CHM 424 EXAM 3 CRIB - COVER PAGE FALL

Transcription

1 CHM 424 EXAM 3 CRIB - COVER PAGE FALL 26 There are six numbered pages with ive questions. Answer the questions on the exam. Exams done in in are eligible or regrade, those done in pencil will not be regraded. 1 coulomb = charges 1 amp = 1 coulomb per 1 second 1 volt = 1 joule per 1 coulomb ε = C 2 J -1 m -1 h = J s = J K -1 c = m s -1 1 amu = g =.694 cm -1 K -1 N A = mol -1 E = hν A = log ( T ) σln A = 2.3 A = εlc τ= 1 t I = Ie φ τ = ix + q[ Q] ix q [ Q ] 1 τ = + ix φ = + ix ( λ, λ ) = ( λ ) Δλ ( λ ) 1 ( λ ) φ F( λ ) Δλ E ( λ ) D( λ ) I I E T ex em s ex ex ex ex ex em em em em em mm ν= ΔE 1 μ= 1 2 h = 2 π μ E m1+ m2 v h 1 = v + 2π μ 2 dθ = dλ m acos θ m dl dλ = dθ dλ N N 1 ΔE = exp T sinθ m = mλ a 2λ Δ θ = Na cosθ m

2 CHM 424 EXAM 3 CRIB FALL 26 Name Score /15 3 pts. 1. Provide the ollowing deinitions or acts at 6 pts each. a. What aspect o an absorption spectrophotometer limits the maximum absorption that can be measured? stray light in the monochromator limits the minimum measurable intensity b. In an experiment, molar absorptivity is computed using absorbance measured at the top o a pea. What happens to the computed value as the bandwidth o the spectrophotometer is increased? as the bandwidth is increased the pea becomes squatter and atter, thus the absorption and computed absorptivity decrease c. Why is the optical path in an absorption spectrophotometer NOT arranged in the order source, sample, monochromator, and detector? The lamp will photolyse the sample i all its radiation impinges on the sample. When the ordder is source, monochromator, sample, only a small range o wavelengths impinge on the sample. d. Give the molecular ormula or one example o a molecule that has no inrared active vibration. F 2 e. For each set o molecules circle the compound which would absorb at the longest UV/visible wavelength. set 1 set 2 H CH 3 OH Exam 3 Crib, 26, page 1

3 3 pts. 2. Perorm the ollowing calculations at 1 pts each. a. The quantum yield o a molecule is 1. The luorescence lietime is 1 ns. What is the rate constant or luorescence,? i φ = 1, << ix + q [Q] then τ = 1/ or = 1/τ = 1 9 s -1 b. How ar does an FTIR mirror have to move to produce a spectral resolution o 2 cm -1? a resolution o 2 cm -1 corresponds to an optical path length change o 1/2 cm -1 =.5 cm the mirror moves hal this distance,.25 cm. c. An absorption transition shows three vibronic bands at 385, 435, and 5 nm. What is the requency (in cm -1 ) o the vibration? 385 nm = 25,974 cm nm = 22,988 cm -1 5 nm = 2, cm -1 The vibrational requency is the dierence between any two o these, e.g. ~3, cm -1 Exam 3 Crib, 26, page 2

4 3 pts. 3. Answer the ollowing two homewor questions or 15 pts each. a. Atomic absorption doesn't use molar absorptivities because the atomic sample is rarely dissolved in solution. However, it is o academic interest to see how much stronger atomic absorption is than molecular absorption. To this end, convert an atomic cross section o 1-13 cm 2 to ε in L mole -1 cm ( ) σ ε = = = b. Given that the vibrational requency o H 2 is 4,4 cm -1, estimate the vibrational requencies o HD and D 2. Assume that the bond orce constant is the same or all three compounds. ν ν HD DD.5 = 4, 4 = 3, = 4, 4 = 3, Exam 3 Crib, 26, page 3

5 3 pts. 4. Shown below is the excited state diagram and orbital diagram or a molecule, lie phenol or aniline, that has a lone pair o electrons. Only the highest occupied π and lowest unoccupied π* orbitals are shown. The lone pair is denoted by l. In the table below the igure, show how many electrons are in each orbital or each labeled energy level. Use arrows to eep trac o electron spin. Also, give the quantum number, v, associated with each vibronic level. The vibronic level o interest cuts through the middle o the label. Labels Î, Ö and reer to S ; Ï and Ð to S π6π* ; Ñ and Ò to S l6π* ; Ó to T l6π* ; and, Ô and Õ to T π6π*. label v =? π* l π Exam 3 Crib, 26, page 4

