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Guide to Instructors
Table of Contents
Page #
Introduction 1 Instructor Vignettes 1 Identifying the Problem 2 Sampling 2 Sample Preparation 8 Gas Chromatography 12 Mass Spectrometry 17 Method Validation 23
Introduction
The materials comprising the Lake Nakuru project are designed in a modular fashion, allowing them to be used in different classes in a variety of ways. The vignettes by instructors who have used these materials show many approaches to using these materials in a variety of classroom settings. For example, one use for these materials could be as a capstone assignment in an Instrumental Analysis course. Alternatively, the modular nature of the materials could allow use of only the sampling unit in a Quantitative Analysis or an Environmental Chemistry course.
Instructor Vignettes
Heather Bullen Northern Kentucky University Anna Cavinato Eastern Oregon University Alanah Fitch, Loyola University – Chicago Cindy Larive University of California – Riverside Rick Kelly East Stroudsburg University David Thompson Sam Houston State University Thomas Wenzel, Bates College
The remainder of this Instructor’s Guide addresses each unit providing a brief overview of how the content might be used, supplemental lecture content that might be helpful to the students before starting the module, and answers to the questions posed within each section.
Identifying the Problem
This module briefly lays out the scientific problem – what in Lake Nakuru is killing the flamingos? Like most real scientific questions, this one is complex. To our knowledge the “real” reason is not known and there may in fact be a variety of contributing factors. Even if only one or two of the modules will be used in your course, it would probably be useful to start with the Identifying the Problem unit to provide a context for the other sections.
Identifying Possible Analysis Methods
It is useful to have students explore the possible analytical methods that might be used to measure the chemicals involved in each of the three hypotheses for flamingo death.
If this exercise is used toward the end of an analytical chemistry course, students may be asked to go back through each method that was covered in the course and explain whether or not it might work for the analysis of the species in question.
An alternative is to ask students to go to the scientific literature and find possible methods for the analysis. In this format, it is probably best to divide the class into groups and give each group one of the three hypotheses. After completion of the assignment, each group can report their findings to the rest of the class. This can lead to a useful discussion of the strengths and weaknesses of the various methods they identify. This discussion may turn up methods that are not covered or emphasized in the course and lead to the introduction of other analysis methods that are usually not discussed.
Sampling
After discussing the Identifying the Problem module and the first page of the Sampling section, students can work in groups on the questions that are provided. The answers to the questions are provided below.
Q1. Pick eight random samples from the grid laid out above. How do you ensure you sampling is random?
Q4. Each of the previous grids is an example of one of these cases. Can you identify which sample is which?
Sample B exhibits constitutional heterogeneity while Sample A exhibits distributional heterogeneity.
Q5. How does distribution heterogeneity affect accuracy and precision?
The answer to Q5 is included with the answer to Q6 below.
Q6. How does constitutional heterogeneity affect accuracy and precision?
When you overlay the sampling scheme with Sample A and Sample B neither sampling scheme is effective, as shown below. The analyte in sample “A” was sampled once (35) as it was in Sample “B” (53). With either sample, unless the sampling scheme could take this heterogeneity into account the accuracy and precision of the measurement would be compromised.
A. Distributional Heterogeneity B. Constitutional Heterogeneity
Q7. Do you see a scenario where distribution heterogeneity could be magnified by mixing and/or sampling?
Sampling is often by weight or by “grab”. In this case settling may alter the sample composition.
Q8. What is the advantage of implementing judgmental sampling over random sampling if one knows the point source for the discharge an analyte into a system?
The advantage is that you can get larger number of relevant samples which should decrease the standard deviation of the average value measured for that sample. The cost should also decrease.
Q9. Assume you have a chosen a selective sampling plan to evaluate pollution from a point source into a lake. Use the diagram below and words to describe your sampling plan.
If we use a purely random grid over the lake we will be unable to tell what the effect of the point source is because we would have sampled only once at the source.
A selective method might be the following:
A sample is taken at the point source and for comparison a sample is taken at a distance from the point source.
Q10. Use a grid design (as we have previously done) to show how you would conduct systematic sampling (regular intervals in space and time) of the pollutant. Is there an advantage to what you might learn using this sampling method? What are the disadvantage(s)?
Q14. Can you think of any control studies you might want to include when compositing samples?
You might want to retain portions of the grabs and analyze them separately.
Q15. Does the EPA Method 525.2 suggest a particular sampling method?
The EPA Method does not describe a particular sampling method because each sampling plan must be designed for the specific analytical question being addressed.
