QUALitative Analysis Qualitative analysis in GC and GC-MS is the identification of analytes in a mixture. Qualitative analysis answers the question, "What is in this mixture?" Using GC-MS: Qualitative analysis of compounds is simple using a mass detector (MS), since the mass ion detected represents the molecular weight of the analyte. In addition, the fragmentation pattern observed in the mass spectrum can indicate the structure of the molecule. Using GC: If you do not have an MS detector (we do!), the retention time of a standard can be compared to that of an unknown to make a qualitative assignment. A more rigorous method of qualitative identification of a compound as follows; a member of a homologous series can be determined by comparing the capacity factor (k') of a series of standards and the k' value of the unknown. When more than one homologous series is being considered, a more positive identification can be made by measuring k' on two separate columns with different retention properties. For example, a non-polar column such as poly-silicone (column A) and a polar column such as diethylene-glycol succinate (column B) can be used. A linear plot of k' for the polar column versus k' for the non-polar column can be obtained for each homologous series. For example, an unknown could be identified as a member of a homologous series by its position on one of the lines of the A versus B plot shown below (Figure \(\PageIndex{1}\)). QUANTtitative Analysis Quantitative analysis tells us how much of an analyte is in a mixture. In an ideal GC or GC-MS analysis, the peak height and area under the peak are proportional to the amount of analyte injected onto the column. A peak’s area is determined by integration, which usually is calculated by the instrument’s computer. If two peak are resolved fully, the determination of their respective areas is straightforward. Overlapping peaks, however, require a choice between one of several options for dividing up the area shared by the two peaks (Figure 4.3.15 ). Which method we use depends on the relative size of the two peaks and their resolution. In some cases, the use of peak heights provides more accurate results [(a) Bicking, M. K. L. Chromatography Online, April 2006; (b) Bicking, M. K. L. Chromatography Online, June 2006]. For quantitative work we need to establish a calibration curve that relates the detector’s response to the analyte’s concentration. If the injection volume is identical for every standard and sample, then an external standardization provides both accurate and precise results. Unfortunately, even under the best conditions the relative precision for replicate injections may differ by 5%; often it is substantially worse. For quantitative work that requires high accuracy and precision, the use of internal standards is recommended (see below). Another important consideration is the variation in detector sensitivity for different substances that may lead to inaccurate quantitative results. To overcome this problem, a response factor should be determined for the substance of interest relative to the internal standard. The molar response factor (r) for a component \(\mathrm{C}_{\mathrm{n}}\) is defined as: \[\mathrm{r}_{\mathrm{C}_{\mathrm{n}}}=\frac{\text { peak area }}{\text { moles of } \mathrm{C}_{\mathrm{n}} \text { injected }}\] Molar response factors must be determined from a calibration curve. Using standard solutions, known amounts of the analyte(s) are injected into the instrument, and the peak height and area are determined. By adding a known quantity of internal standard to every standard and to the sample, injecting it into the chromatogram, measuring peak areas, and incorporating response factors, the quantity of analyte may be determined. Internal Standards If injection volumes vary or if the concentration may change easily because the solvent is volatile, we can still create a calibration using an internal standard as a reference. The internal standard is a compound added to every solution that is used as a reference to compare to the analyte’s signal. The species, which we call an internal standard , must be different than the analyte. Because the analyte and the internal standard receive the same treatment, the ratio of their signals is unaffected by any lack of reproducibility in the injection procedure or changes in concentration due to evaporation. If a solution contains an analyte of concentration