final solution

final solution - Chemistry 420 – Fall 2008 Name: _______...

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Unformatted text preview: Chemistry 420 – Fall 2008 Name: _______ KEY ______________________ Prof. Ryan C. Bailey Final Exam (250 points total) 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 – Final exam Page total: _____ / 65 Easy street: Circle the correct answer (5 points each) 1. Decreasing the slit width of a monochromator decreases or increases the wavelength resolution. 2. Decreasing the slit width of a monochromator decreases or increases the optical throughput. 3. Stimulated emission is coherent or incoherent. 4. The main utility of double‐beam spectrometers is to eliminate matrix interference effects: true or false. 5. Phosphorescent emission has a longer or shorter lifetime than fluorescence. 6. Fluorescence is inherently less sensitive than absorbance spectrometry: true or false. 7. Chemical ionization or electron impact would be the best choice in order to maximize the amount of parent ion observed in a mass spectrum. Short answer: The following questions can be adequately answered in a few bullet points or sentences. 8. (10 points) Photomultiplier tubes are often used for very sensitive detection in the ultraviolet and visible region of the electromagnetic spectrum. Briefly explain how a PMT works and why it can have such high sensitivity. Photons strike a photocathode which leads to the release of electrons. Electrons are then directed towards a dynode and impact results in the production of more electrons. This process is continued across a series of biased dynodes. The end result is a highly amplified signal that is proportional to the number of photons that originally struck the photocathode. The PMT is able to have such high sensitivity, therefore, because of the amplification of signal. 9. (10 points) As we discussed in class, band broadening can be modeled using the van Deemter equation. Write out the van Deemter equation. What do each of the terms represent (names of the terms is sufficient)? H = A+ B + C su + CM u u A is the multipath term B/u is the longitudinal diffusion term CSu is the stationary phase mass transfer term CMu is the mobile phase mass transfer term 10. (10 points) When analyzing hexane (C6H14) and dodecane (C12H26) on a non‐polar gas chromatography column, which molecule would you expect to have a shorter retention time? Why? Under these conditions, hexane will have the shorter retention time. Given that the polarities of the column stationary phase and analyte are well‐matched (they are here) then lower boiling point analytes elute faster. The six‐ carbon hexane (69 oC) has a lower boiling point that the twelve‐carbon dodecane (216 oC). Page 2 of 10 Chem 420 – Fall 2008 – Final exam Page total: _____ / 30 11. (10 points) An aqueous solution of a 50 kDa protein is contaminated with a trimer of the same material having three times the mass (150 kDa/trimer unit). Based upon this information alone, which liquid chromatography technique would you first try to use in the purification of the sample? Using this technique, which species would you expect to elute from the column first? Why? You should use size‐exclusion chromatography. The trimer (being larger) will elute first. Smaller molecules are able to fit into more of the small pores of the porous stationary phase. This means that they are effectively removed from the mobile phase flow for a longer period of time and thus take longer to elute. Larger molecules, like the trimer (compared to the monomer) cannot access as many of the pores and end up traversing the column more rapidly. 12. (10 points) Briefly, describe electroosmotic flow and the significance it has in standard capillary electrophoresis. Silica capillaries are commonly used in CE which renders the walls negatively charged at any pH above 3 (usually the case). A layer of cations from the mobile phase forms a loosely held double layer that neutralizes the interface. These cations are mobile in the presence of the applied electrophoretic field and migrate towards the negatively biased electrode. The cation movement effectively drags the bulk solvent and all other species with it and thus neutral and anionic also can be analyzed by CE on account of electroosmotic