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Madeleine Stewart
CHEM 145 AB
5 November 2013
Experiment 4
Molar Mass of a Low-Boiling Liquid
Purpose:
The purpose of this lab was to determine the molar mass of a low-boiling liquid
by converting it to a gas, which could then be described using the equation PV=nRT. The
method, invented by French chemist Jean-Baptiste André Dumas, involves heating a flask
containing a volatile liquid in a boiling water bath. The liquid evaporates, pushing the air
molecules out of the flask. After the entire unknown has evaporated, the flask is capped and
removed from the heat, preventing air from being sucked back into the flask. The unknown is
allowed to cool to its liquid state, and the flask, cap, and liquid are then massed. This
demonstrates how properties such as boiling point and vapor pressure can be used to determine
the molecular properties of a substance. The boiling point is determined by the strength of
intermolecular forces. Being a volatile liquid, the unknown has week intermolecular forces and
boils at a relatively low temperature. When the gas is cooled, it returns to its liquid phase,
although some molecules remain in the gaseous phase so that the system is at equilibrium in
terms of pressure.
Procedure:
The procedure in the lab manual was followed exactly.
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Data:
Unknown #4
Mass of flask, cap, and air (g)
51.900
Masses of Flask, Cap, and Vapor
Trial
Mass of Flask, Cap, and Vapor (g)
1
2
3
4
51.924
51.914
51.915
51.897
Barometric Pressure (kPa)
101.59
Sample Temperature (°C)
100.5
Water (Room) Temperature (°C)
24.0
Mass of flask, cap, and water (g)
176.894
Calculations:
Volume of Flask (L)
0.12533
Mass of Air in Flask (g)
0.14855
Density of Air in Flask (g L-1)
1.1853
Mass of Completely Empty Flask
51.751
The volume of the flask was found by determining the mass of water in the flask with the
equation,
Mass Water = (Mass of Flask Filled with Water) – (Mass of Flask Filled with air).
The mass of the water was found to be 124.994 g.
The volume of the flask was then found using the equation,
Volume = (mass water) / (density water) and was found to be 0.12533 L
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Density of Air (20% O2 and 80% N2) at 101.59 kPa was found in five steps:
1.)
Moles of air were calculated using
nair = PV/RT, where P = 101.590 kPa, V = 0.12533, R = 8.3145 L kPa
mol-1 K-1 , T = 297 K and found to be 0.0051560 moles
2.)
The approximate moles of oxygen and nitrogen gas in the flask were found using
the mole fraction equations,
nO2 = nair * XO2 and nN2 = nair * XN2 , where nair = 0.0051560 moles, XO2
= 0.20, and XN2 is 0.80 and were found to be 0.0010312 and 0.0041248
respectively
3.)
The masses of oxygen and nitrogen gas in the flask were found using the
equations,
massO2 = nO2 * MO2 and massN2 = nN2 * MN2 , where nO2 = 0.0010312
moles, MO2 = 31.999 g mol-1, nN2 = 0.0041248 moles, and MN2 = 28.014 g
mol-1 , and were found to be 0.032997 g and 0.11555 g respectively.
4.)
The mass of the air in flask was found by adding the masses of the gases in the
flask:
massair = massO2 + massN2, where massO2 = 0.032997 g and massN2 =
0.11555 g, and was found to equal 0.14855 g
5.)
The density of the air in the flask was found using the equation,
ρair = massair / Vflask , where massair = 0.14855 g and Vflask = 0.12533 L
and was found to equal 1.1853 g L-1.
The mass of the flask without air or water was found using the equation,
massflask+cap = massflask+air+cap – massair , where massflask+air+cap = 51.900 g and massair =
0.14855 g and was found to equal 51.751 g
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Mass of Unknown Vapor
Trial
1
2
3
4
51.924
51.914
51.915
51.897
0.17300
0.16300
0.16400
0.14600
Molar Mass of Unknown Gas (g mol-1)
42.20
39.76
40.00
35.61
Avg. Molar Mass (g mol-1)
39.39
Mass of Flask, Cap, and Vapor (g)
Mass of Vapor (g)
The mass of the unknown vapor was found using the equation,
massunknown = massflask+cap+vapor – massflask+cap , where massflask+cap = 51.751 g
The number of moles of unknown gas in the capped flask was found using the equation,
nvapor = PV / RT, where P = 101.590 kPa, V = 0.12533, R = 8.3145 L kPa mol-1 K-1 ,
and T = 373.5 K, and was found to be 0.004100 moles
The molar mass of the unknown was calculated using the equation,
MM = massvapor / nvapor, where nvapor = 0.004100
Molar Masses of Ethyl Acetate, Isopropanol, Acetone, Ethanol, Methanol (g mol-1)
Ethyl acetate (C4H8O2)
88.106
Isopropanol (C3H8O)
60.096
Acetone (C3H6O)
58.080
Ethanol (C2H6O)
46.058
Methanol (CH4O)
32.042
The two closest molecular masses from the above table are those of ethanol (46.06 g mol-1)
and methanol (32.04 g mol-1).
The percent error was calculated using the equation,
% error = |calculated MMavg – actual MM| / actual MM * 100%, where calculated
MMavg was 39.39 g mol-1
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Compound
Percent Error (%)
Ethanol
14.48
Methanol
22.94
Results and Conclusions:
The lab proved how the molar mass of a volatile liquid could be determined with
some accuracy by heating it, which allows it to be defined by the ideal gas equation, and
then letting it cool, so that the sample can be massed. The data gave a molar mass of 39.39
g mol-1, which had a high percent error when compared to the two closest compounds
(ethanol and methanol). The measurements were relatively precise, but less accurate.
Loss of accuracy may have been due to water droplets present inside the rubber
policeman when massing or to some gas escaping while the gas was removed from heat,
before being capped. Water droplets in the rubber policeman would have increased the
measured mass, and increased the experimental molar mass. Escaping gas would have
made the experimental molar mass lower because less liquid would condense and the mass
of that liquid would have been lower. These sources of error underscored the need for
careful measurements and data collection in the lab.
Questions:
1.)
The results were not very accurate (the percent error was between 14.48
and 22.94%). They were relatively precise, as all measurements fell within
0.027 grams of each other.
2.)
Methanol has a slightly different odor than ethanol. Smelling (wafting) both
methanol and ethanol could help the observer to identify the substance. Both
methanol and ethanol are colorless liquids.
3.)
Humid air would have a higher density because it would have more water
vapor per unit volume than dry air. This would give it more mass per unit
volume, making it more dense.
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4.)
The decrease in temperature would also decrease the pressure, which would
suck air molecules into the flask. The mass would then include the masses of
the air molecules and not be useful for calculating the molar mass of the
unknown.
5.)
The compound must have a lower boiling point than water so that it
evaporates before the water in the boiling water bath evaporates. It must
also be a liquid at room temperature so that it will condense and can be
massed. Finally, it must behave relatively ideally in room conditions and at
373 K (the boiling point of water).