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Extraction of malonitrile using different portions of ether

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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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 ...

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.