Q16. What sample size does EPA Method 525.2 suggest for analysis of pesticides in water? Why?
The EPA method suggests 1 L. The method does not say why but presumably the concentrations of the organochlorine pesticides are low and using 1 L helps create a “composite” sample of sufficient quantity that the target analytes can be extracted and measured at levels above the instrumental limit of detection.
Q17. An analytical method (not necessarily the Lake Nakura project) has a percent relative sampling variance of 0.10% and a percent relative method variance of 0.20%. The cost of collecting a sample is $20 and the cost of analyzing a sample is $50. Propose a sampling strategy that provides a maximum percent relative error of ±0.50% (α = 0.05) and a maximum cost of $700.
error = ValueExperimental − ValueKnown = X − μ
relative error = X^ −μ
= X
% relative error = 100 X^ −μ
⎥ =^100
X
If there is no bias then the error is equal to the standard deviation, s and
% relative error = 100
s
μ
⎦^ ⎥
To answer this question requires reading Harvey, specifically eq. 7.
X −μ = error = t
ssamp^2
n samp
+ smeth
2
n samp n rep
In this equation μ is the true value and t is the value for the t test which depends on the number of samples and the confidence level. The value of α is 0.05. Since we are going to be limited by the cost, a rough guess of the sample size is:
=
Begin the iterations assuming t(10,0.05)=2.228 (from t table).
n=10. You have 10 samples you can work with to get the maximum percent relative error of 0.5%. The percent relative error is
The total variance associated with a measurement is
σ= σ s + σ s + x
where …..σx is the variance from any other contributing factor. So we can write
σ total^2 = 0.5 = σ sampling^2 + σ method^2 = σ sampling^2 + 0.
This sets the allowed variance associated with sampling at 0.3.
Q18. The auto sampler on the GC-MS you will be using for pesticide analysis has 200 vial locations. How will you choose your representative samples? Here is a picture of Lake Nakuru. Design your sampling plan. Think about random, systematic, clustering, etc. sample strategies. Will you take grab samples or pool samples together?
One way to use this question is to have the students discuss these questions in groups and put together a final plan. The plans could either be presented to the class or turned in as a graded written assignment.
3. Sample Preparation
Students who have already carried out project-based labs involving extensive sample preparation will likely be familiar with many of the concepts in this module. If this is not the case for your students, this section could be good preparation for a laboratory project or serve as a surrogate for actually conducting such a study. Although the focus of this section is necessarily on the preparation of aqueous samples for GC-MS analysis, many of the concepts are relevant to analysis of trace organic compounds in general. Students will likely need some guidance in answering questions 1-3 and may need to refer to both the EPA method and the Identifying the Problem module, especially to answer Q3. One way to handle these questions is to discuss the issues in class, without necessarily arriving at the correct answers. The remainder of the module is fairly self-contained and students should be able to work through the rest of the questions in small groups or as a homework assignment. Once they have completed this Sample Preparation module, they should be better able to answer questions 1-3.
well. Octadecyl phases are non-polar. Bare silica has a polar surface so it is doubtful that the organochlorine pesticides will adsorb to this material.
Q7. Discuss the following aspects of the solvent you would choose to elute the organochlorine pesticides.
The polarity of the eluting solvent must be such that it readily dissolves the adsorbed pesticides. In this case it ought to be relatively non-polar.
If the eluted sample will be directly injected into the gas chromatograph, then the solvent needs a low boiling point so that it elutes from the chromatographic column prior to the pesticides. Also, the volume of organic solvent used for the elution usually dilutes the analyte too much so that it needs to be concentrated before injection into the gas chromatograph. This concentration step is facilitated by having a solvent with a low boiling point.
The volume of eluting solvent needs to be large enough to remove all of the compound from the solid phase. But if the volume is too large, that will lengthen the time needed to concentrate the sample for injection into the gas chromatograph-mass spectrometer.
Q8. How would you remove any residual water in the organic phase?
Add a suitable drying agent such as sodium sulfate, similar to what students have likely done in the past in organic chemistry.
Q9. How would you reduce the volume of the organic extract that contains your pesticide?
Students will likely think of evaporation with heating as a means of reducing the solvent. A common procedure is to blow an inert gas such as nitrogen over the sample to reduce the volume.
Q10. What concerns might you have in the solvent reduction step?
