1. Instrumental Analysis (I) HPLC Tutorial 8

2. Graded presentation Students in groups of 4-5 individuals are asked to prepare a presentation (weight=5% of the theoretical course assessment) during the session of the last tutorial of chromatography due from 25 to 28 November. Each group will be asked to give a 10 min talk about the presentation and has to be prepared for students’ and instructor’s questions and discussions. Each group will have a topic according to the coming slide. The assessment criteria are: 2 CriterionGrades (out of 10) Choice of the topic2 Content of the presentation2 Comprehension of the content by members of the group 2 Quality of the presentation2 Ability to answer students’ and instructor’s questions 2

3. TOPICS FOR PRESENTATIONS due after next tutorial (last tutorial of chromatography) 5% OF THE TOTAL GRADE GroupTopic 1Protein analysis by HPLC 2Types of stationary phases in Chiral Chromatography 3UPLC 4Monolithic columns and new trends in stationary phases 5Reversed stationary phases 6Softwares in chromatography 7Theories 8Analysis of water/fat soluble vitamins 9Gas chromatography applications 10Supercritical fluid chromatography applications

4. Today's Objective Describe principles of separation in chromatography. Estimate the tailing factor T f. Assess chromatograms visually by comparison. Calculate chromatographic parameters from given data. Assess several chromatographic experiments by calculating parameters from given data Solve a gradient elution problem. Solve an upscaling problem. Next time, the instructor will see your presentations!!! The details of the presentation requirements and assessment are in the last slide of lecture 6.

5. Tailing factor T f = a+b / 2a T f = 1 T f > 1 T f < 1

6. Useful Equations in Chromatography time I t R1 tMtM t R ' = t R1 - t M t R2 t R2 ' = t R2 - t M : partition coefficient : R F value (in planar chromatography) : capacity factor Separation factor

7. Useful Equations in Chromatography Linear flow rate (u x )= L / t M

8. Compare Chromatograms # 1 Chromatograms of compounds A and B were obtained at the same flow rate with two columns of equal length. 1.Which column has more theoretical plates? 2.Which column has a larger plate height? 3.Which column gives higher resolution? 4.Which column gives a greater relative retention? 5.Which compound has a higher capacity factor? 6.Which compound has a greater partition coefficient? 1 2 1 neither B B

9. Simple Numerical Problems -1 # 2 A chromatography column with L = 10.3 cm and ID = 4.61 mm is packed with a stationary phase that occupies 61.0% of the volume. The linear flow rate = 17.4 cm min -1. 1.How long does it take for solvent to pass through the column? 2.Find the retention time for a solute with a capacity factor of 10.0. t M = L / u x = 10.3 cm / [17.4 cm min -1 ] = 0.592 min. k ' = t' R / t M = (t R -t M ) / t M => t R = k' t M + t M = t M (1 + k') = 0.592 min x 11 = 6.51 min.

10. Assess a Chromatographic System -1 # 3 A chromatogram with ideal Gaussian bands has t R = 9.0 min and w 0.5 = 2.0 min. 1.How many theoretical plates are present? 2.Find the plate height if the column is 10 cm long. N = 5.54 t R 2 / w 0.5 2 = 5.54 * 9 2 min 2 / 2 2 min 2 = 112. H = L / N = 10 cm / 112 = 0.89 mm.

11. Predict the Peak Width # 4 A band having a width of 4.0 mL and a retention volume of 49 mL was eluted from a chromatography column. N.B. Assume that the only band spreading occurs on the column itself. What width is expected for a band with a retention volume of 127 mL? from the known: N = 16 t R 2 / w minutes 2 = 16 V R 2 / w volume 2 = 2401 is the number of plates for this column in this experiment. for the unknown: w unk 2 = 16 V R-unk 2 / N = 16 * 127 2 mL 2 / 2401 w unk = 10.4 mL. Chart parameters like retention, peak width, etc. can always be expressed in units of volume, time or cm resp. mm. If you apply rules make sure that you have all of them in suitable dimensions.

12. Assess a Chromatographic System -2 # 5 Two compounds with partition coefficients of 0.15 and 0.18 are to be separated on a column with V m / V s = 3.0. Calculate the number of theoretical plates needed to produce a resolution of 1.5. N = [6 * 1.2/0.2 * 1.055/0.06] 2 = (400689 =) 4 x10 5.

