2. a) R N and L N so R L or L R 2.

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1 1. Use the formulae on the Some Key Equations and Definitions for Chromatography sheet. a) 0.74 (remember that w b = 1.70 x w ½ ) b) 5 c) (α always refers to two adjacent peaks) d) 1.0x10 3 e) 0.1 mm f) 1.53 min (for a substance that is not retained by the stationary phase, t S = 0 so t R = t M ). a) R N and L N so R L or L R. So, if we want to increase R by a factor of (from 0.75 to 1.5) we must increase the length of the column by a factor of. So the required column length is 4 x 15 = 60 m. ii) The retention time is directly proportional to the column length so the retention time with the 60 m column will be four times greater than with the 15 m column; ie 60 min. (Improving the resolution comes at the cost of increased analysis time.) k i) R or, put another way 1 k so k 1 k k and k and the required k = 10.0 R R 1 k 1 k k 1 k 1 ii) We know that t R = t M + t S and that t S = t' R = k x t M t M remains unchanged (1.5 min), so with k = 10, we get t' R = 15 min So, the new retention time is 16.5 min (ie t' R + t M ) (Again, improved resolution takes a longer time.) i) Look at the equation for k (= K/). Do K (the distribution constant) or (the phase ratio) depend on the length of the column? So, k does not depend on the length of the column.

2 ii) iii) The phase ratio (β) is the ratio of mobile phase volume to stationary phase volume. If there is more stationary phase, there will be more retention and k will increase. Do K (the distribution constant) or (the phase ratio) depend on the mobile phase velocity? No. So does k depend on the velocity? No 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: i. Retention time If the column is longer, then everything will take more time to get through it and retention times will increase. ii. Peak area The peak area depends on how much of the analyte gets to the detector (which depends on the analyte concentration in the sample) and this does not depend on the length of the column so peak are will not change. iii. Plate height Plate height is a measure of the column efficiency and is influenced by the processes contributing to band broadening. Look at the rate theory of chromatography: none of the factors contributing to band broadening mention the column length so H is not affected by column length. iv. Peak height If the peak area is not affected by the column length, can the peak height be affected? Yes, if the width of the peak changes for a given area, a wider peak will be a shorter peak. If the column length is increased, then the plate number (N) increases and so does the retention time (t R ). Look at the relationship between N and t R and you will see that w ½ must increase as column length increases so peak height will decrease. b) How will increasing the stationary phase film thickness affect: i. Retention time Increasing the film thickness will increase the volume of stationary phase in the column, which will retain the analyte more so retention time will increase. ii. Capacity factor Increasing the film thickness will increase the volume of stationary phase in the column, which will decrease the phase ratio (β) and so increase the capacity factor so capacity factor will increase. iii. Plate height

3 Look at the rate theory and the contributions to band broadening. The film thickness has an effect on the resistance to mass transfer with thicker films causing slower mass transfer so plate height will increase. iv. Peak area The peak area will not be affected. c) How will increasing the mobile phase velocity affect: i. Retention time The retention time will decrease. ii. Capacity factor Both t R ' and t M will increase the capacity factor will not change. iii. Plate height Trick question? Look at the van Deemter curve and you will see that if the mobile phase velocity is below the optimum, then increasing the velocity will decrease the plate height but if the mobile phase velocity is at or above the optimum, then increasing the velocity will decrease plate height. d) How will increasing the analyte concentration affect: i. Retention time No change (remember what you saw from the results of the GC and practicals that you did.) ii. Plate height No change iii. Peak width No change iv. Peak area Increasing the analyte concentration will increase the peak area. 4. Detector answers a) determination of alcohols and aldehydes in Scotch whisky GC alcohols and aldehydes are volatile so they can enter the gas mobile phase in GC. FID alcohols and aldehydes are organic compounds. b) determination of sucrose and glucose in a fermentation broth HPLC sucrose and glucose are not volatile (GC can only be used for substances that RID sucrose and glucose do not have any conjugated double bonds and they are not rigid compounds nor are they ionic so that leaves refractive index. c) determination of large molecular weight polycyclic aromatic hydrocarbons Polycyclic aromatic hydrocarbons (PAHs) consist of two or more aromatic rings stuck together. Since aromatic rings have a conjugated C=C system and are not flexible (can

