exam 1 solution - Chemistry 420 – Fall 2008 Name:

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Unformatted text preview: Chemistry 420 – Fall 2008 Name: _______________________________ Prof. Ryan C. Bailey Exam #1 (175 points total) This exam is 8 pages long Due at the end of class 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. Key Chem 420 – Fall 2008 – Exam #1 Name: _______________________________ Appetizers—Responses for questions 1‐4 can be answered in just a few sentences 1. (15 points) How does slit width impact an optical measurement made with a grating monochromator? In general, what is the optimal slit width? (Note: I am not looking for a numerical answer, but rather the general objective) Decreasing the slit width of a monochromator increases the resolution (decreases the spectral bandwidth) (5 pts) Decreases the slit width also reduces the light throughput (5pts) In general, one should use the largest slit width capable of providing the required resolution for the measurement (5 pts) 2. (10 points) In atomic emission spectroscopy, why is an inductively coupled plasma source less susceptible to interfering chemical species and matrix effects than a flame source? The plasma is significantly hotter (2‐3 times) than the highest temperature fuel‐oxidant flame. At these higher temperatures, atomization is much more complete (7 pts) and thus there is a reduction in interference and matrix effects. Additionally, the inert environment of the plasma (no oxidant) prevents oxide formation (3 pts), which in turn increases the lifetime of the element. Key Page 2 of 8 Chem 420 – Fall 2008 – Exam #1 Name: _______________________________ 3. (15 points) Atomic line widths are broadened by four effects: the uncertainty effect, the Doppler effect, pressure broadening, and electric and magnetic field effects (not discussed in class). Pick two of the three we discussed and briefly explain their origin. Of the two you choose, which has the largest contribution to line broadening? Uncertainty : The finite (and short) lifetime of the transition states leads to uncertainty in the spectral frequency. ΔνΔt ≥ 1 Doppler: The wavelength of radiation emitted or absorbed b y a rapidly moving object decreases or increases as the object moves toward or away from the transducer. Atoms in these measurements are moving. Pressure : Collisions between absorbing or emitting species cause small changes in energy levels and this spectral broadening Order: Uncertainty < Doppler < (or equal to) Pressure (5 pts each) 4. (10 points) Stimulated emission is a key process in achieving lasing. Briefly describe stimulated emission, mentioning several key differences from spontaneous emission. Stimulated emission is induced by collisions (interactions) with photons of the same energy as spontaneous emission. (5 pts) Stimulated emission is precisely in phase (time) and occurs in the same directions as the colliding (interacting) photon. (Coherent). Spontaneous emission is not in phase (incoherent) (5pts) Page 3 of 8 Key Chem 420 – Fall 2008 – Exam #1 Name: _______________________________ Main Course—Questions 5‐7 are numerical problems and their solutions require calculations 5. (25 points) This past weekend, I weighed my son Marty on our bathroom scale multiple times in succession and obtained the following weights: 23.8, 24.3, 24.5, 23.7, and 24.1 lbs. Determine the mean and standard deviation of the results. (Remember to show all work—magically appearing answers will not be accepted.) Mean = 23.8 + 24.3 + 24.5 + 23.7 + 24.1 120.4 = = 24.08 lbs (4 sig figs) 5 5 Key (10 pts for mean 3 pts for sig figs) Standard Deviation = (23.8-24.08)2 + (24.3-24.08)2 + (24.5-24.08) 2 + (23.7-24.08) 2 + (24.1-24.08) 2 = 0.11 = 0.33 (2 or 3 sig figs) 5-1 (10 pts for standard deviation 2 pts for sig figs) Page 4 of 8 Chem 420 – Fall 