Lab 9: Molar Mass of a Volatile Liquid

Only help on the Problems in the lab manual are available at this time.
PROBLEMS (page 86)

1. A 0.678 gram sample of gas occupies 0.214 L at standard conditions; what is the molar mass of the gas?
Since this is at standard conditions (0°C and 1 atmosphere of pressure, which is the same as 760 torr), you can do a more simpler calculation. You can assume a mole of this gas (or any gas) has a volume of 22.4 liters. In other words you know it is 1 mole per 22.4 liters (or 22.4 liters per mole). Since the question is asking for molar mass, we need the final units to be grams per mole. That guides us in how to set up the units (the dimensions). We need the 0.678 grams in the numerator because out final answer of "grams per mole" has grams in the numerator. We need moles in the denominator. Therefore, we need to use the ratio of "22.4 liters per mole" because that puts moles in the denominator. In "22.4 liters per mole" the liters are in the numerator. To cancel those, we use the data of 0.214 L in the denominator. So that's essentially 0.678 grams per 0.214 liters, which gets changed to grams per mole (molar mass)

 
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Mass of gas
divide by liters
Changes "per liter" into "per mole"
  
molar mass of gas
2
0.678
grams
    
22.4
 Liters
=
???
 grams
3
0.214
 Liters
1
 mole     mole

Note that the answer should be in 3 significant figures because mass and volume were in 3 sig figs.
2. A sample gas occupies 250. mL at 37°C and 730 torr; what volume would the gas occupy at standard conditions?
If this 250. mL was at 0°C and 760 torr, then at standard conditions, it would still be 250. mL. However, 37°C is warmer than standard conditions and 730 torr is less pressure than standard conditions. So you need to decide if the change in conditions will cause the volume to go up or down. If temperature goes down from 37°C to 0°C, then the gas molecules are moving slower and they bang against with less force causing them to shrink more (less volume). To find out by how much we have to turn Celsius temperatures into Kelvin to get them on the proper scale. 0°C is 273K and 37°C is 310K. Now we see their perspective sizes. To make the volume go down, it makes sense to use the fraction of 273K/310K.
      Going from 730 torr to 760 torr means the gas will feel more pressure, causing the molecules to squeeze together more (less volume). The exact amount depends on a fraction based on the starting and final pressures. For example if the pressure started at 380 torr and then went to 760 torr, then there would be twice the pressure causing the gas to shrink to half its volume at standard volume of 760 torr. So multiplying by the fraction of 380 torr/760 torr (which reduces to 1/2) will mathematically decrease our volume of 250. mL to 125. mL. In this case, however, pressure increase isn't as much. It's only going from 730 torr to 760 torr. So a fraction of 730 torr/760 torr, will reduce the volume by 730/760=0.9605. That reduces the volume from 250. mL to 240. mL.
 
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Volume of gas
Adjust for going to an increased pressure (Boyle's Law)
Adjust for going to a lower temperature
(Charles Law)
  
Volume of gas at standard conditions
(760 torr, 273K)
2
250.
mL
730
 torr
273
 K
=
???
 mL
3
760
 torr
310
 K      

The other way to do the above calculations is to use the gas laws. Boyle's Law is PV=constant. Two different condions are also equal to the same constant. So they can be written as equal to each other. So that is P1V1=P2V2. So you could put the starting conditions of 730 torr into P1 and 250. mL into V1. You would then put the final conditions of 760 torr into P2, and V2 would be the final unknown volume. So the formula would be written as
730 torr x 250. mL = 760 torr x V2

You would then solve for V2. Notice when you do, you get 250.mL x 730 torr/760 torr, which is what the above spreadsheet is showing.

