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Solution Manual for Forensic Chemistry, 2nd Edition

Solution Manual for Forensic Chemistry, 2nd Edition simplifies even the toughest textbook questions with step-by-step solutions and easy explanations.

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    Preface ii Chapter 1 Introduction 1 Chapter 2 Foundations 5 Chapter 3 Quality Assurance and Quality Control 20 Chapter 4 Reporting Defensible Uncertainty and Obtaining Representative Samples 39 Chapter 5 Chemical Fundamentals: Partitioning, Equilibria, and Acid/ Base Chemistry 47 Chapter 6 Instrumentation 59 Chapter 7 Drugs as Physical Evidence: Seized Drugs and Their Analysis 68 Chapter 8 Forensic Drug Analysis: Selected Drug Classes 77 Chapter 9 Drugs in the Body 84 Chapter 10 Forensic Toxicology 92 Chapter 11 The Chemistry of Combustion and Arson 96 Chapter 12 Explosives 103 Chapter 13 Firearms and Associated Chemistry Evidence 111 Chapter 14 The Chemistry of Colors and Colorants 115 Chapter 15 The Chemistry of Polymers 121 Chapter 16 Forensic Analysis of Inks and Paints 125 Chapter 17 Chemical Analysis of Materials: Paper and Fiber 131

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    Solutions Manual for Forensic Chemistry ii Preface This solutions guide accompanies the second edition of the text. For the instructor, this manual should help clarify concepts and principles as applied in the homework. For the student, this manual should be the last book opened . Why? In my experience, students tend to use the solutions manual in a less than optimal way. For example, here is the wrong way to use a solutions manual: 1. Obtain the assigned homework from the instructor. 2. Open the solutions manual and copy the results for each problem, cleverly altering them so that it does not appear that you in fact used the solutions manual. * 3. Quit. 4. When test time comes around, study the solutions manual and hope to survive. This is not learning; this is jockeying to do well on a test. Now, here is the correct way to use the solutions manual: 1. Obtain the assignment, also doing extra problems that are similar to those assigned. 2. Work the problem, making sure to reinforce concepts as you go. For example, if the problem involves an acid/base extraction, refer back to that section of the text and refresh your memory. 3. Check and honestly critique your work; focus on “honestly.” 4. Do the next problem. 5. When finished with the assignment and related problems, open the solutions manual. 6. Check your work. 7. Close the solutions manual and put it far away from you. 8. Redo the problems you missed. 9. Return to step 4 and repeat. 10. When test time comes, you are ready. * FYI, you are kidding yourself if you think this works.

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    Solutions Manual for Forensic Chemistry iii Notes: Frequent mention is made of the following references; they are abbreviated within: “Clarke’s Handbook” refers to: Galichet, L. Y., et al., ed. Clarke ’ s Analysis of Drugs and Poisons, Vol 1. and Vol. 2 London: Pharmaceutical Press, 2004. (Volume 2 contains monograms on the individual drugs and is referenced most frequently.) PDR refers to: Physicians ’ Desk Reference, 58 th ed. Montvale, NJ: Thomson PDR, 2004. Physicians ’ Desk Reference for Nonprescription Drugs and Dietary Supplements , 22 nd ed. Montvale, NJ: Medical Economics — Thomson Healthcare, 2001. CRC Handbook refers to: Handbook of Chemistry and Physics , 84 th ed. Boca Raton, FL: CRC Press, 2003 – 2004. Merck refers to: O ’ Neil, M. J., et al., ed. The Merck Index: An Encyclopedia of Chemicals, Drugs, and Biologicals . Whitehouse Station, NJ: Merck Research Laboratories — Merck and Co., 2001. For illustrative purposes and in spreadsheet printouts, extra significant digits may appear in intermediate calculations. This is to avoid rounding errors and make the process and procedure used clear. Errata: The solutions manual was completed after the first printing of the book went to press. Errors in the questions that were found are noted in the solutions manual and corrections applied; subsequent printings will have the corrections applied.