C A and an internal standard of concentration C IS , then the signals due to the analyte, S A , and the internal standard, S IS , are \[S_A = k_A C_A \nonumber\] \[S_{IS} = k_{SI} C_{IS} \nonumber\] where \(k_A\) and \(k_{IS}\) are the sensitivities (aka response factors) for the analyte and the internal standard, respectively. Taking the ratio of the two signals gives the fundamental equation for an internal standardization. \[\frac {S_A} {S_{IS}} = \frac {k_A C_A} {k_{IS} C_{IS}} = K \times \frac {C_A} {C_{IS}} \label{5.12}\] \(K\) is the relative sensitivity (relative response factor) for the analyte compared to the internal standard. Because \(K\) is a ratio of the analyte’s sensitivity and the internal standard’s sensitivity, it is not necessary to determine independently values for either \(k_A\) or \(k_{IS}\) Single Internal Standard In a single-point internal standardization, we prepare a single standard that contains the analyte and the internal standard, and use it to determine the value of K in Equation \ref{5.12}. \[K = \left( \frac {C_{IS}} {C_A} \right)_{std} \times \left( \frac {S_A} {S_{IS}} \right)_{std} \label{5.13}\] Having standardized the method, the analyte’s concentration is given by \[C_A = \frac {C_{IS}} {K} \times \left( \frac {S_A} {S_{IS}} \right)_{samp} \nonumber\] A sixth spectrophotometric method for the quantitative analysis of Pb 2 + in blood uses Cu 2 + as an internal standard. A standard that is 1.75 ppb Pb 2 + and 2.25 ppb Cu 2 + yields a ratio of ( S A / S IS ) std of 2.37. A sample of blood spiked with the same concentration of Cu 2 + gives a signal ratio, ( S A / S IS ) samp , of 1.80. What is the concentration of Pb 2 + in the sample of blood? Solution Equation \ref{5.13} allows us to calculate the value of K using the data for the standard \[K = \left( \frac {C_{IS}} {C_A} \right)_{std} \times \left( \frac {S_A} {S_{IS}} \right)_{std} = \frac {2.25 \text{ ppb } \ce{Cu^{2+}}} {1.75 \text{ ppb } \ce{Pb^{2+}}} \times 2.37 = 3.05 \frac {\text{ppb } \ce{Cu^{2+}}} {\text{ppb } \ce{Pb^{2+}}} \nonumber\] The concentration of Pb 2 + , therefore, is \[C_A = \frac {C_{IS}} {K} \times \left( \frac {S_A} {S_{IS}} \right)_{samp} = \frac {2.25 \text{ ppb } \ce{Cu^{2+}}} {3.05 \frac {\text{ppb } \ce{Cu^{2+}}} {\text{ppb } \ce{Pb^{2+}}}} \times 1.80 = 1.33 \text{ ppb } \ce{Pb^{2+}} \nonumber\] Multiple Internal Standards A single-point internal standardization has the same limitations as a single-point normal calibration. To construct an internal standard calibration curve we prepare a series of standards, each of which contains the same concentration of internal standard and a different concentrations of analyte. Under these conditions a calibration curve of ( S A / S IS ) std versus C A is linear with a slope of K / C IS . Although the usual practice is to prepare the standards so that each contains an identical amount of the internal standard, this is not a requirement. A seventh spectrophotometric method for the quantitative analysis of Pb 2 + in blood gives a linear internal standards calibration curve for which \[\left( \frac {S_A} {S_{IS}} \right)_{std} = (2.11 \text{ ppb}^{-1} \times C_A) - 0.006 \nonumber\] What is the ppb Pb 2 + in a sample of blood if ( S A / S IS )samp is 2.80? Solution To determine the concentration of Pb 2 + in the sample of blood we replace ( S A / S IS ) std in the calibration equation with ( S A / S IS ) samp and solve for C A . \[C_A = \frac {\left( \frac {S_A} {S_{IS}} \right)_{samp} + 0.006} {2.11 \text{ ppb}^{-1}} = \frac {2.80 + 0.006} {2.11 \text{ ppb}^{-1}} = 1.33 \text{ ppb } \ce{Pb^{2+}} \nonumber\] The concentration of Pb 2 + in the sample of blood is 1.33 ppb. In some circumstances it is not possible to prepare the standards so that each contains the same concentration of internal standard. This is the case, for example, when we prepare samples by mass instead of volume. We can still prepare a calibration curve, however, by plotting \((S_A / S_{IS})_{std}\) versus C A / C IS , giving a linear calibration curve with a slope of K . You might wonder if it is possible to include an internal standard in the method of standard additions to correct for both matrix effects and uncontrolled variations between samples; well, the answer is yes as described in the paper “Standard Dilution Analysis,” the full reference for which is Jones, W. B.; Donati, G. L.; Calloway, C. P.; Jones, B. T. Anal. Chem. 2015 , 87 , 