Chapter 8: Sampling, Standardization, and Calibration. Contents in Chapter 08. Analytical Samples and Methods Sampling The Least Squares and Calibration Curve Calibration Methods 1) External Standard Calibration - Matrix Effect 2) Standard Addition Method 3) Internal Standard
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Contents in Chapter 08 • Analytical Samples and Methods • Sampling • The Least Squares and Calibration Curve • Calibration Methods 1) External Standard Calibration - Matrix Effect 2) Standard Addition Method 3) Internal Standard 5. Figures of Merit for Analytical Methods
Analytical Samples and Methods • Classification ofanalysesby sample size • Classification ofconstituenttypesbyanalytelevel
Continued • For example: Determining Hg in the ppb to ppm range in a 1 μL (≈ 1 mg) sample of river water would be a micro analysis of trace constituent. * Interlaboratory error as a function of analyte concentration:
2. Sampling • Classification of sampling plan (Ref/Harvey) - Random sampling - Judgmental sampling - Systematic sampling - Systematic–judgmental sampling - Stratified sampling - Convenience sampling • Random sample: A sample collected at random from the target population. (provides and unbiased estimate of the target population) - using random number table - generating from spreadsheet
Statistics of sampling 1) Sampling variance versus overall variance so2 = ss2 +sm2 so2: overall variance ss2: variance of sampling process sm 2: variance of analytical method
Example: A quantitative analysis for an analyte gives a mean concentration of 12.6 mg/kg. The standard deviation for analytical method (sm) is found to be 1.1 mg/kg, and that due to sampling process (ss) is 2.1 mg/kg. • What is the overall variance for the analysis? • By how much does the overall variance change if sm is improved 10% to 0.99 mg/kg? • By how much does the overall variance change if ss is improved 10% to 1.9 mg/kg ? Solution: • so2 = sa2 + ss2 = (1.1)2 + (2.1)2 = 5.6 • so2 = (0.99)2 + (2.1)2 = 5.4 • so2 = (1.1)2 + (1.9)2 = 4.8 • In general, ss2 is larger than sm2 • More significant improvement in the so is focus on sampling problems.
Example: The following data were particular drug concentration in a animal-feed formulations. The data on the left were obtained under conditions in which random errors in sampling and the analytical procedure contribute to the overall variance. The data on the right were obtained in a repeatedly analysing a single sample. Determine the overall variance and the contributions from sampling and the analytical procedure. sm2 so2
Solution: From left data: so2 = 4.71x10–7 From right data: sm2 = 7.00x10–8 Sampling variance: ss2 = so2 – sm2 = 4.71x10–7– 7.00x10–8 = 4.01x10–7
2) Drawing particles with two types particles (binomial distribution) Application example: nA: analyte containing nB: analyte free *
Example 1: A mixture contains 1% KCl particles and 99% KNO3 particles. What is the number of particle required for a sampling relative standard deviation of 1% for KCl ? Solution:
Example 2: A mixture contains 1% KCl particle and 99% KNO3 particle. The average density of the particles is 2.108 g/cm3, and the average diameter of the particle is 0.152 mm. What is the sample mass equivalent to total 9.9x105 particles of the mixture? Solution: For solid sample: Finer particles require less mass of sample to reach same RSD of sampling operation
Continued For two types particles with different percentage of active ingredientand different density: dA: density of type A particle dB: density of type B particle d: the average density of the particles PA: the percentage of the particle contain a higher amount of active ingredient PB: the percentage of the particle contain a less amount of active ingredient P: the overall average percent of active ingredient
Choosing a sample size (mount) Ks = m (r 100)2 m: mass of the collected sample r : percent relative standard deviation of analyte measurement due to sampling Ks: Ingamells sampling constant
