SKOOG SOLUCIONÁRIO CAPÍTULO 5

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Fundamentals of Analytical Chemistry: 8th ed. Chapter 5 
Chapter 5 
5-1 (a) Constant errors are the same magnitude regardless of sample size. Proportional 
errors are proportional in size to sample size. 
(b) Random error causes data to be scattered more or less symmetrically around a mean 
value. Systematic error causes the mean of a data set to differ from the accepted value. 
(c) The mean is the sum of the results in a set divided by the number of results. The 
median is the central value for a set of data. 
(d) The absolute error of a measurement is the difference between the measured value 
and the true value. The relative error is the absolute error divided by the true value. 
5-2 (1) Random temperature fluctuations causing random changes in the length of the metal 
rule; (2) uncertainties from moving and positioning the rule twice; (3) personal judgment 
in reading the rule; (4) vibrations in the table and/or rule; (5) uncertainty in locating the 
rule perpendicular to the edge of the table. 
5-3 (1) Instrumental errors 
(2) Method errors 
(3) Personal errors 
5-4 (1) The analytical balance is miscalibrated; (2) after weighing an empty vial, fingerprints 
are placed on the vial while adding sample to the vial; (3) a hygroscopic sample absorbs 
water from the atmosphere while placing it in a weighing vial. 
5-5 (1) Incorrect calibration of the pipet; (2) temperature different from calibration 
temperature; (3) incorrect filling of the pipet (overshooting or undershooting the mark). 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 5 
5-6 Systematic method errors are detected by application of the method to the analysis of a 
standard reference material having one or more analytes at exactly known concentrations. 
5-7 Constant and proportional errors. 
5-8 (a) (– 0.4/700) × 100% = – 0.06% 
 As in part (a) 
(b) – 0.09% 
(c) – 0.2% 
(d) – 1% 
5-9 (a) First determine how much gold is needed to achieve the desired relative error. 
 (– 0.4/– 0.2%) × 100% = 200 mg gold 
Then determine how much ore is needed to yield the required amount of gold. 
 (200/1.2%) × 100% = 17,000 mg ore or 17 g ore 
(b) 7 g ore 
(c) 4 g ore 
(d) 3 g ore 
5-10 (a) (0.04/50.00) × 100% = 0.08% 
 As in part (a) 
(b) 0.4% 
(c) 0.16% 
(d) 0.1% 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 5 
5-11 (a) (– 0.4/40) × 100% = – 1.0% 
 As in part (a) 
(b) – 0.23% 
(c) – 0.10% 
(d) – 0.07% 
5-12 mean = 0106.001063.0
3
0105.00104.00110.0
≈=




 ++
 
Arranging the numbers in increasing value the median is: 
 0.0104 
 0.0105 ← median 
 0.0110 
The deviations from the mean are: 
 
0004.00106.00110.0
0001.00106.00105.0
0002.00106.00104.0
=−
=−
=−
 
mean deviation = 0002.000023.0
3
0004.00001.00002.0
≈=




 ++
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 5 
 (b) Using a spreadsheet 
 A B C 
1 Data Deviation 
2 
 24.53 0.174 
3 
 24.68 0.024 
4 
 24.73 0.026 
5 
 24.77 0.066 
6 
 24.81 0.106 
7 
 
8 Mean = 24.704 0.0792 
9 Median = 24.73 
10 
 
11 Spreadsheet Documentation 
12 B8=AVERAGE(B2:B6) 
13 B9=MEDIAN(B2:B6) 
 
14 C2=ABS(B2-$B$8) 
 
15 C8=AVERAGE(C2:C6) 
 
Tabulating the results for (a) through (f) gives: 
 Mean Median Deviation from the Mean Mean Deviation 
(a) 0.0106 0.0105 0.0002, 0.0001, 0.0004 0.0002 
(b) 24.70 24.73 0.17, 0.02, 0.03, 0.07, 0.11 0.08 
(c) 190 189 2, 0, 4, 3 2 
(d) 4.54×10-3 4.52×10-3 0.02×10-3, 0.07×10-3, 0.09×10-3, 
0.06×10-3, 0.01×10-3, 0.04×10-3 
0.05×10-3 
(e) 39.59 39.64 0.24, 0.02, 0.34, 0.09 0.17 
(f) 859 862 9, 3, 10, 10, 6 8