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    Solutions Manual for Forensic Chemistry 1 Introductio n C C H H A A P P T T E E R R 1 1 From the chapter 1. Compare and contrast the adversarial system and the scientific method. List the strengths and weaknesses of both in the context of criminal and civil law. The scientific method is based on experiment and observation, postulations of relationships, iterative testing, and drawing conclusions based on the results. The adversarial system is based on arguments from opposing parties. The scientific method does not by design pit one side against another. Certainly, there are arguments in science, but the arguments are resolved by experiment and observation. The process may take decades, but if the data is consistent and reproducible, conclusions are drawn and consensus is reached. Adversaries argue based on differing viewpoints, interpretations, and agendas. Science can be used to support or refute interpretations of events and evidence, but the outcome depends on the strength of the argument and the skill of those making it. The adversarial system is well-suited to situations in which social and human (for lack of a better term) issues are involved. A person may have committed murder, but the reasons and circumstances are critical to the deliberation. Science, on the other hand, seeks to elucidate natural laws and to apply them to derive new knowledge. Ideally, there is no social element; gravity does not take into account mitigating circumstances, for example. This process is essential for understanding the laws of the universe since consistency and reproducibility are required to derive them. It is not an ideal system for dealing with human beings. 2. During a Daubert hearing, what entity ultimately decides on admissibility? The judge, who acts as the gatekeeper.

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    Solutions Manual for Forensic Chemistry 2 3. What role does peer review play in science and in the law? Compare and contrast. Argument before a judge, jury, or lawyers is loosely analogous to peer review. The appeals process could also be placed in the category since professionals of the law review the work done by other professionals. Peer review is more obvious and central in science where it is an integral part of the dissemination process. Scientific findings are submitted to peer-reviewed journals where editors assign qualified reviewers to comment on and judge the work. They may accept it as is, request more information or work, or recommend that it not be published. Publication in a peer-reviewed journal is validation of acceptance by those with similar skills, background, and understanding of the mechanism of science; materials published in non-peer-reviewed journals or sources generally are not given the same weight or credibility as peer-reviewed data. 4. Describe how a preponderance of inclusive circumstantial evidence can become conclusive in the eyes of a jury. The Wayne Williams case is the perfect example of this phenomenon. Had a few dog hairs been found on victims, such a finding would place under suspicion only those people who owned or might have come in contact with dogs, a large percentage of the population. It was the unique combination of hairs and fibers that was compelling. Consider your own home. What types of hairs and fibers would be found there? How would that compare with a friend’s environment? You might both have the same blue carpet if you live in the same apartment building or dorm, but the hairs to be found in your two rooms would be different, as would fibers from clothing and articles of furniture. Link enough of such observations together and it becomes clear that your apartment is a much different environment than even one next door. Integrative 1. A great scientist can still be a terrible forensic scientist; a person who gives wonderful testimony can be a terrible forensic scientist. Comment on these observations and the implication for forensic chemistry. Forensic science and forensic chemistry require understanding and skills in comparison. It is not always an easy skill to acquire nor is the forensic mindset necessarily a natural one. In addition, a

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    Solutions Manual for Forensic Chemistry 3 forensic scientist must be able to apply advanced techniques and knowledge to the analysis of evidence, yet present the results clearly and concisely to an audience without a comparable body of knowledge or experience. Thus, forensic scientists must be good communicators and good teachers. A great research scientist might be able to communicate with peers, but may struggle to distill the complexities of his or her work such that anyone with a high school education could grasp the concepts. Conversely, a person may be able to concisely and brilliantly present findings that are c ompletely wrong; the skill of the presentation coupled with the audience’s lack of background can result in acceptance of incorrect information. A forensic chemist must be the master of the chemistry and of the forensic aspects — to speak the truth (derived from science) in the public forum. 2. Can jurors ask questions of expert witnesses? Comment on your findings regarding this issue. Except under rare circumstances, they cannot. They also are not usually allowed to take notes; however, practices are changing in some states. For example, in Arizona, jurors may submit questions for expert witnesses to the judge, who screens the questions and then will ask those that are probative and appropriate. In some cases, jurors are also allowed to take notes and to take these notes into deliberation. How jurors interact with expert witnesses and how scientific evidence is presented is becoming an increasingly difficult issue, given the complexity of scientific testimony. Complicating this issue is the general lack of scientific background in jurors, judges, and lawyers. It is hard to imagine reforms not occurring as the complexity of testimony continually increases. Food for thought 1. Is the analysis of drugs using instruments such as mass spectrometers and infrared spectrometry based on comparison? Arguably yes. Identifications are made by comparing data to that stored in a library.