13. Simple Numerical Problems -2 # 6 An open tubular column (OTC) with L = 30.1 m and ID = 0.530 mm is coated on the inside wall with a layer of stationary phase that is 3.1 m thick. Unretained solute passes through in 2.16 min, whereas a particular solute has a retention time of 17.32 min. Find the capacity factor for the solute and the fraction of time spent in the stationary phase. Find the partition coefficient C s /C m for this solute. k ' = t' R / t M = (t R -t M) / t M = (17.32 min – 2.16 min) / 2.16 min = 7.02. k' = t s / t m => t s = k' * t m = 7.02 x 2.16 min = 15.16 min. fraction in stat. phase = 15.16/17.32 = 0.875

14. K = C s / C m k ' = K V s /V m => K = k' * V m / V st = k' * (L*A mob )/(L*A st ) = k' A mob / A st A st = A total - A mob = D 2 /4 - D mob 2 /4 = /4 [(530 x10 -6 m) 2 – (523.8 x 10 -6 m) 2 ] = 5.13 x10 -9 m 2 K = k' * A mob / A st = 7.02 * 2.15 x10 -7 m 2 / 5.13 x10 -9 m 2 = 294.

15. A Problem on Gradient Elution # 7 A mixture of 14 compounds was subjected to a RP gradient separation going from 5% to 100% acetonitrile (CH 3 CN) with a gradient time of 60 min. All peaks were eluted between 22 and 41 min. 1.Is the mixture more suitable for isocratic or gradient elution? 2.If the next run is a gradient, select the starting and ending %age acetonitrile and the gradient time. t G = 60 min, ∆t = 41-22 = 19, ∆t / t G = 19/60 = 0.317 Use gradient elution if ∆t / t G > 0.25 Use isocratic elution if ∆t / t G < 0.25 equation of the graph %age ACN vs. time: %ACN = (100-5) % / (60 min) * X min + 5% %ACN at 22min:%ACN = 95/60 %/min * 22 min + 5% = 40% %ACN at 41min:%ACN = 95/60 %/min * 41 min + 5% = 70% Thus, a suitable gradient would be from 40 to 70% ACN over 60 mins.

16. A Problem on Peak Shape # 8 Suppose that an HPLC column produces Gaussian peaks. The detector measures absorbance at 254 nm. A sample containing equal moles of compounds A and B was injected into the column. (AUC = 1.064 x h x w0.5) What is the peak height of B? compound > V parameters AB  254 2 x 10 4 M -1 cm -1 1 x 10 4 M -1 cm -1 peak height 100 mm ? peak half-width 10 mm12 mm

17. AUC = 1.064 x w 0.5 x h. AUC a  . => AUC A / AUC B =  A /  B  AUC B = AUC A x  B /  A. AUC A = 1.064 * 100 mm * 10 mm = 1064 mm 2. AUC B = AUC A x  B /  A = 1064 mm 2 x 1/2 = 532 mm 2. h B = AUC B / (1.064 x w 0.5B ) = 532 mm 2 / (1.064 x 12 mm) = 41.6 mm.

18. "Rules of Thumb" for Up- (or Down-)Scaling 1.Keep column length constant (if separation is o.k.). 2.Cross-sectional area of column  mass of analyte: 3.Sample volume applied to column  mass of analyte. Suppose you want to separate a mixture by chromatography without having a suitable instruction. So in order to find a suitable method you try several systems with small samples of the mixture. Suppose there is an instruction telling you how to separate 100 mgs of your mixture by a chromatographic method. In fact, you have only 20 mgs of this mixture to be separated. CASE ICASE II What do you have to consider if you want to apply the found method to your mixture?

19. A Problem on Upscaling # 9 A chromatographic procedure separates 4.0 mg of unknown mixture on a column with a length of 40 cm and a diameter of 0.85 cm. 1.What size column would you use to separate 100 mg of the same mixture? same length (40 cm), diameter: m B / m A = r B 2 / r A 2 = d B 2 / d A 2 d B 2 = d A 2 * m B /m A = 0.85 2 cm 2 * 100 mg/4 mg = 18.06 cm, d B = 4.25 cm

20. Separation Principles # 10 Match terms in the 1 st list with the characteristics in the 2 nd list. 1.adsorption chromatography 2.partition chromatography 3.ion-exchange chromatography 4.molecular exclusion chromatography 5.affinity chromatography A.Ions in mobile phase are attracted to counterions covalently attached to stationary phase B.Solute in mobile phase is attracted to specific groups covalently attached to stationary phase C.Solute equilibrates between mobile phase and surface of stationary phase D.Solute equilibrates between mobile phase and film of liquid attached to stationary phase E.Different-sized solutes penetrate voids in stationary phase to different extents. C D A E B