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SOLUTIONS
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MANUAL
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Suzanne Bell
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Complete Solutions Manual for Instructors
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and Students
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© Suzanne Bell
All rights reserved. Reproduction or distribution without permission is prohibited.
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©MEDGEEK
, Solutions Manual for Forensic Chemistry, 3e
by Suzanne Bell (All Chapters)
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Solutions Manual for End of Chapter Questions
Forensic Chemistry, 3rd edition
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Chapter 1
1. A standard of Pb2+ for a gunshot residue analysis using atomic absorption is prepared by first
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dissolving 1.0390 g dried Pb(NO3)2 in distilled water containing 1% nitric acid. The solution is
brought to volume in a class A 500.00-mL volumetric flask with an uncertainty of ±0.20 mL.
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The solution is then diluted 1:10 by taking 10.00 mL (via an Eppendorf pipet, tolerance ±1.3 μL)
and diluting this in 1% nitric acid to a final volume of 100.00 mL in a volumetric flask with a
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tolerance of ±0.08 mL. The balance has an uncertainty of ±0.0002 g.
An unusual feature of this question is that the molecular weight of Pb is listed as 207.2 g/mole
which must be considered in rounding. This point is made in the text as well.
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a. Using conventional rounding rules, calculate the concentration of the final solution of Pb2+, in
ppm.
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The calculations should be done together, and the results rounded at the end. The only exception
to this is with the formula weight for the lead nitrate, which is rounded separately as an
addition/subtraction:
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Formula weight of Pb(NO3)2 = 207.2 + (2 x 14.01) + (2 x 3 x 16.) = 331.22
Formula weight of Pb(NO3)2 = 207.2 + (2 x 14.00674) + (6 x 15.9994) = 331.2
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, To obtain the concentration in ppm (mg/L), the molarity is calculated and then converted to the
final value. Note that the moles of Pb in the sample are equal to the moles of Pb(NO3)2 since
there is a 1:1 mole ratio of lead in lead nitrate:
1.0390g Pb ( NO3 )2 1 mole Pb g 1000 mg
207.2 Pb
Pb ( NO3 )2 1 mole Pb ( NO3 )2
g
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331.2 mole g
mole
0.5000 L
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molarity (moles/L) → ppm (mg/L)
= 1300.0 ppm
This solution is diluted by a factor of 1:10 (10.00 mL to 100.00 mL) to yield the final solution
concentration of 130.0 ppm.
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The result is rounded to four significant figures, the least number present in any of the measured
values.
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b. Determine the absolute and relative uncertainties of each value obtained in part a.
Units must match so convert ul to mL
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0.20mL
Large vol flask: = 4.00x10-4
500.00mL
0.08mL
Small vol flask: = 8.00x10-4
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100.00mL
0.0002g
Balance: = 1.92 x10-4
1.0390g
1 mL 0.0013mL
Eppendorf: 1.3ul = = 1.30x10 −4
1000uL 10.00mL
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c. Report the results as a range by the propagation-of-error method.
This is best done as one calculation to avoid rounding errors:
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Uncertainty = ( uncert.balance ) + ( 500mL flask ) + ( Eppendorf ) + (100ml flask )
2 2 2 2
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, 0.0002 2 0.2 2 0.0013 2 0.08 2
+ + + = 8.54x10−7 = 9.24x10 −4
1.0390 500.00 10.00 100.00
9.24x10−4 (131.0 ppm) = 0.12 = 0.1 rounded
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Concentration of Pb2+ = 131.0 ppm ± 0.1 ppm
d. Comment on your findings and why this case is unique.
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As mentioned in the text, lead is one of the few elements with a molecular weight generally
reported to only one decimal place. Therefore, this weight may limit significant figures in some
calculations. In addition, the value for oxygen is also reported to one decimal, 16.0, in the table
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included in this book. However, additional decimals can be obtained using other tables. This is
generally not the case with lead.
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2. If an outlier is on the low side of the mean, as in the example in the chapter, could a one-tailed
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table be used?
Possibly but this is not recommended. Two-tailed tests are the default choice, and you would
only use a one-tailed test if you were sure that the outlier could only be low or only be high.
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What you cannot do is try a two-tailed test, and then decide to try a one-tailed test and pick the
one you like the best. Unless you have a definitive reason ahead of time, use a two-tailed test.
Question 7 shows an example where a one-tailed test is used.
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3. If replicate measurements are made of a homogenous parent population, the value of the
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sample standard deviation and the population standard deviation will converge. Why?
As the number of samples selected from a population increases, the better this subset reflects the
composition of the parent population. Think of the reverse situation. If you test all the samples
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