1.  The word “chromatography” originated from two Greek words, chroma which means “color” and graphy which means “writing”.  Chromatography was founded in 1906 by a Russian botanist, Micheal Tswett, when he separated plant extracts on a column packed with finely divided calcium carbonate. Chromatography 1

2.  Chromatography is a method to separate two or more compounds in a mixture based on the differences in the property of each individual substance.  The properties include polarity, solubility, ionic strength, and size.  Chromatography is today one of the most useful analytical methods for separation, identification, and quantitation of chemical compounds. Purpose of Chromatography 2

3.  In chromatography, the compounds are physically separated by distributing themselves between two phases: (a) a stationary phase (b) a mobile phase which flows continuously across the stationary phase. Principles Injector Detector was located after the column or stationary phase Mobile phase flow Injected or applied sample Packing material 3

4.  The stationary phase can be (a) a solid packed into a column (b) a solid coating the surface of a flat, plane material (c) a liquid supported on a solid (d) a liquid supported on the inside wall of an open tube  The mobile phase can be a gas, a liquid or a supercritical fluid. Principles 4

5.  The mixture to be analysed is introduced onto the stationary phase and mobile phase carries the components through it.  Each individual analyte interacts with the two phases in a different manner. Principles (cont.) Injector Detector Mobile phase flow t1t1 t2t2 t3t3 Time Interaction with stationary phase Most Least 5

6.  Because analytes differ in their affinity for the stationary phase vs the mobile phase, each analyte exhibits different migration and elution patterns and thus a mixture of analytes can be separated and quantified.  The tracing of the output signal vs time or mobile phase volume is called a chromatogram. Principles (cont.) Time Signal 6

7. Classification of elution chromatographic techniques 7

8. Adsorption Chromatography Eluent flow Legend: Active site Analyte 8

9. Partition Chromatography Solute dissolved in liquid phase coated on surface of solid support 9

10. Affinity Chromatography + + + + Matrix Ligand Immobilized ligand Ligand Sample Complex Impurities Complex Purified sample 10

11. Size Exclusion Chromatography 3 2 1 Signal Volume 1 2 3 Pore size Stationary phase particle 11

12. Chromatographic Techniques Chromatography GasLiquidSupercritical Fluid Solid Liquid IEC Solid GLCGSC SFCBPC LSC LLCTLCSEC GLCGas-Liquid Chromatography GSC Gas-Solid Chromatography SFCSupercritical Fluid ChromatographyLLC Liquid-Liquid Chromatography TLCThin Layer chromatographyLSC Liquid-solid Chromatography BPCBonded Phase ChromatographyIECIon Exchange Chromatography SECSize Exclusion Chromatography Mobile phase Stationary phase 12

13. Chromatographic definitions Retention time, t r - time elapsed between injection of sample and emergence of peak maximum Mobile phase time, t m - time required for an injected solvent molecule or any other unretained compound to traverse the column - also known as t o (column dead volume or column void volume) Adjusted retention time, t’ r - difference between analyte retention time and column void volume, given by t’ r = t r - t o 13

14. Chromatographic definitions Capacity factor, k’ - A measure of how long a sample is retained with respect to the dead volume. Given by k’ = (t r - t o )/t o Column efficiency, N - A measure of the broadening of the sample peak as it passes through the column. Given by N = 16 (t r ’ /w b ) 2 or N = 5.54 (t r ’ /w h ) 2 where w b is the width of peak at base peak w h is the width of peak at half height 14

15. Chromatographic definitions Height equivalent to a theoretical plate (HETP), H - used to compare column efficiency of different lengths, L. H = L /N Resolution, R s = degree of separation of two components leaving the column. 15

16. General Theory of Chromatography 16

17. Properties of a Gaussian peak.  is standard deviation, w i is width at the inflection point, w h is the width at half height, and w b is the width at the baseline intercept. The plate theory Tangents to points of inflection Inflection point Fraction of peak height w h = 2.354   w i = 2  w b = 4  1.213 1.000 0.882 0.607 0.500 0.134 0 17

18. Schematic diagram showing important parameters in chromatography. t o is retention time of unretained compound, t R1 and t R2 are the retention times for component 1 and component 2, respectively. Chromatogram Injection 10% of peak height 18

19. Resolution d w1w1 w2w2 Data: d = 1.8 cm w 1 = 0.8 cm w 2 = 0.9 cm R s = 2 --------------- = ------------- = 2.1 d 3.6 cm (w 1 + w 2 ) 1.7 cm 19

20. Separation factor or selectivity factor  = ( t r (B) - t o ) ( t r (A) - t o ) = t’ r (B) t’ r (A)  = k’ (B) k’ (A) or 20

21. Separation factor or selectivity factor t’ r (B) 2.8 min = ________ = _________ = 1.4 t’ r (A) 2.0 min 123 0 4 time (min) Injection  Air peak t’ r (B) t’ r (A)   21

22. Band broadening It is due to finite rate at which several mass transfer process occur during migration of solute down a column. Non-equilibrium theory - movement of solute through the column treated as a random walk. Random walk - progress of a molecule through column is a succession of random stops and starts about a mean equilibrium concentration or band centre. 22

23. Band broadening When solute desorbed from stationary phase and transferred to mobile phase, it moves more rapidly than the band centre. Three factors contributing:-  Eddy diffusion (A)  Longitudinal molecular diffusion (B)  Rate of mass transfer (C) 23

24. Typical pathways for two solute molecules during elution. The distance travelled by molecule 2 is greater than that travelled by molecule 1. Thus, molecule 2 arrieves at B later than molecule 1. Eddy diffusion (1) (2) A B 24

