Activity 7: : Key Read: Kenkel (p 299 - 389) and Harris, (Chip 22 - 25)

____ / ____ Score Name (last)______(first)______

For all questions, show your complete analysis and provide an explanation or justification.

Basic Chromatography 1. Consider a mixture of compound A, a somewhat polar liquid, and compound B, a somewhat nonpolar liquid. Tell which liquid, A and B,

would emerge from a chromatography column first under the following conditions and why:

a) A polar liquid mobile phase and a nonpolar liquid stationary phase

Compound A would emerge first because polar mixture components tend to dissolve more in the polar mobile phase and thus will

come through the column with the mobile phase and emerge first. B, being nonpolar, will tend to remain behind in the stationary

phase.

b) A nonpolar liquid mobile phase and a polar liquid stationary phase

B, since it is nonpolar it will emerge first since it will tend to dissolve more in the nonpolar mobile phase. This situation is opposite

to condition of part a.

2. The retention factor (or capacity factor), K of a compound is defined as K = ms/mM, that is K is the ratio of the masses of the

compound in equilibrium in the two phases. Show, from the information given in the corresponding chromatogram, that the

expression used k = (tR - tM) / tM is equivalent to this. Remember that for a given compound the relation between the retention

time tR, the time spent in the mobile phase tM (hold-up or dead time) and the time spent in the stationary phase ts, is as follow: tR

= tM + ts

All the mass flows through the column and is eluted at about tr so tr • u mT. The only mass that is actually moving at any

point in time is that in the mobile phase. Therefore, all the mass in the mobile phase (if it isn’t exchanging) will elute at tm, so tm• u mM. Mass in the stationary phase in a 2-phase system is the rest of the mass, so ms = mT – mm.

m (t ⋅ u ) - (t ⋅ u ) t - t m = m - m ∝ (t ⋅ u) - (t ⋅ u) so K ' = s = r M = r M s T M r m m t ⋅ u t M M M

Using the chromatogram example: Recall that Wi = 1.698 FWHM = 4σ 5.3 (4 ⋅ 3.6) m peak height t - t 360 - 60 360 - 60 K ' = s = ≅ 5.3 and using retention times, K ' = r M = = = 5 m 2.4 (4 ⋅1.5) t 60 60 M peak height M A

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3. The efficiency of a column is measured by a parameter called plate height (H, mm), which is related to the gas

flow rate (u = mL/min) by the van Deemter equation: H = A + B/u + Cu, where A, B and C are constant. Prepare a spreadsheet

with a graph showing values of H as a function of u for u = 4, 6, 8, 10 20, 30,40,50, 60, 70, 80, 90 and 100mL / min. Use the

values, A = 1.65 min, B = 25.8, and C = 0.0236.

i) From your plot, what is the optimum flow rate for this column.

ii) What is the optimum pate height (H , mm) for this column.

iii) What is the units for the constants B and C?

4. Solvent passes through a column in 3.0 min, but solute requires 9.0 min.

i) Calculate the capacity factor, k'. 2.0

ii) What fraction of time is the solute in the mobile phase in the column? 0.33

iii) The volume of stationary phase is one-tenth of the volume of the mobile phase in the column (Vs = 0.10 Vm).

Find the partition coefficient, K, for this system. 20.0

Gas Chromatography

3 5 An injection of 3.00 microL of an methylene chloride, CH2Cl2 (density = 1.327 g/mL) gave a peak size of 3.74 cm . The injection of

3.0 microL of an unknown sample (density = 1.174 g/mL) gave a methylene chloride peak size of 1.02 cm3. Calculate a response

factor for methylene chloride and the percent of methylene chloride in the sample. Response Factor = Peak size/quantity

of pure sample injected = 9.74 cm3/[(3•10-3 cc * (1.327 g/cc) * (1000mg/g)] = 0.939 cm3/mg

3 3 Quantity of Analyte = Peak size / response factor = 1.02cm / 0.939 cm /mg = 1.0857 mg CH2Cl2

-3 % of Analyte = Quantity of Analyte / Total quantity injected = 1.0857 mg CH2Cl2 / [(3•10 cc * (1.174 g/cc) * (1000mg/g)] = 30.84%

Response factor = 0.939 cm3/mg, % methylene chloride = 31%

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6. A chromatogram reveals a resolution factor of 1.5 between two neighboring peaks, 1 and 2. If K2 = 5 and a = 1.05, and knowing that

the retention time of the compound 2 is 5 min, then:

a) Calculate the hold-up time of this chromatogram.

b) What is the width of the second peak at half-height, if the scale of the chromatogram is such that 1 min corresponds to 1cm? Use one of the following equations-

tr(2) - tR(1) 1 α - 1 k2 N α - 1 k2 -k1 R = 1.177 ; R = N2 ⋅ ⋅ ; R = ⋅ ⋅ δ1 + δ2 4 α 1 + k2 2 α k1 + k2 + 2

t = 5 min K = 5 t = hold-up time € r 2 m

t - t 5 - t K = r(2) m = 5 = m → t = 5/6 min = 0.833 min 2 t t m m m 1 α - 1 k 1 1.05 - 1 5 R = N ⋅ ⋅ 2 ; 1.5 = N ⋅ ⋅ 4 α 1 + k 4 1.05 1 + 5 2

5.55 t2 N = 151.2 → N = 22861 N = r → w = 5.55 t2 /N = 5.55 (5)2/22861 = 0.078 cm w 1/2 r 1/2

7. What is the optimal flow rate for the. best separation of solutes?

Minimum plate height is at 33 ml/min

The van Deemter eqn contains 3 terms describing three band broadening mechanism.

i) Which term is zero (0) for an open tubular column? Explain.

Multile path term (A) is zero for open tubular column.

ii) What is the physical meaning of the A, B and C term.

A- Mmultiple path term

B- Longitudianal diffusion C- Equilibrium time

iii) For the graph shown, estimate the values of A, B and C.

Note that uo is the flow rate (mL/min) in the equation shown.

A = 1.8 mm B: 1.75 = B / Uo = B / 20mL min-1 : B = 35 mL mm min-1 C: 1.8 = C Uo = C • 70mL min-1 : C = 0.0257 mL-1 mm min

A = 1.65 min

B = 25.8 mn•mL/min

C = 0.0236 mm•min/mL

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High Performance Liquid Chromatography

8 What type of HPLC should be chosen for each of the following separation application?

a) All mixture components have MW less than 2000, are molecular and polar, and are soluble in nonpolar organic solvents.

Normal and reverse phase HPLC

b) Mixture components have formula weights varying from very large to rather small and are nonionic.

Size exclusion HPLC

c) Mixture components have formula weight less than 2000, are molecular and polar, and are water soluble.

Normal and reverse phase HPLC

9. What is the order of of the following acids from HPLC column whose stationary phase is of the type C18 (Bonded Reverse-

Phase Chroatography) while the mobile phase is a formate buffer with a concentration of = 200 nM at a pH = 9.

Mixture

1. Linoleic acid: CH3(CH2)4CH=CHCH2CH=CH(CH2)7CO2H

2. Arachidic acid: CH3(CH2)18CO2H

3. Oleic acid: CH3(CH2)7CH=CH(CH2)7CO2H

First: pKa = 7.05 Linoleic acid: CH3(CH2)4CH=CHCH2CH=CH(CH2)7CO2H

Second: pka = 7.64 Oleic acid: CH3(CH2)7CH=CH(CH2)7CO2H

Third: pka = 9.29 Arachidic acid: CH3(CH2)18CO2H

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