What is a chemical separation? Examples: 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
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Mass SpectroscopyChemical Separations
Collection of a pure product
Isolation for subsequent analysis for either quantification or identification
What is it?Chemical Separations
Distilled SpiritsChemical Separations
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 mixturesChemical Separations
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? compounds will boil away and can be collected.
What is brandy?Distillation
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. 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.
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.
Liquid – Solid 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.
Liquid - LiquidExtraction
We want to partition a solute between two immiscible phases. 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.
Oil and vinegar
Ethyl ether and water
Octanol and water.
The two solvent should have low miscibility
Have different densities – avoid emulsion formation.Solvent Extraction (Liquid-Liquid)
Replace concentration with moles over volume and http://oceanexplorer.noaa.gov/explorations/02sab/background/products/media/chroma1.htmllet q equal the fraction in the aqueous phase.m will be the moles of solute in the entire system
p is the term for the fraction in the organichttp://oceanexplorer.noaa.gov/explorations/02sab/background/products/media/chroma1.html
p + q = 1
GivingIf it does not end up in the aqueous phase it must be in the organic phase.
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.Sample Problem
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.Example
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???What if our compound can dissociate or participate in some other equilibrium?
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?Dissociation
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:
B) 0.1 N NaOH (aq)
C) 0.1 N HCl (aq)
D) I never wanted to work here anyhow.So What, Why is this useful.
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.Separation
System I another. What if we are try to separate two compounds, A and B
Da = 32 Db = 0.032 (A ratio of 1000)
Let's recall our equations
q (fraction in aqueous) = 1 / (DVr + 1)
p (fraction in organic) = DVr / (DVr + 1)
Vr (volume ratio) = Vo / VaExample
p another. What if we are try to separate two compounds, A and Ba = 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 I
D another. What if we are try to separate two compounds, A and Ba = 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.
pb = 1*1 / ( 1*1 +1) = 1/2 = 0.5
What do we get for purity of compound a now?
purity = 0.999 / (0.999 + 0.50) = 0.666
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?????
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)-0.5How can we get around this issue?
So let us look at our two cases and see which will give us the optimum values.
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.Revisit the two cases
Case II the optimum values.
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.Case II Revisited
What is our purity for this system? the optimum values.
pa = 1000*0.032 / (1000*0.032 + 1) = 32/33 = 0.97
pb = 1*0.032 / (1*0.032 + 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.Purity for Case II
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.Can we improve this purity?
Let's look at the numbers. phase.
Da = 32 Db = 0.032 Vr = 1
pa = 0.97 pb = 0.03
qa = 0.03 qb = 0.97
Let’s prepare a table.Back Extraction Case I Example
Before Shaking phase.
Fresh Aqueous Phase
0Initial conditionsprior to starting back extraction
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
(0.03)(0.97)Now we extract – shake shake shake
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)Now what is the purity for A in the organic phase???
After second Back Extraction phase.
0.0009*0.97Let’s 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.
Still a viable option for preparative work.
For separations it has been replaced by HPLC
Called Craig Counter Current Extraction.
Special glassware is used.Can We Expand This?Why Would We Want to?
Starting Conditions phase.
After One Equilibrium
Transfer Step 1
Now we do Transfer 2
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.
We transfer again.
Transfer Step 3
Shake Again Equilibrium 4 equilibrium.
See a trend????
How about a binomial expansion? equilibrium.
(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 1Craig CCE
Or the formula equilibrium.
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. equilibrium.
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!Craig CCE
What if we collect the last two tubes?? equilibrium.
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 CCE
r equilibrium.max = 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(1)
w = 4s = 4(npq)0.5
Resolution would be
R = Drmax/w = Drmax/4s
R = nDp/(4(npq)0.5) = n0.5Dp / 4(pq)0.5Final Formulas(2)