flow. 13. (10 points) Does a double focusing mass analyzer have better or worse resolution than a stand alone magnetic sector mass spectrometer? Why? How does this work? Double focusing mass analyzers have better resolution than stand along magnetic sector mass spectrometers. This is because they use an electrostatic energy filter through with only ions having the same kinetic energy to transit. By narrowing the KE distribution of ions, the magnetic sector, which is the second half of the double focusing instrument, can provide better resolution. Page 3 of 10 Chem 420 – Fall 2008 – Final exam Page total: _____ / 20 14. (20 points) Fill in the blanks for the six radiative and non‐radiative transitions numbered in the figure below. 1. Vibrational relaxation 2. Intersystem crossing 3. Absorbance 4. Fluorescence 5. Phosphorescence Page 4 of 10 Chem 420 – Fall 2008 – Final exam Page total: _____ / 39 The following questions require a longer answer (compared to those immediately above, but not to exceed one paragraph) or perhaps a sketch. 15. (24 points) X‐ray photoelectron, Auger electron, and Energy‐dispersive X‐ray spectroscopies all utilize beam‐sample interactions to probe the chemical composition of a material/surface. Compare and contrast XPS, AES, and EDX in terms of the type of beam used, the type of particle generated from the beam‐sample interaction, and the physical mechanism by which that particle is produced, for each analytical technique. Type of beam used to interrogate surface: (2.5 points each) XPS x‐ray AES electron beam EDX electron beam (from electron microscope) Type of particle generated: (2.5 points each) XPS electron AES electron EDX x‐ray photon Physical mechanism for particle generation: (3 points each) XPS An x‐ray beam interacts with and displaces a core electron from the surface material. The kinetic energy of that ejected electron is measured. AES An electron beam displaces a core electron from the surface material. An electron from a higher energy level falls to fill the void and the difference in energy is released in the form of an emitted electron. It is this emitted electron that is measured in AES. EDX An electron beam displaces a core electron from the surface material. An electron from a higher energy level falls to fill the void and the difference in energy is released in the form of an x‐ray photon. 16. (15 points) Frictional (lateral) force and phase imaging are two operational modes of atomic force microscopy that allow chemical information to be extracted from tip‐surface interactions. Describe the main differences between frictional force and phase imaging in terms of the general imaging mode (how the tip contacts the surface) and method by which differential tip‐surface interactions are transduced by the instrument (how does the instrument “observe” these interactions). Which would you choose if you had to image a very fragile sample? (5 points) Frictional (lateral) lateral force microscopy is a version of contact mode imaging. In this experiment, chemical interactions between the tip and sample lead to a tortional stress (twisting) of the cantilever and thus a lateral (in the plane parallel to the sample) deflection of the laser spot reflected off the back of the cantilever. [Something to this effect] (5 points) Phase imaging is a version of tapping or intermittent contact mode imaging. The microscope tranduces chemical interactions by measuring the lag or acceleration in tip resonance frequency to the drive frequency. [Something to this effect] (5 points) Phase imaging would be the best choice for imaging a very fragile sample because it is an intermittent contact imaging mode. Page 5 of 10 Page total: _____ / 16 17. (16 points total; 4 points each) a. An unknown sample is analyzed on a normal phase HPLC column with a low‐polarity mobile phase and the resulting chromatogram shown on the right. Based upon the order of elution, rank the analytes in order of increasing polarity. A>B>C A is the most polar, followed by B, and C is the least polar. b. Sketch the expected chromatogram (still on a normal phase column) if the solvent polarity was increased. Sketch should show that the elution is faster, but the peaks move closer to one another. For example: c. Would the order of elution change on a reversed phase