AP Chemistry Lab No. 9
Determination of the Molar Mass of Volatile Liquids
by
Veronica Martin
Hayley Gallo
for
Mr. Weiss
AP Chemistry AP1
November 11, 2014
Pre-Lab Questions
1. A determination of the molar mass of methyl alcohol (CH3OH) yielded the following data.
Temperature of boiling water bath
99.5o C
Barometric pressure
738 mm Hg
Temperature of room temperature water
24.0o C
Density of room temperature water
0.9973 g/mL
Trial 1
Mass of empty pipette
1.557g
Mass of piet and condensed methyl alcohol
1.571g
Mass of pipet and water
16.001g
Mass of condensed methyl alcohol
0.014g
Mass of water in filled pipet
14.444g
Volume of pipet
14.483mL
Molar mass of methyl alcohol (experimental)
28.69 g/mol
Molar mass of methyl alcohol (theoretical)
32.05 g/mol
Using the data, fill in the rest of the table. The volume of the pipet is equal to the volume of
water inside the pipet. Use the relationship of mass and density to determine this volume. Once
the volume of the pipet is determined, equation 3 in the Background section can be used to
calculate the molar mass of methyl alcohol. Compare this value to the actual molar mass of
methyl alcohol.
Mass of condensed methyl alcohol
1.571g - 1.557g = 0.014g
Mass of water in filled pipet
16.001g - 1.557g = 14.444g
Volume of pipet
14.444g x
1 mL = 14.483 mL
1
0.9973g
Mass of methyl alcohol (experimental)
molar mass (g/mol) = mass(g) x R x T(K)
P(atm) x V(L)
molar mass (g/mol) = (0.014g)(0.082057 L x atm/K x mol)(372.5K)
(1.03 atm)(0.014483 L)
Molar Mass (experimental) = 28.69 g/mol
2. The following mistakes were made when carrying out the experiment. What effect does each
have on the calculated molar mass? Be specific. For example, too large because…
a. Only part of the pipet was immersed in the boiling water, so the temperature in part of
the pipet was less than that of the water bath.
If the temperature was less than the water bath in some places because only part of the
pipet was immersed in the boiling water, the molar mass calculated would become lower.
This is because the ideal gas law temperature value would be too low, therefore lowering
the overall calculation.
b. The mass of the condensed liquid was not determined quickly. Instead, the pipet was
allowed to stand for a while before immersing it in room temperature and then massing
the pipet.
If the pipet was allowed to cool before immersing it in the room temperature water, the
value determined as the molar mass would be lower. The pipet is placed in the water in
order to change the state of the gas. If it was allowed to cool by itself, it might not fully
change, resulting in a different “n” value for the ideal gas law. Therefore, a lower number
of moles would produce a lower molar mass.
Purpose:
To determine the molar masses of various volatile liquids
Procedure:
Equipment/
Materials:
- ~6mL of Chemical 1 (not labeled)
- Tap Water
- Balance, milligram (.001-g precision)
- 1 250-mL beaker
- beral-type pipets, super jumbo, narrow stem, 6
- hot plate
- permanent marker
- pliers
- thermometer (LabQuest)
Data:
Temperature of boiling water bath
- ~6mL of Chemical 3 (not lableled)
- Distilled Water
- Barometer
- 2 450-mL beakers
- boiling stones (pieces of zinc)
- paper towels
- rubber tubing
- test tube clamp
`
97 o C
Barometric pressure
Temperature of room temp. water
Density of water at room temperature
746.3 mm Hg
20 o C
0.998207 g/mL
Data
Chem
Jumbo
Pipet #1
ical
Jumbo
Pipet #2
Table
#1
Chem
Jumbo
Pipet #3
Jumbo
Pipet # 4
ical
#3
Jumbo
Pipet #5
Jumbo
Pipet #6
Mass of
empty
pipet (g)
1.9523
1.9723
1.9379
1.9866
1.9374
1.9037
Mass of
pipet and
water (g)
15.8548
15.8576
15.8518
15.8623
15.8247
15.8023
Mass of
water in
filled pipet
(g)
13.9025
13.8853
13.9139
13.8757
13.8873
13.8986
Volume of
pipet (mL)
13.9270
13.9102
13.9389
13.9006
13.9122
13.9236
Data Table
Trial 1
Pipet #1
Pipet #2
Pipet #3
Chemical 1
Mass of pipet and
condensed chemical
1.9732
1.9929
1.9586
Mass of condensed
chemical (g)
0.0209
0.0206
0.0207
Molar mass of
chemical (g/mol)
46.40
45.79
46.05
Pipet #4
Pipet #5
Pipet #6
Chemical 3
Mass of pipet and
condensed chemical
(g)
2.0139
1.9646
1.9250
Mass of condensed
chemical (g)
0.0273
0.0272
0.0213
Molar mass of
chemical (g/mol)
60.73
60.46
47.31
Observations:
For the initial boiling of the water, it took longer than anticipated. Eventually, when the heat was
turned to 500o on the hot plate, there was a steady temperature of 97oC. There were multiple
attempts to stretch and cut the pipets, but they ended up becoming similar sizes, lengthwise.
