exam 2 solution - Chemistry 420 – Fall 2008 Name:

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Unformatted text preview: Chemistry 420 – Fall 2008 Name: _______KEY_______________________ Prof. Ryan C. Bailey Exam #2 (175 points total) This exam is 6 pages long Due at the end of class—If you finish early, please remain in your seat so as not to disturb others around you. Please box all answers Show all of your work for each answer. Partial credit is awarded based upon the work that you show. We can’t give credit if we don’t know what you did. Be careful to use correct significant figures and propagate errors through mathematical operations, when appropriate. Recommendation: Read the entire exam first and come up with a strategy to maximize your accumulated points. You do not have to work the problems sequentially. Chem 420 – Fall 2008 – Exam #2 Name: ____KEY__________________________ 1. (5 points) Experimentally, the minima in a van Deemter plot in LC are not usually observed. Why? Flow rates at van Deemter minima are too low to be practical, and thus aren’t typically observed. 2. (5 points) van Deemter plots for GC typically have broader minima than LC. Why? The B/u term is more important in GC because diffusion rates are higher (faster) in the gas phase. This results in broadening of the van Deemter plot. 3. (15 points) Describe the “General Elution Problem” and then list the common methods of reducing this limitation in gas chromatography, liquid chromatography, and supercritical fluid chromatography. General Elution Problem: It is difficult, if not impossible, to optimize separation conditions for every component of a mixture. Gas chromatography Temperature programming Liquid chromatography Gradient elution or solvent programming Supercritical Fluid chromatography Pressure programming 4. ( 10 points) Briefly describe how size exclusion chromatography works? Do larger or smaller molecules elute first? In size exclusion chromatography, the stationary phase is a porous packing material. Small analytes can enter many of the pores, and are effectively removed (shielded) from the mobile phase flow while inside. Larger analytes cannot fit into the pores on account of their size (excluded) and as a result experience almost no retention. Thus, larger molecules elute first. Page 2 of 7 Chem 420 – Fall 2008 – Exam #2 Name: ____KEY__________________________ 5. (20 points) Matrix‐assisted laser desorption ionization is often coupled with time‐of‐flight detection. Separately describe how both a MALDI ion source and a TOF detection system work. Then justify, based upon performance attributes, why MALDI + TOF is a natural combination. Why would a MALDI‐quadrupole pair not make as much sense? MALDI − Sample immobilized in matrix material (usually crystalline) − Matrix absorbs laser pulse and explodes, putting the sample into the gas phase and ionizing. TOF − Pulses of ions are introduced into a field‐free drift tube − Separation on account of different velocities. If all ions have same kinetic energy then lower mass ions travel faster and reach detector first (KE=1/2 mv2) MALDI‐TOF is a good combination because MALDI is good at generating gas phase ions from high MW samples and TOF has a virtually unlimited mass range. A MALDI‐quadrupole configuration would not make as much sense because a quad has a limited m/z range, and therefore doesn’t really take full advantage of the strength of MALDI. 