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Unformatted text preview: Chemistry 420 Problem set #3 (80 points total)—Key Due at the beginning of class October 29 No late assignments will be accepted. Assignments are due at beginning of class—starts at 12 Noon! For full credit assignments must be in box when the lecture begins. Any assignment turned in between noon and 12:05 PM will automatically be deducted 5 points. Assignments will NOT be accepted for credit after 12:05 PM. Score = 0 No name, no grade. Show all of your work for each answer. Partial credit will be given and NO credit will be awarded if no work is shown. Be careful to use correct significant figures. 1. (12 points, Skoog 26‐14) The following data are for a liquid chromatographic column Length of packing Flow rate VM VS 24.7 cm 0.313 mL/min 1.37 mL 0.164 mL A chromatogram of a mixture of species A, B, C, and D provided the following data: Retention time, minutes Nonretained 3.1 A 5.4 B 13.3 C 14.1 D 21.6 Width of Peak Base (W), minutes — 0.41 1.07 1.16 1.72 Calculate: a) the number of plates from each peak. b) the mean and the standard deviation for N. c) the plate height for the column. (a) 2 points for setup, 1 point for each answer, ‐0.5 for sig figs ⎛t ⎞ N = 16 ⎜ R ⎟ ⎝W ⎠
2 ⎛ 5.4 ⎞ 3 for A: N = 16 ⎜ ⎟ = 2775 or 2.8x10 or 2800 plates ⎝ 0.41 ⎠ ⎛ 14.1 ⎞ = 2365 or 2.36x103 or 2360 plates for C: N = 16 ⎜ 1.16 ⎟ ⎝ ⎠
2 2 ⎛ 13.3 ⎞ 3 for B: N = 16 ⎜ ⎟ = 2472 or 2.47x10 or 2470 plates ⎝ 1.07 ⎠ ⎛ 21.6 ⎞ for D: N = 16 ⎜ = 2523 or 2.52x103 or 2520 plates 1.72 ⎟ ⎝ ⎠
2 2 (b) 1 point for each setup, 1 point for each answer, ‐0.5 for sig figs N= 2775 + 2472 + 2365 + 2523 = 2534 or 2.53x103 plates 4 4 1 SD = ( 2775  2534 )2 + ( 2472  2534 )2 + ( 2365  2534 )2 + ( 2523  2534 )2 = 174 or 2x10
2 (c) 1 point for setup, 1 point for answer, ‐0.5 for sig figs H= L 24.7 cm = = 9.75x103 cm N 2534 plates 2. (10 points, Skoog 26‐15) From the data in Problem 1 (Skoog 26‐14), calculate for A, B, C, and D a) the retention factor. b) the distribution constant. (a) 2 points for set‐up, 0.75 points for each answer, ‐0.25 for sig figs k= tR − tM tM 13.3 − 3.1 = 3.3 3.1 21.6 − 3.1 = 6.0 for D: k = 3.1 for B: k = for A: k = 5.4 − 3.1 = 0.74 3.1 14.1 − 3.1 = 3.5 for C: k = 3.1 (b) 2 points for set‐up, 0.75 points for each answer, ‐0.25 for sig figs ⎛ VM ⎞ t R − tM ⎛ VM ⎞ ⎛ 1.37 ⎞ ⎜ ⎟ = k⎜ ⎟ = k⎜ ⎟ = 8.35k tM ⎝ VS ⎠ ⎝ 0.164 ⎠ ⎝ VS ⎠ for A: K = 8.35 ( 0.74 ) = 6.2 for B: K = 8.35 ( 3.3) = 28 K= for C: K = 8.35 ( 3.5 ) = 29 for D: K = 8.35 ( 6.0 ) = 50 3. (14 points, Skoog 26‐16) From the data in Problem 1 (Skoog 26‐14), calculate for species B and C a) the resolution. b) the selectivity factor α. c) the length of column necessary to separate the two species with a resolution of 1.5. d) the time required to separate the two species on the column in part (c). (a) 2 points for setup, 1 point for answer, ‐0.5 for sig figs 2 ⎡ ( t R )C − ( t R ) B ⎤ ⎦ Rs = ⎣ WC + WB Rs = 2 [14.1 − 13.3] 1.16 + 1.07 = 0.7 or 0.72 (b) 2 points for setup, 1 point for answer, ‐0.5 for sig figs αC ,B = ( tR )C − tM ( t R ) B − tM = 14.1 − 3.1 = 1.08 ( 0.926 also acceptable ) 13.3 − 3.1 (c) 2 points for setting up calculation for N2, 1 point for performing N2 calculation (sig figs not important here), 1 point for setting up column length, 1 point for answer, ‐0.5 for sig figs in final answer ( Rs )1 ( Rs )2 = N1 N2