One is that there is some loss of the analyte because some of it has evaporated as well. A second is that some of the compound decomposes. Evaporation under a nitrogen stream avoids the use of higher temperatures and is often used to reduce loss and decomposition of the analyte relative to an evaporation procedure involving heating.
Q11. What would be the final extraction concentration be if you plan to use EPA Method 525.2? Can you detect this level in the lake?
The LOD is given as 0.083 μg/L (ppb) for the sample injected into the GC-MS. The lake water is 150 ng/L or 150 ppt. The extraction step calls for a liter of water that is eventually reduced into a volume of 1 mL of organic solvent. This represents a 1,000-fold increase in the
concentration compared to the original sample of the lake (extract concentration is 150 ppb). This concentration is well above the detection limit. For quantitative purposes, it is usually recommended that the sample concentration be ten times the LOD or 0.83 μg/L (ppb). The sample is well above the value needed to complete a quantitative analysis of DDT in the lake.
Q12. What does it mean to determine the precision of the pesticide analysis? What does it mean to determine the percent recovery of the pesticide?
Precision is a measure of the reproducibility of the extraction process from run to run.
The percent recovery refers to the amount (typically a mass) of the compound that is removed from the solid phase sorbent compared to the amount that was introduced onto the solid phase in the sample. Percent recoveries are evaluated for a standard for which you know exactly how much you introduced to the solid phase.
Q13. Suppose a solid-phase extraction procedure provided a low percent recovery. What could you change in an attempt to raise this value?
There are three possibilities for why the percent recovery is low.
The first possibility could be that the sample was run through the solid phase sorbent at too high a flow rate such that the compounds did not have sufficient time and contact to adsorb to the surface. Correcting this would involve slowing the flow of the sample through the solid phase.
A second possibility is that the surface of the solid does not sufficiently match the polarity of the compounds you are trying to adsorb and the compounds are not being retained. In this case you would need to use a different solid phase. Also, many solid phase materials require a pre- treatment of the surface before using them for adsorption. If the pre-treatment was not performed, the adsorption might be compromised.
A third possibility is that the eluting solvent does not sufficiently match the polarity of the compounds being eluted. In this case, you would need to switch the solvent.
Another possibility that students might suggest would be that the sample could overload the capacity of the solid phase sorbent, however, this is very unlikely with environmental samples as contaminants are typically present in trace amounts.
Q14. Is the solid used in the cartridge effective for removing the pesticides from water?
Yes. All of the % recoveries in Table 8 are in the range of 98-118 percent, so the solid did remove the pesticides.
Q15. How is it possible to get a percent recovery of over 100?
Every step in the process is subject to error. To determine a percent recovery, you must introduce a known concentration onto the cartridge. There is error in the volume you introduce such that the volume is larger than expected.
these topics in a traditional lecture format, one could also consider the companion active learning module on separations as a way to teach these concepts.
Q1. Create a graph of the boiling points of a series of linear alkanes as a function of mass. Can you separate these compounds by boiling point? Describe the trend you observe in the graph?
-
0
50
100
150
200
0 20 40 60 80 100 120 140 160
m.w.
B.P. oC n-alkanes
The graph is almost a linear curve in which the boiling point increases regularly with molecular weight.
Q2. Figure 1 contains a portion of chromatogram showing the elution of a sample of n -alkanes from a column held at a constant temperature of 87 oC. Estimate the approximate time at which C16 will appear?
An examination of the chromatogram shows that each subsequent n-alkane in the homologous series has approximately double the retention time of the previous one. Since C15 takes about 12 minutes, C16 should take about 24 minutes.
Q3. Using the data in Table 2, create a graph of the boiling points as a function of molecular weight. Is the trend in this graph the same as you observed for the series of n -alkanes? If not, what do you think makes it different?
-
0
50
100
150
200
250
300
350
400
0 50 100 150 200 250 300 350 400 450 m.w.
B.P. oC
n-alkanes
Pesticides
No. The graph is quite different and there is no overall correlation between molecular weight and boiling point.
Q4. Examine the molecular structures of the compounds in Table 2. These compounds can be divided into three general types. Which ones would you group together into the three different types? Within a compound group, is there a relationship between structure, molecular weight and boiling point? Do you think that the compounds in Table 2 can be separated based on their boiling points alone?
Group 1 – Compounds 1-4. The compounds in group 1 are all isomers with the same boiling point. These definitely cannot be separated on the basis of boiling points.
Group 2 – Compounds 5-8. There is no agreement between weights and boiling points. Compound 8 is the heaviest of the group but has the lowest boiling point.
