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227 Qa Qc

  1. 1. CHAPTER 1 ANALYTICAL CHEMISTRY OVERVIEW Analytical Chemistry can be defined as the marriage of qualitative and quantitative analyses, that is, the identification of an analyte (the substance sought for) and then the determination of the concentration of that analyte. Simply put, qualitative analysis answers the question “What is in the sample?” and quantitative analysis answers the question “How much is in the sample?”. Usually, the emphasis in analytical chemistry is the determination of the quantity of a pure substance in a sample. Clinical Analytical Chemistry is concerned with the identification and quantitation of substances related to living organisms, especially human beings. This sub-discipline is concerned with the sampling and analysis of body fluids, such as urine, blood serum and blood plasma. Most clinical chemistry laboratories are located in hospitals and medical centers; although today some are located in stand-alone testing facilities. Modern physicians are very dependent on the analytical results that are provided by the clinical laboratory. Health care specialists order many lab tests in an effort to ascertain what illnesses or conditions exist for any particular patient. The lab tests may also rule out the presence of certain diseases, and are also used to track the levels of certain medications in the body. Life-and-death decisions may be based on these clinical laboratory results. For these reasons, great care must be exercised when taking samples, preparing the samples, analyzing the samples, and then interpreting the results. This course will provide a foundation for these very important steps in the clinical analytical chemistry discipline. Please examine Figure 1, on page 2. This figure provides a flow chart of the analytical process. Note how many steps are required for a successful analysis of any material for any component. Problems or carelessness at any step in this process may well compromise the validity of the final result. Understanding of the steps involved in successfully obtaining and then analyzing a sample will be of great benefit to all those who are involved in the healthcare industry. 1
  2. 2. Figure 1: Process Flow for Analytical Chemistry Collect sample Preserve Deliver Sample to lab Log In Sample custody Sample Sample Preparation Sample Analysis Record Data Unacceptable? Re-analyze sample Evaluate Sample Data Acceptable? Write Lab Report There are eight main steps in the analytical chemistry process: 1. The sampling step. Before a sample is analyzed, it must be obtained. Selection of an analytical sample from the gross sample requires that the sample be representative of the whole sample. Representative means that the portion of the gross sample that is collected and analyzed should provide the same results as other portions of the gross sample collected and analyzed. If one is sampling a patient’s urine for protein, for example, the urine sample must be representative. The physician’s assistants give specific directions for obtaining such a sample. Some samples are, by their nature, homogeneous, or well- mixed. Obtaining blood from a living patient usually represents a homogeneous sample. Some samples may not be homogeneous, and the sampling personnel must take 2
  3. 3. additional measures to attempt to achieve this condition. Taking several samples over a period of time (then mixing) may provide an acceptable degree of homogeneity. 2. Sample Preservation. After a sample is successfully obtained from a source (usually a living being), the sample must be preserved before delivery to the lab. Because sample can undergo chemical and even physical changes soon after sampling, steps must be taken to preserve the sample so that it reaches its destination essentially unchanged. Preservation may be as simple as refrigeration of the sample, or some preservation agent may have to be added to the samples just after it is collected (such as acid or base or a chemical salt). 3. Sample Delivery. Samples must be delivered to the testing facility in an expedient manner. Many samples, even when preserved, begin to degrade or decay within 24 hours. Many samples are shipped by Federal Express or some other rapid service. 4. Sample Log-In. Once the sample is received at the testing facility, it must be registered (logged) as having been received. Sloppy receiving procedures cause samples to be mixed up or otherwise poorly identified. Incorrect tests may be performed, which may result in costly re-sampling and re-analyses. 5. Sample Preparation. Once properly received, samples often must be sub-sampled to obtain a quantity that can be tested in the lab. Most analyses today are accomplished through the use of sophisticated instrumentation, and only a few milliliters or milligrams of sample are required. The aliquot (sub-sample) must also be representative. In the case of blood specimens obtained for clinical tests, a specimen of whole blood is collected in a glass tube (a vial). When the specimen is allowed to stand for several minutes, the soluble protein fibrinogen is converted by the coagulation mechanism to fibrin, which forms a clot that entraps blood cells. After the clot forms, it shrinks and squeezes out a straw-colored liquid which is known as serum. The serum contains all the constituents of whole blood except the fibrinogen. An aliquot of the unclotted whole blood (obtained by the addition of an anticoagulant [anti-clotting agent] such as heparin) may be used as the sample, or the blood cells may be separated by centrifugation. The supernatant fluid obtained from centrifugation is known as the plasma. However, plasma may still contain a small quantity of fibrinogen, (which could clot) so serum is often used instead. The preparation step is not always simple. A sample is not usually introduced directly into an instrument or analyzed as-is. It must often be prepared for the analysis steps. The analyte of interest must be amenable to the analysis- that is, it must be in a form where it can be measured. Preparation may be as simple as adjusting the pH of the sample or as complicated as extracting the analytes of interest through a series of extraction and separation steps. Today, many of these extractions are automated, so there is not a lot of intensive hands-on work required. Once the analytes are separated / isolated, the analyte may require purification so no interferences are present during the analysis. A common 3
  4. 4. interference is the presence of serum proteins, which can be removed by precipitation and filtration. Most of the lab investigations we will perform require a minimum of sample preparation. 6. The Sample Analysis / Measurement Step. Once the sample is properly prepared, actual chemical analysis can be performed. This step involves the measurement of some chemical or physical property of the analyte which can then be transformed into a quantitative value (concentration). It is most common to measure a chemical property of the analyte- such as absorbance of some form of energy (atomic absorbance, UV or IR absorbance), reactivity with certain substances (as in a titration), retention on a separation column (GC or HPLC), and similar techniques. Most of these techniques yield relative responses that must be converted to absolute responses. This is done by comparison to standards, or by calculation of results using stoichiometric factors. 6a. The Calculation Step. The results must be in the correct units, and must be interpreted with the proper statistical considerations (error range, for example). The accuracy of a given procedure is determined through the analysis of standards and reference check samples. The weakest link in performing calculations deals with the issue of significant figures or digits. The precision of a result is directly related to the number of significant figures contained in that result. There are two components to significant figures. The first component deals with the precision of the instruments used in the analysis. The second component deals with the actual values themselves. Let’s look at these in more detail. Precision of instruments. Instruments not only include actual analytical instrumentation, but also include measuring tools such as balances, glassware, and thermometers. The following table provides some typical precision values for various instruments: Instrument typical precision Platform balance ± 0.5 g Triple beam balance ± 0.01 g Top-loading semimicro balance ± 0.001 g Analytical Balance ± 0.0001 g 10-ML graduated cylinder ± 0.1 mL 100- ML graduated cylinder ± 0.2 mL 25-ML buret ± 0.01 mL 50-ML buret ± 0.02 mL 10-ML pipette ± 0.01 mL 110o C Thermometer ± 0.2 0C 2. Actual values involved in calculations. In general, the significant figures that are associated with a result are based on the least amount of significant figures associated with the calculations that lead to that result. These calculations can be based on the masses of materials, volumes of materials, the instrument readings themselves, and other factors. 4
  5. 5. Guidelines to determining significant figures: 1. Zeros between non-zero digits are always significant. For example, in the number 302045, there are 6 significant figures. 2. Leading zeros (in front of non-zero digits) are never significant. They may be used to locate a decimal point, but are not significant. For example, in the value 0.000345, there are only three sig. figs. 3. Trailing zeros may or may not be significant. The value 7000 mL may have four sig.figs, or may have one sig. fig. if it can be written as 7 x 103 mL. Some scientists place a line over the significant zeros- 7000 means there are 4 sig. figs. 4. If a value is greater than 1, then all zeros to the right of the decimal point are significant. For example, 5.000 contains 4 sig. figs. 4.0605 contains 5 sig. figs. If the value is less than one, than zeros following nonzero digits are significant. For example, 0.0034500 contains 5 sig. figs. 5. In the addition or subtraction of a set of values, the number of significant figures is governed by the value containing the least amount of sig. figs. after the decimal point. For example, given this series of operations: 101.27 + 2.336 – 10.5 + 0.2973 = 93.4 Three significant figures are determined as the correct precision because of the 10.5 value, which contains 1 sig. fig. after the decimal point. 6. In the multiplication and division of a series of values, the precision of the answer is determined by the value with the lowest number of significant figures. For example, in this series of operations: (2.8 * 108.7)/(0.0373 * 5298.3), the correct result is 1.5 because 2.8 only contains 2 sig. figs. 7. Logarithmic terms: Quantities such as pH should be expressed with the same number of sig. figs. to the right of the decimal point as the number of sig. figs. of the non- exponential digits. For example, a solution with a [H+] of 6.6 x 10-11 has a pH of 10.18 (2 sig. figs to the right of the decimal point) because 6.6 has 2 sig. figs. If the [H+] is 3.75 x 10-5 then the pH will be 4.426 (3 sig. figs to the right of the decimal point) because 3.75 has 3 sig. figs. Conversely, for a solution of pH 10.7, the [H+] = 2 x 10-11 because 10.7 has only one sig. fig after the decimal point. 7. Data Reporting What happens after results are obtained and checked? That depends on who needs the data and in what form they need it. A simple e-mail or fax or phone call may be all that is required. However, for many institutions and programs, a written report may be 5