Example: The following data were obtained for determination of the amount of inorganic ash in a breakfast cereal. (assume the Sm << Ss) • Determine sampling constant Ks. • What is the amount of a single sample needed to give a relative standard deviation for sampling of ±2.0%. • Predict the relative standard deviation and the absolute standard deviation if a single sample of 5 g are collected. (assume same average of percentage ash)
Solution: (a) Measurement mean = 1.298 Ss = 0.03194 r = 2.46% Mean of m = 1.0007 Ks = m (r 100)2 = (1.0007)(2.46)2 = 6.06 g
Number of replicate to be analyzed Assuming Ss >> Sa, the number of defined mass samples (ns) to reach a desired overall error (e):
Regression line (Calibration curve) ŷi yi: Experimental value ŷi: Calculated by y = mx + b Residual error = yi–ŷi yi Total squares of residual error = Σ(yi–ŷi)2 3. The Least Squares and Calibration Curve • Method of Least Squares A mathematical optimization technique that attempts to find a "best fit" to a set of data by attempting to minimize the total squares of residual error, between the fitted function and the data. A linear least squares for example: Assumptions in method of least squares: • error in y >> error in x • Uncertainties (standard deviation) in all y values are same
2) Linear Calibration Curve The linear relationship between the amount of analyte and a method’s measuring signal: Smeas = kCA + b (i.e., y = mx + b) (established by the method of least squares) Smeas: Signal response by sample CA : Analyte concentration k: Slope between Smeas and CA b: y-intercept of the linear equation
3) Establishing Calibration Curve Linear equation: y = mx + b
Example: Establish a linear calibration curve by following data: Solution:
4) Example of Applying Calibration Curve Following are results of spectrophotometric analysis of protein standard: 982_Ch04_Calibration_Curve_Application_Demo.xls • Reject datum of (0.392) (it is an outlier)
Reject data of 25.0 μg protein standard (out of the linear range)
剩下 14 個數據建立 linear equation m = 0.01628 b = 0.10400 y = (0.01628)x + 0.10400 iv) 計算 unknown sample 中分析物含量 Example: 已知 calibration curve: y = (0.01628)x + 0.10400 樣品分析結果 吸收值為 0.406，樣品中 protein 含量為多少？ Solution:
4. Calibration Methods 1) External Standard Calibration • A calibration curve, y = ax + b, established by standard containing known amount of analyte. • External standard calibration is feasible if the standards and the sample’s matrix has no effect on the value of the slope of the calibration curve.
Example: A spectrophotometric method for the quantitative determination of Pb2+ levels in blood yields an absorbance of 0.474 for a standard whose concentration of Pb2+ is 1.75 μg/L. How manyμg/L of Pb2+ occur in a sample of blood if its absorbance is 0.361? Solution: Ans: 1.33 μgL–1
Example: A spectrophotometric method for the quantitative determination of Pb2+ levels in blood gives a linear normal calibration curve for which Sstand = (0.296 μg –1L)CS+ 0.003. What is the Pb2+ level (in ppb) in a sample of blood if Ssamp is 0.397? Solution: Ans: 1.33 μgL-1
– Matrix Effect • Matrix: Everything in the unknown sample other than analyte • Matrix effect: The change in the analytical signal caused by matrix. • Value of slope versus matrix effect: Analyzing groundwater ClO4– for example Signal of sample Incorrect Actual
Solutions for matrix effect: - Matrix matching: Adjusting the matrix of an external standard so that it is the same as the matrix of the samples to be analyzed. • Standard addition: By addingaliquots of standard solution into sample, and resolving analyte concentration by adequate algebra. Cautionary for Standard Addition Method: • Amount of spiked analyte (standard) should close to the amount of analyte in the sample • Volume of the spiked standard should be designed to avoid the matrix difference. • Drawback of standard addition: It can not avoid the signal from blank.
