principios de analise instrumental 6ed skoog resolucoes

analyte ions are separated from electrons and negative 
ions by a suitable field and are accelerated into the mass spectrometer itself. 
11-6. Lasers are used for sampling in ICPMS by exposing a solid sample to an intense, pulsed 
laser beam, which rapidly vaporizes (ablates) the sample. The resulting vapor is carried 
into the ICP torch where atomization and ionization occurs. The resulting gaseous 
mixture then enters the mass spectrometer for ion analysis. 
11-7. Isobaric interferences are encountered when the isotopes of two elements have the same 
mass. A second type of spectroscopic interference occurs from molecular species that 
have the same mass as that of an analyte ion. A third type of interference is from matrix 
species that combine with the analyte and reduce the analyte signal as a result. 
11-8. An internal standard is often used in preparing calibration curves for ICPMS in order to 
compensate for random error from instrument drift and noise, torch instabilities, and 
some matrix effects. The internal standard chosen should be an element that is normally 
absent from the sample, but one that has an atomic mass and ionization potential close to 
that of the analyte. 
 
 
Principles of Instrumental Analysis, 6th ed. Chapter 11
 
 3
11-9. In an isotope dilution experiment, a known weight of the analyte that has been prepared 
from a non-naturally occurring isotope of one of the elements in the analyte is added to 
the sample to be analyzed and mixed thoroughly until homogenization is assured. A 
known weight of the sample is then taken and a fraction of the highly purified analyte is 
isolated and weighed. The amount of enriched analyte is then determined by a mass 
spectrometric measurement. As shown in Section 32 D-1, these data then permit the 
calculation of the percentage of the analyte in the original sample. 
11-10. In a glow-discharge mass spectrometric analysis, the sample is made part of the cathode 
of a glow-discharge atomizer, such as that shown in Figure 8-12. Here, the solid sample 
is bombarded by a stream of argon ions that have been accelerated through a potential of 
5 to 15 kV. Analyte atoms are sputtered from the cathode surface and converted by 
collisions with electrons and argon ions into positive ions. These are then passes into the 
mass spectrometer for analysis. 
11-11. In secondary-ion mass spectrometry, the sample is bombarded with a stream of 5- to 20-
eV positive ions, such as Ar+ or Cs+. The ion beam is formed in an ion gun. The impact 
of the beam on the surface of the sample results in secondary ion formation containing 
analyte ions, which then pass into the mass spectrometer for analysis. 
 
 
Skoog/Holler/Crouch Chapter 12
Principles of Instrumental Analysis, 6th ed. Instructor’s Manual
 
 
 1 
CHAPTER 12 
12-1. λ0 = 12398/V Equation 12-2 
 λ0 = 12398/(90 × 103) = 0.138 Å 
12-2. Equation 12-2 gives the minimum voltage needed to produce an emission line of 
wavelength λ. 
 Thus, solving for Vmin
 Vmin = 12398/λ 
 (a) For U, Kβ1 = 0.111 Å and Lβ1 = 0.720 Å Table 12-1 
 For U Kβ1 Vmin =12398/0.111 = 1.12 × 105 V or 112 kV 
 For U L β1 Vmin =12398/0.720 = 17.2 kV 
 Similarly, 
 
 λ for Kβ1, Å Vmin for Kβ1, kV λ for Lβ1, Å Vmin for Lβ1, kV 
(a) For U 0.111 112 0.720 17.2 
(b) For K 3.454 3.59 No Lβ1 line ---- 
(c) For Rb 0.829 15.0 7.075 1.75 
(d) For W 0.184 67.4 1.282 9.67 
 
12-3. Here we assume that a linear relationship exists between the atomic number Z for the 
elements in question and the square root of the frequency ν of their Kα lines (see Figure 
12-3). This linear relationship can then be used to determine the frequencies and 
wavelengths of the other elements as shown in the spreadsheet. 
 
 
Principles of Instrumental Analysis, 6th ed. Chapter 12
 
 2
 
Principles of Instrumental Analysis, 6th ed. Chapter 12
 
 
 
2
 
12-4. Proceeding as in Pb. 12-3, we construct the following spreadsheet: 12-4. Proceeding as in Pb. 12-3, we construct the following spreadsheet: 
 
 
 
Principles of Instrumental Analysis, 6th ed. Chapter 12
 
 3
12-5. ln P0/P = μMρx Equation 12-3 
 ln (1/0.478) = 49.2 cm2/g × 8.9 g/cm3 × x 
 x = 0.738/(49.2 × 8.9) = 1.69 × 10–3 cm 
12-6. μM = WKμK + WIμI + WHμH + WOμO Equation 12-4 
 (a) 
 M
11.00 39.1 126.9 89.00 16.0 = 16.7 + 39.2 + 0 + 1.50
100.00 166.0 166.0 100.00 18.0
μ ⎛ ⎞ ⎛× × ×⎜ ⎟ ⎜⎝ ⎠ ⎝
⎞⎟⎠ 
 = 4.92 cm2/g 
12-7. ln P0/P = (1/0.965) = μMρx = 2.74 × 2.70x 
 x = 0.0356/(2.74 × 2.70) = 4.82 × 10–3 cm 
12-8. ln P0/P = ln(1/0.273) = μMρx = μM × 0.794 × 1.50 
 μM = 1.298/(0.794 × 1.50) = 1.09 
 (a) μM = 1.09 = 2 × 39.2 
2I
W
 = 0.0139 or 1.39% I
2I
W 2
 (b) μM = 1.09 = 2 × 39.2 + 
2I
W ETOH
24 160.70 + 0.00 + 1.50
46 46
W ⎛ ⎞× ×⎜ ⎟⎝ ⎠ 
 1.09 = 78.4 + 0.887W
2I
W ETOH
 + W
2I
W ETOH = 1 
 Solving the two equations yields 
 = 2.62 × 10
2I
W –3 or 0.262% 
 
