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

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Discussion Overview

The discussion revolves around the extraction of malonitrile from an aqueous solution using ether. Participants explore the calculations involved in determining the amount of malonitrile that can be recovered through different extraction methods: three 100 mL portions of ether versus one 300 mL portion of ether.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Homework-related

Main Points Raised

  • Post 1 presents the initial problem and suggests using a variable x to represent the weight of malonitrile extracted into the ether layer, establishing a concentration ratio based on the solubility in ether and water.
  • Post 3 reiterates the initial suggestion and emphasizes that the necessary information is available for solving the problem.
  • Post 4 provides detailed calculations for both extraction scenarios, arriving at specific weights of malonitrile recovered: 21.1 g for three 100 mL portions and 18.0 g for one 300 mL portion.
  • Post 5 confirms that the calculations presented in Post 4 appear to be correct.

Areas of Agreement / Disagreement

Participants generally agree on the correctness of the calculations presented, although the discussion does not explicitly address any potential disagreements regarding the methodology or assumptions involved.

Contextual Notes

The discussion relies on assumptions about the solubility of malonitrile in ether and water, as well as the ratios used in the calculations. The implications of these assumptions on the accuracy of the results are not fully explored.

chiefy
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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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