You can also use Charlie's Law, which is Volume = a constant times the temperature. That can be arranged as volume divided by temperature equals a constant. And two of these that both equal the same constant can be written as equal to each other. That can be written as:
V1/T1=V2/T2

For the starting conditions you would substitute 250. mL for V1 and 310K as T1. The final condition has 273K as T2. So the formula then looks like:
250.mL/310K = V2/273K

To solve for V2 multiply both sides by 273K. That gives us:
250.mLx 273K/310K = V2

Notice, this is the way it is written in the spreadsheet above. So both ways are good and you should use both techniques just to check yourself.
3. A 5.00 mL sample of an unknown liquid is vaporized in a flask having a volume of 285 mL. At 100°C, 0.4168 g of the vapor exerts a pressure of 740 torr. Calculate the gram molar mass of the unknown liquid.
This problem is closer to what happened in the experiment. The "gram molar mass" is the "grams per mole" of the liquid. The problem tells us the grams (0.4168 g) but not the moles. We know that one mole of gas occupies 22.4 liters if the pressure is 760 torr and the temperature is 0°C. However, here we have 285 mL of volume, a lower pressure of 740 torr, and a higher temperature of 100°C. In problem 1 above, we used the grams divided by volume (liters) and multiplied that by 22.4 liters per mole to get our molar mass; however, that assumed the gas was at standard temperature and pressure (STP). One strategy of doing this problem is to change the volume of 285 mL at this non-standard pressure (740torr) and non-standard temperature (100°C) to the volume at STP. So that is like Problem 2. Once we have the volume at STP, we can do the same as in Problem 1.
 
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Volume of gas
Adjust for going to a higher pressure (gas shrinks)
Boyle's Law
Adjust for going to a lower temperature (gas shrinks)
Charles Law
  
Volume of gas at standard conditions (STP)
(760 torr, 273K)
2
285
mL
740
 torr
273
 K (0°C)
=
???
 mL
3
760
 torr
373
 K (100°C)      
4                  
5
Mass of vapor
Divide by STP volume from H2
Changes "per mL" into "per mole"
  
molar mass of the unknown gas (g/mole)
6
0.4168
 g    
22400
 milliliters
=
???
 g
7    
???
 mL
1
 mole     mole
In cell E6 above, the 22.4 liters was converted to 22,400 mL so that the mL in D7 would cancel. The answer in H2 goes into C7. Notice that milliliters cancel leaving just grams per mole (molar mass). Technically, the answer would be just 1 significant figure because the temperature was given as 100°C. which has only 1 significant figure. Also, the pressure of 740 torr is only 2 sig figs. However, I think the intention was to have 3 sig figs or better for all data, allowing you to report 3 sig figs in the final molar mass. Also, note the 5.00mL mentioned has not needed because most of the evaporates away. Only the 0.4168 grams is used in the calculation.
4. Calculate the density of oxygen gas at 50.°C and 750. torr.
This problem requires you to look up the molar mass of oxygen gas (O2). One source says that is 31.9989 g/mole. You also know that one mole of gas takes up 22.4 liters at STP. Density is mass divided by volume. So 31.9989g/22.4 liters would be the density of oxygen gas at STP. Of course, you would divide by 22.4 to get the density in "grams per 1 liter." However, this oxygen is not at STP so the volume of one mole of O2 is not 22.4 liters. However, above you learned that you can adjust the volume according to the deviations from STP. The setup from Problem 3 adjusted the volume, so that would be a good place to start.
 
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Volume of 1 mole of gas at STP
Adjust for going to a lower pressure (gas expands)
Boyle's Law
Adjust for going to a higher temperature (gas expands)
(Charles Law)
  
Volume of O2 at 50.°C and 750 torr
2
22.4
Liters
760
 torr
323
 K (50°C)
=
???
 Liters
3
750
 torr
273
 K (0°C)      
4                  
5
Mass of 1 mole of O2 gas
Divide by volume from H2
  
Density in grams per liter
  
6
31.9989
 g    
=
???
grams
    
7    
???
 Liters  
1
 Liter    
The decimal point in "50.°C" indicates that this has 2 significant figures (the zero is significant). So the answer (grams quantity) should be set to 2 significant figures.
5. Calculate the density of H2S at STP.
This problem is easier than Problem 4. There are no adjustments due to temperature or pressure. So it's just rows 5, 6 and 7 in the above spreadsheet. Calculate the mass of one mole of H2S and divide by 22.4 liters (which is the volume of one mole of a gas at STP).