Chemical Forums
Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: Winga on July 10, 2004, 01:56:03 PM
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Hi Everybody!
Could anyone explain why we use 2 times of 50mL of a solvent for extraction is better than that of using 1 time of 100mL.
Thank You!
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do your mathematics
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Sorry, I need an explanation without using any calculation.
(e.g. about the volume & of the solvent, concentration...)
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Lets say that your solute is completely soluble in 5 ml of ether. Then why use 2 x 50ml or 1 x 100ml of ether to extract it? Because the extraction procedures are not 100% efficient. Assuming the extraction protocol is 90% efficient then if you do it twice you can get 90% of 90% of the solute out. If you only do it once then you only get 90% of it out. The amout of ether is irrelevant, its a matter of efficiency.
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A further question:
As you say the solute is completely soluble in 5 ml of ether, but if we use 2 ml of ether to do 25 times of extraction, will it affect the original efficiency? (0.9^25)
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of course. The efficiency I speak of assumes your using an amount of ether greater than the amount needed for it to be soluble. The correct chemical name for this behavior isn't efficiency, but something else. I'll try to look it up for you tonight when I get home.
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Sorry, my question is that if we use the amount of solvent which cannot completely dissolve the solute in normal condition, (e.g. 5 ml of ether can completely dissolve 1 g of solute), but I just use 2 ml of ether for each extraction, will the efficiency of each extraction lower than 90%?
Thanks a lot! ;)
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Yes.
and I looked up the theory. It's not called efficiency but percent recovery..
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Sorry, my question is that if we use the amount of solvent which cannot completely dissolve the solute in normal condition, (e.g. 5 ml of ether can completely dissolve 1 g of solute), but I just use 2 ml of ether for each extraction, will the efficiency of each extraction lower than 90%?
But by this equation,
K = [Solvent A]solute / [Solvent B]solute
If the efficiency / percent recovery of each extraction is lower than the original value, will K still constant?