  6. 6. required, especially if the data were generated as a part of a research project. Whatever the end use of the data is, it is imperative that everything is properly recorded, so that the records can be evaluated and the values verified at some later time. 8. The Data Evaluation Step. The data and results should always be reviewed to ensure that the correct analyses were done, that the calculations are correct, and that the units are correct. In the event that the data are found to be incorrect, this situation must be quickly brought to the attention of interested parties, and an effort made to correct the results, or perform the analysis again. In the classroom, incorrect data may only impact your grade; in the medical field, it may impact someone’s life. 6
  7. 7. CHAPTER 2 STATISTICS AND QUALITY ASSURANCE Consider these four diagrams: xx x x xx ● ● x x x x x x ● ● x x x x What can you say about the accuracy and precision of each of the diagrams, if the center black dot is the “true” answer? These four diagrams represent various conditions of accuracy and precision. What types of conditions could have caused these patterns? 1. Determinate (systematic) errors: a. The cause and magnitude of error can be determined. b. The errors are consistent and are about the same magnitude. c. The errors are skewed to one side, either positive or negative. 7
  8. 8. d. These errors affect the accuracy of the analysis. Determinate errors are errors that are repeated, and are caused by a consistent reagent, instrumental, or operator (analyst) malfunction. These types of errors can be discovered and corrected. Correction may include the re-making of all reagents and solutions, re- calibrating all instrumentation, re-training of the analyst, or a combination of these steps. Indeterminate errors cannot be explained or accounted for. They are random, biased in various directions, and not repeated often enough for an analysis to be conducted. The causes may include almost anything, from intermittent operator error, random instrument malfunctions, causes outside of the laboratory (such as power fluctuations), or local environmental contamination that cannot be isolated. However, if sufficient analyses are performed, a statistical expression of the error can be estimated when the analyses are all performed under the same conditions. Assuming the determinate errors have been minimized as much as possible (so that the level is near zero), the effects of the indeterminate errors can be expressed as a Gaussian function mode- that of a bell-shaped curve. This so-called “normal” distribution is shown below. Gaussian Distribution 25 20 number 15 10 68.3 & 5 0 1 2 3 4 5 686 7 8 9 10 11 value The distribution curve can also be viewed as the distribution of the standard deviation of the population. See graph on next page. The area designated as “68.3 %” represents +/- 1.0 standard deviations from the mean. The area designated as “95.5%” represents +/- 2.0 standard deviations from the mean. The area designated as “99.7%” represents +/- 3.0 standard deviations from the mean. These values represent the probability that all values within these ranges do belong to the sample population; any values that fall outside these ranges suggests that the value does NOT belong to the sample population, and should be rejected. Of course, the more data points that are gathered, the better this estimation will be. 8
  9. 9. Gaussian Distribution 25 20 number 15 68.30% 10 5 95.50% 0 99.70% 1 2 3 4 5 6 7 8 9 10 11 value In a normal distribution, 68.3 % of the data fall between 0 and ±1 standard deviations from the mean, 95.5 % of the data fall between 0 and ±2 standard deviations from the mean, and 99.7 % of the data fall between 0 and ±3 standard deviations from the mean. Statistics involving sample populations There are several statistical measures that can be applied to populations, and they are used analyze the data. They are known as “descriptive statistics”. A sample is a member of a population- the entire group of possible data points. Samples are discreet sections or units or pieces of the population. Recall that it is nearly impossible to analyze an entire population; we must settle for a sufficient number of samples from that population. It follows that if we can analyze enough samples, our results should follow the normal distribution shown above. The first measure of a sample set is the measure of central tendency. There are two ways to describe the central tendency, the mean and the median. The mean is simply the sum of all the results (xi) divided by the number of results (n), or _ Mean = x= Σi xi n The median is the value that occurs in the middle of the ordered data set- there are the same number of values on both sides of the median. If there is an even number of data points, then the median is the average of the two middle points. It is often more useful to use the median rather than the mean because it may be a better representation of the actual central tendency; the median is not affected by very high or very low values that would influence (skew) the mean in on direction or the other. For example, consider these values for the mass of a nickel: 9
  10. 10. 4.55 g, 5.00 g , 5.10 g, 5.13 g, 5.20 g, 5.24 g, 5.25 g First, it is helpful to PLOT the data, as a simple frequency distribution, or histogram. X x xx x xx____ 4.00 4.25 4.50 4.75 5.00 5.25 Note the majority of the data are clustered between 5.00 and 5.25. Then we calculate the mean of this data set, which is 5.07 g. The median value, however, is 5.13 g. The mean (5.07 g) is influenced by the relatively low value of 4.55 g, whereas the median, 5.13 g, is a better representation for the majority of the data points. The larger the number of samples is, the more closely the mean and median should be to each other. Another useful statistic is the range of the data. The range is defined simply as the distance between the highest and lowest values of the data set. The range is calculated by computing the difference between the highest and lowest values, and expressing the results without a sign. For the previous data, the range is therefore (5.25 g – 4.55 g) = 0.70 g The range is an expression of the spread of the data, which may be used for other statistical functions that will be discussed later. Another more useful expression of the spread of the data is represented by the Gaussian curve- the normal distribution of the data set. The variation of the data is given by the standard deviation of the mean, and is calculated as follows: n s = [ ( Σi=1 (xi – x)2 ) ] 1/2 n-1 xi represents each individual data point and n represents the number of data points in the series. For the nickel weight data, the standard deviation = 0.244 g . “s” is often referred to by it’s Greek letter designation, sigma (σ ). As shown on the previous graphs, the standard deviation can be related to the confidence the user has for any particular data point. In general, if a data point falls with +/- 3 standard deviations from the mean (3 x σ), then that value ha a 99.7 % “chance” of being acceptable. Expressed more statistically, that value most likely belongs to the sample population, and the user can have a 99.7% level of confidence that is does so. Clearly, the farther a value is from the mean, the more standard deviations from the mean it will be (the higher the s value for that particular data point). (Just for information, the Quality Organization known as Six Sigma is based on the total number of standard 10
  11. 11. deviations included in the 99.7 % portion of the data: +3 sigma on one side of the mean and -3 sigma on the other side, for a total of six sigma.) The values associated with each sigma level form “confidence levels” for the user at various percents. If a user wants to have a very tight confidence range, he or she may use the 1 or 2 sigma limits. For most users, a 2 or 3 sigma confidence limit is acceptable, or even desirable. Another measure of precision of a data set is the coefficient of variation. The coefficient of variation (CV) is calculated as follows: CV = [ σ ] * 100% x-bar This statistic is useful in assessing the precision of a test procedure. The CV must be compared to the mean to determine if the CV is good or bad. A 10% CV associated with a small mean may suggest poor precision, whereas this CV associated with a large mean may be quite acceptable. For our nickel example, the CV = 0.244 * 100% or 4.18 % 5.07 4.18% of 5.07 is a fairly small value, but for the US Mint, the tolerances on nickels are much smaller than +/- 4.18 %. Obviously, the CV value must be interpreted within the context of the process or piece of equipment. Another interesting statistic is the mode, which is defined as the value that occurs most frequently. This also helps to determine the variability of a data set. Exercises: 1. Students determine the density of an aqueous solution of NaCl. They are to deliver 10 mL with an instrument of their choice into a preweighed beaker, determine it’s mass, and calculate it’s density. Class A uses a 10 mL graduated cylinder, while Class B uses a 10 ml volumetric pipette. The graph illustrates their results. The true density is 1.14 g.cm3 . 11
  12. 12. The true density is 1.14 g/mL. The class values suggest which of the following is true? a. Class A achieved greater accuracy and precision than Class B. b. Class A had greater accuracy; class B had greater precision c. Class B had greater accuracy; class A had greater precision d Both classes had substantial systematic errors. e. Class B had greater accuracy and class A had a greater standard deviation. Exercise 2 In the lab, you investigate the relationship between the molarity and density of NaCl. 5.0 M NaCl was made up in deionized water, diluted to volume, and 10 mL samples were weighed to determine density. The following graphs show the results for several students and also for the true values. True Values 1.175 1.15 1.125 Density 1.1 1.075 1.05 1.025 1 1 2 3 4 5 M NaCl Group A 1.175 1.15 1.125 1.1 1.075 1.05 1.025 1 1 2 3 4 5 M Na Cl Group B 1.175 1.15 1.125 1.1 1.075 1.05 1.025 1 1 2 3 4 5 M N a Cl 12
