Isolate Malonitrile:Extr w/ Ether, 3x100mL & 1x300mL

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The discussion revolves around the extraction of malonitrile from an aqueous solution using ether. The calculations show that when three 100 mL portions of ether are used, approximately 21.1 g of malonitrile can be recovered. For a single 300 mL ether extraction, the recovery is estimated at 18.0 g. The extraction ratios were calculated using the concentration of malonitrile in both ether and water, confirming the methodology is sound. Overall, the extraction process and calculations appear to be correct.
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Suppose a reaction mixture, when diluted with water, afforded 300mL of an aqueous solution of 30 g of the reaction product malonitrile,CH2(CN)2 which is to be isolated by extraction with ether. The solution of malonitrile in ether at room temperature is 20.0 g per 100 mL, and in water is 13.3 g per 100 mL. What weight of malonitrile would be recovered by extraction with (a) three 100 mL portions of ether; (b) one 300-mL portion of ether. Suggestion: For each extraction let x equal the weight extracted into the ether layer. In case (a) the concentration in the ether layer is x/100, and in the water layer is (30-x)/300; the ratio of the quantities is equal to k = 20/13.3.
 
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any ideas? anyone?
 
what's the problem? You've got most of what you need right here.

Suggestion: For each extraction let x equal the weight extracted into the ether layer. In case (a) the concentration in the ether layer is x/100, and in the water layer is (30-x)/300; the ratio of the quantities is equal to k = 20/13.3.
 
Is this right?

Well, is this right?


1) k = 20/13.3 = 1.5
x/30-x = .50
x = .5(30-x)
1.5x = 15
x = 10g in ether layer
30 – x = 20g in water layer

x = .5(20 – x)
x = 10 -.5x
1.5x = 10
x = 6.67
20 – x = 13.33


x/13.33 – x =.50
x = .50(13.33 – x)
x = 6.67 - .5x
x = 4.44
13.33 – 4.44 = 8.89

a) 21.1g = three 100-mL portions of ether

x/30-x =1.5
x=1.5(30-x)
x =45-1.5x
2.5x/2.5 =45/2.5
x = 18.0

b) 18.0 g = one 300-mL extraction of ether
 
yep everything seems correct
 
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