The results of using the Dumas method for determining the molar mass of methanol are displayed in Table 1. This table shows that the experimental molar mass of methanol, with sources of error, is 26.57g mol-1.
Table 2: Observations of Unknown # 330 in Flask using Dumas Method to Determine Molar Mass
The results of using the Dumas method for determining the molar mass of the unknown is shown in Table 2. The approximate average molar mass of the unknown volatile liquid is 73.91g mol-1.
Calculations: (using data from tables)
Trial # 1 for Methanol – M=mRT/PV
M=(0.1266g)(0.08205L atm/K mol)(371.05K)/(1.38atm)(0.160L)
M= 23.17g mol-1
Trial # 2 for Methanol - M= mRT/PV
M=(0.1647g)(0.08206L atm/K mol)(369.95K)/(1.038atm)(0.149L)
M= 32.33g mol-1
Trial # 3 for Methanol - M=mRT/PV
M=(0.1273g)(0.08206L atm/K mol)(370.65K)/(1.038atm)(0.154L)
M= 24.22g mol-1
Average Molar Mass = (M1+M2+M3)/3
(methanol) = (23.12 g mol-1+ 32.33g mol-1+ 24.22g mol-1)/3
= 26.57g mol-1
Relative Percent Error = |Actual Molar Mass – Experimental Molar Mass| x 100
(methanol) Actual Molar Mass
=|32.04g mol-1 – 26.57g mol-1| x 100
32.04 g mol-1
= 17.07%
Trial # 1 for Unknown - M=mRT/PV
M=(0.3942g)(0.08206L atm/K mol)(370.85K)/(1.038atm)(0.150L)
M=77.05g mol-1
Trial # 2 for Unknown - M=mRT/PV
M=(0.3458g)(0.08206L atm/K mol)(371.25K)/(1.038atm)(0.149L)
M=68.11g mol-1
Trial # 3 for Unknown- M=mRT/PV
M=(0.4196g)(0.08206L atm/K mol)(371.45K)/(1.038atm)(0.161L)
M= 76.53g mol-1
Average Molar Mass = (M1+M2+M3)/3
(unknown) =(77.05g mol-1+68.11g mol-1+76.53g mol-1)/3
= 73.91g mol-1
Relative Percent Error = |Actual Molar Mass (hexane) – Experimental Molar Mass| x 100
(unknown) Actual Molar Mass
=|86.18g mol-1 – 73.91g mol-1| x 100
86.18g mol-1
= 14.24%
The volatile liquid that was used in the lab was a clear liquid that produced clear gas. The gas smelled strongly of alcohol. The pressure of the room, which in turn was the pressure of the flask, was 788.00mmHg or 1.038atm.
Discussion
In this lab we used methanol as a sort of control or comparison to the unknown volatile liquid. Using the formula for molar mass with respect to mass, pressure, temperature and volume methanol had a percent error of 17.07%. Because this percent error was not zero, it shows that the molar mass of the unknown will not be perfect either.
The mass of these liquids were found by weighing the entire assembly and subtracting it from the mass of the entire assembly plus the liquid, see Tables 1 and 2. The pressure of the room was found by using a barometer. The pressure of the room could be used in the equation because the pressure of the room was the same as the pressure inside the flask when the correct amount of the liquid had been vaporized. The temperature of liquid inside the flask would be the same as the temperature of the water into which it was submerged and that is why the temperature of the water was suitable to use for this experiment. The volume needed to be measured of each flask because each flask has a unique volume.
The unknown # 330 was found to be hexane. Although the determined molar mass of the unknown was directly in the middle of the molar mass of iso-propanol and hexane, determining if the liquid has evaporated fully was difficult and it probably was not. Since hexane would take longer to evaporate than iso-propanol, and suspicion is that the liquid was not fully evaporated at 20 minutes or so, the liquid is hexane. The reason why hexane would take longer to evaporate is because it is a bigger molecule, known by the larger molar mass.