2 2534 N2 = 0.72 1.5 ⎡ ⎤ ⎢ 2534 ⎥ ⎥ = 10998 N2 = ⎢ ⎢ ⎛ 0.72 ⎞ ⎥ ⎢ ⎜ 1.5 ⎟ ⎥ ⎠⎦ ⎣⎝ L = HN = 9.75x103 (10998 ) = 107 cm 1x102 and 1.1x102 also acceptable ( ) ( ) (d) 2 points for setup, 1 point for answer, ‐0.5 for sig figs 2 ( tR )1 ( Rs )1 = ( tR )2 ( Rs )2 2 14.1 ( 0.717 ) = ( tR )2 (1.5)2
2 2 ( t R )2 = 14.1 ⎛ ( 0.717 ) ⎜ ⎜ (1.5 )2 ⎝ ⎞ ⎟ ⎟ ⎠ = 61.7 or 62 minutes 4. (6 points, Modified from Skoog 27‐3) What is meant by temperature programming in GC? Why is it frequently used? What general chromatographic problem does it help solve? Temperature programming involves increasing the temperature of a GC column as a function of time. This technique is particularly useful for samples that contain constituents whose boiling points differ significantly. Low boiling point constituents are separated initially at temperatures that provide good resolution. As the separation proceeds, the column temperature is increased so that the higher boiling constituents come off the column with good resolution and at reasonable lengths of time. Temperature programming helps solve the general elution problem in gas chromatography. 5. (2 points, Skoog 27‐20) The same polar compound is gas chromatographed on an SE‐30 (very nonpolar) column and then on a Carbowax 20M (very polar) column. How will K =
cs vary between the two columns? cm The distribution coefficient, K, for a polar compound will be larger on the Carbowax 20M column than on the non‐polar SE‐30 column because the compound will have a stronger interaction with the polar stationary phase, and thus more of the compound will partition into the Carbowax 20M, compared with the non‐polar SE‐30. 6. (10 points, Modified from Skoog 28‐7) Define: a) isocratic elution. b) gradient elution. c) reversed‐phase packing. d) normal‐phase packing. e) ion‐pairing chromatography. (2 points for each part) (a) In an isocratic elution, the solvent composition is held constant throughout the elution. (b) In a gradient elution, two or more solvents are used and the composition of the mobile phase is changed continuously or in steps as the separation proceeds. (c) A reversed‐phase packing is a non‐polar packing that is used in partition chromatography with a relatively polar mobile phase. (d) In a normal‐phase packing, the stationary phase is polar and the mobile phase is relatively non‐polar. (e) In ion‐pair chromatography a large organic counter‐ion is added to the mobile phase as an ion‐pairing reagent. Separation is achieved either through partitioning of the neutral ion‐pair or as a result of electrostatic interactions between the ions in solution and charges on the stationary phase resulting from adsorption of the organic counter‐ion. 7. (12 points, Modified from Skoog 28‐20) Two components in an HPLC separation have retention times that differ by 15 seconds. The first peak elutes in 9.0 minutes and the peak widths are approximately equal. The dead time tM was 65 seconds. (With or without using a spreadsheet—your choice) Find the minimal number of theoretical plates needed to achieve the following resolution Rs values: 0.50, 0.75, 0.90, 1.0, 1.10, 1.25, 1.50, 1.75, 2.0, 2.5. (Note: In this case, do not worry about significant figures, but do remember that the N must be an integer.) 2 points for setting up and performing α calculation (1 each), 2 points for setting up and performing k calculation (1 each), 3 points for setting up equation for N as a function of Rs, and 0.5 points for each answer α= ( tR )2 − tM ( tR )1 − tM = 9.25 − 1.08 = 1.03 9.00 − 1.08 k2 = ( t R )2 − tM
