Homework #5 (26-2, 26-6, 26-7, 26-11, 27-8, 27-19, 27-21, 27-22, 27-25, 28-4, 28-13, 28-15, 28-16, 28-17, 28-18) 26-2. Describe the general elution problem. The general elution problem is the problem that arises when attempting to separate a group of molecules which have very different retention factors. The problem is that no single isocratic or isothermal separation will satisfactorily separate all of the molecules. Conditions appropriate to weakly retained molecules (those with small retention factors) will be inappropriate for the strongly retained molecules (those with large retention factors), and the same is true in reverse. The solution is to use gradient elution or temperature programmed conditions. 26-6. What variables are likely to affect the α value for a pair of analytes? The selectivity factor is equal to the ratio of the two retention factors which will also be equal to the ratio of the distribution coefficients. Anything that changes this ratio will change the selectivity coefficient. Temperature will affect the value of the equilibrium constants (freshman chem), but it will likely only weakly affect the ratio, depending upon the enthalpies. Changing the composition of the stationary phase will affect the distribution coefficients, and in LC changing composition of the mobile phase will also have an impact on the equilibrium constant. 26-7. How can the retention factor for an solute be manipulated? The retention factor can be manipulated by any of the above factors that affect the value of the equilibrium constant (temperature, composition of stationary phase, composition of mobile phase (in LC)), but you can also change the ratio of the volume of the stationary phase to the mobile phase. This can be accomplished any number of ways, for example by changing the thickness of the stationary phase on the packing material or by changing the diameter of the capillary. 26-11. What is gradient elution? Gradient elution is the process of varying the composition of the mobile phase during the separation. 27-8. Describe the principle upon which each of the detectors listed in Problem 27-7 is based. (a) TCD, (b) atomic emission, (c) thermionic, (d) electron capture, (e) flame photometry, (f) photoionization. (a) TCD operates on the principle of differences in thermal conductivity. Gases eluting from the column are passed over a hot element (the temperature of which is closely monitored). As the thermal conductivity of tha gases passing the element change, so does 5mL .695 As a result the corrected retention volumes are V0air = 0.19.1+273.98min 31.the heat transport from the element.695 9.695 131.2)/(21.6mL The corrected retention volumes include an additional pressure term j given by equation 27-5 j = (3/2)((Pi/P)2-1)/((Pi/P)3-1). 21. (e) the retention time and corrected retention volume for methylhexanoate.98min. as the analyte is diluted in the mobile phase. packed with Chromosorb P.0C retention times: air.0mL = 91.1C.93min 31.16min.4mL/min = 9. 1.44mL Vma = 1. a) the average flow rate through the column in mL/min is determined from equation 27-3 on page 702 F = Fm x Tc/T x (P-PH2O)/P F= (25. 4. 748 torr measured outlet flow: 25. (d) the distribution coefficients for the three esters. How will K = cS/cM vary between the two columns? A polar compound will interact attractively with the polar stationary phase more than it will interact with the nonpolar stationary phase.3mL V0mp = 0. column. This detection method is based on the average thermal conductivity and so will not be very sensitive.1psi above room. room. As a result. and its temperature changes.4mL/min = 249.2))((748-18)/748) = 31.10m x 2.3mL = 43.4mL/min b) the retention volumes for air and the three esters are given by equations 27-1 and 2.80 and thus j = 1.0mm.1psi(760torr/14. peak widths at base: 0.5(1.56mL V0ma = 0.695 62. 0. 1. respectively Calculate: (a) the average flow rate through the column.4mL/min = 131. methyl acetate.79 min. the Cs for the SE-30 will be less than the Cs for the carbowax 20M.4mL/min = 62.695 249. where Pi is the inlet pressure Pi = 748torr+ 26. (b) the corrected retention volumes for air and the three esters.02g/mL pressures: inlet 26.93min.803-1) = 0.802-1)/(1. (c) the specific retention volumes for the three components. 18. and therefore K20M > KSE-30.16min 31. 27-21.6mL = 173.0mL V0mb = 0.40g. methyl n-butyrate.3mL/min)((102.3 mL/min temperature: room. The same polar compound is gas chromatographed on an SE-30 (very non-polar) column and then on a carbowax 20M (very polar) column. A GLC column was operated under the following conditions: column: 1.44mL = 6.0mL Vmb = 7. thus by simple multiplication of the retention times by the average flow rate. weight of stationary liquid added. methyl n-propionate.3mL Vmp = 4. density of liquid 1. 7.0+273. 0.0s/60s)min) 31.7psi) = 1349torr and so Pi/P = 1349/748 = 1. (b) atomic emission operates by creating a plasma 27-19. Vair = ((18.39.0s. 102. 