  13. 13. Group C 1.175 1.15 1.125 1.1 1.075 1.05 0 1.025 1 1 2 3 4 5 M N aC l Match the error ( i, ii, or iii) to the most appropriate graph (A, B, C). i. A systematic error occurred. Some NaCl solution was lost while transferring it to the volumetric flask for preparation of the 5 M NaCl stock solution ii. As far as you can tell, only random error occurred. iii. A systematic error occurred. The tip of the pipette had a crack and no volume was retained in the tip after delivery of the samples prior to weighing. Determination of Outliers Frequently, when a series of data are obtained for a given process, one or more results may appear to be markedly different from the main body of data. These data points may be the result of determinate error. When such an error seems to exist, it would seem wise to actually assess this possible errant data point to indeed show that it is errant (an outlier) and should be discarded. As an example, consider again our values for the nickel masses. The lowest value, 4.55 g seems to be significantly different from the other 6 results. How can we determine if this result is actually an outlier? One way to assess a data result is by using the Q test (rejection Quotient), which can be used when the number of results is fairly small (less than 15). The calculation of the Q value is straightforward: 1. Calculate the difference between the suspected result and the next closest result (5.00-4.55 = 0.45 in our example) 2. Calculate the range (0.70 in our example). 3. Divide (1) by (2) to determine the Q value. (0.45/0.70 = 0.64 in our example) 4. Compare the result (the “rejection quotient”) to the following table. If the calculated Q is greater than the value in the table for the number of data points, then the suspect value can be discarded. The table for Q values is on the next page. 13
  14. 14. Table 1. Rejection Quotient (Q) at a varying confidence levels Number of results Q90 value Q95 value 4 0.76 0.83 5 0.64 0.71 6 0.56 0.63 7 0.51 0.57 8 0.47 0.53 9 0.44 0.49 10 0.41 0.47 11 0.39 0.45 12 0.37 0.43 13 0.36 0.41 14 0.35 0.39 15 0.34 0.38 So, for our example, the calculated Q value is 0.64. Looking at the Q table for n=7 (90%), the rejection quotient value is 0.51. Because 0.64 > 0.51, the 4.55 value can be discarded. Even at the 95% confidence level, Q=0.57, and the value can be rejected. A second, more robust test is the Student’s t-test. It is used in conjunction with the standard deviation and the mean of a normal distribution (or a semi-normal distribution) to construct a confidence interval at a desired percent confidence value (such as 95% or 99%). The suspect result is then compared to the resulting confidence interval, and if it is outside the confidence interval, it is rejected. The process for calculating a Student’s t-confidence interval is as follows. 1. Calculate the mean (x) and the standard deviation (s) of the data set. 2. Determine the “degrees of freedom” of the data set, which is defined as one less than the number of data results, or n-1. 3. Calculate a value known as the “error of the mean”: error = s/√n 4. Look up the t-value in the Student’s –t table (on next page) at the desired level of confidence. 5. Calculate the confidence interval as follows: Confidence interval = x ± (t * error) The T-table is on the next page. 14
  15. 15. Table 2. Student’s T Probabilities Conf. Level 80% 90% 95% 98% 99% 99.7% df . . . . . . 1 1.000 3.078 6.314 12.706 31.821 63.657 2 0.816 1.886 2.920 4.303 6.965 9.925 3 0.765 1.638 2.353 3.182 4.541 5.841 4 0.741 1.533 2.132 2.776 3.747 4.604 5 0.727 1.476 2.015 2.571 3.365 4.032 6 0.718 1.440 1.943 2.447 3.143 3.707 7 0.711 1.415 1.895 2.365 2.998 3.499 8 0.706 1.397 1.860 2.306 2.896 3.355 9 0.703 1.383 1.833 2.262 2.821 3.250 10 0.700 1.372 1.812 2.228 2.764 3.169 11 0.697 1.363 1.796 2.201 2.718 3.106 12 0.695 1.356 1.782 2.179 2.681 3.055 13 0.694 1.350 1.771 2.160 2.650 3.012 14 0.692 1.345 1.761 2.145 2.624 2.977 15 0.691 1.341 1.753 2.131 2.602 2.947 16 0.690 1.337 1.746 2.120 2.583 2.921 17 0.689 1.333 1.740 2.110 2.567 2.898 18 0.688 1.330 1.734 2.101 2.552 2.878 19 0.688 1.328 1.729 2.093 2.539 2.861 20 0.687 1.325 1.725 2.086 2.528 2.845 21 0.686 1.323 1.721 2.080 2.518 2.831 22 0.686 1.321 1.717 2.074 2.508 2.819 23 0.685 1.319 1.714 2.069 2.500 2.807 24 0.685 1.318 1.711 2.064 2.492 2.797 25 0.684 1.316 1.708 2.060 2.485 2.787 26 0.684 1.315 1.706 2.056 2.479 2.779 27 0.684 1.314 1.703 2.052 2.473 2.771 28 0.683 1.313 1.701 2.048 2.467 2.763 29 0.683 1.311 1.699 2.045 2.462 2.756 30 0.683 1.310 1.697 2.042 2.457 2.750 40 0.681 1.303 1.684 2.021 2.423 2.704 50 0.679 1.299 1.676 2.009 2.403 2.678 60 0.679 1.296 1.671 2.000 2.390 2.660 70 0.678 1.294 1.667 1.994 2.381 2.648 80 0.678 1.292 1.664 1.990 2.374 2.639 90 0.677 1.291 1.662 1.987 2.368 2.632 100 0.677 1.290 1.660 1.984 2.364 2.626 Using our nickel data again, we will construct a 95% confidence interval to determine if the 4.55 value is really an outlier. 1. the mean (x) and the standard deviation (s) of the data set: 5.07 g and 0.244. 2. Determine the “degrees of freedom” of the data set, or n-1: 7-1 = 6 3. Calculate a value known as the “error of the mean”: error = s/√n = (0.244/√7 ) = 0.0922 15
  16. 16. 4. the t-value in the Student’s –t table at 95% level of confidence: 1.943 5. Calculate the confidence interval as follows: Confidence interval = x ± (t * error) = 5.07 ± (1.943 * 0.0922) or 5.07 ± 0.179 g, which forms a confidence interval of 4.89 – 5.25 g Comparing our low value of 4.55 g to the confidence interval, we see that 4.55 < 4.89, so we can reject the 4.55 g value. This statistic confirms the Q-test result, which should confirm our “gut” feeling that the 4.55 g value does not belong in our data set. The question might arise, When do we use the Q test and when do we use the t-test? The t-test is better for a larger number of data points, and the Q-test is better for a smaller number of data points. There is no firm dividing line, but for our purposes, if n < 10, use the Q-test; if n ≥ 10, use the t-test. Both statistics can be calculated for additional information and confirmation that an apparent outlier is truly an outlier, and should be rejected. SUMMARY: For our nickel mass data, let us summarize what we have learned so far. Mean = 5.07 g Median = 5.13 g Mode= none Range = 0.70 g s = 0.244 CV = 4.18 % Suspect result = 4.55 g Q-test result: Q4.55 = 0.64 (at a 95% confidence level), which is > 0.51, so 4.55 is an outlier; t-test result: for df = 6, confidence interval = 4.89 g – 5.25 g. Therefore, 4.55 is an outlier. Interpretation of results: We have decided that 4.55 grams is an outlier. Now we have to ask ourselves, what does this mean? In collecting our data, did we weight the same nickel on the same balance 7 times? Or did we weight 7 nickels on the same balance? Or did we weigh the same nickel on 7 different balances? Or did 7 different students weigh their own nickel on their own balance? You can see that it makes a lot of difference which of the above questions (possibilities) is actually, in fact, true. What can you conclude about the possible sources of error for EACH of the 4 possibilities? Which ones make the most sense? 16
  17. 17. IN-CLASS EXERCISE: Statistical Data Set A student obtained the following values for the analysis of Vitamin C in a serving of orange juice. All results are in mg/ 250 mL: 67.2 66.5 63.5 70.0 65.1 64.9 69.2 57.2 65.0 66.8 TRUE VLAUE: 65.2 mg/250 mL How do we analyze the data? Plot the data 54 56 58 60 62 64 66 68 70 72 mean median 17
  18. 18. mode range standard deviation and CV “Q” test for outliers Student’s t-test confidence interval Which data points, if any are truly outliers and should be rejected? Percent recovery or percent error 18
  19. 19. HOMEWORK DATA EXERCISE: The following is a sample data set for cholesterol levels in a certain male patient age 53 years. A clinical lab technician analyzed a sample of blood for total cholesterol. She performed 8 replicates of the analysis, each with a different aliquot of the patient’s blood. The results are as follows, all in units of mg total cholesterol/ 100 mg plasma. Replicate value 1. 185 2 201 3 204 4 175 5 192 6 221 7 182 8 175 Calculate the following: mean median mode standard deviation CV range Are there any outliers? Use the “Q” test to determine if there are. Then check your answer with the t-test. Plot the data in the most useful format using Excel. Is this patient’s cholesterol level within the normal range for a person his age? How did you determine your answer? 19
  20. 20. CHAPTER 3 UNITS OF CONCENTRATION One of the most important yet most easily forgotten aspects of performing analytical work is the use of the correct units. Analytical samples can be gases, liquids (solutions), or solids. The substance that is being sought for via analytical chemistry is called the analyte. For most analytical procedures, the analyte is dissolved in a large quantity of another substance, which is often referred to as the matrix. The matrix may be gas, a liquid, or a solid. The analyte which is dissolved in the matrix is also known as the solute. The combination of solute and matrix (or solvent) is the solution. In this course, we will be working mainly with liquid matrices- especially blood plasma or serum and aqueous (water) solutions. There are several units of concentration that are important to the analytical professional. They include the following: Percent Composition (mass/mass, mass/volume, volume/volume) Molarity Normality Grams/ liter Parts per million (ppm or mg/L) Parts per billion (ppb or ug/L) Percent Composition a. mass/mass or weight/weight (w/w) Many times, the user of analytical data will be interested in knowing the analyte concentration in terms of parts per hundred parts of the solution, which is the same as percent. In aqueous solutions, percent composition is based on the fact that 1.00 mL of water exhibits a mass of almost exactly 1.00 grams at temperatures ranging from 4 oC to 30oC. Thus, if we know the mass of an analyte in a given mass of a water-based solution, we can express the units in percent by mass/mass (mass solute or analyte per mass solvent). For example, the preparation instructions for sodium chloride might require a 10% solution by mass/mass. This means that for every 100 grams of solution, 10 grams of salt are dissolved in 90 grams of solvent (water). You would then dissolve 10 grams of solid, pure NaCl in 90 grams (100 mL) of water. If a liter of solution were required, then 100 grams of NaCl would be dissolved, and brought to a final volume of 1.0 liter with DI water. Some scientists refer to mass/mass solutions as weight/weight solutions (abbreviated w/w). b. mass/volume (or weight/volume, w/v) The most common method of solution preparation in the clinical field is by mass/volume. This method of preparation requires a certain mass of solute to be added to the solvent until a certain final volume is reached. For example, if we were to prepare a 10% by mass/volume solution of NaCl, we would weigh out 10.0 grams of solid NaCl and add 20
