CHEM340 Tutorial 4: Chromatography

Transcription

1 CHEM340 Tutorial 4: Chromatography 1. The data in the table below was obtained from a chromatogram obtained with a 10 cm liquid chromatography column. Under the conditions used, the compound uracil is not retained at all by the stationary phase Calculate: Component A Component B t M (min) t R (min) w ½ (min) a) the resolution (R) between the two peaks, b) the capacity factor (k) for component B. c) the selectivity coefficient ( ) for components A and B, d) the plate number (N) for the column according to component B, e) the plate height (H) for the column, f) the retention time for uracil. 2. In chromatography, the resolution between two adjacent peaks can be described by: R N α 1 k 2 4 α 1 k 2 N (plate number) is a measure of the column efficiency, (the selectivity coefficient) depends on the chemical structure of the substances involved and k 2 (the capacity factor for the second of the two peaks) depends largely on the temperature of the column (for GC) or the composition of the mobile phase (for HPLC). Resolution is usually sufficient when R = 1.5. A 15 m GC column with a plate number (N) of gives a resolution of 0.75 between two peaks, with the second peak eluting after 15 min. Assuming that the plate height does not vary with column length and that and k 2 remain constant: a) What length of column would be required to give a resolution of 1.5? b) What would be the retention time of the second peak? The resolution between two peaks is 1.1, the value of k 2 is 2.0 and t M is 1.5 min. a) What value of k 2 would be required to increase the resolution to 1.5? b) What would be the retention time of the second of the two peaks? For each of the following factors, state with a reason whether it affects the value of k. a) Column length b) Phase ratio (β) (in GC) c) Mobile phase velocity

2 3. In gas chromatography with an open tubular column, there are several experimental variables which can have an effect on the resultant chromatogram: a) How will increasing the column length affect: ii. Peak area iii. Plate height iv. Peak height b) How will increasing the stationary phase film thickness affect: ii. Capacity factor iii. Plate height iv. Peak area c) How will increasing the mobile phase velocity affect: ii. Capacity factor iii. Plate height d) How will increasing the analyte concentration affect: ii. Plate height iii. Peak width iv. Peak area 4. For each of the following applications, choose a technique (GC) or liquid chromatography (LC) and a detector (from the list above). In each case, give a brief (one sentence) reason. a) determination of alcohols and aldehydes in Scotch whisky b) determination of sucrose and glucose in a fermentation broth c) determination of large molecular weight polycyclic aromatic hydrocarbons d) determination of anions in boiler feed water e) determination of very low concentrations of semi-volatile chlorinated pesticides in water Page 2 of 6

3 plate height (mm) 5. The two chromatograms below were obtained using a non-polar HPLC stationary phase and an acetonitrile/water mobile phase for the separation of three compounds in a mixture Chromatogram A Chromatogram B One of the chromatograms was obtained using 12% acetonitrile in water and the other was obtained using a 6% acetonitrile in water mobile phase. The resolution in Chromatogram B is much better than it is in Chromatogram A there is almost no separation evident in Chromatogram A and all three substances are co-eluting. a) For each of the following changes, state with reasons whether it could have led to the increase in resolution seen in going from Chromatogram A to Chromatogram B: i. Increase in plate number (N). ii. Increase in selectivity factor ( ) iii. Increase in capacity factor (k). b) Which mobile phase is stronger (i.e. leads to shorter retention times) the 12% acetonitrile or the 6% acetonitrile? Explain briefly. c) Which mobile phase was used to obtain Chromatogram B? 6. The graph below shows two plots of plate height against mobile phase velocity (commonly known as van Deemter curves). They were both obtained using the same open tubular column with two different carrier gases (He and N 2 ). van Deemter curves for open tubular GC column using He and N 2 as carrier gas B A mobile phase velocity (cm/s) a) For curve B, what is the optimum mobile phase velocity? b) Consider curve B: in what range of mobile phase velocities is resistance to mass transfer (i.e. the C term ) the major contributor to band broadening, as measured by the plate height (H)? Page 3 of 6

