Materials:
Materials
Ethyl Alcohol (Ethanol) (C2H5OH)
Acetone (CH3COCH3)
Isopropyl Alcohol ((CH3)2CHOH)
Distilled water
Tap water
15-mL Beral-type pipets
4-decimals electric balance
Hot plate with magnetic stirrer
Magnetic bar
250-mL beaker
150-mL beaker
600-mL beaker
Ring stand
Thermometer
Scissors
Paper towel
Permanent marker
Plastic tubing pipet holder
Universal clamp
Boiling stones
10-mL graduated cylinder
25-mL graduated cylinder
Buret clamp
Procedures:
Trial 1
9 mL
9 mL
9 mL
Trial 2
9 mL
9 mL
9 mL
2L
1.2 L
18 pipets
1 balance
2 plates
3 bars
2 beakers
3 beakers
5 beakers
3 ring stands
3 thermometers
1 pair
3 blocks
4 markers
4 tubing pipet holders
3 clamps
90 mL
2 cylinders
2 cylinders
3 clamps
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1.) Add about 30 mL of boiling stones to the 600-mL beaker
2.) Add about 500 mL of distilled water to the beaker
3.) Put it on a hot plate with magnetic stirrer
4.) Start heating the water using heat level of 9
5.) Pull the thin part of the Beral-type pipet
6.) Keep pulling until it stretch, creating a capillary tube where it’s pulled
7.) Try to leave as much thin part as possible
8.) Cut the capillary part of the pipet, leaving it about 1 cm long on the side of the bulb head
9.) Repeat step 5.) to 8.) for two more pipets
10.) Label the pipets as 1, 2, and 3
11.) Mass each pipet
12.) Using the graduated cylinder, measure 3 mL of ethyl alcohol
13.) Transfer the alcohol to pipet number 1
14.) Repeat step 12.) and 13.) for pipet number 2 and 3
15.) Insert the tube part of all three pipets through a plastic tubing pipet holder
16.) Clamp the plastic tubing
17.) Put the bulb part of the pipet in the heated water
18.) Put the thermometer in the beaker in a way that the thermometer does not touch the side of
the beaker
19.) Add about 500 mL of tap water to another 600-mL beaker
20.) Wait until all the volatile liquid (ethyl alcohol) had vaporized
21.) As soon as all the liquid had vaporized, check the temperature
22.) Transfer the pipets to the tap water beaker
23.) Check the temperature of the tap water
24.) After a minute, take the pipets out of the water
25.) Gently dry each pipet with paper towel
26.) Mass each pipet
27.) Rinse the inside of each pipet with distilled water
28.) Fill the pipets completely with water
29.) Mass each water-filled pipet
30.) Squeeze all the water from inside the pipet to a 25-mL graduated cylinder
31.) Record the volume of water
32.) Repeat step 30.) and 31.) for the other two pipets
33.) Repeat step 5.) to 32.) using acetone
34.) Repeat step 5.) to 32.) using isopropyl alcohol
35.) Repeat step 1.) to 34.) for one more trial
Results: The longer the thin tube part of the Beral-type pipet was left, the easier it is to get them
through the plastic tubing pipet holder because the bulb part of the pipet would not be pushed against
each other. The boiling stones caused the water to blur from the stone dust. After heating the pipet for
a while, the top of the hot plate turn yellow. The brown color of the permanent marker turned to blue
after it was heated along with the pipets in the water bath. Some of the permanent ink was washed
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away with heated water. The ink is easy to be washed with the volatile liquid. All the volatile liquid feels
cold when touched and exposed to air. Some liquid take longer to vaporize than the other.
