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#1 2008-02-12 00:32:43

orgopete
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Partition of Y in ether in extraction question

Compound Y has a solubility of 5.0 g/100 mL in water and 25.0 g/100 mL in ether.  What weight of compound Y would be removed from a solution of 3.0 g of Y in 100 mL of water by a single extraction with 100 mL of ether?

Kp = (wt in ether/wt in water) = 25 g/5 g = 5
x = amount in ether
3-x = amount in water;
Set the distribution of the 3 g sample equal to the ratio (Kp) in ether and water.
(x/100 mL)/[(3-x)/100 mL] = 5 [Cancel the denominators and solve for x]
x/(3-x) = 5
x = 5 * (3-x) = 15 - 5x
6x = 15
x = 15/6 = 2.5 g  in ether
3- x = 0.5 g in water


What weight of compound Y would be removed from the original water solution if two extractions of 50 mL of ether were used?
Use the same strategy as above, except now the solvent ratios are a factor. Because less ether is used, each ether fraction will contain less total compound than if 100 mL were used. (I changed the order in which I wrote the ratio by putting the ratio (5) first. It is still amount in ether/amount in water = 5. Note the trick I used to cancel out the volumes.

5 = (x/50 mL)/[(3-x)/100 mL]

We need to change the denominators so they can cancel.
Multiplying x/50 mL times 1 or (2/2) will not change its value, so (x/50 mL) * (2/2) = (2x/100 mL)

5 = (2x/100 mL)/[(3-x)/100 mL]
5 = 2x/(3-x)
x = 2.14 g in ether
3-x = 0.85 g in water
5 = 2x/(0.85-x)
x = 0.61 g
3-(2.14 + 0.61) = 0.24 g in water


How much ether must be used to remove 95% from water in a single extraction?
Note how the same strategy is used again, but with the volume (y) being the unknown value. The ratio of ether to water is used, but with the amount of ether unknown.

3 g * 95% = 2.85 g in ether; 3 g - 2.85 g = 0.15 g in water
y = mL of ether
5 = (2.85/y)/(0.15/100 mL); multiply RH side by (100 mL/100 mL)
5 = (285 mL/y)/0.15
5 * 0.15 * y = 285 mL
y = 285/(5*0.15) = 380 mL

I used (2/2) to change the denominator. One can use other equivalent strategies, such as dividing numerator and denominator of the second term by 2. If you are trying to compare odd volumes, this will also work, but you must be more careful with the algebraic changes associated with the calculations. For example, let me convert the 100 mL to 50 mL by dividing by 2 (100 mL/50 mL = 2). I am writing it as multiplying by (0.5/0.5) to fit the formatting of the forum posts.
(0.5/0.5) * (3-x)/100 mL = 0.5*(3-x)/50 mL or ((3-x)/2)/50 mL
Multiplying or dividing either term by 1 will not change its value. We are simply finding a common denominator.

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