2008 – Exam #1 Name: _______________________________ 6. (30 points) I have constructed a simple photometer consisting of only a flashlight, a simple interference filter that transmits light at 649 nm, a silicon photodiode, and a handheld multimeter. Shining the flashlight first through the filter and then through a 2.54 cm pathlength cuvette filled with an aqueous solution containing 396 μg/L of compound Q (791.99 g/mol), I measure a voltage of 0.572 V on the multimeter. Replacing the sample with the blank, a voltage of 1.189 V was measured. Determine the following: a. What is the transmittance of the sample of Q? T= P 0.572 V = = 0.481 (3 sig figs) Po 1.189 V Key (7 pts) b. What is the absorbance of the sample of Q? A = − log T = log P = log(0.481) = 0.318 (2 or 3 sig figs) Po (7 pts) c. What is the molar absorptivity of compound Q? c= 396μg 1g 1mol ⋅6⋅ = 5.00 × 10−7 M 1L 10 μg 791.99g A 0.318 = = 2.50 × 105 M -1cm -1 (3 sig fig) bc 2.54cm ⋅ 5.00 × 10−7 M A = ε bc → ε = (7 pts) d. Now, let’s use this previous sample of Q as a standard to determine the concentration of an unknown aqueous sample of compound Q (Unknown Q). A 99.00 mL aliquot of Unknown Q was combined with 1.00 mL of deionized water and photometric measurement gave an absorbance of 0.194. A second 99.00 mL aliquot of Unknown Q was combined with 1.00 mL of the standard and photometric analysis gave an absorbance of 0.333. What is the concentration of Unknown Q in units of mol/L? cx = Ax ⋅ cs ⋅ Vs 0.194 ⋅ 5.00 × 10−7 M ⋅1.00mL 9.70 × 10−8 M = = = 7.19 × 10−9 M (3 sig figs) ( Ax + s − Ax ) ⋅ Vx (0.333 − 0.194) ⋅ 99.00mL 13.5 (9 pts) Page 5 of 8 Chem 420 – Fall 2008 – Exam #1 Name: _______________________________ 7. (40 points) The fundamental transition between the ground and first vibrational Isotope Isotopic mass (g/mol) 1 excited state for H35Cl is characterized by an infrared absorption signal at 2990.97 H 1.00783 ‐1 2 cm . Using the constants at right, determine the following: H 2.01410 3 H 3.01605 35 a. What is the force constant of the H—35Cl stretch in kg/s2 (same as N/m)? Cl 34.96885 37 (10pts) Cl 36.96590 μH-35Cl = 1.00783g/mol ⋅ 34.96885g/mol ⋅ 1.00783g/mol + 34.96885g/mol 1 6.022141× 1023 atoms 103 g ⋅ mole kg = 0.979587 6.022141× 1026 = 1.62664 × 10−27 kg Key ⇒ k = ( 2π cν ν =⎜ ⎟ ⎝ 2π c ⎠ μ ⎛1k )2 μ = ⎛ 2π (2.99792 × 1010 ⎜ ⎝ cm ⎞ kg N ⋅ 1.62664 × 10−27 kg = 516.319 2 = 516.319 s⎟ m s ⎠ (6 sig figs) (10 pts) b. Assuming that the force constant does not change, what is the frequency (in wavenumbers) of the fundamental (v=0 1) 1H—37Cl stretch? (10 pts) μH-35Cl = 1.00783g/mol ⋅ 36.96590g/mol ⋅ 1.00783g/mol + 36.96590g/mol 1 6.022141× 1023 atoms 103 g ⋅ mole kg = 0.979587 6.022141× 1026 = 1.62912 × 10−27 kg kg ⎛ ⎞ 516.319 2 ⎟ 1 ⎛1⎞k ⎜ s =⎜ = 2988.70cm-1 ν =⎜ ⎟ ⎟ ⎝ 2π c ⎠ μ ⎜ 2π ⋅ 2.99792 × 1010 cm ⎟ 1.62912 × 10−27 kg ⎜ ⎟ s⎠ ⎝ c. (6 sig figs) (10 pts) H Cl and 3H35Cl differ in mass by less than 0.03%, however, they have significantly different IR absorption spectra. Calculate the difference in energy (in Joules) between the fundamental vibrational transitions of 1H37Cl and 3H35Cl. (20 pts) 1 37 kg ⎛ ⎞ 516.319 2 ⎟ 1 ⎛1⎞k ⎜ s =⎜ = 1776.56cm-1 ν =⎜ ⎟ ⎟ ⎝ 2π c ⎠ μ ⎜ 2π ⋅ 2.99792 × 1010 cm ⎟ 4.61060 × 10−27 kg ⎜ ⎟ s⎠ ⎝ μH-35Cl = 3.01605g/mol ⋅ 34.96885g/mol ⋅ 3.01605g/mol + 34.96885g/mol 1 6.022141× 10 23 atoms 10 g ⋅ mole kg 3 = 2.77657 6.022141× 1026 = 4.61060 × 10−27 kg ΔE = 2988.70cm-1 − 1776.56cm-1 = 1212.14cm -1 E = hcν = 6.62607 × 10−34 Js ⋅ 2.99792 × 1010 cm ⋅1212.14cm -1 = 2.40785 × 10−20 J s (6 sig figs) (20 pts) Page 6 of 8 Chem 420 – Fall 2008 – Exam #1 Name: _______________________________ Dessert—Choose your own adventure: Select either problem 8 or 9 to work for credit. In the event that you have extra time, you may choose to complete the other question for up to 10 points of extra credit. If you work on both, you MUST indicate which problem is to be considered for regular (not extra) credit. Failure to indicate your selection will result in our grading problem 8 for regular credit. 