Determination of Molar Mass of Ethanol (Ethyl Alcohol) Trial 1
Temperature of boiling water bath (K)
362.15
Barometric pressure (atm)
0.9949
Temperature of room temp. water bath (K)
298.15
Density of water at room temperature (g/mL)
0.9970479
(http://en.wikipedia.org/wiki/Density)
Mass of empty pipet (g)
Mass of pipet and water (g)
Mass of water in filled pipet (g)
Volume of pipet (mL)
Ethyl Alcohol
Mass of pipet and condensed
ethyl alcohol (g)
Mass of condensed ethyl alcohol
(g)
Molar mass of ethyl alcohol
(g/mol)
Average molar mass of ethyl
alcohol (g/mol)
Pipet 1
1.8037
16.6232
14.8195
14.8634
Trial 1
Pipet 2
1.7560
16.6447
14.8887
14.9328
Trial 2
Pipet 3
1.7468
16.6284
14.8816
14.9257
Trial 3
1.8353
1.7906
1.7772
0.0316
0.0346
0.0304
63.72
69.24
60.87
64.61
Determination of Molar Mass of Ethanol (Ethyl Alcohol) Trial 2
Temperature of boiling water bath (K)
356.15
Barometric pressure (atm)
0.9949
Temperature of room temp. water bath (K)
298.15
Density of water at room temperature (g/mL)
0.9970479
(http://en.wikipedia.org/wiki/Density)
Mass of empty pipet (g)
Mass of water in filled pipet (g)
Volume of pipet (mL)
Ethyl Alcohol
Mass of pipet and condensed
ethyl alcohol (g)
Mass of condensed ethyl alcohol
(g)
Molar mass of ethyl alcohol
(g/mol)
Average molar mass of ethyl
alcohol (g/mol)
Pipet 1
1.7310
15
15
Trial 1
Pipet 2
1.7501
16
16
Trial 2
Pipet 3
1.8514
16
16
Trial 3
1.7544
1.7760
1.8882
0.0234
0.0259
0.0368
45.85
47.57
67.60
53.67
Determination of Molar Mass of Acetone Trial 1
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Temperature of boiling water bath (K)
Barometric pressure (atm)
Temperature of room temp. water bath (K)
Density of water at room temperature (g/mL)
351.15
0.9949
298.15
0.9970479
(http://en.wikipedia.org/wiki/Density)
Mass of empty pipet (g)
Mass of pipet and water (g)
Mass of water in filled pipet (g)
Volume of pipet (mL)
Acetone
Mass of pipet and condensed
acetone(g)
Mass of condensed acetone(g)
Molar mass of acetone(g/mol)
Average molar mass of acetone
(g/mol)
Pipet 1
1.8216
16.8438
15.0222
15.0667
Trial 1
Pipet 2
1.7806
17.0067
15.2261
15.2712
Trial 2
Pipet 3
2.0765
16.8786
14.8021
14.8459
Trial 3
1.8582
1.8389
2.0880
0.0366
70.39
0.0583
110.62
0.0155
30.25
70.42
Determination of Molar Mass of Acetone Trial 2
Temperature of boiling water bath (K)
332.15
Barometric pressure (atm)
0.9949
Temperature of room temp. water bath (K)
298.15
Density of water at room temperature (g/mL)
0.9970479
(http://en.wikipedia.org/wiki/Density)
Mass of empty pipet (g)
Mass of water in filled pipet (g)
Volume of pipet (mL)
Acetone
Mass of pipet and condensed
acetone(g)
Mass of condensed acetone(g)
Molar mass of acetone (g/mol)
Average molar mass of acetone
(g/mol)
Pipet 1
1.7950
15
15
Trial 1
Pipet 2
1.8586
15
15
Trial 2
Pipet 3
1.7680
14
14
Trial 3
1.8249
1.8892
1.8228
0.0299
54.64
0.0306
55.91
0.0548
107.29
72.61
Determination of Molar Mass of Isopropyl Alcohol Trial 1
Temperature of boiling water bath (K)
356.15
Barometric pressure (atm)
0.9949
Temperature of room temp. water bath (K)
298.15
Density of water at room temperature (g/mL)
0.9970479
(http://en.wikipedia.org/wiki/Density)
Mass of empty pipet (g)
Mass of pipet and water (g)
Pipet 1
1.9244
16.9468
Pipet 2
1.7985
16.7753
Pipet 3
1.7719
17.0548
