Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: xangelofxdeathx on February 28, 2007, 11:50:40 PM
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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! :)
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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.
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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?
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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.
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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.
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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).
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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.
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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.
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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. ;)
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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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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.
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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.
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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!
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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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% 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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Your numbers are correct, but in this case irrelevant as you are using percent yield wrong; In my opinion, as your number makes no sense when taking into account the purpose of this distillation.
% yield:
"percent yield: the actual yield of an experiment divided by the theoretical yield and multiplied by 100."
Say you collect 80mL of distillate, and you stopped collecting it immediately when the B.P. started changing. Theoretically, if your apprati and procedures where done to perfection, the maximum amount of ethanol is 76.8 mL. If you then determine you have 70mL of ethanol in your distillate, your percent yield from the maximum theoretical yield you could achieve in that aliquot is 91.15%.
What you are calculating with is what I would call the "total theoretical yield", and what I am saying would be called the "experimental theoretical yield". And in this type of situation, the "experimental theoretical yield" makes more sense, to me, as it provides you directly with statistics of the efficiency of your experiment. Your "total theoretical yield" only makes sense if you originally tried to completely separate out entire compound. Typically in distillation, unless you have very little of the compound, you stop short in order to maintain purity.
I think it is nothing more then semantics and representation of results. I do not see the method you are trying to use worth any statistical merit in this case; because with the equipment he was using, and more then likely the outlined procedures, his goal was to collect some distillate and calculate the efficiency of his distillation apparatus or fractionating column. That makes more sense, to learn about the trials and errors and problems of distillation. Not just boil the solution until you have passed the critical point and you are now adding an extra 25mL of water to your distillate to get those last 4mL of ethanol.
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You can not just decide to change what % yield means just beacuse you don't like the answer it gives.
As I've said all along the question was about % yield not purity or efficency.
You are confusing them and trying to combine them together into some kind of bastard child you call "experimental theroretical yield".
Stop making up stuff to try and prove your point.
Yield is Yield and Purity is Purity they are not interchangeable terms.
Yes in real life distillation you can either get good yield or good purity and under different circumstances you want one over the other.
I work in industry and have done for 12 years since i got my PhD what do you you think distillation is used for in industry?
Sometimes the material you distill off is the one you want to purify in which case purity is the most important and you are prepared to scarifice yield to get the purity you want.
In another process you may have say a wet solvent that you are drying by distilling out water in this case removing all the water is critical and you don't care if you lose extra solvent to get all the water out.
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You can not just decide to change what % yield means just beacuse you don't like the answer it gives.
Stop making up stuff to try and prove your point.
I am not.
If you collect 50 mL of distillate, and the maximum possible ethanol you could have in that distillate is 48mL, but you only get 40mL of ethanol you only obtained a 83% yield.
You are welcome to disagree with me all you want, as I disagree with you. There is no need to get all defensive though, nobody is attacking you personally; and I apologize if it came off as I was, I was not. If you would listen to what I said instead of trying to prove your self right, you would see my point has a logical sense to it. I have already stated it was my opinion that what I am saying makes more sense in this case. Keyword, opinion. Not that I am empirically correct.
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Ok say you have 100ml of 50% ethanol
if you collect 20ml of 80% ethanol you'd say that was 83% yield but i'd say it was 32%
if you collect 40ml of 80% ethanol you'd say that was 83% yield but i'd say it was 64%
if you only colllected 10ml of 96% ethanol you'd say it was 100% yield while i'd say it was 19%
How are your numbers correct as a yield?
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Please, both of you: define your yields in terms of symbols so that they can be easily compared :)
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Ok say you have 100ml of 50% ethanol
if you collect 20ml of 80% ethanol you'd say that was 83% yield but i'd say it was 32%
if you collect 40ml of 80% ethanol you'd say that was 83% yield but i'd say it was 64%
if you only colllected 10ml of 96% ethanol you'd say it was 100% yield while i'd say it was 19%
How are your numbers correct as a yield?
They are just as correct as yours.
Of that first 20 mL collection of distillate, theoretically I can not have anymore then the 19.2 mL of pure ethanol. We are saying the exact same thing....the exact same thing. We are only representing it differently. I personally think mine makes more sense, in a situation like this to refer to theoretical yield as the maximum possible yield for the volume of distillate collected.