 
Principles of Instrumental Analysis, 6th ed. Chapter 12
 
 4
12-9. nλ = 2d sin θ Equation 12-6 
 (a) For topaz, d = 1.356 Å 
 For Fe Kα at 1.76 Å 1 × 1.76 = 2 × 1.346 sin θ 
 θ = 40.46° and 2θ = 80.9° 
 The following data were obtained in the same way. 
 (a) Topaz 
(d = 1.356 Å) 
(b) LiF 
(d = 2.014 Å) 
(c) NaCl 
(d = 2.820 Å) 
Fe (1.76 Å) 80.9 51.8 36.4 
Se (0.992 Å) 42.9 28.5 20.3 
Ag (0.497 Å) 21.1 14.2 10.1 
 
12.10. Using the same method used in Solution 12-9, we find 
 (a) 2θ = 135° (b) 99.5° 
12-11. Vmin = 12398/λ Equation 12-2 
 (a) Vmin = 12398/3.064 = 4046 V or 4.046 kV 
 Likewise for the others, 
 (b) 1.323 kV (c) 20.94 kV (d) 25.00 kV 
12-12. Following the least-squares procedure outlined in Appendix 1, Section a1D, the 
spreadsheet shown was constructed. The unknown was found to contain 
 0.149 ± 0.002 %Mn 
 
 
Principles of Instrumental Analysis, 6th ed. Chapter 12
 
 5
 
 
 
Skoog/Holler/Crouch Chapter 13
Principles of Instrumental Analysis, 6th ed. Instructor’s Manual
 
 
 1 
CHAPTER 13 PROBLEM SOLUTIONS 
13-1. (a) %T = 10–A × 100% = 10–0.038 × 100% = 91.6% 
Proceeding in the same way, we obtain 
(b) 11.0% (c) 39.9% (d) 57.4% (e) 36.7% (f) 20.3% 
13-2. (a) A = – log T = – log (15.8/100) = 0.801 
Proceeding in the same way, we obtain 
(b) 0.308 (c) 0.405 (d) 0.623 (e) 1.07 (f) 1.27 
13-3 (a) %T = 10–0.038/2 × 100% = 95.7% 
Similarly, 
(b) 33.2% (c) 63.2% (d) 75.8% (e) 60.6% (f) 45.1% 
13-4. (a) A = – log (2 × 15.8/100) = 0.500 
Similarly, 
(b) 0.007 (c) 0.103 (d) 0.322 (e) 0.770 (f) 0.968 
13-5. 4
6.23 mg KMnO
 = c
3
410 g KMnO
L
−
×
4mg KMnO
4
4
1 mol KMnO
158.034 g KMnO
× 5= 3.942 10 M−× 
A = – log T = – log 0.195 = 0.710 
ε = A/(bc) = 0.710/(1.00 × 3.942 × 10–5) = 1.80 × 104 L mol–1 cm–1
13-6. 
5.24 mg A
 = c
100 mL
310 g A−×
mg A
1000 mL× 1 mol A
L 335 g A
× 4= 1.564 10 M−× 
A = – log T = – log 0.552 = 0.258 
ε = A/(bc) = 0.258/(1.50 × 1.564 × 10–4) = 1.10 × 103 L mol–1 cm–1
 
 
Principles of Instrumental Analysis, 6th ed. Chapter 13
 
 2
13-7. (a) A = εbc = 9.32 × 103 L mol–1 cm–1 × 1.00 cm × 3.79 × 10–5 M = 0.353 
(b) %T = 10–A × 100% = 10–0.353 × 100% = 44.4% 
(c) c = A/εb = 0.353/(9.32 × 103 × 2.50) = 1.52 × 10–5 M 
13-8. (a) A = εbc = 7.00 × 103 L mol–1 cm–1 × 1.00 cm × 3.49 × 10–5 M = 0.244 
(b) A = εbc = 7.00 × 103 L mol–1 cm–1 × 2.50 cm × 3.49 × 10–5/2 M = 0.305 
(c) For (a), %T = 10–0.244 × 100% = 57.0%; For (b) %T = 10–0.305 × 100% = 49.5% 
(d) A = – log T = – log (0.570/2) = 0.545 
13-9. ε = 7.00 × 103 L mol–1 cm–1 
Fe
mg Fe
 = 7.9c 3 g Fe 10
L mg