1. 11 Analytical Chemistry CHEM 2310 Prof. Monzir S. Abdel-Latif

2. خطة مساق كيمياء تحليلية مدرس المساق: أ.د. منذر سليم عبد اللطيف عنوان المدرس: مبنى المختبرات الجديد ، غرفة 630 الساعات المكتبية: سبت اثنين أربعاء 12-10 2

3. وصف المساق (حسب الدليل): المفاهيم الأساسية في الكيمياء التحليلية، والمفاهيم العامة للاتزان الكيميائي والذي يشمل الأحماض والقواعد، أسس التحليل الوزني والحجمي ومعايرات الأحما ض والقواعد ومعايرات الترسيب، ومعايرات المعقدات، ومعايرات الأكسدة والاختزال، وطرق التحليل المعتمدة على الخواص الكهربائية. 3

4. الأهداف العامة للمساق )أو مخرجات المساق:( امتلاك المهارات الأساسية النظرية في التحليل الكيميائي التقليدي ، بما في ذلك تقييم ومعالجة البيانات بطريقة حسابية وإحصائية مناسبة ، التعرف على طرق التحليل الوزني والحجمي والأسس النظرية التي يقوم عليها كل منهما ، دراسة الأنواع المختلفة من الكواشف ومجال عمل كل نوع منها ، تحضير المحاليل المختلفة بما فيها المحاليل المنظمة والحسابات الكيميائية لكل ما سبق. 4

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6. أسلوب تدريس المساق: المحاضرة والنقاش في بعض الأحيان مراجع المساق: أ. المرجع الرئيس : Analytical Chemistry, 2004, Gary D. Christian, 6th Ed. ب. المراجع الإضافية : 1. Analytical Chemistry, Principles and Techniques, Larry Hargis, Prentice Hall. 2. Arts and Science of Chemical Analysis, Christie Enke, Wiley and Sons. 3. Principles of Analytical Chemistry, Miguel Valcarcel, Springer. 4. Fundamentals of Analytical Chemistry, Skoog, Holler, and West, Brooks/Cole ج. المعلومات الواردة على صفحة المساق الالكترونية 6

7. التقييم 3 امتحانات ساعية: 50% الامتحان النهائي (شامل): 50% ____________________________________________ * ينص البند (4) من مادة (8) في النظام الأكاديمي على أن : تشمل الأعمال الفصلية لكل مساق الامتحانات التحريرية و الشفهية و التقارير و البحوث و الأعمال المخبرية والتطبيقية على أن يخصص له 50 % كحد أقصى من العلامة النهائية للمساق وفق ما يحدده مجلس الكلية. ** ينص البند (6) من مادة (8) في النظام الأكاديمي على أن" يكون الامتحان النهائي شاملاً للمنهج كله ". 7

8. 88 Areas of Chemical Analysis and Questions They Answer Quantitation: How much of substance X is in the sample? Detection: Does the sample contain substance X? Identification: What is the identity of the substance in the sample? Separation: How can the species of interest be separated from the sample matrix for better quantitation and identification?

9. 99 Chapter 1: Introduction This course is a quantitative course where you have seen some qualitative analysis in general Chemistry Lab and will also encounter the topic in other classes. In addition, Analytical Chemistry can be classified as Instrumental or Classical (wet Chemistry).

10. 10 In this course, we will cover the classical methods of Chemical Analysis. However, it should not be implied that the term classical means something old, which is studied like history, but rather the term means understanding the basics of Chemical Analysis that were eventually laid down long time ago. Some of the classical methods still serve as the standard methods of analysis, till now.

11. 11 The analytical process involves a sequence of logical events including: 1. Defining the problem This means that the analyst should know what information is required, the type and amount of sample, the sensitivity and accuracy of the results, the analytical method which can be used to achieve these results, etc...

12. 12 2. Obtaining a representative sample It is very important to collect a representative sample for analysis. This could be appreciated if, for example, an ore is to be analyzed to decide whether the ore concentration in a mine or mountain can be economically produced. One should take several samples from different locations and depths, mix them well and then take a sample for analysis.

13. 13 3. Preparing the sample for analysis Most analytical methods require a solution of the sample rather than the solid. Therefore, samples should be dissolved quantitatively and may be diluted to the concentration range of the method. 4. Chemical separations The sample may contain solutes which interfere with the determination of the analyte. If this is the case, analytes should be separated from the sample matrix by an accepted procedure.

14. 14 5. Performing the measurement This implies conducting the analytical procedure and collecting the required data. 6. Calculations The final event in the analytical process is to perform the calculations and present the results in an acceptable manner.

15. 15 RANGE The size of the sample can be used to describe the class of a method where: A method can be described as meso if the sample size is above 100 mg or 100 microliters. A semimicro method describes a sample size from 10 to 100 mg or 50 to 100 microliters. When the sample size is in the 1 to 10 mg or less than 50 microliters, the method is said to be a micro method A sample size less than 1 mg denotes an ultramicro method.

16. 16 An analyte in a sample can be classified as a major constituent if it constitutes more than 1% of the sample or a minor in the range from 0.1 to 1.0 %. It is classified as trace if it constitutes less than 0.1%. Analyze versus Determine These terms are sometimes misused. Always use the term ‘analyze’ with the sample while use the term ‘determine’ with specific analyte. Therefore, a sample is “analyzed”, while an analyte is “determined”.