tM = 9.25 − 1.08 = 7.56 1.08 ⎛ α ⎞ ⎛ 1 + k2 ⎞ N = 16 Rs2 ⎜ ⎟ ⎟⎜ ⎝ α − 1 ⎠ ⎝ k2 ⎠
2 2 N tR1 tR2 tM 9.00 minutes 9.25 minutes 1.08 minutes 1.031579 1.03 7.538462 7.54 Rs 0.50 0.75 0.90 1.0 1.10 1.25 1.50 1.75 2.0 2.5 N 5476 or 5.5x103 12321 or 1.2x104 17742 or 1.8x104 21904 or 2.2x104 26504 or 2.65x104 34225 or 3.42x104 49284 or 4.93x104 67081 or 6.71x104 87616 or 8.8x104 136900 or 1.4x105 (if sig figs computed for α) 6.0x103 or 6,000 1.4x104 or 14,000 2.0x104 or 20,000 2.4x104 or 24,000 2.9x104 or 29,000 3.8x104 or 38,000 5.4x104 or 54,000 7.4x104 or 74,000 9.7x104 or 97,000 1.5x105 or 150,000 α α (sig figs calculated) k2 k2 (sig figs calculated) 8. (4 points, Skoog 30‐1) What is electroosmotic flow? Why does it occur? Electroosmotic flow occurs when a mobile phase in a capillary tube is subjected to a high potential difference between one end of the tube and the other. For a silica tube, the flow is generally away from the positive electrode towards the negative. The flow occurs because of the attraction of positively charge species toward the negative silica surface. This layer of positive charge is mobile and is attracted toward the negative electrode carrying with it the mobile phase molecules. 9. (4 points, Skoog 30‐5) A certain inorganic cation has an electrophoretic mobility of 4.31x10−4 same ion has a diffusion coefficient of 9.8 x10 −6 cm2 . This Vs cm2 . If this ion is separated from other cations by CZE with a s 50.0 cm capillary, what is the expected plate count N at applied voltages of a) 5.0 kV? b) 10.0 kV? c) 30.0 kV? 1.75 points for set up, 0.75 points per answer (a‐c), ‐0.5 for sig figs (a) ( 4.31x10 ) (5000) = 1.1x10 N = = 2D 2 ( 9.8x10 )
μeV
4 6 5 (b) N = (c) ( 4.31x10 ) (10000) = 2.2x10 2D 2 ( 9.8x10 ) μ V ( 4.31x10 ) ( 30000 ) N= = = 6.6x10 2D 2 ( 9.8x10 )
μeV
4 = 5 6 4 e 5 6 10. (6 points, Skoog 30‐6) The cationic analyte of Problem 9 (Skoog 30‐5) was separated by CZE in a 50.0 cm capillary at 10.0 kV. Under the separation conditions, the electroosmotic flow rate was 0.85 mm/s toward the cathode. If the detector were placed 40.0 cm from the injection end of the capillary, how long would it take in minutes for the analyte cation to reach the detector after the field is applied? 2 points for setting up equation and calculating electrophoretic velocity (1 each), 2 points for total velocity, and 2 points for final time in minutes. ‐1 point for sig figs and ‐1 point if answer is not in units of minutes. 2 ⎛ −4 cm ⎞ ⎜ 4.31x10 ⎟ (10000 V ) Vs ⎟ μeV ⎜ ⎝ ⎠ velectrophoretic = = = 0.0862 cm/s L 50 cm total velocity = velectrophoretic + velectroosmotic = 0.0862 cm/s + 0.085 cm/s =0.171 cm/s time = 1 min distance 40 cm = 3.9 min ⋅ = velocity 0.171 cm / s 60 s ...
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This note was uploaded on 04/08/2010 for the course CHEM 420 taught by Professor Bailey during the Spring '08 term at University of Illinois at Urbana–Champaign.
 Spring '08
 BAILEY
 Chemistry

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