Extraction of malonitrile using different portions of ether
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Suppose a reaction mixture, when diluted with water, afforded 300 mL of an aqueous solution of 30 g of the reaction product malononitrile, CH2(CN)2, which is to be isolated by extraction with ether. The solubility of malononitrile in either at room temperature is 20.0 g per 100 mL, and in water is 13.3 g per 100 mL. What weight of malononitrile 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 these quantities is equal to k = 20/13.3.
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Solution Summary
This solution provides information on extraction of malonitrile from water using ether. We have used 2 different extraction solutions, a) three 100-mL portions of ether and b) one 300-mL portion of ether. It is known that the amount of malonitrile recovered in two cases is different. This is what calculated in this solution.
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k = 20/13.3 = 1.5
a.) Three 100-mL portions of ether are used
x/100 / (30-x)/300 = 1.5
x/1 / (30-x)/3 = 1.5
3x/ (30-x) = 1.5
3x = 1.5 (30-x)
3x= 45 - 1.5x
4.5x = 45
x= 30/4.5
x = 10g in ether layer
30 - x = 20g in water layer
Next, remaining malonitrile in water is 20g, therefore when we do next extraction with 100ml ...
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