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Mass of water in filled pipet (g)
Volume of pipet (mL)
Isopropyl Alcohol
Mass of pipet and condensed
isopropyl alcohol(g)
Mass of condensed isopropyl
alcohol(g)
Molar mass of isopropyl alcohol
(g/mol)
Molar mass of isopropyl alcohol
(g/mol)
15.0224
15.0669
Trial 1
14.9768
15.0211
Trial 2
15.2829
15.3282
Trial 3
1.9792
1.8690
1.8502
0.0548
0.0705
0.0783
106.89
137.94
150.13
131.65
Determination of Molar Mass of Isopropyl Alcohol Trial 2
Temperature of boiling water bath (K)
354.15
Barometric pressure (atm)
0.9949
Temperature of room temp. water bath (K)
298.15
Density of water at room temperature (g/mL)
0.9970479
(http://en.wikipedia.org/wiki/Density)
Mass of empty pipet (g)
Mass of water in filled pipet (g)
Volume of pipet (mL)
Isopropyl Alcohol
Mass of pipet and condensed
isopropyl alcohol(g)
Mass of condensed isopropyl
alcohol(g)
Molar mass of isopropyl alcohol
(g/mol)
Molar mass of isopropyl alcohol
(g/mol)
Pipet 1
1.7551
16
16
Trial 1
Pipet 2
1.7933
16
16
Trial 2
Pipet 3
1.8280
15
15
Trial 3
1.8107
1.8521
1.8849
0.0556
0.0588
0.0569
101.56
107.40
110.86
106.61
Molar Mass Table
Volatile Liquid
Literature
Molar Mass of
the Volatile
Liquids (g/mol)
Average Molar
Mass of Trial 1
(g/mol)
Average Molar
Mass of Trial 2
(g/mol)
Average Molar
Mass of Both
Trials (g/mol)
64.61
70.42
53.67
72.61
59.14
71.515
46.07
58.08
128.37
123.13
131.65
106.61
119.13
60.10
198.22
Percent Errors
(%)
(Wikipedia)
Ethyl Alcohol
Acetone
Isopropyl
Alcohol
Analysis:
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Mass of water in filled pipet = (Mass of water and pipet) – (Mass of pipet)
Mass of water in filled pipet equals mass of water and pipet minus the mass of pipet
Trial 1 (Ethanol Pipet 1): 16.6232 – 1.8037 = 14.8195 g
Trial 1 (Ethanol Pipet 2): 16.6447 – 1.7560 = 14.8887 g
Trial 1 (Ethanol Pipet 3): 16.6284 – 1.7468 = 14.8816 g
Trial 1 (Acetone Pipet 1): 16.8438 – 1.8216 = 15.0222 g
Trial 1 (Acetone Pipet 2): 17.0067 – 1.7806 = 15.2261 g
Trial 1 (Acetone Pipet 3): 16.8786 – 2.0765 = 14.8021 g
Trial 1 (Isopropyl alcohol Pipet 1): 16.9468 – 1.9244 = 15.0224 g
Trial 1 (Isopropyl alcohol Pipet 2): 16.7753 – 1.7985 = 14.9768 g
Trial 1 (Isopropyl alcohol Pipet 3): 17.0548 – 1.7719 = 15.2829 g
Volume of the pipet = (Mass of water in filled pipet) ÷ (Density of water at room temperature)
Volume of the pipet equals mass of the water in filled pipet divide by density of water at room
temperature
Trial 1 (Ethanol Pipet 1): 14.8195 ÷ 0.9970479 = 14.8634 mL ethanol
Trial 1 (Ethanol Pipet 2): 14.8887 ÷ 0.9970479 = 14.9328 mL ethanol
Trial 1 (Ethanol Pipet 3): 14.8816 ÷ 0.9970479 = 14.9257 mL ethanol
Trial 1 (Acetone Pipet 1): 15.0222 ÷ 0.9970479 = 15.0667 mL acetone
Trial 1 (Acetone Pipet 2): 15.2261 ÷ 0.9970479 = 15.2712 mL acetone
Trial 1 (Acetone Pipet 3): 14.8459 ÷ 0.9970479 = 14.8459 mL acetone
Trial 1 (Isopropyl alcohol Pipet 1): 15.0224 ÷ 0.9970479 = 15.0669 mL isopropyl alcohol
Trial 1 (Isopropyl alcohol Pipet 2): 14.9768 ÷ 0.9970479 = 15.0211 mL isopropyl alcohol
Trial 1 (Isopropyl alcohol Pipet 3): 15.2829 ÷ 0.9970479 = 15.3282 mL isopropyl alcohol
Mass of condensed ethyl alcohol = (Mass of pipet and condensed ethyl alcohol) – (Mass of pipet)
Mass of condensed ethanol equals mass of pipet and condensed ethanol minus mass of the pipet