For the first chemical, when fully immersed in the boiling water bath, there was a clear steam
coming out of the capillary tip of all three (#1, 2, 3) pipets. Chemical #3 did not have this
observation. Additionally, when filling the pipets completely with water, they were filled to the tip
of the fine capillary to the best of the experimenter’s ability.
Conclusions/
Questions:
1. Determine the mass of condensed, volatile vapor for each pipet trial and for each liquid. Enter
these values in the data table.
Pipet One: 15.8548 - 1.9523 = 13. 9025
Pipet Two: 15.8576 - 1.9723 = 13.8853
Pipet Three: 15. 8518 - 1. 9379 = 13.9139
Pipet Four: 15.8623 - 1.9866 = 13.8873
Pipet Five: 15.8247 - 1.9374 = 13.8873
Pipet Six: 15.8023 - 1.9037 = 13.8986
2. Use the CRC Handbook of Chemistry and Physics to determine the density of water at the
temperature of the room temperature water bath used in this experiment. Enter this density
value in the Data Table. Use this value and the mass of water in each filled pipet to calculate the
volume of each pipet.
Pipet One
13.9025g x
1 mL
= 13.9270 mL
1
0.998207g
Pipet Two
13.8853g x 1 mL
1
0.998207g
Pipet Three
13.9139g x 1 mL
1
0.998206g
Pipet Four
13.8757g x 1 mL
1
0.998206g
Pipet Five
13.8873g x 1 mL
1
0.998206g
Pipet Six
13.8986g x 1 mL
1
0.998206g
= 13.9102 mL
= 13.9389 mL
= 13.9006 mL
= 13.9122 mL
= 13.9236 mL
3. Determine the mass of the condensed volatile liquid for each run. Enter these values in the
Data Table.
Pipet One
1.9732 - 1.9523 = 0.0209
Pipet Two
1.9929 - 1.9723 = 0.0206
Pipet Three
1.9586 - 1.9379 = 0.0207
Pipet Four
2.0139 - 1.9866 = 0.0273
Pipet Five
1.9646 - 1.9374 = 0.0272
Pipet Six
1.9250 - 1.9037 = 0.0213
4. Calculate the molar mass of the liquid used in each run and the average of the three runs for
each volatile liquid.
Pipet One
(0.0209g)(0.082057L x atm/K x mol)(370 K)
(.9819 atm)(0.013927L)
M=46.40 g/mol
Pipet Two
M = (0.0206g)(0.082057L x atm/K x mol)(370 K)
(.9819 atm)(0.01391L)
M=45.79 g/mol
Pipet Three
M = (0.0207g)(0.082057L x atm/K x mol)(370 K)
(.9819 atm)(0.01393L)
M=46.05 g/mol
Average = 46.08 g/mol
Pipet Four
M = (0.0273g)(0.082057L x atm/K x mol)(370 K)
(.9819 atm)(0.0139L)
M=60.73 g/mol
Pipet Five
M = (0.0272g)(0.082057L x atm/K x mol)(370 K)
(.9819 atm)(0.01391L)
M=60.46 g/mol
Pipet Six:
M = (0.0213g)(0.082057L x atm/K x mol)(370 K)
(.9819 atm)(0.01392L)
M=47.31 g/mol
Average: 56.16 g/mol
5. Volatile liquids with lower boiling points often give better results than those with higher boiling
points. Suggest a reason for this.
When there is a liquid with a lower boiling point, it is much easier to get the particles in a
gaseous state. Therefore, the mass values prove to be more accurate, and therefore better and
more accurate molar mass results.
6. What effect would vapor condensation in the neck of the 15-mL Beral-type pipets have on the
reported molar mass? How large an error might this introduce?
If vapor was to condense in the neck of the pipet, the vapor contains molecules that would not
get accurately measured in experiment. Therefore, they would not be accounted for when
determining the molar mass value. This would be a significantly large error because it would
completely affect the molar mass.
7. Some liquids have enough attraction between molecule to form dimers. (Dimers are
molecules formed from the combination of identical molecules, A + A → A2.) What effect would
this have on the experimental molar mass?
Dimers would not affect the molar mass. Although the two atoms combined to form one
molecule, the masses would still be the same because they are the same gaseous molecule.
The volume would not change because it is the volume of the pipet, not the molecules. The
molar mass value would not change.