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Molar Mass of a Volatile Liquid

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Lisa Liu

on 25 November 2013

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Transcript of Molar Mass of a Volatile Liquid

Molar Mass of a Volatile Liquid
Lisa Liu & Sandhiya Ravichandran

Pre-Lab Question #1
What is the equation used to find the molecular weight of a gas?

MW=dRT/P
Pre-Lab Question #2
For each part of the equation, describe where you will gather that variable during the lab.

For density, the grams of the gas will be found when we subtract the final mass of the flask in each trial from the initial mass in each trial, and the volume will be found when we fill the flask with water and then pour the water into a graduated cylinder. Temperature will be determined in step 10 when we put a thermometer in the boiling water, and pressure will be determined in step 10 as well when we record the pressure from the barometer.
Pre-Lab Question #3
Trial 1
Objectives
The purpose of this lab was to find the molar mass of a volatile liquid through the use of the formula MW=dRT/P.
Trial 2
Mass of flask
and foil
Liquid #
Observations
Temperature of
boiling water
Barometric
pressure of room
Mass of flask after
liquid has vaporized
3
120.1517 g
120.2273 g
liquid is yellow tinted
produced visible gas, looked like water vapor
105.7 degrees Celsius
105.8 degrees
Celsius
71.03 mm Hg
71.03 mm Hg
120.2273 g
120.1778 g
Pre-Lab Question #4
If 2.51 grams of the vapor of a volatile liquid is able to fill a 498 mL flask at 100. degrees Celsius and 775 mm Hg, what is the molecular weight of the liquid?
d=m/v d=2.51g/0.498L =
5.04g/L
775 mm Hg
760 mm Hg
1 atm
= 1.02 atm
MW=dRT/P
MW= (5.04g/L)(0.08205 L atm/mol K)(373K)/(1.02 atm)
100. + 273 = 373 K
=151 g/mol
Pre-Lab Question #5
Why is a vapor unlikely to behave as an ideal gas near the temperature at which the vapor would liquefy?
Gases behave best at high temperatures (not lower ones at which they would liquefy) because there is less interaction between particles, as they move too fast.
Procedures
Step 1:
Set up lab station as shown on the right (but without putting the flask in yet). Add marble chips to the water and let it boil, then adjust the flame so that it boils steadily without bubbling over the beaker.
Step 2:
Mass a dry, empty 250 mL Erlenmeyer flask with a foil cover. Obtain an unknown liquid and record its identification number. Add 5 mL of the liquid to the flask and cover it tightly with the foil again.














Unknown
Marble chips






Procedures
Step 3:
Puncture a small hole in the foil cover. Immerse the flask in the water, clamping the neck so that it does not float.
Step 4:
When it appears that all the liquid has vaporized, continue to heat for 1-2 minutes. Remove from the water bath and set on the lab bench to cool to room temperature.




Procedures
Step 5:
While the flask cools, take the temperature of the boiling water and the barometric pressure of the room. Record these values.
Procedures
Step 6:
When the flask has cooled, mass it.

Step 7:
Repeat procedures for second trial, but do not clean out the flask in between. The mass of the flask after the second sample of unknown liquid is vaporized should agree with the first determination within 0.05g. If not, do a third trial.





Procedures
Step 8:
Remove the foil from the flask and clean out. Fill the flask with water and then pour the water from the flask into a 500 mL graduated cylinder. Read the water level and record (ours was 314 mL). Calculate the molar mass.
Post-lab Question #1
It was important that the flask be completely dry before the unknown liquid was added so that the water present would not vaporize when the flask was heated. The typical drop of liquid water has a volume of approximately 0.05 mL. Assuming the density of water is 1.0 g/mL, how many moles of water are in one drop of it and what volume would this drop occupy when vaporized at 100. degrees Celsius and 1 atm?
0.05 mL water
1 mL water
1.0 g water
18.02 g water
1 mol water
=
0.003 mol water
PV=nRT
100. + 273 = 373K
-> V=nRT/P
V=(0.003 mol)(0.08206 L atm/mol K)(373K) / (1 atm) =
0.09 L







Post-lab Question #2
Determine whether each of the following errors would make the final calculated value of the molecular mass too high, too low or would cause no change. Explain your reasoning.
a) All of the liquid did not vaporize.
b) The temperature of the boiling water was 2.0 degrees Celsius above the temperature recorded.
a) If there was still liquid left, it would add to the final mass. Consequently, density would be higher, and as density and molecular weight are directly proportional, molecular weight would be too high.
b) If the actual temperature was higher, then the molecular weight would be higher as well, since temperature and molecular weight are directly proportional. The final value of molecular mass would then be too low.
Conclusion
In this lab, we vaporized a gas and gathered data to find its molecular weight with the formula MW=dRT/P.

The final mass of our second trial was lower than it was when we started, although we did not notice because it was still within 0.05g of the first one. This may have been because of errors we made in the first trial. For example, if we hadn't properly vaporized all of the liquid then, it may have had a heavier final mass than it should have, which would have resulted in too high of an initial mass for the second trial. Alternatively, we may have heated the flask in trial 2 for too long and let the gas escape from the opening (we didn't notice any condensation after it cooled), thus making the final mass of trial 2 too low.

To improve this lab, we would find a way to make the water boil faster, perhaps by covering it at the beginning and then keeping a close eye on it.
Liquid is vaporizing.
Trial 1 Calculations
MW=dRT/P
120.2273g-120.1617g=0.0756g
0.314L
314 mL
Volume of flask
=0.241g/L
T=105.7+273.15=378.9K
71.03 mmHg
760 mmHg
Step 8:
Remove the foil from the flask and clean out. Fill the flask with water and then pour the water from the flask into a 500 mL graduated cylinder. Read the water level and record (ours was 314 mL). Calculate the molar mass.
Procedures
d=
1 atm
P=
=0.09346 atm
MW=
(0.241g/L)(0.08206L atm/mol K)(378.9K)
0.09346 atm
=
80.2 g/mol
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