  21. 21. enough water to bring the final volume to 100.0 mL using a volumetric flask. This would then be a 10% (w/v) saline solution. c. volume/volume (v/v) Percent by v/v is usually used when one liquid is being dissolved in another liquid, such as alcohol in water. A 5% solution of ethyl alcohol in water would require that the clinician measure out exactly 5.0 mL of ethyl alcohol and then placed in enough water to make exactly 100 mL of solution. Again, the use of a volumetric flask is very helpful in preparing these types of solutions; one has to only measure out the correct volume of the solute and place it into the volumetric flask, then carefully add water until the meniscus is resting on the 100 mL line on the neck of the flask. meniscus Molarity A common unit of concentration used in many analytical laboratories is molarity (M). Molarity is equivalent to moles (grams solute/molecular or atomic weight of solute) per one liter of solution. Molarity is almost exclusively used with aqueous solutions. A 1.0 M solution of dried sodium chloride would be prepared by dissolving 58.45 grams of NaCl in some deionized water and then adjusting the final volume to exactly 1.0 L (in a volumetric flask). Normality Normality (N) is based on the equivalent weight of a species. Equivalents are used to compare combining ratios of elements or compounds, especially acids and bases. Electrolyte compounds (salts of sodium, potassium, chloride, and bicarbonate) are often described by the number of equivalents. An equivalent is most easily defined as the molecular mass of a substance (usually and acid or a base) divided by the number of reactable hydrogen or hydroxide units provided by that substance upon reacting with another species. One gram-equivalent weight is therefore defined as the mass of an element or compound that is chemically equivalent in reacting power with one gram of a H+ or (OH-) ion. For salts or other non-acid/base compounds, the gram equivalent weight is the molecular weight divided by the charge of the ion in question. Here are some examples: 21
  22. 22. Substance Formula Mass Equivalent Mass HCl 36.45 g 36.45 g (one H+) NaOH 40.00 g 40.00 g ( one OH-) H2SO4 98.08 g 49.04 g (2 H+ ions) H3PO4 98.00 g 32.67 g (3 H+ ions) NaCl 58.45 g 58.45 g (one Na+1 ion) Na2SO4 142.00 g 71.00 g (two Na+1 ions) One gram equivalent weight can represent a fairly large quantity of the compound in question. In clinical situations, we may prefer to use milliequivalents (mEq), which are 1/1000 of an equivalent. We then would use mg as our mass unit rather than grams. As an example, the normal concentration of sodium ions in human serum is 140 mEq/L, or 0.140 Eq/L. The Normality of a solution is defined as one equivalent mass per liter of solution. For solutions that consist of species with only one reactable H+ , OH- or one ion with a +1 charge, the Normality equals the Molarity. In cases where there are multiple H or OH ions, or where there are multiple metal ions, or metal ions with a charge greater then +1, the Normality is most easily calculated by multiplying the Molarity by the number of ions or the charge. For example, a 2.00 M solution of H2SO4 is also 4.00 N because of the presence of 2 H+1 ions. A 0.25 M solution of CaCl2 is also 0.500 N because of the presence of one Ca+2 ion. (Can you prove this?) Grams per Liter (g/L) Sometimes it is important to express concentrations as a unit of mass per volume, especially when there is a substantial quantity of solute dissolved in the solvent. Environmental chemists often prefer the unit of grams per liter (g/L), or milligrams per liter (mg/L), or even micrograms per liter (μg/L). In each case, the analyst can have an immediate idea as to the physical amount of solute in the solvent. In clinical applications, the solvent is usually human blood or other bodily fluid, such as urine. In these cases, it is more customary to express the concentration in units of mg/DL, or milligrams per deciliter (0.1 L). It is then easy to extrapolate to the amount of a species in the entire body, since the “average” human body contains anywhere from 5.2 – 8.3 DL of blood/ kg of body weight. In a 70 kg person (male or female), there would be about 470 DL of blood (4.7 L or around 9 pints). Many routine blood test results are reported in the units of mg/DL, such as blood sugar, creatine, cholesterol, calcium and so on. Thought Question: How many mg/DL of Chloride ion is represented by a 0.130 M solution of NaCl ? (Answer given later!) 22
  23. 23. Parts per million (ppm), parts per billion (ppb) These units are derived from the grams/L and milligram/L units just described. Consider 1.0 mg/L. A milligram is 1/000 of one gram, and one liter of water weighs just about 1000 g. Therefore, a milligram of solute in a liter of aqueous solvent represents (1/(1000)/ 1000) or 1/106 parts of solute per part of solvent, or 1 part per million (ppm). A microgram/Liter represents 1/109 parts of solute per part of solvent, or one part per billion (ppb). (Can you prove this?) These units are used mostly for trace analyses in analytical chemistry and are often used by the Environmental Protection Agency when they promulgate safe drinking water regulations that specify the maximum allowable concentrations of potential pollutants. Other units: Some other somewhat less common units exist as well. They include: Molality (mole solute/kg solvent) Mole fraction (moles solute/total moles of solute + solvent)* 100 SUMMARY: Concentration can be expressed by these units: Unit Symbol Definition Relationship Molarity M Moles solute/liter of solution M = moles/ liter Normality N Equivalents solute/ liter N = equiv./Liter solution % by weight/weight % w/w Ratio of the weight of the % w/w = gsolute (* 100%) solute to the total wt. of solute gsol + gsolv + solvent % by weight/volume % w/v Ratio of wt. of solute to total % w/v = gsolute (* 100%) volume of solution Vol sol’n % by volume/volume % v/v Ratio of volume of solute to % v/v = Vsolute (* 100%) total volume of solution Vsol + Vsolv Molality M Moles solute/kg solvent M = moles solute/kg solvent Mole fraction X Ration of moles of solute to X= nsolute total moles in the solution nsol + nsolv Parts per million ppm Milligrams solute/ liter of ppm = mg/L or mg/kg or ug/g solution (or mg solute/kg sample) Parts per billion ppb Micrograms solute/liter of ppb = ug/l, ug/kg or ng/g or ng/ solution or mg solute/kg ML sample Dilutions: In analytical chemistry, sample concentrations are often much too high for an instrument to measure, due to an instruments analytical range, also called the dynamic range or linear range- the highest value that can be measured. To compensate for this limitation, samples often have to be diluted many times to reduce the analyte concentration to the 23
  24. 24. point where it is within the analytical range of the instrument. This reduction in analyte level is known as a dilution. Sometimes, dilution may occur as a sample is prepared, but additional dilutions may still be necessary. Dilutions must be accounted for when calculating the final result in a sample. If a sample was diluted 10 times, and the final results for glucose is 9.8 mg/dL, then this value must be multiplied by a factor of 10 to obtain a sample result of 98 mg/dL. Dilutions are usually accomplished through the use of volumetric glassware- volumetric pipettes and volumetric flasks. These items must be clean to insure a correct dilution has been made. Dilution instructions sometimes may state the parts of solute to be added to parts of solvent. For example, a ten-fold dilution may be stated as a 1:9 dilution- one part solute plus 9 parts solvent, (which is a 10 times dilution). If you have done titrations you are familiar with the formula for calculating the concentration of an unknown species. The analogous formula can be used to calculate the concentration of a species when a dilution has occurred, or to calculate how much of a dilution must be made to achieve a certain concentration when we are interested in diluting concentrated solutions (such as acids or bases). The general formula is V a * Ca = V b * Cb where a is the concentrated solution and b is the diluted solution. V and C are the volume and concentration of the solutions. For example, we have a solution of 2.0 M sulfuric acid and we wish to make 500 ML of 0.50 M H2SO4. Using the formula, we can substitute in the known values for the variables and solve for Va: Va * 2.0M = 500mL * 0.50 M Va = 125 mL. Therefore, we would measure out 125 mL of the 2.0 M solution of H2SO4 and dilute it to 500 mL with DI water. Now, try the following problems for homework: 1. How would you prepare one liter of a 4.000 M solution of ferric sulfate, pentahydrate? 2. How would you prepare a 15 % m/v solution of potassium chloride? 3. How would you prepare 500 mL of a 0.35 M solution of sulfuric acid, given that concentrated sulfuric acid is 18.0 M? (NOTE: the specific gravity of H2 SO4 is 1.835) 4. If 30 mL of absolute ethyl alcohol are dissolved in 50 mL distilled water, what is the % concentration of the alcohol, by volume? 24
  25. 25. 5. 100 grams of Calcium Hydroxide are dissolved in 2.00 L of deionized water. What is the Normality (N) of the solution that is formed? 6. A 100 mL sample of liquid waste contains 0.0045 grams of uranium. How many ppm of uranium are in the solution? 7. 1.0 gram of hazardous waste contains 15 nanograms (ng) of lead. How many ppt does this represent? How many ppb? How many ppm ? 8. You have 450 mL of 2.5 M nitric acid. How would you prepare 2.0 L of 0.10 M HNO3 ? 9. You analyzed a solution for sodium. You had to dilute the original solution by 5 to digest it, then you took 1.0 mL of that solution and diluted it to 1.0 L. The instrument result was 1.25 ppm. What is the sodium level in the original solution? 25
