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# Molar Mass of a Volatile Liquid

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by on 1 December 2015

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#### Transcript of Molar Mass of a Volatile Liquid

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

MW=dRT/P

2) Read through the lab procedures. For each part of the equation you wrote above, describe where you will gather that variable during the lab.

d (density)- grams of liquid added divided by the volume of the container
R (constant)- 0.08206 L atm/mol K
T (temperature)- temperature of the boiling water in Kelvin using a thermometer.
P (pressure)- barometric pressure of the laboratory in atmospheres

3) Create a data table to hold the data for this lab.

Post-Lab Questions
1. It was important that the flask be completely dry before the unknown liquid was added so that water present would not vaporize when the flask was heated. A typical single drop of liquid water has a volume of approximately 0.05mL. Assuming the density of the water is 1.0g/mL, how many moles of water are in one drop of liquid, and what volume would this amount of water occupy when vaporized at 100 degrees Celsius and 1 atm?
Post Lab Questions Continued
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.
-Because liquid is more dense than gas, the calculated density would have been higher when compared to the density of only the gas. The molecular weight would have been higher (if there was liquid), thus the calculated molecular weight would have been higher if all of the liquid in the flask did not vaporize.
b)The temperature of the boiling water was 2.0 degrees Celsius above the temperature needed.
-If the temperature that we recorded was 2 degrees Celsius lower than the actual temperature, then the molecular weight given would be smaller than the actual molecular weight; multiplying by a smaller number will obviously give you a smaller number.
Molar Mass of a Volatile Liquid
By: Lamya Elnihum and Nour Hilal

Observations Data Table
4) If 2.51 g of the vapor of a volatile liquid is able to fill a 498-mL flask at 100.° C and 775 mm Hg, what is the molecular weight of the liquid? What is the density of the vapor under these conditions?
MW=dRT/P
MW=?
d= 2.51g/0.498 L≈5.04 g/L
R= 0.08206 L atm/mol K
T= 100+273=373K
P= 775/760≈1.02 atm
MW= (5.04)(0.08206)(373)
(1.02)
MW=151g/mol
5) Why is a vapor unlikely to behave as an ideal gas near the temperature at which the vapor would liquefy?

An ideal gas has zero interactions between its particles. When a gas approaches condensations, the molecules have strong attractions for one another.
Procedure
Then, we covered two 250 mL flasks with foil and poked small openings
Initial Volume: 122.5373g
First, we allowed the water to boil while we obtained our supplies.
Next, we obtained 5 mL of unknown #2
We then added the unknown to the flask and placed the foil back on top of the rim.
We immersed the flask containing the unknown into the boiling water.
When it appeared that all the unknown liquid had vaporized, we continued to heat the flask for 1-2 more minutes.
As the flask cooled, we measured the exact temperature of the boiling water.
Temperature of boiling water:
94.4° C
We dried and massed the flask with the condensed vapor and foil cover.
Mass: 123.2653g
Objective:
Determine the molar mass of the unknown liquid and identify the unknown liquid.
Possible unknowns: Ethyl acetate, acetone, Cyclohexane, methanol, Tert-Butyl alcohol.

Barometric pressure in the lab: 749 atm
We then removed the flask, set it on the lab bench, and allowed the flask to cool to room temperature.
The flask was now cooled to room temperature and the vapor had condensed.
Trial #2
Next, we emptied the liquid into the organic waste container.
To measure exact volume, we filled the flask completely with water.
Then we poured the water into a 500mL graduated cylinder.
Volume of flask #1: 280.7 mL
Initial mass: 125.112g
5 mL of unknown #2
Lowered flask into the boiling water bath.
Once all the liquid vaporized, we allowed it to boil for 2 more minutes.
Then we set it on the lab bench to cool to room temperature.
Temperature: 93.5° C
Barometric pressure: 748 atm
We then sped up the cooling process with tap water.
Final mass of flask: 125.8110 g
Volume of the flask: 280.1 mL
Then we discarded the liquid into the organic waste container.
Molar Mass of the Liquid
Calculate the molar mass of the liquid. Show all work.
MW=dRT/P (Trial 1)
MW=?
m=123.2653-122.5373=0.728g
d= 0.728g/0.280 L≈2.60 g/L
R= 0.08206 L atm/mol K
T= 94.4+273=367.4K
P= 749/760≈0.986 atm
MW= (2.60)(0.08206)(367.4)
(0.986)
MW=79.5g/mol
122.5373
123.2653
749
94.4
280.7
125.112
125.8110
748
280.1
93.5
MW=dRT/P (Trial 2)
MW=?
m=125.8110-125.1120=0.699g
d= 0.699g/0.280 L≈2.50 g/L
R= 0.08206 L atm/mol K
T= 93.5+273=366.5K
P= 748/760≈0.984 atm
MW= (2.50)(0.08206)(366.5)
(0.984)
MW=76.4g/mol
0.05 mL water 1.0g water 1 mol water
1 mL water 18.02 g water
=0.003 mol water
PV=nRT
P= 1 atm
V = ?
R = 0.08206
T = 100+273=373
n =0.05/18.02 = 0.003
V=nRT/P
V=(0.003)(0.08206)(373)
(1)
V = 0.08 L
Conclusion
We identified our unknown as Tert-Butyl Alcohol.