6. (20 points) Describe the process of generating gas phase ions from the sample in both electron impact and chemical ionization ion sources. What is the main difference in the sample ions generated? What common limitations do both of these gas phase ion sources share? Electron impact Chemical ionization − Sample is thermally vaporized − Sample is thermally vaporized − Ionized by stream of electrons − Ionized by collisions with ions Differences in ions generated: EI is a “hard” ion source and causes significant fragmentation. CI is a relatively “soft” ion source. Less fragmentation and more parent ion. Common limitations: Both of these gas phase ion sources are only amenable to solutes that are volatile and thermally stable. Page 3 of 7 Chem 420 – Fall 2008 – Exam #2 Name: ____KEY__________________________ 7. (35 points) Consulting the example liquid chromatogram to the right, which was obtained on a 27.25 cm column, compute the following parameters relevant to separations based upon the event time points provided. a. Based upon the signal observed for Component A, calculate the number of theoretical plates. ⎛t ⎞ NA = 16 ⎜ R ⎟ ⎝ WA ⎠ 2 Component A Dead or void time (unretained solvent front) 5 3.8 0 40 3.3 4. 0 5 WA = 2.40 .75 12 .20 17 Component B WB = 3.10 .55 3.95 5.65 1 1 11 10 15 .75 18 20 25 30 ⎛ 12.75 ⎞ NA = 16 ⎜ ⎟ = 452 theoretical plates (3 sig figs) ⎝ 2.40 ⎠ 2 Time (minutes) b. Calculate the plate height of the column based upon the number of plates determined in part (a). L L H = H N 27.25 H= = 6.03 ⋅ 10 −2 cm 6.035 ⋅ 10 −2 cm also acceptable (3 or 4 sig figs acceptable) 452 N= c. Calculate the retention factor for Component A. t R − tM tM 12.75 − 3.85 kA = = 2.31 (3 sig figs) 3.85 kA = d. Calculate the resolution for the separation of Component A and Component B. 2 ⎡ ( t R )B − ( t R ) A ⎤ ⎦ Rs = ⎣ WA + WB Rs = 2[17.20 − 12.75] 2.40 + 3.10 = 1.62 (3 sig figs) e. Calculate the selectivity factor for this separation. α= ( t R )B − t M ( tR ) A − t M 17.20 − 3.85 = 1.50 (3 sig figs) α= 12.75 − 3.85 Page 4 of 7 Chem 420 – Fall 2008 – Exam #2 Name: ____KEY__________________________ 8. (10 points) The protein Akt1 is an important mediator in the apoptotic signaling pathway and perturbations to its signaling are known to contribute to several types of cancer. Notably, the function of Akt1 is significantly altered when it is phosphorylated. The mass of Akt1 is 59.1 kDa (59,100. Da; treat this as an exact number with 5 significant figures). Protein phosphorylation results in an addition of HPO3, which is measured via mass spectrometry as a mass increase of 80 Da relative to the native (non‐phosphorylated) protein. What resolution needed to resolve two peaks? m 59100 + 59180 mavg = = 59140 Δm 2 59140 R= = 739.25 ; 7x102 , 740, and 7.4x102 also accepted 80 R= 9. (20 points) Referring back to problem 10: if we performed the analysis of Akt1 and phospo‐Akt1 on a Reflectron MALDI‐TOF instrument having a 1.00 m drift tube operated with a 2.00 kV accelerating voltage. What would be the difference in flight times for the +1 ions of Akt1 and phospho‐Akt1? (Note: Remember to be aware of units: A Da is a g kg ⋅ m2 , a V is a 3 , and a C is an A ⋅ s .) mol s ⋅A Since it is a reflectron configuration, L=2.00 m = 9.8138x10 −23 kg 23 6.022141x10 ⋅ 1000 m 59180 g/mol = 9.8271x10 −23 kg t f =L m (kg ) phospho‐ Akt 1 = 23 2zeV 6.022141x10 ⋅ 1000 t fAkt 1 =2.00 m ⋅ 9.8138x10−23 kg ⎛ kg ⋅ m 2 2 ⋅ 1 ⋅ 1.602177x10 −19 A ⋅ s ⋅ ⎜ 2.00 x103 ⎜ s3 ⋅ A ⎝ ⎞ ⎟ ⎟ ⎠ = 7.826423689x10 −4 s (7.83x10‐4 s if rounding here) m (kg ) Akt 1 = 59100 g/mol ( ) ⎞ ⎟ ⎟ ⎠ = 7.831725213x10 −4 s (7.83x10‐4 s if rounding here) t fphospho−Akt 1 =2.00 m ⋅ 9.8271x10 −23 kg ⎛ kg ⋅ m 2 2 ⋅ 1 ⋅ 1.602177x10 −19 A ⋅ s ⋅ ⎜ 2.00 x103 ⎜ s3 ⋅ A ⎝ ( ) Difference = Δt f = 7.831725213x10−4 ‐7.826423689x10 −4 = 5.301524x10‐7 s ≈ 0.53 μsec or because I did not engineer the sig figs correctly: Δt f = 7.83x10 −4 ‐7.83x10 −4 = 0 s in the end, we did not take off for signifcant figures for the final part of question 9 Page 5 of 7 Chem 420 – Fall 2008 – Exam #2 Name: ____KEY__________________________ Acc. Chem. Res. 2004, 37, 53‐59. 