4 c) In the region of the van Deemter curve where resistance to mass transfer dominates (i.e. where eddy diffusion and longitudinal diffusion can be ignored) would a gas with large diffusion coefficients give more or less band broadening than a gas with small diffusion coefficients? Explain briefly. d) Denser gases have lower diffusion coefficients. Which of the curves shown above was obtained using He as carrier gas? e) Curve B has a lower plate height at the optimum mobile phase velocity than does curve A. However, for most applications the gas for curve A would be preferred and the velocity would be set to about 30 or 40 cm/s. What advantage would there be for using a mobile phase velocity above the optimum? What would be the disadvantage? What advantage is there to having a less steep van Deemter curve? 7. The following two chromatograms were obtained using two different stationary phases; column a) had a methyl(95%) phenyl(5%) siloxane stationary phase (ZB-5) which is a non-polar phase, whereas column b) had a poly(ethylene glycol) Carbowax stationary phase (ZB-WAX) which is a polar phase. a) What effect would you expect changing the polarity of the stationary phase to have on: i) K ii) iii) iv) Peak area Plate height α b) The hydrocarbon mixture used for chromatograms a) and b) consists of two groups of compounds; alkanes and aromatics. Which group is most affected by the change to a more polar stationary phase? Why? c) Using the more polar stationary a) ZB-5 column 30 m x 0.32 mm x 0.25 µm 1= pentane, 2 = solvent, 3 = benzene, 4 = heptane,, 5 = toluene, 6 = ethylbenzene,7 = m-xylene, 8 = p-xylene, 9 = o-xylene, 10 = decane, 11 = dodecane b) ZB-WAX column 30 m x 0.32 mm x 0.50 µm 1= pentane, 2 = heptane, 3 = solvent, 4 = benzene, 5 = decane, 6 = toluene, 7 = ethylbenzene, 8 = p-xylene, 9 = m-xylene, 10 = dodecane, 11 = o-xylene phase has improved the resolution of the xylene isomers. What accounts for the improved resolution a change in N, α or k? Explain briefly. Page 4 of 6

5 d) The boiling points of some of the compounds are: compound pentane heptane toluene m-xylene p-xylene decane dodecane b. p 35 o C 98 o C 110 o C 138 o C 138 o C 174 o C 216 o C i. Look at the order of the boiling points and at the elution order in the chromatograms. ii. What property of the compounds determines the elution order on the non-polar column and why are m-xylene and p-xylene not separated? iii. What causes the elution order to change when the mixture is separated on the (polar) ZB- WAX column? 8. The following chromatogram shows the separation of some triazine herbicides by reversed phase HPLC (i.e. a non-polar stationary phase is used with a more polar mobile phase). A C- 18 column (15 cm x 4.6 mm) was used with a mixture of acetonitrile (33%) and 0.1 M sodium acetate (67%) as the mobile phase. Separation of herbicides by reversed phase HPLC 1 = simazine, 2 = atrazine, 3 = prometon, 4 = ametryn, 5 = propazine, 6 = prometryn, 7 = terbutryn Varian Chrompack Application note 1201-HPLC a) Which compounds will be more strongly retained on a reversed phase HPLC column polar compounds or non-polar compounds? b) Based on the chromatogram, which is the less polar compound, terbutryn or simazine? Justify your answer. c) The peaks for prometryn (6) and terbutryn (7) are much broader than those for simazine (1) and atrazine (2). Explain briefly why this is so. d) To make the prometryn and terbutryn peaks sharper and elute earlier, the mobile phase could be made stronger for reversed phase HPLC making the mobile phase stronger means Page 5 of 6

6 making it less polar so that it can better dissolve the non-polar analytes. To make the mobile phase stronger, would you increase or decrease the percentage of acetonitrile. Why? e) By changing the percentage of acetonitrile in the mobile phase are you changing the selectivity factor, α? (Hint: are you changing the chemical structure of the mobile phase?) f) By changing the percentage of acetonitrile in the mobile phase are you changing the capacity factor, k? (Hint: are you changing the average time an analyte spends in the stationary phase?) g) If you make the mobile phase stronger, what would happen to the separation between ametryn and propazine? How could you overcome that problem? h) These triazine herbicides can also be separated by normal phase HPLC in this case the mobile phase is less polar than the stationary phase. For example, a silica stationary phase could be used with a hexane/ethanol mobile phase. In normal phase chromatography, in which order would simazine and terbutryn elute? Explain briefly. i) In normal phase HPLC with a hexane/ethanol mobile phase in order to increase retention times and increase resolution would you increase or decrease the percentage of hexane in the mobile phase? Explain briefly.. Page 6 of 6