39 .2 = 1.9 mL/g= 61.1mL/g Vg-mp = (273.329 Using the slope is 0. By using equation 26-8 on page 680.2/(102. From the data in problem 27-21.357 (from excel) you can obtain the last line of values.0+273. (b) α values for each adjacent pair of compounds. the corrected retention volume will be V0mh = jFtr-mh = (0.40g = 19. With a retention time of 2150s.8 249.5 (e) these three esters represent a homologous series.02g/mL) (102+273.9767 1738 1820 1612 1723 7. (a) The k’ can be readily determined from the retention times.40Vg = 1. First make a table of the log corrected retention times and the carbon number and then determine the slope and intercept. k’ = tr-tm/tm values from this calculation are shown in the attached table.40 g/mL 86. the average number of theoretical plates and plate height for the column.3mL-6.16 7. (d) the resolution for each adjacent pair of compounds. and as such you would expect that the log of the corrected retention times would depend upon the carbon number in a linear way. The selectivity coefficient in the table represents the selectivity between that compound and the previous compound in the table.5 Kmb = 1.39 methylbutyrate 0.2)) (91.365 2.4 47.40Vg and so for the esters Kma = 1.6 457.6 475.40Vg = 1.2976 458 25.56mL)/1. These values were computed using the equation 26-17 on page 683 N = 16 (tr/Wb)2.2 = Vg (1.5mL-6. (c) The number of theoretical plates is measured for each of the peaks and is shown in the table along with the average value.56mL)/1. N 2 3 4 6 tr(s) 118.19 methylpropionate0.2/(102.87 2.8mL/g d) the distribution coefficient depends upon the specific retention volume and the density of the stationary phase and the temperature according to equation 27-7 on page 702 K = Vg ρs Tc/273.2)) (43.98 4.661 3.40g = 43.93 tr sec 18 119 250 476 tr' sec k' ! N Rs 0 0 101 5.52 6.3 1.2)/273.0mL-6.4 tr min 0.40g = 86.43 1.8 231.0+273. Values for these are shown in the attached table below. (d) The resolution for each adjacent pair of compounds was determined by using equation Wb min water methylacetate 0.9mL/g Vg-mb = (273.2/(102.8 2132 Log(tr’) 2.8 mL/g= 121.40 g/mL 43.7 Kmp = 1.003 2.0+273.4 23.c) the specific retention volumes are calculated via equation 27-6 on page 702 Vg = (273.79 average Wb sec 11.6 232 12. (c) the average number of theoretical plates and plate height for the column.695) (31. calculate: (a) k’ for each compound.4mL/min) (2150s/60s) min = 782mL 27-22.8 2150 tr’(s) 100.40g/mL 19.2)) (173.1 mL/g= 26.2/Tcolumn)(V0R-V0m)/W and so for air this will be zero Vg-ma = (273.56mL)/1.3286 and the intercept is 1.40Vg = 1. (b) The α values for each adjacent pair of compounds can be readily determined by simply plugging the retention times into equation 26-10 on page 680 a = k’A/k’B. (d) increasing the flow rate.second equation on page 689. This decreases the mobile phase mass transfer contribution to the plate height. (d) Increasing the flow rate. and thus makes a tighter more confined plug of sample at the start of the column. you should note. The total . So increasing the injector temperature will improve efficiency and decrease the plate height. 27-25. Increasing the injection port temperature (assuming no degradation of the analytes or solvent) will enhance evaporation. (f) Decreasing the column temperature. (e) reducing the particle size of the packing. (b) Decreasing the rate of sample injection. In my calculation I used the average of the peak widths as the width. The values next to a compound represent the resolution between that compound and the previous compound on the list. This produces the opposite effect of slow injections. and this will improve efficiency leading to smaller plate heights. (c) Increasing the injection port temperature. However. and therefore does not represent a modification of the plate height (as described by the vanDeemter equation) it will nonetheless lower the efficiency and increase the calculated plate height. Reducing the packing particle size decreases the distance that the analyte needs to travel in the mobile phase to gain access to the stationary phase. (a) increasing the weight of the stationary phase relative to the packing weight. which will increase the plate height due to increasing the stationary phase component of the mass transfer term. (b) decreasing the rate of sample injection. while this has nothing to do with the column. Decreasing the rate of sample injection will cause all of the peaks to become broadened. (a) Increasing the weight of the stationary phase relative to the packing weight. because it makes the sample vaporization as quick as possible. and this will reduce efficiency by increasing both of the mass transfer components to the plate height. What would be the effect of the following on the plate height of a column? Explain. Increasing the flow rate from zero will first improve decrease the plate height. (c) increasing the injection port temperature. Decreasing the column temperature will decrease the extent of longitudinal diffusion. (e) Reducing the particle size of the packing. (f) decreasing the column temperature. Again. as in the previous case this improvement is unrelated to the column and has nothing to do with the van Deemter equation. This is essentially the same as making the stationary phase film thickness greater. and then above the optimum flow rate (minimum plate height) it will increase the plate height. decreasing diffusion will also reduce the mass transfer of the analyte