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Unformatted text preview: Name: ________________________________ 10 questions @10 points each 1] Describe chemical ionization in mass spectrometry. What type of information is it likely to give about the analyte? 2] What is the electro‐osmotic effect in capillary electrophoresis? Why do all species migrate to one electrode? Do they migrate to the cathode or the anode? What is the general order of migration time for cations, anions, and neutrals? Why are neutrals poorly resolved? 3] When considering the van Deemter equation, why does HPLC require small column packing particles? 4] What is gradient elution and how does this differ from an isocratic one? What advantage does gradient elution have over isocratic separations? When using a C‐18 stationary phase is it more beneficial to increase or decrease m.p. composition polarity during elution? 5] What is meant by a bulk property detector? Give an example of an HPLC detector that is based on bulk properties and one that is not. 6] Why do capillary columns predominate in analytical GC? 7] What is temperature programming in GC? How does it gain an advantage over single T separations? 8] What is the electron capture detector? Explain its basis for operation, why is N2 necessary? What types of species are detected with the ECD? 9] Generally, it is thought by many chromatography dilettantes that twice the column length will give you twice the separation “power”. Comment on why this is false. 10] A GC‐FID analysis was conducted on a soil sample containing pollutant X. The following separations were conducted: Injection 1 21.1 ppm Toluene Internal Standard 33.4 ppm X Injection 2 21.1 ppm Toluene Internal Standard unknown concentration X What is the concentration of X in the sample? tr(minutes) 10.11 14.82 10.05 14.77 peak area 36,242 45,997 38,774 39,115 Chem 454 – Exam 3 – April 23, 2008 1 ANSWERS 1] Describe chemical ionization in mass spectrometry. What type of information is it likely to give about the analyte? See for example: http://en.wikipedia.org/wiki/Chemical_ionization 2] What is the electro‐osmotic effect in capillary electrophoresis? Why do all species migrate to one electrode? Do they migrate to the cathode or the anode? What is the general order of migration time for cations, anions, and neutrals? Why are neutrals poorly resolved? http://en.wikipedia.org/wiki/Capillary_electrophoresis 3]When considering the van Deemter equation, why does HPLC require small column packing particles? H = A + B/u + Cu In the v‐D equation the MT effects predominate, i.e. Cu. Increasing the surface area/bulk ratio of the s.p. is a way to great improve the MT between the two phases. This requires small diameter supports for the s.p. The cost is the pressure required to squeeze the m.p. through the space between the smaller diameter particles. 4] What is gradient elution and how does this differ from an isocratic one? What advantage does gradient elution have over isocratic separations? When using a C‐18 stationary phase is it more beneficial to increase or decrease m.p. polarity during elution? Gradient elutions vary the m.p. solvent composition and polarity during separation. This has an advantage over isocratic separations where solvent compostions are kept constant. A gradient elution will allow for the separation of a large variety of species with a broad spectrum of polarities with a much shorter times than isocratic ones. Generally it’s best that polarity decrease during separation when using a C‐18 s.p. If a non‐polar m.p. is used at the beginning of the elution, there will be no retention between the solutes and the C‐18 s.p. 5] What is meant by a bulk property detector? Give an example of an HPLC detector that is based on bulk properties and one that is not. See problem 28‐7 h. Measure a physical property of the m.p. Example – UV‐vis absorbance, fluorescence techniques are examples of bulk property detectors. Electrochemical detectors are not, since they are based on redox exchanges with solutes near the electrode surface. 6] Why do capillary columns predominate in analytical GC? Again this gets back to the v‐D eqn. The B/u, longitudinal diffusion term contributes most to band broadening in the gas phase. Capillary columns allow for the unobstructed and therefore faster flow of the gaseous m.p. over their packed counterparts. 7] What is temperature programming in GC? How does it gain an advantage over single T separations? Chem 454 – Exam 3 – April 23, 2008 2 By going from colder to warmer temperatures , it is possible to add another dimension separation of solutes beyond the chromatographic ones. This is based on boiling point differences. Generally the initial T is below that of the solutes species and slowly ramped up. See also problem 27‐3. 8] What is the electron capture detector? Explain its basis for operation, why is N2 necessary? What types of species are detected with the ECD? See http://www.instrumentalchemistry.com/gasphase/pages/ecd.htm Nickel‐63 source emits energetic electrons collides with N2 (introduced as make‐up gas or can be used as carrier gas) producing more electrons: Ni‐63 => e‐ e‐ + N2 => 2e‐ + N2+ The result is a constant current that is detected by the electron collector (anode). Chem 454 – Exam 3 – April 23, 2008 3 As an analyte flows through past the Ni‐63 source electron capture is possible by electron‐withdrawing species: A + e‐ => A‐ Current decreases as a result of e‐ capture by analyte. This is one of the few instances in which a signal is produced by a decrease in detectable phenomenon. Sensitive to electron withdrawing groups especially towards organics containing –F, ‐Cl, ‐Br, ‐I also, ‐CN, NO2 9] Generally, it is thought by many chromatography dilettantes that twice the column length will give you twice the separation “power”. Comment on why this is false. Remember that Rs α L1/2. See chapter 26 pages 776‐782. So 2x the column length increase resolution by 1.4. Also remember that B/u effects increase with separation time and 2x will increase t by 2x. Also, using a longer column uses more m.p. and decreases experimental throughput. 10] A GC‐FID analysis was conducted on a soil sample containing pollutant X. The following separations were conducted: Injection 1 Injection 2 21.1 ppm Toluene Internal Standard 33.4 ppm X 21.1 ppm Toluene Internal Standard unknown concentration X tr(minutes) 10.11 14.82 10.05 14.77 peak area 36,242 45,997 38,774 39,115 What is the concentration of X in the sample? 39,115*(36,242/39,115)*(33.4ppm/45,997) = 26.3 ppm Chem 454 – Exam 3 – April 23, 2008 4 ...
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This note was uploaded on 02/16/2010 for the course CHEM 589 taught by Professor -1 during the Fall '02 term at Pennsylvania State University, University Park.

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