vacuum distillation question?

[mubaby]

I'm stuck on question 40 from TBR's chemistry section 7 passage 6.

Here's the part of the passage talking about taking the aliquots:
The researcher reduces the pressure above the mixture and collects the vapor in successive aliquots of 5mL. A total of twenty-five 5mL aliquots are collected before the amount of solution remaining in the flask becomes too small to measure. The samples are labeled 1 through 25 in the order in which they were collected.
Acetophenon BP 203C, MW 120, Pvapor 12 torr
2-propanol BP82C, MW 60, Pvapor 48torr

The question: What is observed when comparing Aliquot 5 with Aliquot 10?
A. Aliquot 10 has a greater total vapor pressure and larger mole fraction of acetophenone than Aliquot 5.
B. Aliquot 10 has a smaller total vapor pressure and lower mole fraction of acetophenone than aliquot 5.
C. The total vapor pressure is greater above aliquot 10 than aliquot 5, while the mole fraction of acetophenone is greater above aliquot 5.
D. The total vapor pressure is greater above aliquot 5 than aliquot 10, while the mole fraction of acetophenone is greater above aliquot 10.

Their answer: D - The vapor pressure of pure acetophenone is less than the vapor pressure of 2-propanol at all temperatures below the boiling point of acetophenone. This means that 2-propanol evaporates faster than acetophenone. With each successive aliquot, there is a larger percentage of acetophenone. This means that the mole fraction of acetophenone is greater above aliquot 10 than above aliquot 5. This eliminates choices B and C. Because aliquot 5 contains a greater percentage of 2-propanol, there is greater vapor pressure above aliquot 5 than above aliquot 10.


What I'm confused about is...what's the point of vacuum distillation? I thought it was to be able to collect the compound with the lowest boiling point. If the aliquots taken have more and more acetophenone (the compound with the higher boiling point), I thought that would defeat the purpose?

Please help! I'm so utterly confused...

---

Members don't see this ad.

 

[stsa84]

The point of distillation is to separate compounds based upon their different bp's. The only difference in vacuum vs. regular distillation is that vacuum distillation is used to separate liquids that boil above 150 F. The reduced pressure in the flask lowers the bp of the liquids, so the distillation can take place under milder (i.e. less destructive) conditions.

The first aliquots collected in this problem are primarily 2-propanol, since it boils at a lower temperature. As successive aliquots are collected, most of the 2-propanol has already been collected out of the sample, so the later aliquots are primarily acetophenone. This is a normal vacuum distillation...the lower boiling liquid is collected off first, and the higher boiling liquid later.

 

[exquisitemelody]

Oh, I guess that makes sense. Thanks

 

[Integrated MCAT Course]

stsa84's explanation is incomplete. The solution component boils out if its vapor pressure is higher than the external pressure. However, you can't use the vapor pressure of the pure liquid or the order of boiling points of the pure liquids. This is a Rault's Law question - The partial vapor pressure of a component in a mixture equals the vapor pressure of the pure component multiplied by its mole fraction in the mixture. The one that leaves first will be the one where the product of mole fraction and vapor pressure is highest.

The point of distillation is to separate compounds based upon their different bp's. The only difference in vacuum vs. regular distillation is that vacuum distillation is used to separate liquids that boil above 150 F. The reduced pressure in the flask lowers the bp of the liquids, so the distillation can take place under milder (i.e. less destructive) conditions.

The first aliquots collected in this problem are primarily 2-propanol, since it boils at a lower temperature. As successive aliquots are collected, most of the 2-propanol has already been collected out of the sample, so the later aliquots are primarily acetophenone. This is a normal vacuum distillation...the lower boiling liquid is collected off first, and the higher boiling liquid later.

 

[exquisitemelody]

Thanks. I guess what I'm confused with is what this experiment looks like if you just let the experiment run without taking aliquots. Because then the end product wouldn't be separated. Would someone mind just explaining how distillation works? I think I'm just confused with the whole technique. When do you stop it so that the 2nd compound doesn't evaporate as well? Or...is it controlled by temperature?

 

[stsa84]

Distillation (in general):

Say you have a mixture of 2 liquids, one of which boils at 60 C and the other of which boils at 100 C. You slowly heat the mixture, and the lower boiling liquid will evaporate when the temp reaches 60 C. Through a variety of techniques, this vaporized liquid can be collected. Separation complete. You have the collected liquid that vaporized at 60 C, and the liquid that would boil at 100 C is left behind in the original container.

Edit: Usually the distillation is continued to vaporize the higher boiling fraction as well, so you end up collecting each liquid in a separate, clean vial. This may provide a higher purity of the second liquid, as well as giving you the boiling point of the second liquid to use for identification purposes if needed.

 

[exquisitemelody]

Ohhhh. Okay...yeah, later in the passage, it does mention that 25 aliquots were taken until the remaining solution in the flask was too little to measure. That means everything was heated, which would mean that 2nd compound would in fact evaporate and be collected. * hits self * So I'm guessing in a real situation in which you actually want to separate them, you'd stop before the 2nd compound would evaporate, correct? I mean, the temperature you'd run the experiment at would be kept below the boiling point of the 2nd compound.

 

[stsa84]

You could stop it between the boiling points, having collected the lower boiling fraction in another vial and leaving the higher boiling fraction in the original container. When I've done distillations however, you usually continue on to the second boiling point and collect that fraction as well. The idea is probably for purity's sake (any impurities in the original mixture may be left behind in the original flask), as well as for identification purposes. If the liquids involved are of unknown identity, obtaining a bp for each one is a great identification tool.

 

[Integrated MCAT Course]

Here's an image that may help. When the mole fraction of hexane and octane are equal, hexane boils first. However, if the solution is more than eighty percent octane, octane is what will distill out.

 

[californiamed]

Okay, it seems TBR worded this poorly as I'm having issues with this as well.
The last parag of the passage states: Researcher reduces the pressure above th mixxture and collects the vapor in successive aliquots of 5mL. A total of 25 5mL aliq are collected before the amt fo solun reamining in flask becomes too smal to measure.

Acetophenon BP 203C, MW 120, Pvapor 12 torr
2-propanol BP82C, MW 60, Pvapor 48torr

The question: What is observed when comparing Aliquot 5 with Aliquot 10?
A. Aliquot 10 has a greater total vapor pressure and larger mole fraction of acetophenone than Aliquot 5.
B. Aliquot 10 has a smaller total vapor pressure and lower mole fraction of acetophenone than aliquot 5.
C. The total vapor pressure is greater above aliquot 10 than aliquot 5, while the mole fraction of acetophenone is greater above aliquot 5.
D. The total vapor pressure is greater above aliquot 5 than aliquot 10, while the mole fraction of acetophenone is greater above aliquot 10.

Answer: D. Why isn't it B? The point of distillation is that the more VOLATILE component gets evaporated into the "successive aliquots", right? SO shouldn't aliquot 10 have a lower mole fract of acetpehnone? Maybe i'm misreading, but the aliquots collect the vapors of each aliquots before it, right?

Side note: Question 43's answer states that Aliquot 1 from th emixture has 11.1% acetophenone and 88.9% 2-proponal, thus confirming what I previously reasoned..

 

[californiamed]

Okay, I re-read the chapter, and in one section it says OVER time, for distillation, there's an increase in the more non-volatile solution due to less effective condensation cycles. Why is this so?

 

[drechie]

--oops.