Chemical Separations

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Chemical Separations

• What is a chemical separation?
• Examples
• Filtration
• Precipitations
• Crystallizations
• Distillation
• HPLC
• GC
• Solvent Extraction
• Zone Melting
• Electrophoresis
• Mass Spectroscopy

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Chemical Separations

• What is the object of the separation.
• Collection of a pure product
• Isolation for subsequent analysis for either
  quantification or identification
• Analysis
• How Much?
• What is it?

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Chemical Separations

• Major Industries
• Petroleum Distillation
• Distilled Spirits

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Chemical Separations

• Petroleum is a mixture of hydrocarbons.
• The larger the molecular weight the less
  volatile.
• So we must separate into various molecular weight
  fractions (different boiling points)
• The results are still complex mixtures

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Chemical Separations

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Distillation

• As heat is added to the system the lower
  volatility compounds will boil away and can be
  collected.
• In the spirits industry the low boilers are call
  foreshots (75 EtOH)
• The high boilers are called feints
• Congeners - Chemical compounds produced during
  fermentation and maturation. Congeners include
  esters, acids, aldehydes and higher alcohols.
  Strictly speaking they are impurities, but they
  give whisk(e)y its flavour. Their presence in the
  final spirit must be carefully judged too many
  would make it undrinkable.

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Distillation

• What is whiskey?
• What is brandy?

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Interesting Facts

• Bourbon - US whiskey made from at least 51 corn,
  distilled to a maximum of 80 abv (160 proof) and
  put into charred new oak barrels at a strength of
  no more than 62.5 abv.
• Organic whisk(e)y - That made from grain grown
  without chemical fertilizers, herbicides and
  pesticides.
• Tennessee whiskey - As bourbon, but filtered
  through a minimum of 10 feet of sugar-maple
  charcoal. This is not a legal requirement, but is
  the method by which Tennessee whiskies are
  currently produced.

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Interesting Facts

• Malt whisky - Whisky made purely from malted
  barley.
• Angels' share - A certain amount of whisk(e)y
  stored in the barrel evaporates through the wood
  this is known as the angels' share. Roughly two
  per cent of each barrel is lost this way, most of
  which is alcohol.
• http//www.whisky-world.com/words/index.php

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Solvent Extraction

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Replace concentration with moles over volume and
let q equal the fraction in the aqueous phase

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Define a new term for the ratio of the volumes of
the phases

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We can do a little algebra and find an expression
for q
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Since if it does not end up in the aqueous phase
it must be in the organic.

• p is the term for the fraction in the organic
• p q 1
• Giving

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Sample Problem

• You have 100.0 mL of an aqueous solution that is
  100.0 mM in compound C.   This solution is
  extracted with 50.0 mL of diethyl ether and the
  aqueous phase is  assayed and it is found that
  the concentration of compound C that remains is
  20.0 mM.  What is the equilibrium constant for
  this extraction system.

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Solution

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We can do multiple extraction from the aqueous
phase.

• We end up with the following expression for what
  is left in the aqueous phase.

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Example

• How many extractions would be required to remove
  99.99 of aspirin from an aqueous solution with
  an equal volume of n-octanol? 
• Since 99.99 must be removed the decimal fraction
  equivalent of this is  0.9999.  This leaves
  0.0001 in the aqueous phase.  Since we have equal
  volumes then Vr is 1.00.
• We are able to find from the Interactive Analysis
  Web site that K for Aspirin is 35.5.  We plug
  these values into the q equation and the power is
  the unknown.

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Solution
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What if our compound can dissociate or
participate in some other equilibrium?

• A compound such as aspirin is a carboxylic acid.
  We can represent this as HA.
• Do we expect the ion A- to be very soluble in the
  organic phase???

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Dissociation

• So if we have dissociation then less will go into
  the organic phase.
• Kp is the ratio of concentration of aspirin (in
  the un-dissociated form) in each phase. This
  ratio will always be the same.
• How do we account for the ion formation?

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Distribution Coefficient

• Where C is the formal concentration of the
  species.
• Ca HA A-
• Dc will vary with conditions
• For this compound what is that condition?

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Dc

• Since the ion is not very soluble in the organic
  phase then we may assume that the dissociation
  will not happen in that phase.
• This gives us the expression to the right.

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Acid Equilibria

• What is the equilibrium?
• Ka

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With a little algebra

• So if you know Kd and Ka then you can determine
  Dc as a function of H (pH)
• However if H is much larger than Ka then Dc
  will equal Kd. If the H is close in value to
  Ka then D will be related to the pH
• Plotting this we get.

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So What, Why is this useful.

