Solutions for Undergraduate Instrumental Analysis, 8th Edition Bruno (All Chapters included)

SOLUTIONS MANUAL FOR

UNDERGRADUATE
INSTRUMENTAL ANALYSIS
8TH EDITION
by



Thomas J. Bruno
James W. Robinson
Eileen M. Skelly Frame
George M. Frame II



Complete Chapter Solutions Manual
are included (Ch 1 to 17)


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** All Chapters included

, 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 × 106 rounds
off to 7.8 × 106.

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;
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, (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-

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