SOLUTIONS MANUAL FOR

SOLUTIONS MANUAL FOR

UNDERGRADUATE

INSTRUMENTAL ANALYSIS

8TH EDITION

by Thomas J. Bruno James W. Robinson Eileen M. Skelly Frame George M. Frame II (All Chapters) 1 / 4

3 Chapter 1 Answers 1.1 (a) See Section 1.3.3.1 (b) determinate error 1.2 (a) See Section 1.3.2 (b) No 1.3 (a) See Section 1.3.2 (b) Analyze a known material, such as a National Institute of Standards and Technology Standard Reference Material. Analyze the sample using a different procedure that is known to be reliable (i.e., a “standard method”).

1.4 Uncertainty is a quantity associated with the result of a measurement that characterizes the dispersion of the values that could reasonably be attributed to the measurement, and ideally includes all factors that are incorporated in the uncertainty budget, including any calibration or standardization.Uncertainty ideally incorporates all of the factors that can influence dispersion about the measured value (that is, the uncertainty budget). Precision is a description of the dispersion about a central or average value of a measurement. Error is the difference (by subtraction) between the reference result (or accepted true result) and the measured result of an analysis.

1.5 (a) The population standard deviation, σ, is the difference between the maximum and one inflection point of a Gaussian distribution of answers. Mathematically, Eqn. 1.11 defines it.(b) 95.5 % should fall within ± 2σ of the mean.

1.6 s = 0.10; σ = 0.097 = 0.10 (rounded off to 2 decimal places) 1.7 mean = 0.1022; average deviation = sum of the absolute deviations divided by the mean = (0.0001 + 0.0003 + 0.0003 + 0.0001)/0.1022 = 0.008; s = 0.0002; % RSD = 0.2 %; absolute error of the mean = | 0.1022 – 0.1026| = 0.0004; relative error of the mean = (0.0004/0.1026) = 0.004.

1.8 4; 3; 2 (assume that the terminal zero is significant); 1; 3; 6 (assume that the terminal zero(s) is/are significant).

1.9 6.483 rounds off to 6.5; 2.9879 rounds off to 3.0 (assume 1.0 has 2 sig. figs.); 7.77316 × 10 6 rounds off to 7.8 × 10 6 .

1.10 Using all data points: (a) mean = 2.17; (b) s = 0.12; (c) 2.17 ± 0.24; (d) 2.51 seems high. Without using 2.51, recalculate a, b and c. New (a) mean = 2.13; (b) s =0.03; 4s = 0.12 and mean + 4s = 2.25.Therefore 2.51 can be considered as a suspected outlier. Remind students that there is no test that indicates a result can or should be rejected, only that it may be an outlier. (c) 2.13 ± 0.06. The second set of values for a, b and c are correct.

1.11 (a) See Section 1.2.3; (b) See Section 1.2.4 1.12 (a) See Section 1.3.2; 2 / 4

4 (b) precision (Section 1.3.3.2) 1.13 (a) No, 2.13 + 0.06 = 2.19, which is less than 2.5 ppm.(b) Yes, the lower limit is 2.13 – 0.6 = 2.07 ppm, which is greater than 2.00.

1.14 0.70 – 2(0.09) = 0.52 ppm Cu. The patient’s serum is within the 95 % CL for normal serum Cu based on the test method (and well within the 99 % CL; 0.70 – 3(0.09) = 0.43 ppm). Treatment should not be started. The doctor could have another serum Cu test done, along with a urine Cu test, to provide more information, or should ask the laboratory if a method for serum Cu with a smaller standard deviation (better precision) is available.

1.15 3.30 × 10

-2

1.16 68.3 %

1.17 Using the Student’s t table (Table 1.9) and 7 degrees of freedom, 1.18 (a) 99.7 %; (b) 95.5 % 1.19 Drift or flicker noise and some types of noise from the surroundings (building vibrations, power lines) are frequency dependent. Decreasing the temperature of instrument components can decrease thermal noise, a type of white noise.

1.20 (a) S/N = 1.50/0.299 =5.02; (b) S/N = 100 is a 20 fold improvement over the current S/N. To improve the S/N by a factor of 20, 400 measurements must be averaged, since (400) 1⁄2 = 20.

1.21 Figure 1.13 (b) S/N ≈ 4.3; Figure 1.13 (c) S/N ≈ 1.5. (b) is approximately 3 times greater than (c), therefore 9 measurements must be aver- aged.

1.22 From Table 1.9, t for (N–1)=5 is 2.57. The calculated t of 3.31 > 2.57; therefore a significant differ- 3 / 4

5 ence between the two methods exists. The ICP method appears to have a bias, giving lower results than the standard XRF method.

1.23 The area under the curve between the maximum and one inflection point.

1.24 See Section 1.1.4.1 1.25 (a) 3; (b) 2.52; (c) 2.7 × 10 3 ; (d) 2.600; (e) 2.73 1.26 (a) See attached plot Problem 1.26 Plot Nitrate Ion mg/LAbsorbance

0.00 0.000

1.00 0.042

2.00 0.080

5.00 0.198

7.00 0.281

Slope 0.03988235 0.040 Intercept 0.00055294 0.00

CORREL 0.999923 1.00

(b) y = 0.040x + 0.00; (c) from Eq. 1.23-1.25 and the DEVSQ function in Excel, s m = 0.023; s b

= 0.091;

(d) 0.9999 = 1.00 1.27 (a) and (b) See Table 1.1

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