Group 3 – Compounds 9-17. Several of these compounds do not have data because the compounds decompose before the boiling point. For those that do, there is no correlation between the boiling point and the molecular weight.
Final conclusion is that these compounds cannot be separated on the basis of boiling point alone. It will be necessary to select a stationary phase that has different interactions with the compounds to produce different partition coefficients.
Q5. Use the information in Table 3 above to find the column recommended for a boiling point- based separation of linear alkanes. What is the chemical composition of the columns suggested for the separation of alkanes?
Methyl polysiloxane. This phase only has methyl groups on a siloxane polymer.
Q6. From Table 3 and Figure 2 what are the compositions of the DB-5 and DB-1 stationary phases? Why is the DB-5 phase a better choice for separation of the organochlorine compounds listed in Table 2?
during the chromatogram. Compounds with low boiling points stay on the column longer and are better separated. Compounds with high boiling points move through the column faster as the temperature is raised and have shorter retention times and more narrow peaks than they would in an isothermal run.
Q11. Look at the maximum temperature at which the stationary phases in Table 3 are stable. What is the maximum operating temperature for the DB-5 column?
325 oC.
Q12. What would an ideal chromatographic peak look like? What would a chromatographic peak look like if you overloaded the column? Why is it undesirable to overload a column?
An ideal peak exhibits a symmetrical Gaussian shape. If you overload the column, you will either get peaks that exhibit fronting or tailing, both of which are a form of asymmetry of the peak. The specifics of whether you observe fronting or tailing are rather complex and depend on particular parameters of the system. Whichever one you observe, peaks that show tailing or fronting will lead to a loss of column efficiency and a reduction in the resolution between adjacent chromatographic peaks. It can also be more difficult to achieve accurate quantitation with asymmetrical peaks.
Q13. A 1 L water sample containing 0.1 μg/L of DDT was concentrated to ~1 mL by the sample preparation procedure and 1 μL injected splitless onto a 30 m X 0.25 mm id fused silica capillary column coated with a 0.25 μm bonded film of DB-5.MS. Calculate the ng of material injected. Is this value greater or less than the capacity of the column?
The original concentrated sample (1 mL) contains 0.1 μg or 100 ng.
1 μL (1/1000) of the 1 mL sample is injected into the GC in a splitless mode.
(100 ng) x (1/1000) = 0.1 ng was injected into the instrument.
This is well under the column capacity of 50-100 ng.
Q14. Evaluate the GC data below for benz[a]anthracene and chrysene and determine if the GC column requires maintenance.
Height of benz[a]anthracene peak = 12 mm
Height of chrysene peak = 31 mm
Average height of the two peaks = 21.5 mm
Height of valley between peaks = 5 mm
(5 mm/21.5 mm) x (100) = 23.26%
This is less than 25% so the column does not require maintenance.
5. Mass Spectrometry
Much like the chromatography section, the focus of this unit is on the practical aspects of mass spectrometry measurements and data interpretation. The module introduces the basic components and there is a short discussion of electron ionization, however, the module does not explain how the various mass analyzers work or how the ions are detected. Students will likely benefit from a lecture on these topics before completing the module, although this knowledge is not required to answer the questions. The discussion of both quadrupole and ion trap mass analyzers in EPA method 525-2 opens the possibility of dissecting both mass analyzers in lecture and discussing their relative merits.
Q1. What important information would you need to unambiguously identify a pesticide like DDT in the presence of other similar compounds?
You would need to know its retention time in the separation and have a match of the mass spectrum of the unknown peak to a standard of DDT.
Q2. Why do you need ions instead of neutral species in mass spectrometry?
Ions can be removed from a gaseous mixture using an electric field whereas neutral species cannot. Ions also will interact with a magnetic or electric fields used in the mass analysis portion of the instrument.
Q3. Mass spectrometers operate under high vacuum conditions? Why?
To insure that the ion has a long enough lifetime to be analyzed by the mass analyzer. If the ion collides with another molecule, it is likely to pick up an electron, be fragmented or be deflected from the ion beam.
Q4. What is an advantage of having a large number of fragment ions present in the mass spectrum? Are there any disadvantages?
Large numbers of fragment ions allow for a more definitive match of the mass spectrum of an unknown compound to the known compound. The disadvantage of having a molecule form too many fragments is that the most intense peak (called the base peak) in the mass spectrum is used for quantitative analysis. A large amount of fragmentation may reduce the intensity of the most intense peak.