6 3 pts. 5. Fluorescence is oten used to quantiy at the picomolar level. This is most oten done by preparing standards and creating a calibration graph. The chie technical impediment to this approach is developing a protocol to prepare accurate standard solutions this dilute. A urther need is to convince yoursel that a solution labeled M is indeed that concentration. That is, you have not lost or gained analyte using your protocol. The ollowing is the general aproach. A microbalance is used to weigh 1 ± 1 μg. This mass is dissolved in 1 ml o ethanol to mae a stoc solution. The stoc is then diluted into a series o 1 ml volumetric lass using 1, 2,..., 1 ml pipettes to create concentrations extending one decade. For each o these solutions obtain a 1-old dilution by pipetting 1 ml into 1 ml o ethanol. Repeat this last step to get into the picomolar range. Consider such an experiment using rhodamine B as the luorophore. The molecular mass o rhodamine B is 479 g mol -1. The molar absorptivity at the absorption maximum is 11, M -1 cm -1. The calibration graph is shown on the next page. The slope o the calibration curve is photons M -1. Ater that long preamble, here is the question! Propose a method or demonstrating that the calibration curve concentration labels are accurate. This proo will indirectly rely on the molar absorptivity and will be general (that is, not restricted to rhodamine B). Your method can only use an absorption spectrophotometer, the same luorimeter that yielded the picomolar calibration graph, balance, pipettes, and volumetric lass. The strategy has two parts: (1) i no solution concentrations are erroneous, the slope o the luorescence calibration curve will be independent o luorophore concentration; (2) i solution concentrations are accurate, the slope o the absorbance calibration curve is the molar absorptivity. First, prepare a decade range o solutions that provide linear calibration curves or both luorescence and absorption. The slope o the absorbance calibration curve can be compared to the molar absorptivity to see i they are accurate. I they are accurate, a luorescence calibration curve obtained or these same solutions will yield the sensitivity o the luorimeter. Since luorimeter sensitivity is independent o concentration, the slope o this calibration curve should be the same as that or the picomolar solutions. There are a couple practical problems. First, the absorbance has to be suiciently high that the calibration curve yields an accurate estimate o the molar absorptivity. This might require that the absorbance values all within the range rom, say, 1. to.1. Second, the concentration o the solutions used to obtain the absorption calibration curve have to be suiciently dilute that the luorescence calibration curve is linear. For rhodamine B, 1 μg diluted into 1 ml produces a solution M. I 1, 2,..., 1 ml o this solution are diluted to 1 ml the resulting solutions will be M to M. In a 1-cm cell the resulting absorbance values will be.23 to.231, and the slope should be a reliable indicator o ε l. Second, the solutions have to have absorbance values small enough to produce a linear calibration curve. In a 1-cm cell the most concentrated solution, M, will have an absorbance o.23. This value is less than the "rule o thumb" given in your notes or luorescence linearity, i.e. 2.3 A <.1. So, the slope o the luorescence curve should be an accurate indicator o instrumental sensitivity. Exam 3 Crib, 26, page 5

7 Callibration Graph 1 75 counts r( x) Molarity, x Knowing that the range o concentrations rom M to M are accurate, the two steps o dilution can be used to obtain solutions rom M to M. These are only o by a actor o ~2 rom the graph, but should be suicient or obtaining a calibration regression curve. I the slope is not the same as the more concentrated solutions a mistae has been made during the dilution procedure. This is the most common place an error is made. Most oten the pipettes are contaminated with small amounts o the luorophore, maing the signal greater than it should be. Also common is the adsorption o the luorophore on the pipette and volumetric walls, maing the signal less than it should be. One additional practical problem (that I don't expect you to now) is the magnitude o the luorescence signal when the high concentration solutions are used. For a sensitivity o photons M -1 and a concentration o M, the signal will be counts. This value is so high it will destroy most photomultipliers. To eep this rom happening the excitation intensity will have to be reduced by about a actor o 1. Obviously the reduction actor has to be accurately nown, usually by calibration in a spectrophotometer that can measure absorbance values up to 3 or more. As you can see, this simple protocol "guarantees" that the x-axis labels are reasonably accurate and you can brag about the limit o detection. Exam 3 Crib, 26, page 6