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Topic: Fractional Distillation (separating mixture of ethanol and water) question!  (Read 50115 times)

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Offline xangelofxdeathx

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Hi, basically one of my labs was to separate a mixture of ethanol and water using fractional distillation. Basically, we heated the mixture so that the alcohol would vaporize at a faster rate and thus, we obtained a more concentrated alcohol solution.

Now one of the lab question is for us to find the % yield (of the distillate), which says actual yield / theoretical yield * 100 on the lab sheet. The problem is our teacher never explained to us how to find the theoretical yield. Could someone show me how I am supposed to find the theoretical yield?

This is my data:

Initial Sol'n

Total Volume = 128.0 mL (ethanol)
Density = 0.962 g/mL
% Alcohol = 25.00%
Proof = 50

Distillate 1

Total Volume = 48.5 mL (concentrated ethanol)
Density = 0.870 g/mL
% Alcohol = 69.00%
Proof = 138

Thanks!  :)

Offline enahs

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Your theoretical yield is the most efficent possible. In many cases for distillation this is ~100% seperation. However, ethanol/water forms an azeotropic mixture ( http://en.wikipedia.org/wiki/Azeotrope ). For ethanol/water the most efficent possible separation is 96% ethanol/4%water.

Offline xangelofxdeathx

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Ah okay, so in this case my theoretical yield would be

.96 * 128.0mL = 122.88 mL?

And thus, my percent yield would be 48.5 mL (Actual Yield) / 122.88 mL (Theoretical) * 100 = 39.5%?

Is this correct?

Offline english

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Ah okay, so in this case my theoretical yield would be

.96 * 128.0mL = 122.88 mL?

And thus, my percent yield would be 48.5 mL (Actual Yield) / 122.88 mL (Theoretical) * 100 = 39.5%?

Is this correct?

Looks right.

Offline DrCMS

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NO NO NO NO.

How the hell can it be right.

You have 128ml of  a solution of 25% ethanol (is that weight/weight weight/vol or vol/vol?)

You got 48.5ml of distilate at 69% ethanol (again is that weight/weight weight/vol or vol/vol?)

Work out how much ethanol was in the starting solution and how much was in the distilate divivde the 2nd number by the first and you have a % yield.

Offline english

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NO NO NO NO.

How the hell can it be right.

You have 128ml of  a solution of 25% ethanol (is that weight/weight weight/vol or vol/vol?)

You got 48.5ml of distilate at 69% ethanol (again is that weight/weight weight/vol or vol/vol?)

Work out how much ethanol was in the starting solution and how much was in the distilate divivde the 2nd number by the first and you have a % yield.

So I forgot the percentages.  Don't lose it.

And yes xangelo you need to clarify whether this is (w/v) or (v/v).


Offline DrCMS

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No you did not just forget the percentages.

Look at his calcluation again they've multiplied the starting volume by 96% and divided the total weight of distilate by this figure and you've said it looks about right

Please explain how its about right?

The figure 96% is not required at all but the starting concentration and final concentrations are.

If you're going to post a reply read and understand what's going on first.

Offline enahs

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Instead of just complaining about what g English did wrong, perhaps you would like to attempt to help the original poster?
But you are wrong, the 96% is used, because in order to calculate the % yield he has to know the maximum possible yield.

xangelofxdeathx
The 96% refers to your distillate. The maximum percentage of ethanol is your distillate can be 96% ethanol.
That means if you originally had 100mL of pure ethanol, and then mixed it with water, by distillation you can not get anymore then 96mL of pure ethanol back.

Offline english

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I'm sorry for the little skirmish, but when I did simple distillation we never used percent composition of ethanol.

The thing is, by that time the majority of us hadn't even had analytical chemistry yet, and since we have so many biology majors taking organic, they tried to keep it simple by leaving that all-too important facet out.

We just calculated the yield directly from density.  Not exactly the most precise thing in the world.   ;)
« Last Edit: March 02, 2007, 12:34:01 AM by g english »

Offline Borek

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That means if you originally had 100mL of pure ethanol, and then mixed it with water, by distillation you can not get anymore then 96mL of pure ethanol back.

