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

  • Thread starter Thread starter chiefy
  • Start date Start date
  • Tags Tags
    Extraction
AI Thread Summary
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.
chiefy
Messages
14
Reaction score
0
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.
 
Chemistry news on Phys.org
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
 
It seems like a simple enough question: what is the solubility of epsom salt in water at 20°C? A graph or table showing how it varies with temperature would be a bonus. But upon searching the internet I have been unable to determine this with confidence. Wikipedia gives the value of 113g/100ml. But other sources disagree and I can't find a definitive source for the information. I even asked chatgpt but it couldn't be sure either. I thought, naively, that this would be easy to look up without...
I was introduced to the Octet Rule recently and make me wonder, why does 8 valence electrons or a full p orbital always make an element inert? What is so special with a full p orbital? Like take Calcium for an example, its outer orbital is filled but its only the s orbital thats filled so its still reactive not so much as the Alkaline metals but still pretty reactive. Can someone explain it to me? Thanks!!
Back
Top