Trial 1 (Pipet 1): 1.8353 – 1.8037 = 0.0316 g
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Trial 1 (Pipet 2): 1.7906 – 1.7560 = 0.0346 g
Trial 1 (Pipet 3): 1.7772 – 1.7468 = 0.0304 g
Trial 2 (Pipet 1): 1.7544 – 1.7310 = 0.0234 g
Trial 2 (Pipet 2): 1.7760 – 1.7501 = 0.0259 g
Trial 2 (Pipet 3): 1.8882 – 1.8514 = 0.0368 g
Mass of condensed acetone = (Mass of pipet and condensed acetone) – (Mass of pipet)
Mass of condensed acetone equals mass of pipet and condensed acetone minus the mass of the pipet
Trial 1 (Pipet 1): 1.8582 – 1.8216 = 0.0366 g
Trial 1 (Pipet 2): 1.8389 – 1.7806 = 0.0583 g
Trial 1 (Pipet 3): 2.0880 – 2.0765 = 0.0115 g
Trial 2 (Pipet 1): 1.8249 – 1.7950 = 0.0299 g
Trial 2 (Pipet 2): 1.8892 – 1.8586 = 0.0306 g
Trial 2 (Pipet 3): 1.8228 – 1.7680 = 0.0548 g
Mass of condensed isopropyl alcohol = (Mass of pipet and condensed isopropyl alcohol) – (Mass of
pipet)
Mass of condensed isopropyl alcohol equals mass of pipet and condensed isopropyl alcohol minus mass
of pipet
Trial 1 (Pipet 1): 1.9792 – 1.9244 = 0.0548 g
Trial 1 (Pipet 2): 1.8690 – 1.7985 = 0.0705 g
Trial 1 (Pipet 3): 1.8502 – 1.7719 = 0.0783 g
Trial 2 (Pipet 1): 1.8107 – 1.7551 = 0.0556 g
Trial 2 (Pipet 2): 1.8521 – 1.7933 = 0.0588 g
Trial 2 (Pipet 3): 1.8849 – 1.8280 = 0.0569 g
Molar mass of ethanol = [(Mass of condensed ethanol) × (Gas constant) × (Temperature of the boiling
water bath)] ÷ [(Barometric pressure) × (Volume of the pipet)]
Molar mass of ethanol equals product of mass of condensed ethanol, gas constant, and temperature of
boiling water bath divide by the product of barometric pressure and volume of the pipet
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Trail 1 (Pipet 1): (0.0316 × 0.0821 × 362.15) ÷ [(1.48195 × 10-2) × 0.9949] = 63.72 g/mol ethanol
Trial 1 (Pipet 2): (0.0346 × 0.0821 × 362.15) ÷ [(1.49328 × 10-2) × 0.9949] = 69.24 g/mol ethanol
Trial 1 (Pipet 3): (0.0304 × 0.0821 × 362.15) ÷ [(1.49257 × 10-2) × 0.9949] = 60.87 g/mol ethanol
Trial 2 (Pipet 1): (0.0234 × 0.0821 × 356.15) ÷ (0.015 × 0.9949) = 45.85 g/mol ethanol
Trial 2 (Pipet 2): (0.0259 × 0.0821 × 356.15) ÷ (0.016 × 0.9949) = 47.57 g/mol ethanol
Trial 2 (Pipet 3): (0.0368 × 0.0821 × 356.15) ÷ (0.016 × 0.9949) = 67.60 g/mol ethanol
Molar mass of acetone = [(Mass of condensed acetone) × (Gas constant) × (Temperature of the boiling
water bath)] ÷ [(Barometric pressure) × (Volume of the pipet)]
Molar mass of acetone equals product of mass of condensed acetone, gas constant, and temperature of
boiling water bath divide by the product of barometric pressure and volume of the pipet
Trail 1 (Pipet 1): (0.0366 × 0.0821 × 351.15) ÷ [(1.50667 × 10-2) × 0.9949] = 70.39 g/mol acetone
Trial 1 (Pipet 2): (0.0583 × 0.0821 × 351.15) ÷ [(1.52712 × 10-2) × 0.9949] = 110.62 g/mol acetone
Trial 1 (Pipet 3): (0.0155 × 0.0821 × 351.15) ÷ [(1.48459 × 10-2) × 0.9949] = 30.25 g/mol acetone
Trial 2 (Pipet 1): (0.0299 × 0.0821 × 332.15) ÷ (0.015 × 0.9949) = 54.64 g/mol acetone
Trial 2 (Pipet 2): (0.0306 × 0.0821 × 332.15) ÷ (0.015 × 0.9949) = 55.91 g/mol acetone
Trial 2 (Pipet 3): (0.0548 × 0.0821 × 332.15) ÷ (0.014 × 0.9949) = 107.29 g/mol acetone
Molar mass of isopropyl alcohol = [(Mass of condensed isopropyl alcohol) × (Gas constant) ×
(Temperature of the boiling water bath)] ÷ [(Barometric pressure) × (Volume of the pipet)]