  26. 26. CHAPTER 4 QUANTITATIVE ANALYSIS – TITRIMETRY QUANTITATIVE ANALYSIS Quantitative analysis is concerned not only with the identification of a substance, but the concentration of that substance. Titrimetry is a useful and fairly simple method of quantitative analysis. A reactive solution (the titrant) is added to a “sample” from a buret. A suitable indicator (which provides a visual clue that the reaction between the titrant and the sample is completed), is added to the sample before the titration is started. As the titration continues, the reactive species in the titrant reacts with the species of interest in the sample. The indicator turns a certain color just after the last molecules of the reactive species in the titrant react with the species of interest in the sample. (The indicator is not the preferred reactive species in the sample- it only reacts AFTER the preferred species has been consumed.) There are four types of reactions that are amenable to Titrimetry. 1. Neutralization Reaction. Example: Acid + Base Salt + water HCl + NaOH NaCl + HOH with phenolphthalein as an indicator 2. Oxidation/ Reduction Titration Example: the Analysis of Vitamin C (Ascorbic Acid) Vitamin C – 2D structure Vitamin C -2D structure – m.p. 190-192 Oc., C6H8O6 Courtesy of UM at Frostburg 26
  27. 27. Overall reaction: I3- + C6H8O6 + H2O ——> C6H8O7 + 3 I- + 2 H+ using starch as the indicator After all the ascorbic acid has been consumed (oxidized) by the tri-iodide ion (I3 -1) the excess iodide ions turn the starch indicator from a cloudy milky white color to a dark purple color. 3. Complex Ion Formation Example: EDTA + Ca +2 EDTA/Ca+2 complex, using Eriochrome Black-T as an indicator The Lewis structure of ethylenediamminetetraacetate ion (EDTA4-), is shown below. EDTA4- forms very stable complexes with most of the transition metals as well as alkali earth metals, such as Calcium. EDTA4- Courtesy of NSTA web site link One ion of Ca+2 reacts with one molecule of the EDTA, so there is a 1:1 mole ratio between the Ca+2 ion and the EDTA. As soon as all the Ca+2 is the test solution has been consumed, the Eriochrome Indicator turns from wine red to a blue color. Precipitation Reaction Example: Analysis for Chloride Ion by titration with Silver Nitrate Cl-1 + AgNO3 AgCl + NO3 -1 using potassium chromate as the indicator 27
  28. 28. As AgNO3 reacts with the chloride ion in the flask (sample), solid AgCl is formed. When all of the Cl-1 ion has been consumed by reacting with the Ag +1 ion, the first excess ions of Ag +1 then react with the KcrO4 indicator, turning the solution in the flask a very dark orange-red. Reactions using silver nitrate are known as argentometric reactions. Titrimetry requires that the amounts of the reacting solution in the buret react with the analyte in the flask (sample) in a stoichiometric relationship. In a valid titration reaction, the moles or equivalents of the titrant must be equal to the moles or equivalents of the analyte. The use of reference solutions (of known composition and concentration) can be used to simplify the calculations by establishing the ratio between the mass of the analyte and moles or equivalents of titrant, often expressed by the volume of titrant required to react with a certain mass of analyte. Once the ratio has been established, a value known as the titer (spelled “titre” in Europe) can be determined, which is usually in the form of mg analyte/1.00 mL titrant. The titer can be used as a sort of conversion factor to determine the analyte concentration in any sample. When the standard solution and the unknown solution containing the analyte are titrated with the same titrant under the same laboratory conditions, the calculations become a simple proportionality: Vtitrant std = Vtitrant analyte Cstandard Canalyte Where V = volume in mL, and C = concentration in the appropriate units. The above equation can be rearranged as follows: Canalyte = Vtitrant analyte * Cstandard Vtitrant standard Example: Problem: A sample of human serum is analyzed for chloride content by argentometric titration. The sample is deproteinized by adding 1.0 mL of the serum sample to a 10 mL volumetric flask. Trichloracetic acid is added to the flask, up to the 10.0 mL line and shaken well. The solution is filtered and then exactly 5.0 mL of the filtrate is titrated with the AgNO3, and it requires 4.98 mL of the titrant to react with all the chloride in the sample and reach the endpoint. 5.00 mL of a standard chloride solution containing 10.0 mEq/L of requires exactly 5.26 mL of titrant to reach the endpoint. What is the chloride concentration in the original human serum sample? 1. Determine the titer of the known solution: 10 mEq Cl -1 requires 5.26 mL titrant. 10 mEq Cl/ 5.26 mL = 1.90 mEq Cl/1.00 mL titrant. This is the titer. 2. Multiply the titer by the volume required for the sample: 1.90 mEq Cl * 4.98 mL titrant = 9.47 mEq Cl in the filtrate 1.0 ML titrant 28
  29. 29. Since the standard was in units of mEq/L, our result is also in the same units- 9.47 mEq Cl -1/L. 3. Looking at how the sample was prepared, 1 mL was diluted to 10 mL, so there is a 10 fold dilution factor to be considered: 9.47 mEq Cl-1 /L* 10 = 94.7 mEq Cl-1 /L in the serum sample. (Since 5 .00 mL of chloride ion standard were used to develop the titer, the 5.00 mL of filtered serum does not enter into the calculations- they effectively cancel out.) Going on- What is the chloride concentration in units of mg/dL? Well, we know that for Cl-1, the number of equivalents = moles, since Cl is a monovalent ion. Therefore, 94.7 mEq/L = 94.7 mmoles/L, or 0.0947 moles of Chloride ion/L. 0.0947 moles Cl-1 /L * 35.5 g Cl-1/ 1.00 mole Cl-1 = 3.357 grams Cl-1 /L are present. 3.357 grams/L = 3357 mg/L, which is equal to 335.7 mg Cl -1/dL. And THAT’S the final answer! 29
  30. 30. SPECTROSCOPY Many centuries ago, scientists discovered that if sunlight was passed through a prism, the white light was dispersed into seven colors. The same phenomenon was evident as sunlight passed through rain drops, forming the rainbow. The reason this happens is due to the fact that ordinary light is actually a combination of many wavelengths. This led scientists to conclude that light is a wave. In the 1800’s Sir Maxwell said that light is one example of electromagnetic radiation- where light consists of packets of energy each oscillating at specific wavelengths, comparable to an alternating electric field. These packets of light, called photons by Einstein, propagate through space. Since these photons do not seem to “run out” of energy, and they move at a constant rate (3.0 * 108 meters/second, denoted by c) early scientists considered light a pure form of energy, and only energy. Since the photon oscillates like a sine wave, it has amplitude (height) and frequency (number of oscillations per unit time, usually seconds). These parameters describe the wavelength- denoted by the Greek letter lambda λ. c = υ* λ or υ=c/λ By the early 1900’s, problems were encountered with the wave theory of light. It could not explain black body radiation (the emission of heat and /or light from substances that absorb light energy, or other forms of energy) or the photoelectric effect (the generation of electricity [the flow of electrons] when a substance is subjected to light), discovered by Millikan. Max Planck was able to explain and define these phenomena by suggesting that light also had a particle nature, and experimentally determined that the energy of a photon is directly proportional to the frequency (υ) of the photon, or E = hυ The constant h is Planck’s constant, or 6.626 * 10-34 J-s. We can use this relationship to determine the energy associated with the light emitted from specific materials, such as the chemical elements. Visible light covers the range of 380 nanometers (nm) at the violet end to around 700 nm at the red end. UV radiation is lower than 380 nm and IR radiation is greater than 700 nm. Spectroscopy in analytical chemistry refers to the measurement of light given off during specific chemical reactions. The light can be UV, Visible or IR. We have instruments capable of measuring light at a wide range of wavelengths. What produces light in chemical species? Simply put, from general chemistry, all atoms posses a given quantity of electrons dispersed across a number of energy levels, each of which contains a discreet amount of energy, or quantum. When an atom absorbs energy, some electrons are excited- that is, they move from one energy level to a higher one- the ground state to the excited state. However, an electron in an excited state ( in a higher energy level) is an unstable state, and the electrons return back to their initial, ground 30
  31. 31. state. When they do this, they release energy, much of which is in the form of photons- light- at specific wavelengths. The pattern of wavelengths that are emitted by various elements are known as the atomic spectrum of the element. These energy levels and energy releases are not random. They are quite predictable. Electrons must absorb a given quantity of energy to move from one state to the next, and when they “relax” they give off a precise, predictable amount of energy. When an electron moves from an excited state back to the ground state, the energy that was absorbed must be dissipated. In the usual interaction of light in the visible spectrum (380-700 nm), this energy is usually dissipated as heat, which is such a small amount it cannot be easily measured. However, some of energy is dissipated as light, which provides the spectra that we can see for most elements. The color that appears is the color that is transmitted from the atoms of the element; the other colors are absorbed by the atom. Spectroscopy is concerned about the transmission or absorption of color. The transmission or absorption of specific wavelengths is measured by a spectrophotometer. Absorption spectroscopy is based on the fact that chemical species absorb light at various wavelengths that are specific for that species. The spectrophotometer is capable of generated light that covers a wide range of wavelengths. It is also capable of measuring absorbance at specific wavelengths. Essential components of a spectrophotometer include the following: Source Slits Lens Monochromators Cells, or other means of sample introduction Detectors Read-out devices Most spectrophotometric techniques produce data that are linear with respect to absorbance versus concentration, whether that concentration is measured in ppm, ppb, molarity etc. In the mid 19th century, scientists developed a relationship that allows us to calculate the concentration of a substance based on its absorbance reading. That relationship is known as Beer’s Law. The law can be stated as follows: A = a*b*c Where A= Absorbance, a = proportionality constant for the species, b = distance light travels through the sample (called the path), c = concentration. 31