between the mobile and stationary phases. It is not labeled but it reads as Rs = (tr)B – (tr)A / W. the temp can also be programmed or changed. which is similar to changing the mobile phase composition.288. really obvious in chiral separations. a solute was found to have a retention time of 29. or gradient elution. if it changes one more than the other then there will be a change in the selectivity. Calculate (a) k’ for the solute and a (b) solvent composition that would bring k’ down to a value of about 10. Of course altering the composition of the stationary phase is also a solution for LC. Alternatively you could change the composition of the stationary phase.71) = 2. because something that changes the retention factor MAY affect the ratio of two retention factors.impact of column temperature will depend on the relative sizes of each of these contributions (B smaller with lower temp) and (C gets larger with lower temp) to the plate height. So 28.1+4.05/1.1) = 2. while an unretained sample had a retention time of 1.05 = 26. aka the selectivity factor. To reduce the value to 10 then 10/26. b) In the case of HPLC. Using table 28-2 on page 743 and the equation for normal phase partition chromatography on page 742 (28-3).1x + 4. The initial value for the polarity index was Pi = 0. Suggest a type of liquid chromatography that would be suitable for the separation of (a) and (b) CH3CH2OH and CH3CH2CH2OH (c) Ba2+ and Sr2+ (d) C4H9COOH and C5H11COOH (e) high-molecular-weight glucosides .5Phexane + 0. 28-4.4.5(0.95 = 0. but first we need to determine the initial value for the retention factor k’i = tr-tm / tm = 29. a) In the case of GC the mobile phase composition is not as important as the column temperature.5Pchloroform = 0. via different isocratic runs. This could simply be via a different temp isothermal run or it could involve temperature programming.1 . (b) liquid chromatography? The selectivity factor can be manipulated in much the same way as the retention factor in the previous problem. but more significant impact will likely come from variation of the mobile phase composition.2& chloroform. 28-15. thereby enhancing the retention of one analyte over the other.1-1.8% hexane with 71.95. to get this polarity index 2.1 – 2 log(10/26. is to add an additive to the mobile phase that complexes or interacts with one of the analytes.71 (quite large).How can the selectivity factor be manipulated in (a) gas chromatography. One last possibility.1.1min.05 min when the mobile phase was 50% by volume chloroform and 50% n-hexane. In a normal phase partition column.1x and thus x = 0.71 = 10(Pi-Pf)/2 thus Pf = 2. 28-13. Changing the column temp can change the equilibria for both analytes. 4. Which solvent. high molecular weight glucosides will be best separated with size exclusion chromatography. b) These small polar (essentially non-ionizable) molecules will be well suited to normal phase partition chromatography c) Atomic ions such as these are best separated with ion exchange chromatography. 28-18. the stationary phase is more polar than the mobile phase. elution will depend on the molecular polarizability. benzene and then n-hexanol.a) These extremely nonpolar polyaromatic hydrocarbons are well suited to adsorption chromatography. ethyl acetate and then nitrobutane In normal phase partition chromatography. if you are separating anions with the analytical column. That is. and thus the more polar compounds will be retained longer than the less polar compounds. 28-16. The primary function of the ion suppressor column is to decrease the ionic conductivity of the mobile phase that often contains a high fraction of supporting electrolyte (in the form of H+ or OH-) 28-17. benzene (b) ethyl acetate. which tells you that smaller retention indices and thus smaller retention times will result when the polarity index for the mobile phase is larger.0min when the mobile phase was toluene. What is a suppressor column and why is it employed? The suppressor column is an ion exchange column that is opposite to the kind of ion column used for the analytical separation. . although ion pair would also be a possibility.6 and CHCl3 is 4. On a silica gel column. then the suppressor would be a cation exchanger. predict the order of elution of (a) n-hexane. For a normal phase separation. carbon tetrachloride or chloroform.1. For molecules with similar polarities. CCl4 is 1. d) These short chain ionizable highly polar molecules are well suited for normal phase and possibly ion pair chromatography (in fact these were also described in the text as an example for ion exclusion chromatography) e) Of the techniques introduced so far. diethyl ether and nitrobutane. So to shorten the retention time you should use a solvent that has a higher polarity index and in this set of solvents that is chloroform. would be more likely to shorten the retention time? From table 28-2 on page 743 you can find the polarity indices for these three solvents toluene is 2. n-hexanol. Recalling that this is normal phase you would use equation 28-3 on page 742. For part (a) the elution order would be first n-hexane. a compound was found to have a retention time of 28. and for part (b) first diethyl ether.