Molar mass of isopropyl alcohol equals product of mass of condensed isopropyl alcohol, gas constant,
and temperature of boiling water bath divide by the product of barometric pressure and volume of the
pipet
Trial 1 (Pipet 1): (0.0548 × 0.0821 × 356.15) ÷ [(1.50669 × 10-2) × 0.9949] = 106.89 g/mol isopropyl
alcohol
Trial 1 (Pipet 2): (0.0705 × 0.0821 × 356.15) ÷ [(1.50211 × 10-2) × 0.9949] = 137.94 g/mol isopropyl
alcohol
Trial 1 (Pipet 3): (0.0783 × 0.0821 × 356.15) ÷ [(1.53282 × 10-2) × 0.9949] = 150.13 g/mol isopropyl
alcohol
Trial 2 (Pipet 1): (0.0556 × 0.0821 × 354.15) ÷ (0.016 × 0.9949) = 101.56 g/mol isopropyl alcohol
Trial 2 (Pipet 2): (0.0588 × 0.0821 × 354.15) ÷ (0.016 × 0.9949) = 107.40 g/mol isopropyl alcohol
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Trial 2 (Pipet 3): (0.0569 × 0.0821 × 354.15) ÷ (0.015 × 0.9949) = 110.86 g/mol isopropyl alcohol
Average molar mass of ethyl alcohol = Σ(Molar mass of ethyl alcohol) ÷ 3
Average molar mass of ethyl alcohol equals the sum of molar mass of ethyl alcohol divide by 3
Trial 1: (63.72 + 69.24 + 60.87) ÷ 3 = 64.61 g/mol ethanol
Trial 2: (45.85 + 47.57 + 67.60) ÷ 3 = 53.67 g/mol ethanol
Average molar mass of acetone = Σ(Molar mass of acetone) ÷ 3
Average molar mass of acetone equals the sum of molar mass of acetone divide by 3
Trial 1: (70.39 + 110.62 + 30.25) ÷ 3 = 70.42 g/mol acetone
Trial 2: (54.64 + 55.91 + 107.29) ÷ 3 = 72.61 g/mol acetone
Average molar mass of isopropyl alcohol = Σ(Molar mass of isopropyl alcohol) ÷ 3
Average molar mass of isopropyl alcohol equals the sum of molar mass of isopropyl alcohol divide by 3
Trial 1: (106.89 + 137.94 + 150.13) ÷ 3 = 131.65 g/mol isopropyl alcohol
Trial 2: (101.56 + 107.40 + 110.86) ÷ 3 = 106.61 g/mol isopropyl alcohol
Average molar mass of both trial = [(Average molar mass of trial 1) + (Average molar mass of trial 2)] ÷ 2
Average molar mass of both trial equals the sum of average molar mass of trial 1 and trial 2 then divide
by 2
Ethanol: (64.61 + 53.67) ÷ 2 = 59.14 g/mol
Acetone: (70.42 + 72.61) ÷ 2 = 71.515 g/mol
Isopropyl Alcohol: (131.65 + 106.61) ÷ 2 = 119.13 g/mol
Percent Error = |(Average molar mass of both trial) ÷ (Literature molar mass of the volatile liquids) – 1|
Percent error equals the absolute value of the average molar mass of both trial divide by the literature
molar mass of the volatile liquids minus 1
Ethanol: |(59.14 ÷ 46.07) – 1| = 28.37 %
Acetone: |(71.515 ÷ 58.08) – 1| = 23.13 %
Isopropyl Alcohol: |(119.13 ÷ 60.10) – 1| = 98.22 %
The hypothesis was not confirmed true. This is because the molar mass of each volatile liquid calculated
from the experiment and the literature molar mass is significantly different from each other. Volatile
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liquids with lower boiling point often give better results than those with higher boiling point. This is
because the volatile liquid would get more energy than it needed to vaporize. Thus, it is quite certain
that all liquid was vaporized. So when the vapor condenses, the mass inside the pipet would be accurate
since there’s no liquid left in the pipet that would be added to the mass of the condensed vapor. If vapor
condenses in the neck of the 15-mL Beral-type pipet, the mass of the condensed vapor would be greater.