  32. 32. When a graph is constructed of the absorbance vs. the concentration of a species, a straight line is usually generated. The slope of the plot gives the value for b. The following is an example plot for a chemical species. Example Calibration Curve (Linear) 0.45 y = 0.0038x + 0.0076 2 R = 0.9998 0.4 0.35 0.3 0.25 ABS trend 0.2 line 0.15 0.1 0.05 0 0 20 40 60 80 100 120 PPM This particular graph is called a calibration curve because it plots known concentration values (on the x-axis) versus the observed absorbance. Note the equation of this plot is that of a straight line, in the y = mx + b format. In analytical chemistry, the measurements of the calibration (known or reference) standards are done under the same conditions as the actual clinical or laboratory samples. Therefore, the terms of “a” and “b” are the same, and cancel out of the equation used for calculating the concentration of the unknown sample: Aref = aref * bref * cref Aunk aunk * bunk * cunk The only values that will change, depending on concentrations and subsequent absorbance, are Aref , Aunk , cref and cunk. As noted above, the “b” and “a” terms all cancel, because they are not dependent on individual reference or unknown sample measurements. Therefore, the above equation simplifies to: Aref = cref Aunk cunk 32
  33. 33. Solving for cunk , we get: cunk = Aunk * cref Aref Beer’s Law can only be used when the optimum wavelength is chosen for the species being measured. At the optimum wavelength, only the species of interest absorbs light; other species will not absorb light at that wavelength. Once can perform a spectral absorbance plot to determine what the optimum wavelength is for any particular species. For example, Fe +2 absorbs best at 510 nm. Modern, automated spectrophotometric instruments, such as Atomic Absorption and Atomic Emission Spectroscopy can automatically scan a species to determine the optimum wavelength, and can set the instrument up to perform the analysis with very little operator. Molecular and Atomic Spectroscopy In general, spectroscopy is concerned with the absorption or emission of electromagnetic radiation by a sample. The molecules or atoms in a particular sample can be excited, bent, stretched, fluoresced, or be fragmented by the energy source. These various effects of electromagnetic radiation have all been utilized to develop techniques for qualitative and quantitative analysis of a wide variety of inorganic and organic species, including biochemical species of clinical (medical) significance. Molecular Spectroscopy There are five commonly utilized methods of analysis that are based on the effects of certain types of electromagnetic radiation on molecular species. Note that each of the methods works by causing some physical effect on the molecule which allows us to detect and quantify it in a sample. In other words, there are no chemical reactions taking place with molecular spectroscopy (except that of dissolution or complexation); the actual molecule stays relatively intact throughout the analysis. Electromagnetic radiation can be broken down into eight distinct regions: Gamma---- x-rays---- UV--- visible--- IR--- microwave---- radio---- long wave High frequency low frequency High energy low energy Short wavelength long wavelength The working UV-Visible wavelengths range from 150-750 nm. Working IR wavelengths range from 2.5 microns to 16 microns (2500 – 16000 nm). A brief summary of each type of analysis follows. A. Infra-red spectroscopy 33
  34. 34. The absorption of IR light causes vibrational energy transitions in molecules. Specific molecules absorb only at specific wavelengths, which makes IR an excellent tool for identifying components of a sample. When the absorbance is plotted against wavelength, a “molecular fingerprint” is obtained. On the next page, an IR spectrum in presented for the molecule benzyl alcohol. IR Spectrum IR spectrum courtesy of http://chipo.chem.uic.edu/web1/ocol/spec/IRex1.htm Note the “stretching” associated with the benzene ring (about 3100 cm-1 and near 1500 cm-1 ), the –OH group (3300 cm-1 ), and aliphatic C-H bonds (about 2900 cm-1 ). B. Mass Spectrometry Mass Spectroscopy is a technique that is based upon the fragmentation of molecules into their component atoms or molecular fragments after being subjected to a beam of electrons. Generally, Mass spectrometry (MS) follows Gas Chromatography, (GC) which is used to separate a mixture of molecules or components. The effluent from the GC is fed into the MS, where the individual compounds are fragmented. In reality, then, the MS serves as a very sensitive and compound-specific detector for GC. The software for MS contains and extensive library of thousands of organic compounds, which is used to match the sample mass spectrum. Each molecule has it’s own, specific fragmentation pattern (a fingerprint) which is reproducible on any MS system. This fingerprint can be used to identify and quantify the compounds present in liquid, solid, and gaseous samples. Here is a photograph of a mass spectrometer unit. 34
  35. 35. Picture of HP6890 Gas Chromatograph with HP5973 Mass Selective Detector, HP7694 Headspace Autosampler and thermal conductivity detector. Gas Chromatography/Mass Spectrometry - GC/MS flowcha–t 35
  36. 36. See a sample mass spectrum on the next page. Courtesy of http://depts.washington.edu/spectral/massspec/GCMSintro/GCMS_4.htmL Notice the “logical losses” from the molecular ion (at 194 amu) to 165, 136, 109, etc. C. Nuclear Magnetic Resonance 36
  37. 37. Nuclear Magnetic Resonance (NMR) is an analytical technique based on nuclear spin energy transitions that occur when molecules are subjected to light at radio wavelengths in a magnetic field. The Nuclei of atoms that are bonded together spin on an axis. Since the nuclei are positively charged, a small magnetic field surrounds the nuclei. When a spinning nucleus is subjected to an external magnetic field, the nucleus and it’s small magnetic field will align with the more powerful field, such as is created with an NMR instrument. The alignment of the nuclear magnetic field will either be in the same direction as the NMR field, or in the opposite direction. The difference in energy states between aligned and opposite nuclear magnetic fields creates a resonance, or the measurable difference that occurs in the radio wave region of the electromagnetic spectrum. The light generated in this region can then be absorbed by other molecules in the magnetic field of the NMR and cause these transitions (directly aligned to oppositely aligned) to occur. It ha been discovered that hydrogen atoms are best measured by this technique, and since nearly all organic molecules contain hydrogen, this technique is useful for many compounds. NMR is more useful for structure determination than for any quantitative measurements. NMR spectra are based on “chemical shifts”, which occur because of the effect of electrons that surround nuclei. The electrons shield the nuclei to measurable extents from the magnetic field, thereby shifting the absorption of the radio wave light. Different hydrogens in a molecule, therefore, are shielded to different degrees, and the magnitude of the shift can be used to identify what type of carbon a hydrogen is bonded to. Below is an NMR spectrum of 2-butanon-4-ene. There are chemical shifts for the hydrogens bonded to a carbon with a carbonyl group (C=O), to a carbon double-bonded to another carbon (C=C), and hydrogens bonded to carbons that are single-bonded to other carbons, as well as terminal carbons (-CH3 ) and internal CH2 groups. 13 C NMR: courtesy of http://chipo.chem.uic.edu/web1/ocol/spec/C13ex3.htm D. UV-Visible Spectroscopy. This technique has been examined previously, so only a very brief review will be presented here. This technique measures the light absorbed by a compound or complex. This light can be in the UV or Visible portions of the electromagnetic spectrum. The degree to which a sample absorbs the light striking the sample is 37
  38. 38. measured at the optimum wavelength at which the species under consideration absorbs light. The amount of absorbance is directly related to the quantity (concentration) of the absorbing species in the sample, per Beer’s law. If the optimum wavelength is not known, a scanning UV-Vis instrument can be used to determine the best wavelength for analytical work. Below is a graph of absorbance versus wavelength for a solution. Courtesy of http://www.santafe.cc.fl.us/chemscape/catofp/measurea/concentr/spec20/spec2wq2.htm For this species, the optimum wavelength (called “lamda-max”) is about 575 nm ( a nanometer is 10-9 meters), which is in the visible range (yellow light is absorbed, so the solution will appear violet to the eye). E. Fluorometry Again, we have discussed this technique, but a review is in order. Some substances emit light, or fluoresce, when subjected to energy. This irradiating energy is usually UV light, and emitted energy is visible light. Both atoms and molecules can absorb UV light and emit visible light. Emission occurs when a species has absorbed energy and is excited- raised to a higher energy level. Excited molecules or atoms seek to return to their ground state as quickly as possible, and do so by various means- molecular collisions and direct emission of energy, in the form of light. Since there are competing mechanisms for a molecule to return to the ground state, the light emitted is at a lower energy than the energy that was absorbed. This results in the emission of energy at a longer wavelength. The UV-Visible spectrometer measures the intensity of the emitted light as percent transmittance, which is mathematically translated to absorbance units, since absorbance is a linear function (% T is a logarithmic function). This equation is used to convert % T to absorbance (A) is 38