Q5. Compare the experimentally determined mass spectrum for DDT in Figure 3 to that from the NIST database in Figure 4. Do you think this is a good match? Why?
Yes. The most intense clusters of peaks do match up with each other. Many of the smaller clusters match as well.
Q6. Table 2 of the EPA method 525.2 lists the quantitation ions for DDT as 235 and 165. Why do you think these ions were selected to determine the amount of DDT in our sample?
Because they are the two most intense peaks in the mass spectrum.
CH 3 Cl+^ CH 2 Cl 2 +
CHCl 3 +^ CCl 4 +
CH 3 Cl+^ CH 2 Cl 2 +
CHCl 3 +^ CCl 4 +
Q9. Now calculate the isotopic distribution expected for the molecular ion (M+) of DDT. How many 35 Cl atoms are in the molecule that comprises the most intense peak?
Entering C 14 H 9 Cl 5 as the molecular ion, the chlorine isotopic distribution pattern shows peaks at 352, 354, 356, 358, 360, and 362 that correspond to chlorine fragments. The peak at 352 would have five 35 Cl atoms. The most intense peak is the 354 peak, so this corresponds to a fragment with four 35 Cl atoms and one 37 Cl atom.
Q10. In a complex environmental sample that has pesticide contaminants at low levels, the GC- MS spectrum extracted for the DDT molecular ion or base peak may have contributions from other compounds, naturally occurring or anthropogenic, that might be confused with DDT. How could the chlorine signature in the mass spectrum be used to help with compound identification?
While many molecules or fragments of molecules might have masses identical to the mass of some of the DDT fragments, the likelihood that these contaminants have the exact same number of chlorine atoms and the exact same pattern for the chlorine isotope distribution is much less likely.
Q11. Propose a structure for the 235 fragment ion of DDT. Use the isotopic distribution calculator to calculate the relative abundances of the ions expected for the structure you propose. How does the predicted spectrum match the intensities in Figure 2?
Usually the easiest way to determine the formula for a fragment is to examine the loss from the molecular ion. Using 352 as the molecular ion for DDT with only 35 Cl isotopes, the loss to 235 involves a mass of 117. One way to get 117 is to lose a CCl 3 group (12 + (3 x 35) = 117). The DDT molecule has a CCl 3 group in it so this loss does make sense. The formula for a fragment with a mass of 235 would then be C 13 H 9 Cl 2. The predicted masses and intensities for the chlorine isotope pattern are 235 (100), 237 (65.7) and 239 (11.1). This agrees well with the pattern shown in the mass spectrum.
Q12. Why was chrysene- d 12 selected as an internal standard rather than chrysene with a natural isotopic abundance of hydrogen nuclei? Would you expect that the chrysene and chrysene- d 12 to be resolved in the GC chromatogram?
While it is unlikely that chrysene would be found in a sample of the water, chrysene is found in some samples in nature and could be a contaminant in the lake water. The isotopically- substituted chrysene- d 12 has a weight 12 units higher than normal chrysene and there is no expectation of finding this compound in samples of natural systems.
Sometimes it is possible to resolve molecules that differ only by isotope distribution in chromatographic separations, but the system has usually been designed specifically to facilitate such a separation. The column and conditions used to separate the pesticides in EPA Method 525.2 would not separate chrysene and chrysene- d 12.
Q13. Section 13.1.2 of EPA 525.2 describes the calculation of the method detection limit (MDL) for each analyte. Use the data in Tables 1 and 2 to calculate the MDL for DDT. How does this value compare with those found in Tables 3-6 of EPA Method 525.2?
According to the EPA Method, “the MDL is defined as the statistically calculated minimum amount that can be measured with 99% confidence that the reported value is greater than zero .”
To calculate the MDL, students will use the variance in the calculated concentrations for the fortified blanks (Table 1) as shown below. Additional information about the calculation of the MDL for EPA methods can be found at http://water.usgs.gov/owq/OFR_99-193/detection.html.
MDL = s × t ( n − 1 , 1 −α= 0. 99 )
For each of the fortified blanks, the first step is to normalize the intensity of the DDT peak relative to that of the internal standard for each measurement, as shown in the Table below. The data in Table 2 is then used to construct a calibration plot, from which the concentration values for each replicate can be calculated. (Note: some attention will need to be made to the fact that concentration values for calibration curve are in mg/L, while MDL should be calculated in values of μg/L. A subtle point is that the internal standard was added to the sample before the pre- concentration at 5.00 μg/L, and since we have normalized to this value, the calculated
