[getchoo0329]

Having a tough time here. dont know where to start

organic compound A has the following solubilities at room temp: 12g A in 100ml Ether; 6g A in 100ml H20 (k=2). calculate which of the following extraction techniques will yield the largest mass of A

a. a solution of 6g A in 100ml H2O is shaken with 100ml ether (one extraction)
b. a solution of 6g A in 100ml H2O is shaken with two 50ml portions of ether (two extractions)

any help is much appreciated

[Arkcon]

Your textbook will have info on how an extraction should be done, and why it's done that way.  *Hint* The answer would be the same, no matter what the setup of this problem was.  That is, the specific values given in this problem don't add anything toward it's answer.

[ARGOS++]

Dear Arkcon;

Could you accept, if I disbelieve your Hint by all basics of Mother Nature, and from the “Nernst Partition Law” (= K- Coefficient)?

In this particular Example the second method outbids the first by 12.5 % relative.
(The second method must “all times” bet the first  –  Sorry!)

For the “Nernst Partition Law”, see on:   "Nernst Partition Law:  Absorption (chemistry)”.

Good Luck!
                    ARGOS⁺⁺

[agrobert]

getchoo0329

So because of the Distribution coefficient unique to each solvent system (ether-water) and substrate (Compound A) you will get a greater mass by extracting twice with equivalent amount of solvent.

[Arkcon]

Dear Arkcon;

Could you accept, if I disbelieve your Hint by all basics of Mother Nature, and from the “Nernst Partition Law” (= K- Coefficient)?

Yeah, I went a little heavy when I said:

*Hint* The answer would be the same, no matter what the setup of this problem was.

I did mean that, following the partition equation, you get better extraction in two steps, than just one, regardless of the solubility in each solvent.  Like I said,

That is, the specific values given in this problem don't add anything toward it's answer.

That all hinges back on this previous question we all batted around for a while:

http://www.chemicalforums.com/index.php?topic=22813.0

and this one that went smoother,

http://www.chemicalforums.com/index.php?topic=23166.0

I seem to remember a worse one, with quite a bit of arguing from the O.P. of that thread, can't find it with a search now.  Oh well.

[ARGOS++]

Dear Arkcon;

But did you recognise that at that time I pointed already out:

http://www.chemicalforums.com/index.php?topic=23166.0

    –––   exactly the same as I did today?


Dear Getchoo0329:  Did you get the same result as I?,  –
or should we do it together to have once a real Example?


Good Luck!
                    ARGOS⁺⁺

[getchoo0329]

Hi guys.

I appreciate all of the help.

I figured the second part will recover more product. I'm having trouble crunching the numbers and I can't find any good examples in my book.

i really am clueless. i have been staring at this problem for hours

[ARGOS++]

Dear Getchoo0329;

I will soon be back with “some” pictures to explain it in more details to you.

Please give me a little time to do it!

Good Luck!
                    ARGOS⁺⁺

[ARGOS++]

Dear Getchoo0329; 

Let’s start:
A.)    From the solubilities K is given as their ratio: K = c_Ether / c_Water = 12.0 g / 100 ml / 6.0 g / 100 ml = 2.00. (- as given in the Question.)
To be more precise we should use Activities, and not Concentrations; but for this Question it is sufficient!
As from the K it is given that the c_Ether must be “all times” the double of c_Ether, we can draw our first picture for the -
B.)    “Single Extraction”:
             Before shaking:       ________________
                                        |                        |
                                        |   a_E = 0.00 g      |     with a = whole amount  = c * v!
                                        |-----------------|
                                        |   a_W = 6.00 g     |
                                        |_______________|
             After  shaking:        ________________
                                        |                        |
                                        |   a_E =  4.00 g     |       =   4.0 g / 100 ml !!
                                        |-----------------|
                                        |   a_W = 2.00 g     |       =   2.0 g / 100 ml
                                        |_______________|              Why?      

             For equal volumes (V) you can easily show, that the whole amount in the Funnel is:
                   A_Total =    c_Ether * V   +  c_Water     * V
                           =    c_Ether * V   +  c_Ether / K * V
                           =    c_Ether * V * (1 + 1/K)
                  And that results in:  c_Ether = 6.0 g / 100 ml / 1.5  =   4.0 g / 100 ml   or:
                                              a_Ether = 4.0 g / 100 ml  * 100 ml  = 4.0 g extracted!

C.)    “Multiple Extraction”: 
             Before shaking:       ________________
                                        |                        |
                                        |   a_E = 0.00 g      |     with a = whole amount  = c * v!
                                        |-----------------|
                                        |   a_W = 6.00 g     |
                                        |_______________|
             After  shaking:        ________________
                                        |                        |
                                        |   a_E =  3.00 g     |       =   6.0 g / 100 ml !!
                                        |-----------------|
                                        |   a_W = 3.00 g     |       =   3.0 g / 100 ml
                                        |_______________|              Why?      

             For non-equal volumes (V_E & V_W) you get for the whole amount in the Funnel:
                   A_Total  =    c_Ether * V_E    +  c_Water      *      V_W 
                             =    c_Ether * V_E    +  c_Ether / K  *      V_W
                             =    c_Ether * V_E    +  c_Ether / K  * 2 * V_E
                             =    c_Ether * V_E    *  (1 + 2/K)
                   And that results in:  c_Ether = 6.0 g /  50 ml /  2.0  =   6.0 g / 100 ml   or:
                                               a_Ether = 6.0 g / 100 ml  * 50 ml  =  3.0 g extracted!

For the second Extraction the identical Formula holds, but you start now with a total amount of only 3.0 g (in Water):
                   A_Total =    c_Ether * V_E    +  c_Water       *      V_W
                           =    c_Ether * V_E    *  (1 + 2/K)
                   And that results in:  c_Ether = 3.0 g /  50 ml /  2.0  =   3.0 g / 100 ml   or:
                                               a_Ether = 3.0 g / 100 ml  * 50 ml  =  1.5.0 g extracted!

So you end up with 3.0 g + 1.5g Compound extracted, what is 12.5% more then by a “Single Extraction”. 
(Please watch carefully the Indices: _E and: _W !! )     

Of course!,  – You can combine both extractions into one Formula and easily expand it for n Extractions! But it's not so illustrative!

I hope this may help you (and all others too) to understand.

Good Luck!
                    ARGOS⁺⁺

P.S.:  I hope I did no mistype!
.