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    Solutions Manual for Forensic Chemistry 4 2. How important is the way in which scientific evidence is presented? Comment on the relative importance of content versus presentation. Why is learning how to testify such an important skill? See Integrative question #1 above. A forensic scientist must be able to communicate findings concisely and clearly to the court so the trier-of-fact can properly weigh the information. Poorly communicated evidence can be as damaging as poor scientific practice.

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    Solutions Manual for Forensic Chemistry 5 Foundations C C H H A A P P T T E E R R 2 2 From the chapter 1. A standard of Pb 2+ for a gunshot residue analysis using atomic absorption is prepared by first dissolving 1.0390 g dried Pb(NO 3 ) 2 in distilled water containing 1% nitric acid. The solution is brought to volume in a class A 500 mL volumetric flask with an uncertainty of +/- 0.20 mL. This solution is diluted 1/10 by taking 10 mL (via an Eppendorf pipette, tolerance +/- 1.3 μL) and diluting this in 1% nitric acid to a final volume of 100 mL in a volumetric flask with a tolerance of +/- 0.08 mL. The balance has an uncertainty of +/- 0.0002 g. a) Using conventional rounding rules, calculate the concentration of the final solution in ppm of Pb 2+ . 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. Using the periodic table in the textbook: Formula weight of Pb(NO 3 ) 2 = 207.2 + (2 x 14.01) + (2 x 3 x 16.) = 331.22 For this calculation, using this periodic table is a problem since the formula weight will needlessly limit the number of significant figures. Thus, a periodic table with more reported digits is needed, even though Pb will remain as is. The new calculation: Formula weight of Pb(NO 3 ) 2 = 207.2 + (2 x 14.00674) + (6 x 15.9994) = 331.2 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(NO 3 ) 2 since there is a 1:1 mole ratio of lead in lead nitrate:

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    Solutions Manual for Forensic Chemistry 6         L 0.5000 g mg 1000 Pb mole g 207.2 NO Pb mole 1 Pb mole 1 NO Pb mole g 331.2 NO Pb 1.0390g 2 3 2 3 2 3                        molarity (moles/L)  ppm (mg/L) = 130.0 ppm The result is rounded to four significant figures, the least number present in any of the measured values. b) Determine the absolute and relative uncertainties of each value. Select the largest and report the results as a concentration range. An easy way to visualize the largest uncertainty at a glance is to convert each one to a “1 part per” expression. For the balance: 0.0002 0.0002g 1 0.0002 = = 1.0390 1.0390g 5195 0.0002 or 1 part per 5195 parts. Notice that this value is unitless. Use the same approach to obtain the relative uncertainties of the other tools: 500 mL volumetric flask = 1 part per 2500 parts 10.00 mL transfer = 7692 parts (remember to convert μL to mL) 100 mL volumetric flask = 1 part per 1250 parts The largest value will dominate the calculation of uncertainty and here that value is the uncertainty attributed to the 100 mL volumetric flask, 1 part in 1250 parts. As a percentage, this is (1/1250) *100 or 0.08%, also unit-independent. Therefore, the uncertainty of the final concentration can be calculated two ways: 130.0/1250 = 0.104 (1 part in 1250 parts, same relative uncertainty as the limiting value from the 100 mL volumetric flask). -or-