8. (30 points) Intentional melamine contamination in some brands of Chinese manufactured pet food (2007) and infant formula (current) has led to the death of many pets and a handful of children on account of acute kidney toxicity. Thankfully, my son no longer drinks formula, but I would like to test my dog’s food to check the melamine level. (Note: Melamine by itself is minimally toxic, however, when it is combined with cyanuric acid, also present in the dangerous products, it can form fatal kidney stones.) Earlier this year, Lin and co‐workers reported a surface‐ enhanced Raman method for determining the concentration of melamine in food samples (J. Food Science, 2008, 73, T129‐T134 and Sens. & Instrumen. Food Qual., 2008, 2, 66‐71). Using a similar method, I measured the Raman scattering intensity at 676 cm‐1 of several standard melamine samples and the plotted the results below. Also included is the Excel regression analysis with slope and intercept (and corresponding errors) in bold font. SERS intensity at 676 cm (arbitrary units) Melamine Raman scattering concentration (μM) intensity (arbitrary units) 8000 -1 Key 7000 6000 5000 4000 3000 2000 1000 0 0 0 10 20 30 40 50 18 1478 2943 4337 5712 7040 y = 140.59x + 73.286 R2 = 0.9996 10 20 30 40 50 60 Melamine concentration (μM) SUMMARY OUTPUT Intercept X Variable 1 Coefficients Standard Error 73.28571429 40.78069772 140.5885714 1.346942115 I took 100 g of dog food, extracted the melamine into 100 mL of the same solvent used to generate the calibration curve, and measured a SERS intensity to be 1737 arbitrary units. a. What is the concentration of melamine in the extracted solution in units of mol/L? SERS Signal-140.6 ± 1.3 [ melamine] + (70 ± 40) [ Melamine] = SERS − (70 ± 40) 1737 − (70 ± 40) 1667 ± 40 = = = 11.9 ± 0.3μM 140.6 ± 1.3 140.6 ± 1.3 140.6 ± 1.3 (5 pts) (10 pts) error1 = 402 = 40 ⎛ 40 ⎞ ⎛ 1.3 ⎞ Error2 = ⎜ ⎟ +⎜ ⎟ = 0.305 ⎝ 1667 ⎠ ⎝ 140.4 ⎠ 2 2 b. What is the mass percentage of melamine (FW=126.12 g/mol) in the sample? Mass = 11.9 ± 0.3μmol 1mole 126.12g × 0.1L = 1.19 ± 0.03μmol ⋅ 6 ⋅ = 1.50 ± 0.04 × 10−4 g L 10 μmol 1mole (5 pts) (10 pts) Mass % = Mass relative 1.50 ± 0.04 × 10−4 g × 100% = × 100% = 1.50 × 104 % Melamine Total Mass 100g Page 7 of 8 Chem 420 – Fall 2008 – Exam #1 Name: _______________________________ 9. (30 points) In 1979, Fernandez and Bastiaans studied the temperature variation at 11 cm above the burner top of an air‐acetylene flame as a function of oxidant‐fuel ratio. (Applied Spectroscopy, 1979, 33, 145‐150.) Notably, they found that a 13% decrease in air‐acetylene ratio caused a drop in flame temperature (at +11 cm) from 2147oC to 2032oC. Consider the 5P3/2 to 5S1/2 atomic emission line for atomic Rubidium (Rb), which appears at 780.0 nm. Assuming that the ratio atoms in the excited and ground state is directly proportional to the emission intensity; calculate the relative intensity of the emission spectra obtained at the two different air‐acetylene ratios. (Note: gj/g0=2 for 5P3/2 to 5S1/2) E= hc Key λ = 6.62607 × 10−34 Js ⋅ 2.99792 × 108 780.00 × 10−9 m m s = 2.5467 × 10−19 J (5 pts) 2147°C 2420K Nj Nb gj go −E ⎛ ⎞ ⎜ ⎟ −3.5467×10−19 J ⎜ ⎟ ⎜ 1.38065×10−23 J ⋅2420K ⎟ ⎜ ⎟ K ⎝ ⎠ 2e I 2420 = = e kT = = 9.790 × 10−4 (5 pts) 2032°C 2305K Nj Nb gj go −E ⎛ ⎞ ⎜ ⎟ −2.5467×10−19 J ⎜ ⎟ −23 J ⎜ 1.38065×10 ⋅2305K ⎟ ⎜ ⎟ K ⎠ 2e ⎝ I 2420 = = e kT = = 6.693 × 10−4 (5 pts) I 2420 9.790 × 10 −4 = = 1.463 I 2308 6.693 × 10−4 or (5 pts) I 2305 6.693 × 10 −4 = = 0.6837 I 2420 9.790 × 10−4 Page 8 of 8 ...
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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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