0.27 Sample Preparation

Transcription

1 0.7 Sample Preparation Problems: 3, 11 Dr. Fred Omega Garces Chemistry 51 Miramar College 1 Sample Preparation 5/5/1

2 Doping Catching the Dopers coming soon this August Sample Preparation 5/5/1

3 Chemical analysis Reliability Confidence in the analysis starts with meaningful samples 3 Sample Preparation 5/5/1

4 Chain of Custody and Sample Storage Besides choosing samples judiciously, sample must be stored such that cross contamination is eliminated. Samples can lose potency after as little as few hours of storage. Samples can also pick up contaminants if storage vials have not been properly cleaned. The chain of custody must be also be observed and documented. 4 Sample Preparation 5/5/1

5 Statistics of Sampling For random errors, the overall variance s o is the sum of the variance of the analytical procedure s a and the variance of the sampling operation s s. total analytical sampling = + variance variance variance s o = s a + s s Uncertainty of sampling can best be understood by the following example: In an analysis of a barrel of powder, the standard deviation of sampling operation is + 4% and the standard deviation of the analytical procedure is +3%. What is the overall standard deviation? s a = 3% s s = 4% : s o = s a + s s : s o = 3% + 4% = s o = = 5 = 5 To what value must the sampling standard deviation be reduced so that the overall standard deviation is + 4%? s 0 = 4% s a = 3% s s =? : s s = s 0 + s a : s s = 4% - 3% = 16-9 s s = = 5 =.6% 5 Sample Preparation 5/5/1

6 Statistics of Sampling For random errors, the overall variance s o is the sum of the variance of the analytical procedure s a and the variance of the sampling operation s s. overall analytical sampling = + variance variance variance s o = s a + s s Uncertainty of sampling can best be understood by the following example: In an analysis of a barrel of powder, the standard deviation of sampling operation is + 4% and the standard deviation of the analytical procedure is +3%. What is the overall standard deviation? s a = 3% s s = 4% : s o = s a + s s : s o = 3% + 4% = s o = = 5 = 5 To what value must the sampling standard deviation be reduced so that the overall standard deviation is + 4%? s 0 = 4% s a = 3% s s =? : s s = s 0 + s a : s s = 4% - 3% = 16-9 s s = = 5 =.6% 6 Sample Preparation 5/5/1

7 Nature of Sampling The uncertainty when selecting a sample for analysis is express in terms of the probability of selecting A over B in a population of n T = n A +n B. p A = possibility of drawing A = n A n A + n B p B = probability of drawing B = n B n A + n B = 1 - p A If n particles are drawn, then the probability of selecting an A particle is np A. The standard deviation of many sampling is known as the binomial distribution standard deviation in sampling operation s = n p p n A B An example of a mixture of 1-mm-diameter particles of KCl and KNO 3 in a ratio 1:99. A sample containing 10 4 particles weights 11.0 g. i) What is the expected number in a sample weighing g? and ii) relative standard deviation of KCl particles. p KCl = ; 104 particles = 11.0 g; 10 6 particles = 1100 g: = np A = particles 100 = 104 Relative standard deviation : s KCl 100 = % n p KCl p KNO3 100 = % 100 = 0.99 % 7 Sample Preparation 5/5/1

8 Nature of Sampling The uncertainty when selecting a sample for analysis is express in terms of the probability of selecting A over B in a population of n T = n A +n B. p A = possibility of drawing A = n A n A + n B p B = probability of drawing B = n B n A + n B = 1 - p A If n particles are drawn, then the probability of selecting an A particle is np A. The standard deviation of many sampling is known as the binomial distribution standard deviation in sampling operation s = n p p n A B An example of a mixture of 1-mm-diameter particles of KCl and KNO 3 in a ratio 1:99. A sample containing 10 4 particles weights 11.0 g. i) What is the expected number in a sample weighing g? and ii) relative standard deviation of KCl particles. p KCl = ; 104 particles = 11.0 g; 10 6 particles = 1100 g: = np A = particles 100 = 104 Relative standard deviation : s KCl 100 = % n p KCl p KNO3 100 = % 100 = 0.99 % 8 Sample Preparation 5/5/1

9 Choosing a Sampling Size Often it is desired to have a specific sampling variance, the question is what sample size must be used? Consider the analysis of radioactive sodium in liver. The tissue is homogenized and contains an average of 37 cps / g with std dev = 13.1%. When the sample size is increase to 1.3 g, the std dev lowers to 5.5%. When the sample size increases to 5.8g, the std dev lowers to.4%. 9 Sample Preparation 5/5/1

10 Choosing the Number of Replicate Analysis Often the overall uncertainty can be reduce by analyzing more samples How many 0.7 g samples must be analyzed to give 95% CI that the mean is known to within +/- 4 % µ - x = e = 4% = 0.04 t = s x = 7% or 0.07 µ - x = e = t s or n = n e where, u is the true population mean x is the measured mean n is the number of samples t is the students t s x is the sampling variance t s e e is the sought -for uncertainty First iteration n = t s e % n = = % Third iteration, t 14 =.150 n = t s e % n = = % Second iteration, t 11 =.09 n = t s e % n = = % Fourth iteration, t 13 =.170 n = t s e % n = = % 10 Sample Preparation 5/5/1

11 The Dissolution Process or Digestion Grinding Dissolution in acid, fusion, ashing 11 Sample Preparation 5/5/1

12 Sample Preparation Series of steps required to transform sample to state suitable for analysis Liquid Extraction with extraction vessel in microwave Supercritical extraction with vessel for high pressure extraction. GC of the product from technique. Solid-Phase extraction: Shown are the steps for extraction process. 1 Sample Preparation 5/5/1

13 Summary of Sample processing in relation to the Analytical Strategy Obtain the sample Representative of Bulk Process the Sample Prep for analysis by weighing and drying then dissolving... Carry out the Analysis Method Obtain weight or Vol Data Prep Referernce Std or Protecting group Std solution or Calib Equib Obrain required Data Work the Data Calcuate the results work up Workup Statistic Analysis Calculate and Report the Results 13 Sample Preparation 5/5/1

14 Lecture Notes Exercise Answers Slide 5 s a = 3% s s = 4% : s o = s a + s s : s o = 3% + 4% = s o = = 5 = 5 s 0 = 4% s a = 3% s s =? : s s = s 0 + s a : s s = 4% - 3% = 16-9 s s = = 5 =.6% Slide 7 1 p KCl = 100 ; 104 particles = 11.0 g; 10 6 particles = 1100 g: = np A = particles 100 = 104 Relative standard deviation : s KCl 100 = % n p KCl p KNO3 100 = % 100 = 0.99 % 14 Sample Preparation 5/5/1