4 you twist a double bond?) PAHs can be detected by fluorescence which means that HPLC would be suitable. Aromatic hydrocarbons are quite stable and can be evaporated without decomposing them, so GC could also be used with FID. d) determination of anions in boiler feed water HPLC using conductivity detector. e) determination of very low concentrations of semi-volatile chlorinated pesticides in water. Semi-volatile should make you think of GC. Then you can use the sensitive ECD (chlorine is electronegative, so the ECD works well for chlorinated compounds). 5. Chromatograms 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). There are two ways to increase N either increase L or decrease H. Has the column length been increased? No. What about decreasing H? Will a decrease in H (ie a decrease in peak width) change the retention times? Can changing the mobile phase composition (ie from 6% acetonitrile to 1 % acetonitrile) change H (look at the rate theory to see what factors affect H). ii. Increase in selectivity factor () The selectivity factor changes if there is a chemical change in the structure of the stationary phase or (for liquid chromatography) the mobile phase. Changing the % acetonitrile in the mobile phase does not change the chemical structure. iii. Increase in capacity factor (k). Acetonitrile is less polar than water so changing the % acetonitrile in the mobile phase changes the polarity of the mobile phase. In chromatography, the analyte is partitioned between the mobile and stationary phases the distribution constant (K) describes this partition. Changing the polarity of the mobile phase will change the value of K it will change the solubility of the analyte in the mobile phase. Changing K will change k, which will change t R and R. b) Which mobile phase is stronger (ie leads to shorter retention times) the 1% acetonitrile or the 6% acetonitrile? Explain briefly. The stationary phase is non-polar (read the bit that comes before the chromatograms). So, less-polar compounds will be the difficult ones to remove from

5 the stationary phase into the mobile phase. So a less-polar solvent will be stronger than a more-polar solvent. The 1% acetonitrile is less polar than the 6% acetonitrile so the 1% acetonitrile mobile phase is the stronger one and will lead to shorter retention times because it is better at getting solutes out of the stationary phase. c) Which mobile phase was used to obtain Chromatogram B? 6% acetonitrile 6. Van Deemter a) For curve B, what is the optimum mobile phase velocity? About 5 cm/s. This is the velocity that gives the best (ie lowest) value of H. (Why is a small value of H good?) b) Consider curve B: in what range of mobile phase velocities is resistance to mass transfer (ie the C term ) the major contributor to band broadening, as measured by the plate height (H)? For the C term H Cu (where u is the mobile phase velocity). So, the range of mobile phase velocities where resistance to mass transfer is the major contributor to band broadening is the range where H vs u has a positive slope. In this case the range is from about 5 cm/s upwards. c) In the region of the van Deemter curve where resistance to mass transfer dominates (ie 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. Consider the rate theory: for the C term, is it better to have fast diffusion or slow diffusion? We want less resistance to mass transfer to give us lower H and less badn broadening so we want fast diffusion and we want to use a gas with large diffusion coefficients. d) Denser gases have lower diffusion coefficients. Which of the curves shown above was obtained using He as carrier gas? Is He less dense or more dense than N? (Have you ever seen a He-filled balloon sink to the ground or does it rise up through the atmosphere?) So, if He is less dense than N will it give smaller or larger diffusion coefficients? (ie is it easier for gas molecules to move through He than through N?) So, will He give lower H? So, curve e) If you look closely, you will see that 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

