Skoog Solution of Chapter 15

Crouch Principles of Instrumental Analysis, 6th ed. Chapter 15 Instructor’s Manual CHAPTER 15 15-1. In a fluorescence emission spectrum, the excitation wavelength is held constant and the emission intensity is measured as a function of the emission wavelength. In an excitation spectrum, the emission is measured at one wavelength while the excitation wavelengths are scanned. The excitation spectrum closely resembles an absorption spectrum since the emission intensity is usually proportional to the absorbance of the molecule. 15-2. a) Fluorescence is the process in which a molecule, excited by the absorption of radiation, emits a photon while undergoing a transition from an excited singlet electronic state to a lower state of the same spin multiplicity (e. g. , a singlet > singlet transition).Phosphorescence is the process in which a molecule, excited by the absorption of radiation, emits a photon while undergoing a transition from an excited triplet state to a lower state of a different spin multiplicity (e. g. , a triplet > singlet transition). (c) Resonance fluorescence is observed when an excited species emits radiation of he same frequency at used to cause the excitation. (d) A singlet state is one in which the spins of the electrons of an atom or molecule are all paired so there is no net spin angular momentum (e) A triplet state is one in which the spins of the electrons of an atom or molecule are unpaired so that their spin angular moments add to give a net non-zero moment. (f) Vibrational relaxation is the process by which a molecule loses its excess vibrational energy without emitting radiation. 1 Principles of Instrumental Analysis, 6th ed. (g) Chapter 15Internal conversion is the intermolecular process in which a molecule crosses to a lower electronic state with emitting radiation. (h) External conversion is a radiationless process in which a molecule loses electronic energy while transferring that energy to the solvent or another solute. (i) I ntersystem crossing is the process in which a molecule in one spin state changes to another spin state with nearly the same total energy (e. g. , singlet > triplet). (j) Predissociation occurs when a molecule changes from a higher electronic state to n upper vibrational level of a lower electronic state in which the vibrational energy is great enough to rupture the bond. (k) Dissociation occurs when radiation promotes a molecule directly to a state with sufficient vibrational energy for a bond to break. (l) Quantum yield is the fraction of excited molecules undergoing the process of interest. For example, the quantum yield of fluorescence is the fraction of molecules which have absorbed radiation that fluoresce.Chemiluminescence is a process by which radiation is produced as a result of a chemical reaction. 5-3. For spectrofluorometry, the analytical signal F is proportional to the source intensity P0 and the transducer sensitivity. In spectrophotometry, the absorbance A is proporti onal to the ratio of P0 to P. Increasing P0 or the transducer sensitivity to P0 produces a corresponding increase in P or the sensitivity to P. Thus the ratio does not change. As a result, the sensitivity of fluorescence can be increased by increasing P0 or transducer sensitivity, but the that of absorbance does not change. 2 Principles of Instrumental Analysis, 6th ed. Chapter 15 5-4. (a) Fluorescein because of its greater structural rigidity due to the bridging –O– groups. (b) o,o’-Dihdroxyazobenzene because the –N=N– group provides rigidity that is absent in the –NH–NH– group. 15-5. Compounds that fluoresce have structures that slow the rate of nonradiative relaxation to the point where there is time for fluorescence to occur. Compounds that do not fluoresce have structures that permit rapid relaxation by nonradiative processes. 15-6. The triplet state has a long lifetime and is very susceptible to collisional deactivation.T hus, most phosphorescence measurements are made at low temperature in a rigid matrix or in solutions containing micelles or cyclodextrin molecules. Also, electronic methods must be used to discriminate phosphorescence from fluorescence. Not as many molecules give good phosphorescence signals as fluorescence signals. As a result, the experimental requirements to measure phosphorescence are more difficult than those to measure fluorescence and the applications are not as large.3 Principles of Instrumental Analysis, 6th ed. 15-7. Chapter 15 4 Principles of Instrumental Analysis, 6th ed. 5-8. Chapter 15 15-9. Q = quinine ppm Q in diluted sample = 100 ppm ? 245 = 196 125 mass Q = 196 mg Q 500 mL ? 100 mL ? = 490 mg Q 10 mL solution 20 mL 3 5 Principles of Instrumental Analysis, 6th ed. 15-10. cQ = A1csVs (448)(50 ppm)(10. 0 mL) = = 145. 45 ppm ( A2 ? A1 )VQ ( 525 ? 448) (20. 0 mL) Chapter 15 145. 45 ppm ? 1 mg quinine 1 g solution ? ? 1000 mL = 145. 45 mg quinine 3 1 mL 1 ? 10 g solution 0. 225 g Q ? 100% = 3. 43% 4. 236 g tablet 15-11. Assume that the luminescent intensity L is proportional to cx, the concentration of iron in the original sample.Then, L1 = kcxVx / Vt where Vx and Vt are the volume of sample and of the final solution, and k is a proportionality constant. For the solution after addition of Vs mL of a standard of concentration cs, the luminescence L2 is L2 = kcxVx / Vt + kcsVs / Vt Dividing the second equation by the first yields, after rearrangement, cx = L1csVs (14. 3)(3. 58 ? 10? 5 )(1. 00) = = 1. 35 ? 10? 5 M ( L2 ? L1 )Vx (33. 3 ? 14. 3)(2. 00) 15-12. Assume that the luminescence intensity L is proportional to the partial pressure of S* . 2 We may then write L = k[S* ] 2 and K = S* ][H 2 O]4 2 [SO 2 ]2 [H 2 ]4 where the bracketed terms are all partial pressures and k and K are constants.The two equations can be combined to give after rearrangement 6 Principles of Instrumental Analysis, 6th ed. Chapter 15 [SO 2 ] = [H 2 O]2 [H 2 ]2 L kK In a hydr ogen-rich flame, the pressure of H2O and H2 should be more or less constant. Thus, [SO 2 ] = k ? L where k? = 1 kK 15-13. The fluorescent center is the rigid quinoline ring, which is rich in ? electrons. 15-14. From Equation 15-7, we can write F = 2. 303 ? f K bcP0 = 2. 303 ? K cP0 ? 0 Dividing both sides by the lifetime ? yields F = 2. 303K bcP0 ? ?0 Since K? , ? , b, ? 0 and P0 are constants, we can write F ? = Kc where K is a compilation of all the constants in the previous equation. 7 Principles of Instrumental Analysis, 6th ed. 15-15. (a) Chapter 15 (b) (c) The corrected fluorescence Fcorr would be Fcorr = F? 0/? , where F is the observed fluorescence, ? 0 is the lifetime for [Cl–] = 0. 00, and ? is the observed lifetime. The results are in the spreadsheet. 8 Principles of Instrumental Analysis, 6th ed. Chapter 15 9