Solution Manual for Forensic Chemistry, 2nd Edition

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PrefaceiiChapter 1 Introduction1Chapter 2 Foundations5Chapter 3 Quality Assurance and Quality Control20Chapter 4 Reporting Defensible Uncertainty and ObtainingRepresentative Samples39Chapter 5 Chemical Fundamentals: Partitioning, Equilibria, andAcid/ Base Chemistry47Chapter 6 Instrumentation59Chapter 7 Drugs as Physical Evidence: Seized Drugs and Their Analysis68Chapter 8 Forensic Drug Analysis: Selected Drug Classes77Chapter 9 Drugs in the Body84Chapter 10 Forensic Toxicology92Chapter 11 The Chemistry of Combustion and Arson96Chapter 12 Explosives103Chapter 13 Firearms and Associated Chemistry Evidence111Chapter 14 The Chemistry of Colors and Colorants115Chapter 15 The Chemistry of Polymers121Chapter 16 Forensic Analysis of Inks and Paints125Chapter 17 Chemical Analysis of Materials: Paper and Fiber131

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Solutions Manual for Forensic ChemistryiiPrefaceThis solutions guide accompanies the second edition of the text. For the instructor, this manualshould help clarify concepts and principles as applied in the homework. For the student, thismanual should bethe last book opened. Why? In my experience, students tend to use thesolutions manual in a less than optimal way. For example, here is the wrong way to use asolutions manual:1.Obtain the assigned homework from the instructor.2.Open the solutions manual and copy the results for each problem, cleverly altering themso 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 thesolutions 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 theproblem involves an acid/base extraction, refer back to that section of the text and refreshyour 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 ChemistryiiiNotes: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. 2London: Pharmaceutical Press, 2004. (Volume 2 contains monograms on the individualdrugs and is referenced most frequently.)PDR refers to:Physicians’Desk Reference,58thed. Montvale, NJ: Thomson PDR, 2004.Physicians’Desk Reference for Nonprescription Drugs and Dietary Supplements, 22nded.Montvale, NJ: Medical Economics—Thomson Healthcare, 2001.CRC Handbook refers to:Handbook of Chemistry and Physics, 84thed. 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, andBiologicals. Whitehouse Station, NJ: Merck Research Laboratories—Merck and Co., 2001.For illustrative purposes and in spreadsheet printouts, extra significant digits may appear inintermediate calculations. This is to avoid rounding errors and make the process and procedureused 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 Chemistry1IntroductionCCHHAAPPTTEERR11From the chapter1. Compare and contrast the adversarial system and the scientific method. List thestrengths 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 basedon arguments from opposing parties. The scientific method does not by design pit one sideagainst another. Certainly, there are arguments in science, but the arguments are resolved byexperiment and observation. The process may take decades, but if the data is consistent andreproducible, conclusions are drawn and consensus is reached. Adversaries argue based ondiffering viewpoints, interpretations, and agendas. Science can be used to support or refuteinterpretations of events and evidence, but the outcome depends on the strength of the argumentand the skill of those making it.The adversarial system is well-suited to situations in which social and human (for lack of a betterterm) issues are involved. A person may have committed murder, but the reasons andcircumstances are critical to the deliberation. Science, on the other hand, seeks to elucidatenatural 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 isessential for understanding the laws of the universe since consistency and reproducibility arerequired to derive them. It is not an ideal system for dealing with human beings.2. During aDauberthearing, what entity ultimately decides on admissibility?The judge, who acts as the gatekeeper.