column? If so, how? Yes. The elution order would be reversed on a reverse phase column. d. What is the effect of reducing the mobile phase polarity on reverse phase separations? Reducing the polarity of the mobile phase in reversed phase separation speeds up elution, but can also lead to a decrease in resolution. Reduced polarity contrast between the stationary and mobile phases tends to speed up separation because the driving interactions for analytes to partition/adsorb to the stationary phase are lessened on account of the lower polarity difference. Chem 420 – Fall 2008 – Final exam Page 6 of 10 Chem 420 – Fall 2008 – Final exam Page total: _____ / 15 Work it out: These questions involve a calculation of one form or another. 18. (15 points) My son Marty reached 16 months of age this week and I wanted to check to see what his weight was. (Recall that I determinedthis earlier in the semester as well, as included on Exam #1). In addition to his growth, he has also become much more active so I had to hold him in my arms and weigh us together, obtaining values of 196.8, 197.3, and 196.6 pounds. After he wriggled out of my arms I weighed myself three times, giving values of 169.9, 169.5, and 169.7 pounds. From this information, determine Marty’s weight and associated error. Note: First determine the mean and standard deviation for the each set of measurements (mean±s.d.) and then remember to propagate the error when determining the difference. Show all work. (3 points each for setting up mean and standard deviation equations, 2 points each for the value of weight from the “both” and “me” measurements, 3 points for taking the difference and 2 points for correctly propagating error; ‐0.5 for sig figs.) MeanBoth = sdBoth = 196.8 + 197.3 + 196.6 590.7 = = 196.9 3 3 3−1 = 0.4 (196.8 − 196.9 )2 + (197.3 − 196.9 )2 + (196.6 − 196.9 )2 Weight of both: 196.9 ± 0.4 pounds 169.9 + 169.5 + 169.7 509.1 = = 169.7 3 3 3−1 = 0.2 MeanMe = sdMe = (169.9 − 169.7 )2 + (169.5 − 169.7 )2 + (169.7 − 169.7 )2 My weight: 169.7 ± 0.2 Marty's weight = Both ‐ Me = (196.9 ± 0.4 ) ‐ (169.7 ± 0.2 ) = 27.2 ± 0.4 (difference should have 3 sig figs) error propagation : error = ( errorA )2 + ( errorB )2 = ( 0.4 )2 + ( 0.2 )2 = 0.4 (1 sig fig) Page 7 of 10 Chem 420 – Fall 2008 – Final exam Page total: _____ / 15 19. (15 points) The vibrational frequency of 12C16O was determined to be 2143.783 Isotope Isotopic mass (g/mol) 12 cm‐1 from infrared vibrational spectrometry. Using this information, determine C 12 (exactly, by definition) 13 the spring constant (in units of N/m) for this vibrational mode and use it to C 13.0033548378 16 predict the IR stretching frequency for 13C16O. O 15.99491461956 (5 points for setting up and manipulating expression for frequency, 5 points for proper value and calculating method for k (3 points for effective mass of species), 5 points for determining the expected value of the infrared vibrational frequency for 13C16O. ) ν= 1 2π c k μ k = ( 2π cν μ 12 C −16 O = k = ( 2π cν )2 ⋅ μ (12 g/mol) ⋅ (15.99491461956 g/mol) (12 g/mol) + (15.99491461956 g/mol) 6.022141x1026 = 1.138500 ⋅ 10 −26 kg 2 ) 2 ⎛ 100 cm ⎞ ⎞ N ⎛ −26 ⋅ μ = ⎜ 2π ⋅ 2.99792 ⋅ 108 ⋅ ⎜ 2143.783 cm‐1 ⋅ ⎟ ⎟ ⋅ 1.138500 ⋅ 10 kg = 1856.50 m ⎠⎠ m ⎝ ⎝ ( ) ν= 1 2π c k μ μ 13 C −16 O = ν 13 C −16 O = (13.0033548378 g/mol) ⋅ (15.99491461956 g/mol) (13.0033548378 g/mol) + (15.99491461956 g/mol) 6.022141x1026 1 2π c = 1.191007 ⋅ 10−26 kg N 1856.50 1 m = ⋅ ⋅ = 2096.00 cm‐1 μ 2π ⋅ 2.99792 ⋅ 108 m/s 1.191007 ⋅ 10 −26 kg k Page 8 of 10 Chem 420 – Fall 2008 – Final exam Page total: _____ / 25 20. Throughout our discussion of atomic emission spectrometry, we highlighted the temperature dependence of the excited state population. This is after all one of the reasons that ICP sources are valuable tools for atomic emission spectrometry. However, we also discussed the Doppler Effect as one source of temperature‐dependent line Δλ v = , where c is the velocity of broadening, whereby the spectral line width, Δλ, can be determined according to: λ c light, and ν is the average speed of an atom, given by: by v = 8kT , where k is Boltzmann’s constant, T is the πm absolute temperature, and m is the mass.Herein lies a potential dilemma—particularly in the analysis of heavier element‐containing samples that have a large number of