Chemical Separations
• What is a chemical separation?
• Examples:
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Filtration
Precipitations
Crystallizations
Distillation
HPLC
GC
Solvent Extraction
Zone Melting
Electrophoresis
Mass Spectroscopy
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?
Chemical Separations
• Major Industries
– Petroleum Distillation
– Distilled Spirits
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
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.
Distillation
• What is whiskey?
• What is brandy?
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.
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
Solvent Extraction
Replace concentration with moles over volume
and let q equal the fraction in the aqueous phase
Define a new term for the ratio of the
volumes of the phases
We can do a little algebra and find an
expression for q
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
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.
Solution
We can do multiple extraction from the
aqueous phase.
• We end up with the following expression for what is
left in the aqueous phase.
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.
Solution
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???
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?
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?
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.
Acid Equilibria
• What is the equilibrium?
• Ka
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.
So What, Why is this useful.
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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.
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.
Example
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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
Case I
• pa = 32*1 / (32*1 + 1) = 0.97
• pb = 0.032*1/ (0.032*1 + 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 %
Case II
• Da = 1000
Db = 1 VR = 1 (Ratio is still 1000)
• pa = 1000*1 / (1000*1 + 1*1) = 1000/1001 = 0.999
• Aha! we got more a into the organic, as we would expect with a
higher D value.
• Now
• pb = 1*1 / ( 1*1 +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!
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) = (Da*Db)-0.5
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.
Case II Revisited
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Case II
Da = 1000 and Db = 1
Vr (opt) = (1000*1)-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.
Purity for Case II
• What is our purity for this system?
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pa = 1000*0.032 / (1000*0.032 + 1) = 32/33 = 0.97
and
pb = 1*0.032 / (1*0.032 + 1) = 0.032/1.032 = 0.03
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Purity of a then is 0.97/ (0.97 + 0.03)
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Which will give us the 97% purity we had for Case I with with the Vr of
1.
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.
Back Extraction
• Called that since you are extracting back into
the original phase.
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
• Let’s prepare a table.
Initial conditions
prior to starting back extraction
.
Before Shaking
Amount A
Amount B
Organic Phase
0.97
0.03
Fresh Aqueous Phase
0
0
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
After Shaking
Amount A
Amount B
Organic Phase
(0.97)(0.97)
(0.03)(0.03)
Aqueous Phase
(0.97)(0.03)
(0.03)(0.97)
Now what is the purity for A in the organic phase???
• Purity = Amount A / (Amount A + Amount B) =
• 0.97*0.97 / (0.97*0.97 + 0.03*0.03) =
• 0.94/(0.94 + 0.0009) = 99.9%
• What is the yield of A (fraction of the total amount that we started
with)
Let’s do it again – Can we improve purity
even more?
After second Back Extraction
Amount A
Amount B
Organic Phase
0.94*0.97
0.0009*0.03
Aqueous Phase
0.94*0.03
0.0009*0.97
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.
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.
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.
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.
Craig CCE
• Rotate Back to Horizontal
Starting Conditions
Tube#
0
Organic
Phase
0
Aqueous
Phase
1
1
2
3
4
0
0
0
0
After One Equilibrium
Tube#
Organic Phase
0
1
2
3
4
0
0
0
0
2
3
4
0
0
0
p
Aqueous Phase q
Transfer Step 1
Tube#
0
1
0
p
Aqueous Phase q
0
Organic Phase
Now here is what is in each tube/phase after equilibrium is reached.
Tube#
0
1
Organic
Phase
pq
pp
Aqueous
Phase
qq
qp
2
3
4
0
0
0
Now we do Transfer 2
Tube#
0
1
2
0
pq
pp
Aqueous Phase q2 pq
0
Organic Phase
3
4
0
0
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
Organic
Phase
pq2
p*2pq
p3
Aqueous
Phase
q3
q*2pq
qp2
We transfer again.
Transfer Step 3
Tube#
Organic Phase
0
1
0
pq2
Aqueous Phase q3 2pq2
2
3
4
2p2q p3
p2q
0
0
3
4
0
0
Shake Again Equilibrium 4
Tube#
0
1
2
3
Organic
Phase
pq3
p*3pq2
p*3p2q
p4
Aqueous
Phase
q4
q*3pq2
q*3p2q
q*p3
4
0
Transfer 4
Tube#
0
1
2
3
4
Organic Phase
0
pq3
3p2q2
3p3q
p4
Aqueous Phase
q4
q*3pq2
3p2q2
p3q
0
See a trend????
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
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!
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
Amount of B
• Purity of A
Yield of A
• Horrible Yield!
p4 or 0.754 = 0.3164
p4 or 0.254 = 0.0039
0.3164 / (0.3164 + 0.0039) = 0.9878
or 98.78%
We collect a fraction of 0.3164 or 31.64%
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.
Craig system n= 200 transfers.
Da of 2.0 and Db of 4.0
pa of 0.666 pb of 0.800.
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 n>24)
• Fr,n = (2p)-0.5(npq)-0.5 exp-[((np-r)^2)/2npq]
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