There are numerous sources of error for this lab and this is a fact because of the relative percent error of the molar mass of ethanol. Some sources of error that could contribute to the molar mass being larger than the actual molar mass is another substance contaminating the liquid making it weigh more. An example of this substance could have been the water vapour from the boiling water bath. Another source of error that would make the molar mass larger was that the pressure inside the flask was not equal to the room pressure and not all of the liquid evaporated. Not letting the flask cool and not having all of the gas condense back into a liquid would contribute to the molar mass being less than it was. The temperature of the liquid is also another source of error because the flask could have not been fully submerged into the water meaning the temperature inside the flask is lower than that of the water and making the molar mass too high. Also the volume of the flask measured could be inaccurate because the liquid used to measure could have spilled, making the volume higher than what was measured and the molar mass lower than its actual molar mass.
Questions
- Density = 0.796g/mL m=(d)(V)
Volume = 4.00mL m=(0.796g/mL)(4.00mL)
Mass =? m=3.184g
Molar mass = 32.04g mol-1
Pressure= 1.038atm n=m/M
ΔTemperature= 370.55K n=(3.184g)/(32.04g mol-1)
R= 0.08206L atm/K mol n=0.09938mol
Volume for gas =?
ΔVolume flask=0.1543L
n=?
V=nRT/P
V=(0.09938mol)(0.08206L atm/K mol)(370.55K)/(1.04atm)
V=2.91L
The volume of the flask was 0.1543L and the volume that the gas can occupy is 2.91L and therefore the volume of 4.00mL was a sufficient amount of liquid to use.
- A) After methanol is added and the flask is covered in foil, liquid methanol, and oxygen gas are present in the flask.
B) After the flask is taken out of the hot water bath the oxygen gas is pushed out and all that occupies to flask is methanol gas.
C) Being weighed after the flask is cooled the oxygen gas is back in the flask along with liquid methanol.
- The factors that could cause the experimental value of the molar mass to be higher than it was were letting the water vapour into the flask, which adds the weight of the water to the mass of the liquid. The solution to this is not to have the water bath boiling as rapidly and tilting the flask at a lesser angle so the vapour will not enter the flask. Another factor of getting a higher molar mass is not letting the methanol fully vaporize so the pressure in the flask is not equal to that in the room and the methanol is still in a liquid form which would add to the mass of the condensation. The solution to this is to leave the flask in boiling longer. Not drying the flask so that the water on the outside of the flask is added to the mass of the assembly is a factor that would make the molar mass higher. The solution to this is to dry the flask completely before weighing it.
- The factors that would cause the calculated molar mass of methanol to be less than it actually is are not letting the flask cool enough so that all of the methanol did not condense and was still in gas form which would make the mass of the liquid lighter. The solution to this would be to wait the same amount of time for each flask so that there is a proper ratio. Letting the flask cool too much could have been another factor for making the molar mass lower than it actually is because the pressure would have been lower. Again this can be solved with letting the flasks cool down for the same amount of time. Measuring errors could have applied as well making the mass or volume of the liquid higher or lower and making the molar mass higher or lower. This could be solved with using more accurate ways of measuring the liquid.
Conclusion
The purpose of this lab, to find the molar mass of an unknown, was accomplished. The molar mass of the substance was 73.91g mol-1, and is the substance hexane. The sources of error greatly affected this lab and the substance that was predicted may not have been accurate due to the percent error. This experiment is not reliable for finding the molar mass of a substance because the molar mass will never be perfect due to the several possible sources of error.
References
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Department of Chemistry 2008 First Year Chemistry Chem 120L Laboratory Manual. University
of Waterloo, Waterloo. pp 16-20.
HyperPhysics Concepts. (2005) “Georgia State University”. October 22, 2008.
< >
Petrucci, Ralph et.al. General Chemistry Principles & Modern Applications, Ninth Edition.
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