emission lines that can begin to spectrally overlap. To further investigate these corroborating phenomena, lets choose the most famous atomic emission line; the 3p‐3s atomic sodium transition at 588.9950 nm (gj=6, g0=2). Use 22.98977 g/mol for the atomic molar mass of sodium. a. (10 points, No penalty for sig figs) Considering the temperatures as exact numbers and neglecting all other sources of line broadening (Doppler Effect only), calculate the line width of the transition measured at 10,000 K (hottest zone within an ICP torch) and at 2100 K (hottest zone within a natural gas‐air flame). mNa = 22.98977 g/mol ⋅ v10,000K = 1kg 1 mol ⋅ = 3.817541 kg 103 g 6.022141 ⋅ 1023 atoms 8 ⋅ 1.38065 ⋅ 10 −23 J/K ⋅ (10,000 K ) ( π ⋅ 3.817541 ⋅ 10−26 kg v10,000K ⋅ λ c ( ) ) = 3034.730 m/s Δλ10,000K = v2,100K = = 3034.730 m/s ⋅ 588.9950 nm = 5.96227 ⋅ 10 −3 nm 2.99792 ⋅ 108 m/s 8 ⋅ 1.38065 ⋅ 10 −23 J/K ⋅ ( 2,100 K ) ( π ⋅ 3.817541 ⋅ 10 −26 kg v2,100K ⋅ λ c ( ) ) = 1390.688 m/s Δλ2,100K = = 1390.688 m/s ⋅ 588.9950 nm = 2.73226 ⋅ 10−3 nm 8 2.99792 ⋅ 10 m/s b. (10 points, No penalty for sig figs) Assuming that the ratio of atoms in the excited versus ground state for each temperature is directly proportional to the emission intensity: calculate the ratio of the emission intensity ⎛ I10,000K ⎞ ⎜ ⎟ at the two temperatures mentioned in part a given the same set of assumptions. ⎝ I2100K ⎠ E= hc λ (6.62607 ⋅ 10 = −34 Js ⋅ 2.99792 ⋅ 108 m/s ‐9 ⎛ E ⎞ )( 588.9950 ⋅ 10 m ) = 3.37260 ⋅ 10 −19 J I10,000K I2,100K ⎛ ⎞ 3.37260⋅10−19 J ⎛ Nj ⎞ ⎜ ⎟ g j −⎜ k (10000K ) ⎟ ⎟ −⎜ ⎠ ⎜ 1.38065⋅10 −23 J/K (10000K ) ⎟ e⎝ ⎜ ⎟ ⎜( ⎟ ) ⎝ ⎠ N0 ⎠10,000K g0 ⎝ e 0.0869204 = = = 9793.16 = = ⎛ ⎞ ⎛ ⎞ E ⎛ Nj ⎞ 8.87562 ⋅ 10−6 3.37260⋅10 −19 J ⎟ −⎜ ⎜ ⎟ g j −⎜ k (2100K ) ⎟ ⎜ ⎟ ⎜ 1.38065⋅10 −23 J/K ( 2100K ) ⎟ ⎠ ⎜( ⎟ ) e⎝ ⎠ ⎝ N0 ⎠2,100K e⎝ g0 c. ( 5 points) Briefly, comment on how this might impact the analysis of heavier elements such as Neodymium and Samarium, which have atomic emission lines at 359.259 nm and 359.260 nm, respectively. (Anything even somewhat reasonable will receive the points) One might have to consider whether it would be more advantageous to take a hit on intensity in order to have a better chance of resolving two emission lies. More insightful: Since the intensity falls off as eT and Doppler broadening as T , you should try to use the highest resolution spectrometer possible because population distribution will kill you very quickly (~2 fold reduction in line broadening, but 104 reduction in emission intensity over the same temperature range). Page 9 of 10 Chem 420 – Fall 2008 – Final exam Page total: _____ / 25 5.62 5.29 21. (25 points) The chromatogram at right was obtained on a reversed phase HPLC column that was 25.0 cm in length, using UV absorbance detection. Unretained compounds elute in 0.25 minutes. Start by listing your estimated times. You won’t be counted against as long as the numbers are reasonable. You should estimate to 0.01 accuracy (i.e., 5.29 for the retention time of B), but you certainly can’t resolve better than that given this plot. a. Calculate the number of theoretical plates for components B and C. Determine the average number of theoretical plates based upon these two components values. ⎛ ( t r )B NB = 16 ⎜ ⎜ WB ⎝ 2 5.19 5.42 5.59 5.80 2 2 ⎞ 5.29 ⎛ ⎞ ⎛ 5.29 ⎞ 3 ⎟ = 16 ⎜ ⎟ = 16 ⎜ ⎟ = 8.5 ⋅ 10 or 8464 plates 2 or 4 sig figs ⎟ ⎝ 5.42 − 5.19 ⎠ ⎝ 0.23 ⎠ ⎠ 2 2 2 ⎛ ( tr )C ⎞ 5.62 ⎛ ⎞ ⎛ 5.62 ⎞ 4 NC = 16 ⎜ = 16 ⎜ ⎟ = 16 ⎜ ⎟ ⎟ = 1.1 ⋅ 10 or 11459 plates 2 or 5 sig figs ⎜ WC ⎟ ⎝ 5.80 − 5.59 ⎠ ⎝ 0.21 ⎠ ⎝ ⎠ 8464 + 11459 = 9962 or 9.75 ⋅ 103 plates Naverage = 2 b. Using the average number of plates you just calculated, determine the average plate height for this separation. H= L 25.0 cm = = 2.51 ⋅ 10−3 cm 3 sig figs N 9962 plates H = 2.56 ⋅ 10‐3 cm if using 9.75 ⋅ 103 for Naverage c. Calculate the retention factor for component C. kC = (tr )C − tm tm = 5.62 − 0.25 = 21 2 sig figs 0.25 d. Calculate the resolution for the separation of components B and C. 2 ⎡( tr )C − ( tr )B ⎤ 2 ( 5.62 − 5.29 ) 2 ( 0.33 ) ⎦= Rs = ⎣ = = 1.5 2 sig figs WC + WB 0.23 + 0.21 0.44 e. Calculate the selectivity factor for the separation of components B and C. α= (tr )C − tm (tr )B − tm = 5.62 − 0.25 = 1.07 3 sig figs 5.29 − 0.25 Page 10 of 10 ...
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