Exam 3 Chem 454 April 25, 2018, Name

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Exam 3 Chem 454 April 25, 2018, Name K = Cs/Cm k = (tr tm)/tm tr = tr tm = tr2 /tr1 = K2/K1 H = L/N = 2 /L Rs = tr/wavg R s t r, B r, A N t t r, B 4 70 total points. 14 questions @ 5 points each.s 1] Sketch the Raman and IR active modes of CO2. 1 2] Assume the source of the Michelson inferometer below is at a single wavelength,. Sketch the response of the detector as movable mirror moves back from /4 to. 2 3] Sketch the Stokes, anti-stokes, and Rayleigh transitions in the following diagram. Which one is associated with the Raman Effect? 3

4] What is population inversion in the lasing effect? 4 5] What are the refractory oxides? Name two examples. Why are they are a problem in AA spectroscopy? 5 6] Rank the following atomic spectroscopy techniques in terms of LOD trends from lowest to highest. Lowest LOD > Intermediate LOD > Highest LOD 6 a) GFAAS Flame AAS >> ICP-AES b) ICP-AES = Flame AAS > GFAAS c) Flame AAS > ICP-AES > GFAAS d) GFAAS ICP-AES > Flame AAS e) ICP-AES > Flame AAS > > GFAAS 7] Label the following diagram in terms of the three major contributions to plate height, H in the van Deemter Equation. What the name of each effect? Draw the total effect. Where is the optimal flow rate? 7 8] Which of the three terms in the van Deemter Equation contributes most to band broadening in GC? How do capillary columns address that effect? 8

9] Contrast the TCD and the FID for GC. Discuss the operational theory of the devices. Discuss which is a universal detector and which might be specific to a class of analytes. Which of the two has a lower LOD? 9 10] A non-polar s.p. is used in a GC separation. Predict the elution order of the following for compounds in terms of shortest to longest retention times. 10 propanol, octanol, methanol & ethanol 11] Sketch the chromatogram expected if the resolution, Rs is reported as 1.0. 11 12] Calculate the increase in resolution Rs, by doubling the column length, L. 12 13] A GC analysis was conducted for formaldehyde using 10 ppm isopropanol as an internal standard (IS). The results are summarized below. tr (mins) signal Injection 1 25 ppm formaldehyde 2.35 4335 10 ppm isopropanol 1.11 3440 Injection 2 unknown formaldehyde 2.37 5032 10 ppm isopropanol 1.13 3122 What is the concentration of formaldehyde in the sample? 13 14] A flame AAS analysis was conducted on 0.1050 g of a catalyst for Pt. This was conducted by dissolving that catalyst in 100.00 ml of conc. HNO3. The solution was divided into two 50.00 ml aliquots and treatment as described along with their spectral absorbances below. abs. Solution A) 50.00 ml aliquot + 1.00 ml blank spike 0.341 Solution B) 50.00 ml aliquot + 1.00 ml spike of 5.32 x 10-3 M Pt salt 0.417 What is the mass percentage of Pt in that catalyst? 14 1 IR Active modes below. The inactive (1340 cm -1 ) is Raman active.

2 3 The Stokes transition is associated with the Raman effect.

4 This is where the population of the excited electronic state is greater than the ground state. Note this that is cannot be achieved by heat but requires incident photons.

5 The refractory oxides are the translucent heat-stable forms of the metal/metalloid oxides that cause light scattering within the flame. This is a non-absorption route for the decrease in the power of the emergent beam, and thus adds to the background. Examples: Al 2O 3, SiO 2, B 2O 3, SnO 2. 6 d) GFAAS ICP-AES > Flame AAS 7 Optimal Flow Rate Total H = A + B/u + Cu Cu, Mass Transfer H, cm A, multiple paths B/u, Linear Diffusion Velocity 8 The linear diffusion or the B/u term of the van Deemter Equation is the largest contributor to band broadening in GC. This is from diffusion being large in the gas phase. As evident from B/u, faster m.p. flow rates minimize this effect. Capillary columns allow for faster flow rates as there aren t obstructions from packing particles. Other advantages include thinner s.p. for faster MT and only one path. 9 TCD thermal conductivity detector. It is based on differences in thermal conductivity between pure gaseous m.p. and when the analytes are in the m.p. In the diagram below, the temperature of the resistive element increases as an analyte migrates between it and the metal block. i Metal Block Metal Block Flow from Column As the m.p. in GC is almost always He, anything else in that gaseous phase increases the T of the heating element. -It is universal but has a relatively high LOD.

FID flame ionization detector. It based on the reduction of radical produce in the flame. The apparatus is below: (-) Current Meter Flame Air H 2 Column Flow The current from the reduction of these radicals (usually CHO + ) is proportional to the analyte concentration. -Specific for organics, lower LOD than TCD. 10 methanol, ethanol, propanol, octanol 11 Remember that there s complete separation with Rs = 1.5, so at Rs = 1.0 they ll be peak overlap. 12 Rs = L Rs = 2 = 1.41, the resolution increases by 1.41 13 5032 * (3440/3122) * (25 ppm/4335) = 32.0 ppm 14 First calculate conc. Pt spike in Solution B. 5.32 x 10-3 (1.00/51.00) = 1.043e-4 M Now plot Signal vs. spike concentration and find slope. Slope = (0.417-0.341) / 1.043e-4 = 7.29e2 line: y = 7.29e2 (x) + 0.341 next find x-int. 0 = 7.29e2 (x) + 0.341 x = 4.68e-4 M Now calc. mass of Pt. this is the conc. of Pt in the 51.00 ml solution mass Pt = (4.68e-4 mol / L) * 0.051 L*195.08 g/mol = 4.66 e-3 g Remember 2 aliquots, 2*4.66e-3 g = 9.32e-3g mass % = (9.32e-3 / 0.1050)*100 = 8.87%