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    Solutions Manual for Forensic Chemistry 7 130.0 * (0.00080 from the percentage) = 0.104. The concentration in ppm using this method of calculating the uncertainty is 130.0 ppm +/- 0.1 ppm or 129.9 – 130.1 ppm. c) Report the result as a range by the propagation of error method. This is accomplished by taking the individual relative uncertainties (unitless) as calculated above and combining them as per equation 2.1:           2 2 2 2 2 flask 100ml Eppendorf flask 500mL balance uncert. e     2 2 2 2 2 0.0002 0.2 0.0013 0.08 e 1.0390 500.00 10.00 100.00                                     Notice that units will cancel out so that the μL uncertainty attributed to the Eppendorf pipette, 1.3 μL, is converted to mL. Solving for the error e yields 9.2 x 10 -4 or 0.09%. This result is close to that we obtained using the largest uncertainty as in part b, but as expected, the value is slightly larger when all uncertainties are taken into account. As a result, the uncertainty in ppm is slightly larger: 131.0. * 0.000924 = 0.121, rounded to the result as 0.1 However, with rounding, the reported range is the same as in part b. d) Comment on your findings and why this case is somewhat unique. 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 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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    Solutions Manual for Forensic Chemistry 8 2. If an outlier is on the low side of the mean as in the example in the chapter, could a one- tailed table be used? A one-tailed table is used when the method would always give a higher value than expected, a concentration for example. In this case, the potential outlier is on the low side of the mean and therefore a one-tailed table would not be appropriate. 3. Find the following reference for Q value tables: Rorabacher, D. B. “ Statistical Treatment for Rejection of Deviant Values: Critical Values of Dixon ’ s ‘ Q ’ Parameter and Related Subrange Ratios at the 95% Confidence Level. ” Analytical Chemistry , 63 1991, 139 – 148. For the trainee, determine what the %cocaine would have to be in the 11th sample for it to be eliminated at the following significance levels: 0.20, 0.10, 0.05, 0.04, 0.02, and 0.01. Present graphically and comment on the findings. This problem can be addressed using Excel® and by realizing that the result will be a range rather than a single value. The data is first summarized and the Q values obtained from the reference, which calls for careful reading of the text and the scenarios described. Here, we wish to find out what a single outlier value will be at either end of the distributed values, with no other outliers of concern. Therefore, the single outlier test and table, designated r 10 in the paper, is appropriate (p. 141). This directs us to the proper table found on page 142, Table 1. The value of n will be 11 with one more included, so the corresponding table values (Q table ) are: CI = 80% 90% 95% 96% 98% 99% 0.332 0.392 0.444 0.460 0.502 0.542 An outlier will be rejected if Q calc > Q table , so the next step is to determine what outlier values, when plugged into the calculation of Q (equation 2.8), will result in a Q calc that just equals Q table . As shown in Figure 2.5, when the data is arranged from low to high, the lowest value is 11.5% and the highest value is 15.0%. Taking the 80% interval (p = 0.20), the calculations are:

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    Solutions Manual for Forensic Chemistry 9     11.5- x = 0.332 15.0 - x 11.5 - x = 0.332[15.0 - x] 11.5 - x = 4.98 - 0.0322x 6.52 - x = -0.332x 6.52 = 0.668x x = 9.76% where [11.5 – x] represents the gap and [15.0 – x] represents the range. To simplify later calculations, this calculation can be solved generically:     T T T T T T T T T low - x = Q high - x low - x = Q high - Q x low - Q high = x - Q x low - Q high = x(1-Q ) low - Q high x = 1-Q Substituting the values of 11.5 for low, 15.0 for high, and 0.332 for Q T yields a value of 9.76, confirming the calculation above. To calculate the high end, the calculation is modified slightly where the gap is represented by [x – 15.0] and the range by [x – 11.5]. Simplifying as above, the generic expression is obtained: T T high - Q low x = 1 - Q