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Solutions Manual for Forensic Chemistry23. 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 appealsprocess could also be placed in the category since professionals of the law review the work doneby other professionals. Peer review is more obvious and central in science where it is an integralpart of the dissemination process. Scientific findings are submitted to peer-reviewed journalswhere editors assign qualified reviewers to comment on and judge the work. They may accept itas is, request more information or work, or recommend that it not be published. Publication in apeer-reviewed journal is validation of acceptance by those with similar skills, background, andunderstanding of the mechanism of science; materials published in non-peer-reviewed journalsor sources generally are not given the same weight or credibility as peer-reviewed data.4. Describe how a preponderance of inclusive circumstantial evidence can becomeconclusive in the eyes of a jury.The Wayne Williams case is the perfect example of this phenomenon. Had a few dog hairs beenfound on victims, such a finding would place under suspicion only those people who owned ormight have come in contact with dogs, a large percentage of the population. It was the uniquecombination of hairs and fibers that was compelling. Consider your own home. What types ofhairs 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 clothingand articles of furniture. Link enough of such observations together and it becomes clear thatyour apartment is a much different environment than even one next door.Integrative1. A great scientist can still be a terrible forensic scientist; a person who gives wonderfultestimony can be a terrible forensic scientist. Comment on these observations and theimplication for forensic chemistry.Forensic science and forensic chemistry require understanding and skills in comparison. It is notalways 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 Chemistry3forensic scientist must be able to apply advanced techniques and knowledge to the analysis ofevidence, yet present the results clearly and concisely to an audience without a comparable bodyof knowledge or experience. Thus, forensic scientists must be good communicators and goodteachers. A great research scientist might be able to communicate with peers, but may struggle todistill the complexities of his or her work such that anyone with a high school education couldgrasp the concepts. Conversely, a person may be able to concisely and brilliantly present findingsthat are completely wrong; the skill of the presentation coupled with the audience’s lack ofbackground can result in acceptance of incorrect information. A forensic chemist must be themaster of the chemistry and of the forensic aspects—to speak the truth (derived from science) inthe public forum.2. Can jurors ask questions of expert witnesses? Comment on your findings regarding thisissue.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 submitquestions for expert witnesses to the judge, who screens the questions and then will ask thosethat are probative and appropriate. In some cases, jurors are also allowed to take notes and totake these notes into deliberation.How jurors interact with expert witnesses and how scientific evidence is presented is becomingan increasingly difficult issue, given the complexity of scientific testimony. Complicating thisissue is the general lack of scientific background in jurors, judges, and lawyers. It is hard toimagine reforms not occurring as the complexity of testimony continually increases.Food for thought1. Is the analysis of drugs using instruments such as mass spectrometers and infraredspectrometry 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 Chemistry42. How important is the way in which scientific evidence is presented? Comment on therelative importance of content versus presentation. Why is learning how to testify such animportant skill?See Integrative question #1 above. A forensic scientist must be able to communicate findingsconcisely and clearly to the court so the trier-of-fact can properly weigh the information. Poorlycommunicated evidence can be as damaging as poor scientific practice.

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Solutions Manual for Forensic Chemistry5FoundationsCCHHAAPPTTEERR22From the chapter1. A standard of Pb2+for a gunshot residue analysis using atomic absorption is prepared byfirst dissolving 1.0390 g dried Pb(NO3)2in distilled water containing 1% nitric acid. Thesolution 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 avolumetric 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 inppm of Pb2+.The calculations should be done together and the results rounded at the end. The only exceptionto this is with the formula weight for the lead nitrate, which is rounded separately as anaddition/subtraction. Using the periodic table in the textbook:Formula weight of Pb(NO3)2= 207.2 + (2 x 14.01) + (2 x 3 x 16.) = 331.22For this calculation, using this periodic table is a problem since the formula weight willneedlessly limit the number of significant figures. Thus, a periodic table with more reporteddigits is needed, even though Pb will remain as is. The new calculation:Formula weight of Pb(NO3)2= 207.2 + (2 x 14.00674) + (6 x 15.9994) = 331.2To obtain the concentration in ppm (mg/L), the molarity is calculated and then converted to thefinal value. Note that the moles of Pb in the sample are equal to the moles of Pb(NO3)2sincethere is a 1:1 mole ratio of lead in lead nitrate:

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Solutions Manual for Forensic Chemistry6L0.5000gmg1000Pbmoleg207.2NOPbmole1Pbmole1NOPbmoleg331.2NOPb1.0390g232323molarity (moles/L)ppm (mg/L)= 130.0 ppmThe result is rounded to four significant figures, the least number present in any of themeasured values.b) Determine the absolute and relative uncertainties of each value. Select the largestand report the results as a concentration range.An easy way to visualize the largest uncertainty at a glance isto convert each one to a “1part per” expression. For the balance:0.00020.0002g10.0002==1.03901.0390g51950.0002or 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 parts10.00 mL transfer = 7692 parts (remember to convert μL to mL)100 mL volumetric flask = 1 part per 1250 partsThe largest value will dominate the calculation of uncertainty and here that value is theuncertainty 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 finalconcentration can be calculated two ways:130.0/1250 = 0.104 (1 part in 1250 parts, same relative uncertainty as the limiting valuefrom the 100 mL volumetric flask).-or-

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Solutions Manual for Forensic Chemistry7130.0 * (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 calculatedabove and combining them as per equation 2.1:22222flask100mlEppendorfflask500mLbalanceuncert.e222220.00020.20.00130.08e1.0390500.0010.00100.00Notice that units will cancel out so that the μL uncertainty attributed to the Eppendorfpipette, 1.3 μL, is converted to mL. Solving for the erroreyields 9.2 x 10-4or 0.09%. Thisresult 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, theuncertainty in ppm is slightly larger:131.0. * 0.000924 = 0.121, rounded to the result as 0.1However, 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 generallyreported to only one decimal place. Therefore, this weight may limit significant figures insome calculations. In addition, the value for oxygen is also reported to one decimal, 16.0, inthe table included in this book. However, additional decimals can be obtained using othertables. This is generally not the case with lead.