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Please, both of you: define your yields in terms of symbols so that they can be easily compared :)
Actual Yield; What you Got (X mL of pure ethanol)
Theoretical Yield; Maximum ethanol you could theoretically obtain with that volume of distillate collected (Y mL of pure ethanol)
% Yield = (X/Y)*100
I just think he is refusing the even admit mine is a theoretical yield; it might not be the theoretical yield he would use, but to me; in this situation it makes more sense. I have at least admitted that his number are correct; that he is not mistaken; just in my opinion I think my representation makes more sense when you take into account the particular application of this experiment.
What he is calculating, while is a percent yield; is more suitably called percent recovery.
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I do refuse you admit you are right as your definition of yield is independant of the starting solution.
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I do refuse you admit you are right as your definition of yield is independant of the starting solution.
That is fine. You are free to choose! But think of it like this.
Percent Yield is defined as (by many people other then me) "( Actual yield of product / Expected maximum yield of product) *100"
Is that not what you define it?
If you have a 50mL ethanol/50mL water solution in your distillation flask, and you collect 20mL of distillate, is your expected yield even remotely possibly 50mL of ethanol? Clearly it is not. If your goal was to obtain ethanol and that 20mL of distillate contained 15 mL of ethanol; where it could only theoretically obtain 19.2, your yield is 78%. Your % recovery from the original 50mL of ethanol is what you are calculating, which in this case would be 30%. Unless you intentionally tried to recollect all 50mL of ethanol, the percent recovery makes not sense in representing data to other people.
These % things are statistics designed to convey efficiency about the process; yours only does that if the goal was to transfer all of your ethanol from the distillation flask to the receiving container.
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Forget the ethanol/water azetrope system for a minute and think of this instead.
100ml of a 50:50 mix of A and B which do not form an azeotrope.
In your system
if you get 50ml of 75% A 25% B the yield is 75%
if you get 1ml of 90% A 10%B the yield is 90%
In a non azetrope system your definition of yield is purity.
As i've said time after time yield and purity are different both are important but they are not interchangable.
% Yield = (amount of distillate collected x purity of this distilate) / (amount of starting mixture x purity of starting material)
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Ok, forget the azeotropic mixture.
100 mL of a 50:50 mix of compound A and B.
You distill it, and you collect 10 mL of the distillate.
10 mL of distillate. Now, how is it theoretically possible in that 10mL of distillate to have 50mL of either A or B? That is how you are defining your percent yield. Your number is a percent recovery. Percent recovery only makes sense if your goal is to recover it all. If my goal was to just make 10mL of compound A, and I succeeded in doing that, my yield is 100%, because in 10mL of distillate I can not have more then 10mL of either compound A or B. 10mL is my maximum possible yield, in 10 mL.
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100 mL of a 50:50 mix of compound A and B.
You distill it, and you collect 10 mL of the distillate.
10 mL of distillate. Now, how is it theoretically possible in that 10mL of distillate to have 50mL of either A or B? That is how you are defining your percent yield.
No, he is not, please reread his formula. 10 mL of pure substance out of 50mL of possible yield will give 20%.
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No, he is not, please reread his formula. 10 mL of pure substance out of 50mL of possible yield will give 20%.
How is it possible to yield 50mL in 10 mL, though? It is not, it is in fact impossible; would you not agree?
That is my point. This is more better termed a percent recovery.
Forget distillation.
I have 20 real diamonds and 20 fake diamonds. I then throw them on the ground and allow you to pick only 10 diamonds, and you of course want to pick all real diamonds. Theoretically it is only possible for you to collect 10 real diamonds. If you collect 9 real diamonds your percent yield is 90%, as (actual/theoretical * 100), because you can not collect more then 10. You, however, only recovered 45% of the total real diamonds.
This might just be a regional terminology problem. But I have never heard what he is trying to calculate a percent yield, just a percent recovery. It makes no sense, as a percent yield, to theorize you can recover more then is actually possible.
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I have 20 real diamonds and 20 fake diamonds. I then throw them on the ground and allow you to pick only 10 diamonds, and you of course want to pick all real diamonds. Theoretically it is only possible for you to collect 10 real diamonds. If you collect 9 real diamonds your percent yield is 90%, as (actual/theoretical * 100), because you can not collect more then 10. You, however, only recovered 45% of the total real diamonds.
Only collecting 10 diamonds from a possible 40 is an arbitary thing. If you collect 10 diamonds and 1 is fake thats 90% purity but as you started with 20 real dimonds your yield is 45%.
How about if you start with 1000000 diamonds and only 1 is fake.
If you collect 10 diamonds and 9 are real is that 90% yield?
If you collect 10 real diamonds from 999999 is that a 100% yield?