Add Vs of [S]i 3) Standard Addition Method i) Single-point standard addition: • Spiked sample • Initial sample [X]i: Analyte conc. of sample Vo: Vol. of sample IX : Signal for [X]i [S]i: Analyte conc. in standard VS: Vol. of spiked standard IS+X: Signal of spiked sample Vo of [X]i Vo of [X]i
Example: A spectrophotometric method for measurong Pb2+ in blood sample yields an absorbance f 0.712. A 5.00 mL blood sample after spiking with 5.00 μL of a 1560 μgL–1 Pb2+ standard, an absorbance of 1.546 is measured. Determine the concentration of Pb2+ in the original sample of blood. Solution: Ans: 1.33 μgL–1
Multiple-point standard addition (Linear Graphic application): Experimental example: Place constant volume (Vo) of sample with conc. of [X]i Add different volume (VS) of standard with conc. of [S]i Dilute to a particular constant volume (Vf)
[X]i: Analyte conc. of sample IX : Signal for [X]i Vo: Vol. of sample (constant) [S]i: Analyte conc. in standard (constant) VS: Vol. of added standard (variable) IS+X: Signal of spiked sample (variable) V: Final vol. after diluting y = b + a x
y = b + a x
Example: Ascorbic acid in orange juice was determined by an electrochemical method. Following data and a linear graph were constructed by eight standard addition. Calculate the concentration of ascorbic acid in the sample.
y = b + a x Solution: Ans: 2.89 mM
4) Internal Standard i) Definition: • Adding known amount of a reference substance (internal standard) other than analyte is added to all blank, analyte standard and samples. • Establish a response factor based on the ratio of standard analyte signal/internal standard signal versus standard analyte conc. /internal standard conc. • Analyte concentration in unknown is then calculated according to the established response factor. • Standard addition: the “standard” is the same substance as the analyte. • Internal standard: The internal “standard” is a different substance from the analyte.
Applying internal standard: • 常用於具有 multichannel dtecting 功能 (e.g., mass) 的分析方法。 • 常用於 Chromatography (層析) 法。 • 克服配製溶液其溶劑為揮發性的問題。 • 克服上機時樣品體積不易控制的問題。 • 克服訊號容易受到操作環境干擾的問題。
iii) Internal standard interpretation 以一個製備樣品其溶劑為揮發性者為例：。 a) 溶劑未揮發 • Analyte b) 之訊號值高於 a) • a) 和 b) Analyte standard 訊號值/Internal standard 訊號值之比例不變 Analyte Standard Internal Standard Signal 300 200 Absorption wavelength (nm) b) 部分溶劑揮發 Analyte Standard Internal Standard Signal 300 200 Absorption wavelength (nm)
iv) Establishing response factor (F) F: Response factor AX: Signal of analyte standard [X]: Conc. of analyte standard AS: Signal of internal standard [S]: Conc. of internal standard v) Calculate analyte in unknown F: Response factor AX: Signal of analyte in unknown [X]: Conc. of analyte in unknown AS: Signal of internal standard [S]: Conc. of internal standard
Example: A spectrophotometric method for the quantitative determination of Pb2+ levels in blood uses Cu2+ as an internal standard. A standard containing 1.75 μgL–1 Pb2+ and 2.25 μgL–1 Cu2+ yields a ratio of AX/AS of 2.37. A sample of blood is spiked with the same concentration of Cu2+, giving a signal ratio of 1.80. Determine the concentration of Pb2+ in the sample of blood. Solution: Ans: 1.33 μgL–1
5. Figures of merit for analytical methods • Linearity: A measure of how well data in a graph follow a straight line, e.g., square of the correlation coefficient R2 required close to 1. • Linear dynamic range: The concentration interval over which linearity, accuracy, and precision are acceptable. • (Calibration) Sensitivity: The change in the response signal per unit change in analytical concentration, thus, the slope of the calibration curve.
Limit of Detection (LOD): The smallest quantity of analyte that is “significant” different from blank, e.g., 99% chance of being greater than the blank. • Only 1% of measurement for a blank are expected to exceed the detection limit. • But, 50% of measurement for a sample containing analyte at LOD will below LOD.
LOD estimation procedure: • Prepare sample whose analyte concentration is 1~5 times the LOD. • Measure 7 replicate samples. • Compute the signal standard deviation (s) of the 7 measurements. • Establish a calibration curve: y = mx +b • LOD (concentration) = 3s/m
Example: Signal from 7 replicate sample with a concentration 3 times the detection limit were 5.0, 5.0, 5.2, 4.2, 4.6, 6.0, and 4.9 nA. The slope of the calibration curve is m=0.229 nA/μM. What is the concentration of the limit of detection? Solution: Ans: 7.3μM