• Well we can now move a solute (analyte) from one
  phase to another.  This can be very useful when
  extracting a compound that has significant
  chemical differences from other compounds in
  solution.  As a matter of fact this has been used
  as an interview question for prospective co-ops
  when I worked in industry.
• The question would go like this.  You have
  carried out a series of reactions and it is now
  time to work up the product which currently sits
  in an organic solution (methylene chloride). 
  Your expected product is a primary amine.  Which
  of the following solutions would you extract this
  methylene chloride solution with to isolate your
  amine.  
• Your choices are
• A)   Toluene.
• B)   0.1 N NaOH (aq)
• C)    0.1 N HCl (aq)
• D)    I never wanted to work here anyhow.
•  

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Separation

• So far we can tell how one compound moves from
  one phase to another. What if we are try to
  separate two compounds, A and B
• Well we might just suspect that if we find a
  solvent system that has different values of Dc
  for each compound we could end up with most of
  one compound in one phase and the other compound
  in the opposite phase. It is not that simple.

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Example

• System I
• Da 32 Db 0.032 (A ratio of 1000)
• Vr 1
• Let's recall our equations
• q (fraction in aqueous)    1 /  (DVr 1)
• p (fraction in organic)          DVr / (DVr 1)
  
• Vr (volume ratio)               Vo / Va

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Case I

• pa    321 / (321 1)    0.97
• pb    0.0321/ (0.0321 1)    0.03
• If we assume that we have equal moles of A and B
  to start then what is the purity of A in the
  Organic Phase?
• Purity   moles A /  (moles A moles B)
• Purity   0.97 / (0.97 0.03)    0.97 or 97
   

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Case II

• Da 1000 Db 1 VR 1 (Ratio is
  still 1000)
• pa    10001 / (10001 11)    1000/1001   
  0.999  
• Aha! we got more a into the organic, as we would
  expect with a higher D value.
• Now
• pb   11 / ( 11 1)   1/2  0.5
• oh-oh
• What do we get for purity of compound a now?
• purity   0.999 / (0.999 0.50)    0.666
• Yuck!

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How can we get around this issue?

• Once we have selected the solvent and pH,  then
  there is little that we can do to change D.    
  What else do we have in our control?????
• Let's look
• p    DVr / (DVr 1)
• Not much here except Vr  and in fact that is the
  key to this problem.  Is there an optimum Vr
  value for the values of D that we have?  Yes!
• Our equation for this is      V r(opt)   
  (DaDb)-0.5

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Revisit the two cases

• So let us look at our two cases and see which
  will give us the optimum values.
• Case I
• Da    32   and Db    0.032  
• V r(opt)   (32 0.032)-0.5      ( 1 )-0.5  1
• So we were already at the optimum.

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Case II Revisited

• Case II
• Da 1000 and Db 1
• Vr (opt)    (10001)-0.5  1000-0.5    0.032
• Which mean that when we do our extraction we will
  extract _______ mL of organic for each _______ mL
  of aqueous.
•  

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Purity for Case II

• What is our purity for this system?
• pa    10000.032 / (10000.032 1)    32/33 
    0.97
• and
• pb    10.032 / (10.032 1)    0.032/1.032
  0.03 
• Purity of a then is 0.97/ (0.97 0.03)
• Which will give us the 97 purity we had for Case
  I with with the Vr of 1.  

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Can we improve this purity?

• If we were to extract again then we would just
  remove the same proportions. We would get more
  compound extracted but it would be the same
  purity.
• What if we were to take the organic phase and
  extract it with fresh aqueous phase. We know
  that one of the two compounds will end up mostly
  in that aqueous phase so we should enhance the
  purity of the other compound in the organic phase.

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Back Extraction

• Called that since you are extracting back into
  the original phase.

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Back Extraction Case I Example

• Let's look at the numbers.
• Da 32 Db 0.032 Vr 1
• pa 0.97 pb 0.03
• qa 0.03 qb 0.97
• Lets prepare a table. 

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Initial conditionsprior to starting back
extraction
.    
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Now we extract shake shake shake

• How much goes to the Aqueous phase
• q        which is 0.03 for A and 0.97 for B
• How much goes to the Organic phase
• p        which is 0.97 for A and 0.03 for B    
   

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Now what is the purity for A in the organic
phase???  

• Purity Amount A / (Amount A Amount B)   
• 0.970.97 / (0.970.97 0.030.03)
• 0.94/(0.94 0.0009) 99.9
• What is the yield of A (fraction of the total
  amount that we started with)

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Lets do it again Can we improve purity even
more?
  Purity A     0.913 / (0.913 0.000027)  
99.997 But our yield has dropped to 91.3,   
there is a price to pay for the added purity.  
 
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Can We Expand This?Why Would We Want to?