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    Solutions Manual for Forensic Chemistry 10 For the 80% confidence interval, the high value cutoff is thus 16.74%. If the forensic chemist obtains a percentage cocaine value of greater than 16.74% or less than 9.76%, that value would be an outlier at the 80% confidence interval. With generic equations in hand, a spreadsheet allows for quick solution and graphing of the results: Low value: 11.5 High value: 15.0 CI: 80 90 95 96 98 99 Table: 0.332 0.392 0.444 0.46 0.502 0.542 Low cutoff: 9.76 9.24 8.71 8.52 7.97 7.36 High cutoff: 16.74 17.26 17.79 17.98 18.53 19.14 Confidence interval Cutoff low Cutoff high 80 9.76 16.74 90 9.24 17.26 95 8.71 17.79 96 8.52 17.98 98 7.97 18.53 99 7.36 19.14 Graphically: Range of Cutoff Values 5.0 7.0 9.0 11.0 13.0 15.0 17.0 19.0 80 85 90 95 100 Confidence Interval % cocaine cutoff Cutoff low Cutoff high Note that as the confidence level increases, the range of values that are accepted (not classified as outliers) increases as well. To be more certain, a bigger range is required. As noted in the text,

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    Solutions Manual for Forensic Chemistry 11 a larger range does not mean the data is “better”; in fact, the data can be less useful because the range is too large. 4. Differentiate clearly between statistical and analytical errors. A statistical error is one that is traceable to normal fluctuations due to random errors, which are small and equally positive and negative. An analytical error is traceable to, and due only to, laboratory procedures. 5. Justify/explain the use of the factor 3 in equation 2.13. The factor 3 represents three standard deviations. The range of +/- 3 standard deviation units includes 99.7% of all possible outcomes, assuming that the underlying distribution is Gaussian. Refer to Figure 2.7. 6. In the quote from NAS report (page 13 of the text), what aspects of NUSAP are specifically addressed? Sources of uncertainty (S, A, P), and the scale (A assessment). “…they should identify, as appropriate, the sources of uncertainty in the procedures and conclusions along with estimates of their scale (to indicate the level of confidence in the results).” 7. A forensic chemist prepares a standard of caffeine in chloroform for use in a quantitative assay. The caffeine is purchased from a reputable supply house and arrives with a certificate stating that it is 99.5% pure or better. The analyst needs to make a stock solution at a concentration near 1 mg/mL. To do so, he obtains a Type A 50.00 mL volumetric flask (+/- 0.05 mL) and uses a microbalance to accurately weight out 49.6 mg. The balance uncertainty is listed as +/- 0.0003g near the 50mg range. The analyst quantitatively transfers the powder to the flask and carefully dilutes the solution to volume. Report the concentration of the solution in mg/mL and the propagated uncertainty associated with it. Be sure to use the proper significant figures. This is set up very much like the example in the chapter: first calculate the concentration, then calculate the propagated uncertainty. Because the caffeine is certified at 99.5% or better, a

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    Solutions Manual for Forensic Chemistry 12 correction was not applied and the value as stated is used. Note that significant figures will be dealt with at the end of the calculation; extras are shown in the intermediate steps as needed for clarity. Calculation of the quantity:   49.6mg caffeine(mg/mL) = = 0.992 50.00mL Propagation of uncertainty: 2 2 -3 t 0 1 0.0003g 0.05mL u = + = 6.13 x10 x 0.992mg/mL = 0.006 mg/mL 0.0496g 50.00mL             Range: 0.992 mg/mL +/- 0.006 mg/mL or 0.986-0.998 mg/mL 8. An analyst proposes a new method for the analysis of blood alcohol. As part of a method validation study, she analyzes a blind sample 5 times and obtains the following results: 0.055%, 0.054%, 0.055%, 0.052%, and 0.056%. a) Are there outliers in the data? We’ll use the Grub bs test since it is recommended by ISO. The questioned point is 0.052 since when the data points are arranged in order, the gap is largest between it and the nearest point: 0.052 0.054 0.055 (2) 0.056 The mean of this data set is 0.0544 and s is 0.00152: G = |0.052 – 0.0544| / 0.00152 or 1.58. The G table value for 4 degrees of freedom is 2.132 at the 95% CI, so the calculated value is less than the table value and the point is retained. There are no outliers. b) Based on the results of the outlier analysis and subsequent actions, calculate the mean and %RSD of the analyst’s results. The mean and s are above; the %RSD = 2.8%.

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