6 the optimum? What would be the disadvantage? What advantage is there to having a less steep van Deemter curve? Does it matter how long it takes for the analysis? ( Time is money ) What would be the disadvantage of working above the optimum velocity? Is it good for H to increase? Larger H means smaller N and decreased resolution which is not good. What advantage is there to having a less steep van Deemter curve? You can get quicker analysis without a big increase in H. If you look at Curve A (the less steep one) at a mobile phase velocity of 30 cm/s (ie twice the optimum) H is only very slightly above the minimum value so you can get fast analysis without losing much resolution. 7. Stationary phase; a) What effect would you expect changing the polarity of the stationary phase to have on: i. distribution constant (K) It would change the value of K depends on the chemical structure of the stationary phase. ii. peak area No change the peak area depends on the concentration of the analyte. iii. plate height No change do any of the three contributions to band broadening depend on the chemical structure of the stationary phase? iv. selectivity coefficient () Change the value of depends on the chemical strucuture of the stationary phase and, if the polarity of the stationary phase has changed, that means that the chemical structure has changed. 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? Aromatics alkanes are non-polar and the only intermolecular forces they are involved in are the non-specific dispersion forces (London forces). On the other hand, the aromatics have the delocalized -electrons, which can give the molecules a very slight polarity. c) Using the more polar stationary phase has improved the resolution of the xylene isomers. What accounts for the improved resolution a change in N, α or k? Explain briefly. It must be α since a change in chemical structure is involved. d) The boiling points of some of the compounds are:

7 compound pentane heptane toluene m-xylene p-xylene decane dodecane b. point 35 o C 98 o C 110 o C 138 o C 138 o C 174 o C 16 o C Look at the order of the boiling points and at the elution order in the chromatograms. What property of the compounds determines the elution order on the non-polar column and why are m-xylene and p-xylene not separated? The boiling point order is the same as the elution order. It should be obvious why the xylene isomers are not separated (look at their boiling points). e) What causes the elution order to change when the mixture is separated on the (polar) ZB-WAX column? With the more polar column, the aromatics can interact with the stationary phase in a way in which the alkanes cannot. (Why? what do the aromatics have that the alkanes do not?) 8. HPLC: a) Which compounds will be more strongly retained on a reversed phase HPLC column polar compounds or non-polar compounds? In reversed phase, the stationary phase is non-polar so non-polar compounds will be more strongly retained. b) Based on the chromatogram in figure 3, which is the less polar compound, terbutryn or simazine? Justify your answer. Terbutryn it is retained longer by the non-polar stationary phase. c) The peaks for prometryn (6) and terbutryn (7) are much broader than those for simazine (1) and atrazine (). Explain briefly why this is so. Under isocratic conditions (ie not changing the proportions of the solvents in the mobile phase) peak width increases with retention time that s what the plate theory tells us (consider the formula used for calculating the plate number from t R and w ½ ). 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 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? Which is the less polar solvent acetonitrile or water? So increase the % acetonitrile (See the answer to Question b) 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?)

8 No, the selectivity factor is not changing. For α to change, you would need to use a chemically different mobile phase eg use a methanol/water mixture instead of acetonitrile/water. 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?) Yes, k is changing the stronger solvent decreases the time that the analyte spends in the stationary phase and k is decreased. g) If you make the mobile phase stronger, what would happen to the separation between ametryn and propazine? How could you overcome that problem? Will the decreased k increase or decrease the resolution between the ametryn and propazine peaks? To overcome the problem, you need to start with a weaker mobile phase (more water and higher values of k) and then increase the strength of the mobile phase (increase acetonitrile and decrease k). In other words, use gradient elution. 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. In normal phase, would you expect the more polar compounds to be retained more or less than the non-polar compounds? Which type of compounds will have more affinity for the polar stationary phase polar or non-polar? 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. To increase retention times, you want to push the analytes out of the mobile phase by making it less like the stationary phase. In normal phase HPLC this means making the mobile phase less polar so increase the percentage of hexane in the mobile phase.

CHEM340 Tutorial 4: Chromatography

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