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Solutions Manual for Forensic Chemistry82. 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, aconcentration for example. In this case, the potential outlier is on the low side of the mean andtherefore 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 Valuesof 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 %cocainewould have to be in the 11th sample for it to be eliminated at the following significancelevels: 0.20, 0.10, 0.05, 0.04, 0.02, and 0.01. Present graphically and comment on thefindings.This problem can be addressed using Excel® and by realizing that the result will be a rangerather than a single value. The data is first summarized and the Q values obtained from thereference, which calls for careful reading of the text and the scenarios described. Here, we wishto find out what a single outlier value will be at either end of the distributed values, with no otheroutliers of concern. Therefore, the single outlier test and table, designatedr10in the paper, isappropriate (p. 141). This directs us to the proper table found on page 142, Table 1. The value ofn will be 11 with one more included, so the corresponding table values (Qtable) are:CI = 80%90%95%96%98%99%0.3320.3920.4440.4600.5020.542An outlier will be rejected if Qcalc> Qtable, so the next step is to determine what outlier values,when plugged into the calculation of Q (equation 2.8), will result in a Qcalcthat just equals Qtable.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 Chemistry911.5- x= 0.33215.0 - x11.5 - x = 0.332[15.0 - x]11.5 - x = 4.98 - 0.0322x6.52 - x = -0.332x6.52 = 0.668xx = 9.76%where [11.5–x] represents the gap and [15.0–x] represents the range. To simplify latercalculations, this calculation can be solved generically:TTTTTTTTTlow - x= Qhigh - xlow - x = Q high - Q xlow - Q high = x - Q xlow - Q high = x(1-Q )low - Q highx =1-QSubstituting the values of 11.5 for low, 15.0 for high, and 0.332 for QTyields 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:TThigh - Q lowx =1 - Q

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Solutions Manual for Forensic Chemistry10For the 80% confidence interval, the high value cutoff is thus 16.74%. If the forensic chemistobtains a percentage cocaine value of greater than 16.74% or less than 9.76%, that value wouldbe an outlier at the 80% confidence interval.With generic equations in hand, a spreadsheet allows for quick solution and graphing of theresults:Low value:11.5High value:15.0CI:809095969899Table:0.3320.3920.4440.460.5020.542Low cutoff:9.769.248.718.527.977.36High cutoff:16.7417.2617.7917.9818.5319.14ConfidenceintervalCutofflowCutoffhigh809.7616.74909.2417.26958.7117.79968.5217.98987.9718.53997.3619.14Graphically:Range of Cutoff Values5.07.09.011.013.015.017.019.080859095100Confidence Interval% cocaine cutoffCutoff lowCutoff highNote that as the confidence level increases, the range of values that are accepted (not classifiedas 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 Chemistry11a larger range does not mean the data is “better”; in fact, the data can be less useful because therange 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 aresmall 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 unitsincludes 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 arespecifically addressed?Sources of uncertainty (S, A, P), and the scale (A assessment).“…they should identify,as appropriate, the sources of uncertainty in the procedures andconclusions along with estimates of their scale (to indicate the level of confidence in theresults).”7. A forensic chemist prepares a standard of caffeine in chloroform for use in aquantitative assay. The caffeine is purchased from a reputable supply house and arriveswith a certificate stating that it is 99.5% pure or better. The analyst needs to make a stocksolution at a concentration near 1 mg/mL. To do so, he obtains a Type A 50.00 mLvolumetric 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 analystquantitatively 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 uncertaintyassociated 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, thencalculate the propagated uncertainty. Because the caffeine is certified at 99.5% or better, a

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Solutions Manual for Forensic Chemistry12correction was not applied and the value as stated is used.Note that significant figures will bedealt with at the end of the calculation; extras are shown in the intermediate steps as needed forclarity.Calculation of the quantity:49.6mgcaffeine(mg/mL)== 0.99250.00mLPropagation of uncertainty:22-3t010.0003g0.05mLu=+= 6.13 x10x 0.992mg/mL = 0.006 mg/mL0.0496g50.00mLRange:0.992 mg/mL +/- 0.006 mg/mL or 0.986-0.998 mg/mL8. An analyst proposes a new method for the analysis of blood alcohol. As part of a methodvalidation 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 Grubbs test since it is recommended by ISO. The questioned point is 0.052 sincewhen the data points are arranged in order, the gap is largest between it and the nearest point:0.0520.0540.055 (2)0.056The mean of this data set is 0.0544 and s is 0.00152:G = |0.052–0.0544| / 0.00152 or 1.58. The Gtablevalue for 4 degrees of freedom is 2.132 at the95% CI, so the calculated value is less than the table value and the point is retained. There are nooutliers.b) Based on the results of the outlier analysis and subsequent actions, calculate the meanand %RSD of the analyst’s results.The mean and s are above; the %RSD = 2.8%.

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