10. (35 points) Fourier transform ion cyclotron resonance mass spectrometry can determine masses of proteins with incredibly high mass accuracy (greater than 1ppb), which in turn leads to ultrahigh resolution (350,000+!). The mass accuracy in FT‐ICR improves with increasing cyclotron resonance frequencies, i.e. more cycles of a given m/z during a fixed measurement period. One emerging application of FT‐ICR is the analysis of crude petroleum samples, termed “Petroleomics.” As can be seen in the FT‐ICR spectrum at right, components with very similar masses, which might be indistinguishable using lower resolution mass analyzers, can be easily resolved. a. Consider the highest mass component in the spectrum above: C16H33+ at 225.25769 Da. Calculate the cyclotron resonance frequencies at 5.6 Tesla (used in obtaining the data in the figure above) and at 12 Tesla (the strength of the magnetic field on several of Prof. Kelleher’s instruments). (Notes: Consider both of the magnetic field strengths as exact numbers, i.e. they should not limit the number of significant figures in the final answer. A Tesla is a zeB 225.25769 g/mol = 3.740492x10−25 kg m (kg )C H + = 23 16 33 m 6.022141x10 ⋅ 1000 ⎛ ⎞ kg ⋅ m2 ⎜ ⋅s ⎟ ⎜ ⎟ s3 ⋅ A 1 ⋅ 1.602177x10 −19 A ⋅ s ⋅ ⎜ 5.6 ⎟ m2 ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ = 2.398666x106 Hz 7 sig figs ωc5.6T = ( ) −25 3.740492x10 kg V ⋅s m2 .) Will the mass resolution be better or worse at the higher magnetic field strength? ωc = ( ) ⎛ kg ⋅ m2 ⎜ ⋅s ⎜ s3 ⋅ A −19 1 ⋅ 1.602177x10 A ⋅ s ⋅ ⎜ 12 m2 ⎜ ⎜ ⎝ ωc12T = 3.740492x10 −25 kg ( ) ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = 5.139999x106 Hz 7 sig figs ( ) Mass resolution is better with higher magnetic field strength. Page 6 of 7 Chem 420 – Fall 2008 – Exam #2 Name: ____KEY__________________________ b. Now suppose that two of the ions in the figure above are present in the same 5.6 Tesla FT‐ICR chamber and both amplified via the AC electric field to an identical cyclotron radius of 2.54 centimeters (consider this an exact number). Calculate the distance traveled by the highest (C16H33+, 225.25769 Da, same as in part (a)) and lowest (C15H13S+, 225.07326 Da) mass ion during the 2.00 second spectral acquisition period. (Note: Consider the acquisition time as an exact number, i.e. it should not limit the number of significant figures in the answer) What is the difference in distance traveled by these two ions during analysis? zeB 225.07326 g/mol m (kg )C H S+ = = 3.737429x10 −25 kg 23 15 13 m 6.022141x10 ⋅ 1000 ⎛ ⎞ kg ⋅ m2 ⎜ ⋅s ⎟ ⎜ ⎟ s3 ⋅ A 1 ⋅ 1.602177x10−19 A ⋅ s ⋅ ⎜ 5.6 ⎟ m2 ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ = 2.400632x106 Hz = ωc C15H13S+ , 5.6T 3.737429x10 −25 kg ωc = ( ) cyclotron radius = 2π r=2π ⋅ ( 2.54 cm ) = 15.959291 cm ⎛ cycles ⎞ ⎛ 15.959291 cm ⎞ 7 7 In 2 seconds, C15H13 S+ : 2.00 s ⋅ ⎜ 2.400632x106 ⎟ ⋅⎜ ⎟ = 7.662477x10 cm, or 7.66x10 cm if rounding here ⎜ ⎟⎜ ⎟ cycle s ⎠⎝ ⎝ ⎠ ⎛ cycles ⎞ ⎛ 15.959291 cm ⎞ 7 7 In 2 seconds, C16H33+ : 2.00 s ⋅ ⎜ 2.398666x106 ⎟ ⋅⎜ ⎟ = 7.656202x10 cm, or 7.66x10 cm if rounding here ⎜ ⎟ cycle s ⎟⎜ ⎝ ⎠⎝ ⎠ Difference = Δdistance = 7.662477x107 − 7.656202x107 = 6.2750x104 cm = 625.70 m ≈ 0.4 miles!!! or because I did not engineer the sig figs correctly: Δdistance = 7.66x107 − 7.66x107 = 0 cm in the end, we did not take off for signifcant figures for the final part of question 10b Page 7 of 7 ...
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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.

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