25. Eddy diffusion Effect of different path lengths due to irregular flow of molecules through packed particles in a column Each molecule see different paths causing the solute molecules to arrive at the column outlet at different times Independent of mobile phase velocity 25

26. Band broadening due to longitudinal diffusion in mobile phase (A) initial band and (B) Diffusion of band with time. Longitudinal diffusion Profile of band conc. Band width Direction of mobile phase 26

27. Longitudinal diffusion Results from tendency of solutes to diffuse from concentration centre of a band to the more dilute regions on either sides. Occur primarily in the mobile phase Longitudinal diffusion inversely proportional to mobile phase velocity Higher velocity provides less time for diffusion to occur: Means no band broadening 27

28. Illustration of the influence of local nonequilibrium on band broadening. Non-equilibrium Mobile phase Stationary phase Interface Flow Equilibrium concentration Actual concentration Legend: 28

29. Concentration profile of analyte at the interface between the stationary phase and the mobile phase: (a) ideal condition before longitudinal movement of mobile phase takes place. (b) Actual condition after longitudinal movement of the mobile phase. Influence of local nonequilibrium on band broadening (a) (b) Analyte concentration Mobile phase Stationary phase Mobile phase Stationary phase New equilibrium Analyte concentration 29

30. Resistance to mass transfer Finite time required for the solutes to transfer in and out of the stationary phase Kinetic lag in attaining equilibrium between two phases Broadening occur when either of these rates is slow Broadening depend on diffusion rate of analyte (time dependent) Broadening worsen with increasing mobile phase flow rate 30

31. Plot of HETP against flow velocity, showing the contributions of A (eddy diffusion), B (longitudinal diffusion) and C (interface mass transfer). Van Deemter plot Flow velocity C B (H min ) C HETP Van Deemter curve 31

32. Van Deemter’s Equation Where;  (A) Eddy diffusion  (B) Longitudinal molecular  (C) Diffusion rate of mass transfer 32

33. Effect of N and  on resolution toto toto toto toto Poor resolution Good resolution due to column efficiency Good resolution due to column selectivity Poor resolution despite adequate column efficiency and selectivity (low k’) 33

34. Resolution, selectivity, and effeciency (a) (b) (c) Initial profile Increased selectivity, N unchanged Selectivity unchanged, N increased Good resolution Poor resolution 34

35. Qualitative Analysis 35

36. Retention time comparison  By using identical column and operating conditions, the identity of the analyte can be determined by comparing the retention time with the standard compound.  The the absence of a peak at the same retention time as that of a standard under identical conditions suggests that either the respective compound is absent in the sample or the compound is present at a concentration level below the detection limit of the method. 36

37. Kovats’ retention index (RI) For linear temperature- programmed GC For isothermal GCPlots of log t R ’ vs C-no. gives a straight line Plots of t R ’ vs C-no. gives a straight line 37

38. Kovats RI The Kovats retention index. The log t R ' values of a series of n- alkanes area plotted against (100 x carbon number). Compound x (log t R ' = 0.6) has a retention index value of 680. Retention Index 38

39. Quantitative Analysis 39

40. Quantitative Analysis: Peak area Measurement of peak area: (a) By triangulation. Peak area  area of the triangle = (h' x w)/2. (b) Peak area  area of the rectangle = h x w 1/2. 40

41. Response factor Response factor of compound X = F X Detector response X 0 5 10 min S 41

42. Response factor Response factor of compound X relative to that of compound S: If the response factors of all compounds in the sample are the same, 42

43. External standard M x Molarity of compound X, M s Molarity of standard S, F x Response factor of compound X, F s Response factor of standard S, V (x) Volume of injected sample X, V (s) Volume of injected standard S, A x Peak area of sample X, A s Peak area of standard S. 43

44. External standard External standard poses a problem with reproducibility. Injection – the volume of injection may vary from one injection to another injection. Detection – the sensitivity of the detector may vary from time to time. 44

45. Internal standard M x Molarity of compound X, M s Molarity of standard S, F x Response factor of compound X, F s Response factor of standard S, V (x) Volume of injected sample X, A x Peak area of sample X, A s Peak area of standard S. N (s) Amount of internal standard S, 45

46. Criteria for internal standards Characteristics of ideal internal standards (a)It is well separated from the components in the analyzed mixture. (b)Its response factor is similar to that of the analyte. (c)Its retention time is close to the that of the analyte to minimize errors in peak area measurement. (d)Its concentration should be similar to that of the analyte. 46

47. Example 1. External standard Consider the following data of a typical quantitative gas chromatographic analysis where a compound, X, is used as the external standard. An injection (1  L) of a mixture containing 10, 12, and 13 ppm of X, Y, and Z, respectively, gave respective peak areas of 515, 748 and 939 arbitrary units. An injection (2  L) of the sample containing compounds X, Y, and Z gave peak areas of 232, 657, and 984 arbitrary units, respectively. Calculate the concentrations of compounds X, Y and Z. 47

48. Example 1. Solution 48

49. Example 2. Internal standard In a chromatographic analysis, a mixture containing 0.0567 M of a compound, X, and 0.0402 mM of a standard, S, gave peak areas of A x = 498 and A S = 526 arbitrary units. An aliquot (10.0 mL) of the unknown was added with 10.00 mL of 0.102 M S and the mixture was diluted to 25 mL in a volumetric flask. The mixture gave a chromatogram with A x = 678 and A S = 588 arbitrary units. Calculate the concentration of X in the unknown. 49

50. Example 2. (Cont.) Detec tor respo nse X 0 5 10 min S Determination of the relative detector response for the standard  Add a known amount of the standard compound into a known quantity (volume or weight) of sample, V x,  Measure the peak area, and calculate the concentration of the analyte using the proper equation. 50

51. Example 2. Solution 51

52. 52