  39. 39. A = log (100/ %T) or A= 2 - log %T As in –v-Visible spectroscopy, a substance can be scanned to determine the optimum wavelength at which the species fluoresces. The relationship is linear between absorbance and concentration, according to Beer’s law. As can be seen, molecular spectroscopy is a valuable tool for analytical chemistry. A vide variety of organic and ionic analytes can be identified and subsequently, concentrations can be determined. Limits of detection can be as low as micrograms per liter (ppb) for many substances. In clinical chemistry, most analyses do not require such low detection limits- often, milligrams per liter (mg/L or ppm) are sufficient, or even grams per liter are low enough for some constituents. Atomic Absorption Spectroscopy In atomic absorption (AA) spectrophotometry, a liquid sample is aspirated into an air- acetylene flame whose temperature is approximately 3140-4940°F. The sample is atomized in the flame. A selected light source emits characteristic frequencies of the atoms of interest. For example, if testing a sample for calcium (Ca), a Ca hollow cathode lamp would be necessary. If the sample contained Ca then the atomized Ca atoms would absorb a portion of the emitted frequencies from the lamp. Absorbance of Ca frequencies by the sample is indicative of the concentration of Ca within the sample. The electronic energy level spacings for atoms are very specific for the element. As a result, the absorption of quantized energy from a monochromatic light beam of the appropriate wavelength can give selective information about the identity and amount of elements (normally metals) in a sample. In AAS, solutions containing metal ions are aspirated into a flame in which they are converted to a free atom vapor. A monochromatic light source is directed through the flame, and the amount of radiation of a specific energy is detected. In this way, the amount of metal present in the original sample can be determined. Recall that when metallic ions, which are dissolved in an aqueous solution, are subjected to heat energy in an air-acetylene flame, the atomized ions are raised to a higher energy state by the absorption of light at specific wavelengths (specific to each element), usually in the visible range. The AA detects and measures the amount of energy absorbed by the ions in the flame. The amount of absorption is directly related to the quantity (concentration) of the ions of that element that are in the solution, again by Beer’s law. These absorption transitions are very rapid- on the order of 10-15 seconds. And, as soon as the atom absorbs energy, then emits energy and returns to the ground state, the atom can once again absorb light energy. This process is exactly opposite to fluorescence, which is based on the emission of light, as previously discussed. The fluorescence process takes considerably longer- on the order of 10-7 seconds. Graphite furnace AA (GFAA) operates by the same principle, except that instead of using a flame, an graphite tube with a “shelf” (L’Vov platform) is used. A single drop of the 39
  40. 40. sample solution is placed on the platform, which is bathed in an inert gas to prevent oxidation. The sample drop is dried at around 100 degrees C, then is ashed at around 800 degrees C (to drive off any organic molecules). Finally, the ashed sample is heated to around 2500 degrees C to atomize the ions, which then absorb light generated by the hollow cathode lamp. After the absorbance has been measured, the tube is heated to 2700 degrees C, which volatilized any remaining sample material. After the tube cools, another sample is then analyzed. Interferences can occur, and are minimized by adding small quantities of matrix modifiers to bind species which would either absorb at the same wavelength as the analyte of interest, or would prevent the analyte of interest from absorbing energy at all. Below is a picture of a state-of-the art GFAA instrument. courtesy of http://www.enveng.ufl.edu/homepp/townsend/Research/Leach/Leaching_TAG_Meeting_00_07_13/sld118.htm Below is a close-up of the graphite furnace section of the GFAA. 40
  41. 41. A magnetic field is induced, which is used to generate the heat needed to operate the GFAA. 41
  42. 42. Spectroscopy take- home assignment Name__________________________________ 1. A clinical lab technician analyzed the blood taken from a three year old male for the lead level in his blood. The boy lives in an old inner-city tenement building that was last painted in 1974. The paint is peeling badly, and it is suspected that the boy is regularly ingesting a significant amount of lead. The following represent the data from the analysis of blood for lead by flame atomic absorption spectroscopy (FAAS). Calibration curve: ppm lead absorbance, in absorbance units 0.00 (blank) 0.005 0.050 0.028 0.100 0.055 0.250 0.128 0.500 0.219 1.00 0.408 a. Using Excel or your statistical calculator, draw the curve. Determine the line of best fit and write the equation for the line. What is the slope of the line? R- squared? b. The analyst then analyzed five separate aliquots of the boy’s blood, all taken from one sampling event. The following absorbance readings were obtained (per analysis of a blood sample initially diluted 1000 times, and then 0.10 mL of that diluted sample digested and brought to a final volume of 100 mL): 0.345 0.299 0.355 0.350 0.339 What concentrations do these absorbances correspond to? Perform a statistical evaluation of the five concentration data points (Mean, median, mode, range, any outliers?). c. The action level for a 3-year old child is 80 ug lead /0.100 L of blood. Is this child ingesting too much lead? 42
  43. 43. ANALYTICAL SEPARATIONS In the current world of chemical analysis, analytical scientists are faced with a wide variety of sample media that require preparation and analysis. The objective of the scientist is to adequately separate the species/analytes of interest from the sample matrix. Insufficient separations can cause matrix interference, which can skew (bias) results in either a positive (high) or negative (low) direction. In some cases, a difficult matrix can cause data to be completely inscrutable, and therefore, of no value to the scientist. Samples can be composed of the following, non-inclusive, types of material: • Blood (serum and plasma) • Urine • Fecal matter • Tissue samples (skin, organ biopsies, bone material, fat, hair…) • Aqueous (water-based) • Soil or sediment • Food matter- milk, meat, starch or vegetable matter, oils… • Air or gas Many of these sample media can be contaminated with pathogens, which require special handling to prevent infection and spread of the pathogen. There are several basic types of separations that can be accomplished. The first type is purification. Purification is the process of separating a particular analyte from a sample matrix by either recrystallization or distillation. Recrystallization involves the concept of solubility. Certain species will dissolve only in hot solutions of a particular solvent. Generally, a “rough” separation is accomplished by isolating the species and some other similar material from a complex matrix. Then, this material is placed in a particular solvent and heated. The species of interest will dissolve in the warm or hot solution, whereas the non-wanted material will not. The heated solution is filtered and the unwanted material is retained, while the dissolved target material is in the filtrate. The filtrate is cooled, and the target analyte precipitates from the cool solution, based on the fact that solubility for most solutes decreases with temperature. This precipitation is the recrystallization process. The solid material then can be filtered, and the target material is now retained on the filter paper. An example is the separation of benzoic acid from salt (NaCl). Salt is soluble in cold water to a far greater degree than benzoic acid. So when a solution containing both materials is cooled to 4 degrees C, the benzoic acid will recrystallize and the salt will remain in the solution. After the benzoic acid is filtered, the salt water can be discarded, or the water can be vaporized to recover the salt. The “crude” benzoic acid is then redissolved in hot water, and recrystallized a second time, which yields a nearly pure produce. A second method of purification is distillation, which was the subject and focus of an earlier laboratory investigation. Distillation is predicated on the fact that various materials have different vapor pressures and therefore, boiling points. A simple distillation set up 43
  44. 44. (such as was done to remove ethyl alcohol from a beverage) is used when there is only one component that is desired for separation in the material to be analyzed. Another example of simple distillation involves the purification of hard water. Upon distillation, “pure” water is produced, and all the inorganic contaminants are left behind in the distillation vessel. However, if one wishes to separate a mixture of organic solvents, simple distillation will not be successful. A more complicated distillation method, known as fractional distillation, is utilized. A fractionating column is placed on top of the distillation vessel. This column is often packed with glass beads or other non-reactive material. This packing provides a large surface area for the continuous heating and condensing of vapors, allowing the operator to adjust the heat to allow a very fine separation of materials with a difference of boiling points of as little as 2 degrees C. (This is the theory behind crude oil fractional distillation.) A second manner in which to accomplish separations in liquid samples is through the use of liquid-liquid extraction. Other names for this process include solvent extraction or solvent exchange. The premise behind this manner of separation is the movement (transfer) of an analyte from one solvent to another solvent in which it is more soluble. For example, the pesticide DDT is still found in some water supplies. DDT is marginally soluble in water, but very soluble in methylene chloride (dichloromethane). If a water sample is mixed with a small volume of methylene chloride, the DDT will be transferred (extracted) into the organic solvent phase from the water phase. The two liquids are immiscible, so after the two liquids are mixed well (by utilizing a separatory funnel), the methylene chloride layer (being more dense than water) settles to the bottom of the separatory funnel and can be easily removed. The analyte of interest (if present) is now dissolved in a solvent, which can be analyzed by HPLC or GC-GC/MS for the quantity of the analyte. Liquid-Liquid extraction is based on the partition coefficient of the system. The ratio (K) of the concentration of the analyte ([A]) in the extraction solvent is divided by the concentration of the analyte in the original sample after the extraction process: [A]extr = K (partition coefficient) [A]orig If K is sufficiently large (greater than 100), then it can be assumed that nearly all of the analyte has been extracted into the solvent, which is then a quantitative transfer. If the K value is less than 100, a more suitable solvent must be used. If a better solvent is unavailable, then the K value can be used to calculate how much (in % or by concentration) analyte will be extracted, and how much will remain in the original solution. This “correction factor” can then be applied to the final analytical results to determine actual concentration in the original sample. Many analytical procedures require the use of special compounds to determine the K of a particular extraction system. These compounds are known as surrogates, and are chemically similar to target analytes. If surrogates recover poorly, it is assumed that the target analytes will not recover (be extracted) well either. (Recovery is defined as [the amount measured in the extract divided by the amount added before extraction], times 100). 44