2321-2327. Marriott and Carpenter report the following data for five replicate injections of a mixture that contains 1% v/v methyl isobutyl ketone and 1% v/v p -xylene in dichloromethane [Marriott, P. J.; Carpenter, P. D. J. Chem. Educ. 1996 , 73 , 96–99]. injection peak peak area (arb. units) I 1 48075 NaN 2 78112 II 1 85829 NaN 2 135404 III 1 84136 NaN 2 132332 IV 1 71681 NaN 2 112889 V 1 58054 NaN 2 91287 Assume that p -xylene (peak 2) is the analyte, and that methyl isobutyl ketone (peak 1) is the internal standard. Determine the 95% confidence interval for a single-point standardization with and without using the internal standard. Solution For a single-point external standardization we ignore the internal standard and determine the relationship between the peak area for p -xylene, A 2 , and the concentration, C 2 , of p -xylene. \[A_{2}=k C_{2} \nonumber\] Substituting the known concentration for p -xylene (1% v/v) and the appropriate peak areas, gives the following values for the constant k . \[78112 \quad 135404 \quad 132332 \quad 112889 \quad 91287 \nonumber\] The average value for k is 110 000 with a standard deviation of 25 100 (a relative standard deviation of 22.8%). The 95% confidence interval is \[\mu=\overline{X} \pm \frac{t s}{\sqrt{n}}=111000 \pm \frac{(2.78)(25100)}{\sqrt{5}}=111000 \pm 31200 \nonumber\] For an internal standardization, the relationship between the analyte’s peak area, A 2 , the internal standard’s peak area, A 1 , and their respective concentrations, C 2 and C 1 , is \[\frac{A_{2}}{A_{1}}=k \frac{C_{2}}{C_{1}} \nonumber\] Substituting in the known concentrations and the appropriate peak areas gives the following values for the constant k . \[1.5917 \quad 1.5776 \quad 1.5728 \quad 1.5749 \quad 1.5724 \nonumber\] The average value for k is 1.5779 with a standard deviation of 0.0080 (a relative standard deviation of 0.507%). The 95% confidence interval is \[\mu=\overline{X} \pm \frac{t s}{\sqrt{n}}=1.5779 \pm \frac{(2.78)(0.0080)}{\sqrt{5}}=1.5779 \pm 0.0099 \nonumber\] Although there is a substantial variation in the individual peak areas for this set of replicate injections, the internal standard compensates for these variations, providing a more accurate and precise calibration. Figure 4.3.16 shows chromatograms for five standards and for one sample. Each standard and sample contains the same concentration of an internal standard, which is 2.50 mg/mL. For the five standards, the concentrations of analyte are 0.20 mg/mL, 0.40 mg/mL, 0.60 mg/mL, 0.80 mg/mL, and 1.00 mg/mL, respectively. Determine the concentration of analyte in the sample by (a) ignoring the internal standards and creating an external standards calibration curve, and by (b) creating an internal standard calibration curve. For each approach, report the analyte’s concentration and the 95% confidence interval. Use peak heights instead of peak areas. Answer The following table summarizes my measurements of the peak heights for each standard and the sample, and their ratio (although your absolute values for peak heights will differ from mine, depending on the size of your monitor or printout, your relative peak height ratios should be similar to mine). [standard] (mg/mL) peak height of standard (mm) peak height of analyte (mm) peak height ratio 0.20 35 7 0.20 0.40 41 16 0.39 0.60 44 27 0.61 0.80 48 39 0.81 1.00 41 41 1.00 sample 39 21 0.54 Figure (a) shows the calibration curve and the calibration equation when we ignore the internal standard. Substituting the sample’s peak height into the calibration equation gives the analyte’s concentration in the sample as 0.49 mg/mL. The 95% confidence interval is ±0.24 mg/mL. The calibration curve shows quite a bit of scatter in the data because of uncertainty in the injection volumes. Figure (b) shows the calibration curve and the calibration equation when we include the internal standard. Substituting the sample’s peak height ratio into the calibration equation gives the analyte’s concentration in the sample as 0.54 mg/mL. The 95% confidence interval is ±0.04 mg/mL. To review the use of Excel or R for regression calculations and confidence intervals, see Chapter 5.5. The data for this exercise were created so that the analyte’s actual concentration is 0.55 mg/mL. Given the resolution of my ruler’s scale, my answer is pretty reasonable. Your measurements may be slightly different, but your answers should be close to the actual values.