Thus, the mass that should have been of the vapor that was in the ratio of its molar volume would be
increase because there was some mass that did not evaporate. This could cause a considerably big error
because the mass of the gas for a certain volume is much lower than the mass of a liquid for a certain
volume because the density of a liquid is greater than gas. Thus, the mass of little amount of liquid could
affect the mass of volume significantly. If the liquids have enough attraction to form dimers, the molar
mass would increase. This is because the molecules combined from two to one so the volume to that
was previously taken by each molecule will be decrease in half. Thus, the molar mass would be doubled
since each molecule now has twice the mass of before.
Conclusion: The hypothesis cannot be verified from the results acquired from the experiment. The
hypothesis did not work possibly because not all liquid was vaporized because the water bath absorbed
a lot of heat so the heat might have been too low to vaporize all the liquid. If not all the liquid was
vaporized, the molar mass would be greater than it should have been. Also, some errors could have
happened during the experiment. The pipets might have been moved to the room temperature water
bath too slowly because there were some hindrance such as the Buret clamp above the universal clamp
which was clamping on to the pipets. This might have caused the air that was pushed out to get back
inside and increase the mass of the vapor. If the mass increased, the molar mass that would also
increased because the molar mass uses the mass in the calculation. Another error that could have
happened was that the vapor might have condensed at the neck of the pipet. The neck of the pipet was
cut so that it was long in order to put it through the plastic tube pipet holder. However, the longer the
neck also meant that it was further away from the heat of the boiling water bath. Thus, there is a higher
chance that the vapor condensed at the neck of the pipet. If the vapor condensed at the neck of the
pipet, the mass would be greater because instead of getting the mass of the volatile vapor, the mass
found would have been of the volatile vapor and the volatile liquid. Since the total volume was the same
whether it is purely vapor or vapor plus liquid, the mass of the vapor plus liquid would be greater
because the liquid is denser than gas. The increase in mass would also results in the increase in molar
mass. Some ways to prevent the errors in the future experiment would be to make sure there would be
nothing to obstruct the transfer of pipet from the boiling water bath to the room temperature water
bath. For example, the Buret clamp holding the thermometer could be put on other ring stand so that
the universal clamp holding the pipets could slide up and down the stand with no hindrance. Another
way to prevent the future errors would be to cut the neck of the pipet just the right amount so that it
would not be too long that it was too far from the water to allow the volatile vapor to condense. The
neck should be measured and cut so that all three pipets could easily slip through the plastic tube and
ended right above the tube
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