  45. 45. If solid samples are to be analyzed, an analogous method is used, called liquid-solid extraction. A solvent is mixed with the solid sample to extract analytes of concern or interest. This mixing can occur in several ways. One of the oldest ways is through the use of the separatory funnel, as with liquid-liquid extractions. However, this method often results in emulsions (a foamy mixture of solvent and sample which will not separate), which leads to poor recovery of the analytes of interest. An apparatus known as the Soxhlet extraction apparatus is used for many liquid-solid extractions. For this type of extraction, a portion of the sample is mixed with a drying agent, such as sodium sulfate, and is then placed in a rigid thimble made of a special paper/cloth material. The thimble does not dissolve in the solvent, so the solvent can wash through the thimble, and extract the analytes of concern. The solvent vessel is on the bottom, usually placed in a heating mantle. The thimble containing the sample is placed in the glass holder on top of the extraction vessel (see the arrow above). On top of the thimble section is a condenser, which prevents the solvent from escaping the system. As the solvent is heated, it boils and rises into the condenser. It then condenses and drips through the sample in the thimble, and fills up the glass enclosure. When the solvent reaches the top of the thimble, it empties into the solvent vessel through the siphon tube (at the right side of the glass thimble holder). In this way, the analytes of interest are incrementally extracted into the solvent. After about 10 or 15 wash cycles, the extraction is terminated, and the solvent is then analyzed in the same manner as the liquid-liquid extraction solvent for the analytes of interest. 45
  46. 46. Once the analytes are in the appropriate solvent and/or form for analysis, the proper method of analysis must be selected. One of the most useful and powerful tools is chromatography. Chromatography is the separation of components based on the degree to which different analytes react physically with two separate material phases. Often, one phase is the “mobile” (moving) phase, and the other is a fixed phase (the “stationary” phase). The mixture containing the analytes of interest is introduced into the mobile phase, which moves through or along the stationary phase. Different analytes are retained on the stationary phase for different lengths of time, which allows separation to occur. Mobile phases can be liquid or gaseous, and the stationary phase can be liquid or solid. Techniques utilizing liquid mobile phases are known collectively as Liquid chromatography (LC), and techniques utilizing a gaseous mobile phase are called gas chromatography (GC). Separation can also be accomplished using an electric field. Different ions are attracted to different degrees in varying directions in an electrical field, and can be separated. This type of analysis is known as electrophoresis. Positive ions move towards the negatively charged electrode (cathode), and the negatively charged ions move towards the positively charged electrode (anode). Obviously, only ions can be separated in this way. Electrophoresis is very useful in separating amino acids and other charged biomolecules, such as DNA. Different sized ions and ions of varying charge can be separated even when they are attracted to the same electrode. The shape of a molecule is also a determining factor in how far an ion will move towards a charged electrode. Electrophoresis can be accomplished using solvent-soaked paper, a slab of gel, or a capillary tube. The carrying (mobile) phase may be organic or inorganic and at a range of pH’s. The loading of a gel electrophoresis investigation is shown above. 46
  47. 47. The white bands represent DNA of a particular size. The arrows are included to point out bands that are legitimate, yet might be overlooked as background noise until you have looked at enough gels to recognize them. The DNA exists in equal amounts, but one fragment is larger than the other • On a molar level, much more DNA of one size is present in that band than in a different band, although the lesser amount may be a larger fragment. http://www.life.uiuc.edu/molbio/geldigest/photo.htmL An example gel electrophoresis result is shown above. As can be seen from this brief summary, there is a wide variety of methods than can be used to separate and then analyze analytes of concern. Much more information on all the techniques introduced here are available in many textbooks and on the internet. 47
  48. 48. CHROMATOGRAPHY One of the most useful tools that analytical chemists have is the gas chromatograph. The use of this instrument to identify and quantify a mixture of unknown organic compounds is known as gas chromatography. The method, known commonly as “GC”, is based on the varying affinities of organic compounds for the stationary phase of a chromatographic column (a long, thin metallic tube coated on the inside with a material for which organic compounds exhibit varying affinities). These varying affinities allow for separation of individual components, often by their boiling points. The sample is “injected” into the instrument, and is carried into and through the column by the “carrier gas” (the “mobile phase”), often helium or nitrogen. Each compound has a unique affinity for the stationary phase that is reproducible under the same conditions. The compounds are separated and “elute” from the end of the column at specific “retention times”, the amount of time that has elapsed from the moment of injection to when the compound is “detected”- sensed by a detector at the end of the column. Separation almost always occurs at elevated temperatures, which is why the column is enclosed in an oven which can be programmed to remain at one particular temperature (isothermal) or the oven can be programmed to elevate the temperature at specific rates over a specific period of time. The more complex the mixture, the more likely the analyst will use a temperature program to attain the best separations possible. The amount, or volume of sample required for this method is very small- often only 1 or 2 microliters (uL- 10-6 L). Because of the very Small volumes needed, this is an ideal method for analyzing specimens from crime scenes, or for other forensic and medical applications. The output of a GC is in the form of chromatographic peaks-which represent the presence and detection of vapors other than the carrier gas. These peaks are recorded by the instrument via a strip-chart recorder, or a software program. This vapor (shown as a peak) represents a particular compound eluting at a particular retention time at a particular concentration. Analysts inject known compounds at known concentrations (the “calibration” of the instrument). Then, after calibration, the sample is injected, and the sample peaks are compared to the known peaks, and identification and quantitation of the compound(s) can be accomplished. This is known as the external standard method of analysis. Most modern GC instruments use a software program that draws the peaks and identifies the retention time, identification, and concentration for each peak. For GC, the sample to be injected must either be an organic liquid or in gaseous form. Solids cannot be introduced into the GC because solid particles will “clog” the column, and prevent any material from passing through the column. Even small amounts of water can damage a column. There is a wide variety of GC instruments, chromatographic columns, detectors, and software packages available today. GC instruments are often connected to a specific type of detector called a mass spectrometer, which breaks the eluted vapors down into 48
  49. 49. molecular or ionic fragments. Each compound has a unique “fingerprint”, or pattern of fragmentation, which makes identification more certain. This system is commonly called “gas chromatography/mass spectrometry” or GC/MS for short. High Performance Liquid Chromatography (HPLC or LC) High-performance liquid chromatography (HPLC) is a form of liquid chromatography (LC) to separate compounds that are dissolved in solution. HPLC instruments consist of a reservoir of mobile phase, a pump, an injector, a separation column, and a detector. Compounds are separated by injecting a plug of the sample mixture onto the column. The different components in the mixture pass through the column at different rates due to differences in their partitioning behavior between the mobile liquid phase and the stationary phase. Instrumentation Solvents must be degassed to eliminate formation of bubbles. The pumps provide a steady high pressure with no pulsating, and can be programmed to vary the composition of the solvent during the course of the separation. Detectors rely on a change in refractive index, UV-VIS absorption, or fluorescence after excitation with a suitable wavelength. The different types of HPLC columns are described in a separate document. Schematic of an HPLC instrument 49
  50. 50. Picture of an HPLC instrument Courtesy of http://elchem.kaist.ac.kr/vt/chem-ed/sep/lc/hplc.htm Applications for HPLC Preparative HPLC refers to the process of isolation and purification of compounds. Important is the degree of solute purity and the throughput, which is the amount of compound produced per unit time. This differs from analytical HPLC, where the focus is to obtain information about the sample compound. The information that can be obtained includes identification, quantification, and resolution of a compound. Chemical Separations can be accomplished using HPLC by utilizing the fact that certain compounds have different migration rates given a particular column and mobile phase. Thus, the chromatographer can separate compounds from each other using HPLC; the extent or degree of separation is mostly determined by the choice of stationary phase and mobile phase. Purification refers to the process of separating or extracting the target compound from other (possibly structurally related) compounds or contaminants. Each compound should have a characteristic peak under certain chromatographic conditions. Depending on what needs to be separated and how closely related the samples are, the chromatographer may choose the conditions, such as the proper mobile phase, to allow adequate separation in order to collect or extract the desired compound as it elutes from the stationary phase. The migration of the compounds and contaminants through the column need to differ enough so that the pure desired compound can be collected or extracted without incurring any other undesired compound. HPLC of Proteins and Polynucleotides Identification of compounds by HPLC is a crucial part of any HPLC assay. In order to identify any compound by HPLC a detector must first be selected. Once the detector is selected and is set to optimal detection settings, a separation assay must be developed. The parameters of this assay should be such that a clean peak of the known sample is observed from the chromatograph. The identifying peak should have a reasonable 50