I don't get it. I would rather say that 100 mL of pure ethanol, once mixed with water, can at best be distilled to 104 mL of 96%. Your approach (100 mL -> 96 mL) seems to deny mass preservation :)
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Offline DrCMS

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Thats my point aswell the 96% ethanol azeotrope is not required for this calculation and if enahs took the time to read my first post they'll see i spelt out how to do the % yield.

Offline enahs

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That means if you originally had 100mL of pure ethanol, and then mixed it with water, by distillation you can not get anymore then 96mL of pure ethanol back.

I don't get it. I would rather say that 100 mL of pure ethanol, once mixed with water, can at best be distilled to 104 mL of 96%. Your approach (100 mL -> 96 mL) seems to deny mass preservation :)

It is just confusion of wording, as I was saying the same thing....only levelizing it back down to 100 (as in, collect only 100mL of distillate). But I do suppose your way makes more sense.

« Last Edit: March 02, 2007, 07:51:15 AM by enahs »

Offline enahs

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Thats my point aswell the 96% ethanol azeotrope is not required for this calculation and if enahs took the time to read my first post they'll see i spelt out how to do the % yield.

No, what you said is just wrong. Unless you distill every bit of the solution, and then calculate the amount of ethanol in it. But that is not the point of distillation, that is just a slow method of transferring one solution into another container.

If you start with 50mL of ethanol and 50mL of water. You then distill it and you only collect the distillate while the B.P. is appropriate. When you stop collecting the distillate, you at most have 96% ethanol in it. Your method (original/result) assumes you can collect 100% of the ethanol. Once you pass that critical point (in this example) of the ~50 mL of distillate, the B.P. will start shifting towards water, and you get more and more water in each drop then you do ethanol.

The only possible way to get 100% of the ethanol back is to distill the entire solution; but the point of distillation is to separate compounds.


You collect a volume of distillate, you then recognize that at the maximum possible efficiency, only 96% of that is going to be ethanol. That is your theoretical yield for that possible volume of distillate. You then calculate how much ethanol you actual have in your distillate (say, using densities). This is then what you use to find your % yield, and determine the efficiency of your distillation apparati and usage.


Your method is a % yield, but it is pointless when applied to distillation like this; as your maximum possible separated compound yield is not what you started with. The only way to get it what you started with is to distill all 100mL of the 50/50mL solution; which is not distillation!


« Last Edit: March 02, 2007, 08:26:51 AM by enahs »

Offline Borek

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Problem lies - IMHO - in the definition of percent yield. Could you all please define percent yield as you see it in the case of a distillation where you start with Vs Cs solution and you end with Ve Ce solution?

Just please post the definitions and - untill all definitions will be in place - stop commenting what others have posted :)
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Offline DrCMS

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% Yield is amount you get at the end of the experiment as a percentage of the theoretical maximum.

If you start with 100ml of 50% ethanol and collect 60ml of 60% ethanol your yield is 72%.
If you start with 100ml of 50% ethanol and collect 70ml of 60% ethanol your yield is 84%.
If you start with 100ml of 50% ethanol and collect 80ml of 60% ethanol your yield is 96%

If you start with 100ml of 50% ethanol and collect 70ml of 70% ethanol your yield is 98%.

If you start with 100ml of 50% ethanol and collect 30ml of 96% ethanol your yield is 57%
If you start with 100ml of 50% ethanol and collect 52ml of 96% ethanol your yield is 100%

If you start with 100ml of 50% ethanol and collect 65ml of 80% ethanol your yield is 100%


The maximun PURITY is 96% but the yield can be anywhere from 0-100%

In reality you either get a high yield or a high purity, not both.
 
Yes the point of distilation is to purify the compound but in doing so you lose yield. 
100% yield from a distillation is pretty much impossible.

Knowing the azeotrope concentrations allows you to calculate your distilation efficency but is irelevent when calculating the yield.

The question asked was how to calculate the yield not the efficency.

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