• Such multiple extraction systems have been
  developed.
• Still a viable option for preparative work.
• For separations it has been replaced by HPLC
• Called Craig Counter Current Extraction.
• Special glassware is used.

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Craig CCE

• Equal amounts of organic (red) and aqueous (blue)
  solvents with the analyte(s) are added to the A
  arm of the tube via port O. Fresh Aqueous Solvent
  is added to each of the tubes down the apparatus.

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Craig CCE

• Rock the system back and forth and to establish
  equilibrium.
• Allow the system to stand for the layers to
  separate.
• Rotate the apparatus counter clockwise about 90o
  to 100o.

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Craig CCE

• Rotate Back to Horizontal

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Starting Conditions
        Tube 0 1 2 3 4
Organic Phase 0        
Aqueous Phase 1 0 0 0 0
After One Equilibrium
         Tube 0 1 2 3 4
Organic Phase p        
Aqueous Phase q 0 0 0 0
Transfer Step 1
        Tube 0 1 2 3 4
Organic Phase 0 p      
Aqueous Phase q 0 0 0 0

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Now here is what is in each tube/phase after
equilibrium is reached.
         Tube 0 1 2 3 4
Organic Phase pq pp      
Aqueous Phase qq qp 0 0 0
Now we do Transfer 2  
         Tube 0 1 2 3 4
Organic Phase 0 pq pp    
Aqueous Phase q2 pq 0 0 0

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Now here is what we have in each tube after the
next equilibrium. The total in each tube times
either p or q as appropriate.
         Tube 0 1 2 3 4
Organic Phase pq2 p2pq p3    
Aqueous Phase q3 q2pq qp2 0 0
We transfer again.   Transfer Step 3    
        Tube 0 1 2 3 4
Organic Phase 0 pq2 2p2q p3  
Aqueous Phase q3 2pq2 p2q 0 0

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Shake Again  Equilibrium 4    
        Tube 0 1 2 3 4
Organic Phase pq3 p3pq2 p3p2q p4  
Aqueous Phase q4 q3pq2 q3p2q qp3 0
Transfer 4    
        Tube 0 1 2 3 4
Organic Phase 0  pq3 3p2q2 3p3q p4
Aqueous Phase q4 q3pq2 3p2q2 p3q 0
  See a trend????
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Craig CCE

• How about a binomial expansion?
• (q    p)n    1
• Powers of the two terms in each tube will add up
  to n
• Coefficients will be found from Pascal Triangle
• 1 1     1 1     2     1 1     3     3    
  1 1     4     6     4     1 1     5     10    
  10     5     1 1     6     15     20     15    
  6     1 1     7     21     35     35     21    
  7     1

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Craig CCE

• Or the formula
• Fr,n    n!/((n-r)!r!) pr q(n-r)
• n is the number of transfer and r is the tube
  number. You start counting at zero!

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Craig CCE

• Let's look at and example for a four tube system.
• Da   3 p   0.75 q
  0.25
• Db  0.333 p 0.25 q 0.75  
  
• What would be the purity and yield of Compound A
  if collected from the last in our above example.
   
• Amount of A               p4   or  0.754     
  0.3164 Amount of B               p4   or 
  0.254      0.0039
• Purity of A         0.3164 / (0.3164 0.0039) 
    0.9878    or 98.78
• Yield of A          We collect a fraction of
  0.3164  or 31.64
• Horrible Yield!

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Craig CCE

• What if we collect the last two tubes??
• Amount of A      p4   and  4p3q  or  0.754  
  4(0.75)3(0.25)     0.3164 0.4219 0.7383
  Amount of B      p4   and  4p3q   or  0.254  
  4(0.25)3(0.75)     0.0039 0.0469 0.0508
• Purity of A    (0.3164 0.4219) / (0.3164
  0.4219 0.0039 0.0469 )    0.9356 or 93.56
• Yield of A              We collect a fraction of
  0.3164    0.4219    0.7383 or 73.83
• Purity still ok and yield is much better.

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Craig system n 200 transfers.  Da of 2.0 and Db
of 4.0 pa of 0.666 pb of 0.800.
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Final Formulas(1)

• rmax    np   nDVr/(DVr 1)
• To find the separation between two peaks we would
  use.
• Drmax (rmax)a - (rmax)b    n(pa-pb)  
• The Gaussian distribution approximation for our
  binomial expansion would be (when ngt24)
• Fr,n   (2p)-0.5(npq)-0.5 exp-((np-r)2)/2npq

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Final Formulas(2)

• The width of the distribution through the system
  would be
• w 4s 4(npq)0.5
• Resolution would be
• R Drmax/w Drmax/4s
• or
• R nDp/(4(npq)0.5)    n0.5 Dp / 4(pq)0.5