17. 17 Chapter 2 Data Handling

18. 18 Accuracy and Precision Accuracy can be defined as the degree of agreement between a measured value and the true or accepted value. As the two values become closer, the measured value is said to be more accurate. Precision is defined as the degree of agreement between replicate measurements of the same quantity.

19. 19 Assuming the correct or accepted value is represented by the center of the circles below; If all values occurred close together within, for example, the red or blue circles, results are precise but not accurate. If all values occurred within the yellow circle, results are both accurate and precise. If results were scattered randomly in one direction of the center, results are neither precise nor accurate.

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21. 21 Example The weight of a person was measured five times using a scale. The reported weights were 84, 83, 84, 85, and 84 kg. If the weight of the person is 76 kg weighed on a standard scale, then we know that the results obtained using the first scale is definitely not accurate. However, the values of the weights for the five replicate measurements are very close and reproducible. Therefore measurements are precise. Therefore, the measurement is precise but definitely not accurate.

22. 22 Significant Figures At the most basic level, Analytical Chemistry relies upon experimentation; experimentation in turn requires numerical measurements. And measurements are always taken from instruments made by other workers.

23. 23 Some information about measurements 1)Examples we will study include the metric ruler, the graduated cylinder, and the scale. 2) Because of the involvement of human beings, NO measurement is exact; some error is always involved. This means that every answer in science has some uncertainty associated with it. We might be fairly confident we have the correct answer, but we can never be 100% certain we have the EXACT correct answer.

24. 24 3) Measurements always have two parts - a numerical part (sometimes called a factor) and a dimension (a unit). Significant figures are concerned with measurements not exact countings.

25. 25 Identifying significant digits The following rules are helpful in identifying significant digits 1. Digits other than zero are significant. e.g., 42.1m has 3 sig figs. 2. Zeroes are sometimes significant, and sometimes they are not. 3. Zeroes at the beginning of a number (used just to position the decimal point) are not significant. e.g., 0.025m has 2 sig figs. In scientific notation, this can be written as 2.5*10 -2 m

26. 26 4. Zeroes between nonzero digits are significant. e.g., 40.1m has 3 sig figs 5. Zeroes at the end of a number that contains a decimal point are significant. e.g., 41.0m has 3 sig figs, while 441.20m has 5. In scientific notation, these can be written respectively as 4.10*10 1 and 4.4120*10 2 6. Zeroes at the end of a number that does not contain a decimal point may or may not be significant. If we wish to indicate the number of significant figures in such numbers, it is common to use the scientific notation.

27. 27 e.g., The quantity 52800 km could be having 3, 4, or 5 sig figs—the information is insufficient for decision. If both of the zeroes are used just to position the decimal point (i.e., the number was measured with estimation ±100), the number is 5.28×10 4 km (3 sig figs) in scientific notation. If only one of the zeroes is used to position the decimal point (i.e., the number was measured ±10), the number is 5.280×10 4 km (4 sig figs). If the number is 52800±1 km, it implies 5.2800×10 4 km (5 sig figs).

28. 28 Exact Numbers Exact numbers can be considered as having an unlimited number of significant figures. This applies to defined quantities too. e.g., 1.The rules of significant figures do not apply to (a) the count of 47 people in a hall, or (b) the equivalence: 1 inch = 2.54 centimeters. 2.In addition, the power of 10 used in scientific notation is an exact number, i.e. the number 10 3 is exact, but the number1000 has 1 sig fig. It actually makes a lot of sense to write numbers derived from measurements in scientific notation, since the notation clearly indicates the number of significant digits in the number.

29. 29 Look at the following example

30. 30 The length is greater than 2.6 cm but less than 2.7 cm, and so the estimated value is 2.62 cm. The measurement can be written as 2.62±0.01 cm or 26.2±0.1 mm. The number 26.2mm contains three significant figures.

31. 31 How do you read it?

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36. 36 Lecture 3 Data Handling Significant Figures, continued…. Math using significant figures

37. 37 Math With Significant Figures Addition and Subtraction In mathematical operations involving significant figures, the answer is reported in such a way that it reflects the reliability of the least precise operation. Let's state that another way: a chain is no stronger than its weakest link. An answer is no more precise that the least precise number used to get the answer. Let's do it one more time: imagine a team race where you and your team must finish together. Who dictates the speed of the team? Of course, the slowest member of the team. Your answer cannot be MORE precise than the least precise measurement.

38. 38 For addition and subtraction, look at the decimal portion (i.e., to the right of the decimal point) of the numbers ONLY. Here is what to do: 1) Count the number of significant figures in the decimal portion of each number in the problem. (The digits to the left of the decimal place are not used to determine the number of decimal places in the final answer.) 2) Add or subtract in the normal fashion. 3) Round the answer to the LEAST number of places in the decimal portion of any number in the problem.

39. 39 Find the formula weight for Ag 2 MoO 4 given the following atomic weights: Ag = 107.870, Mo = 95.94, O = 15.9994. The number with the least number of digits after the decimal point is 95.94 which has two digits for expression of precision. Also, it is the number with the highest uncertainty. The atomic weights for Ag and O have 3 and 4 digits after the decimal point. Therefore if we calculate the formula weight we will get 375.6776. However, the answer should be reported as 375.68 ( i.e. to the same uncertainty of the least precise value.

40. 40 Multiplication and Division The number having the least number of significant figures is called the KEY NUMBER. The LEAST number of significant figures in any number of the problem determines the number of significant figures in the answer. In case where two or more numbers have the same number of significant figures, the key number is determined as the number of the lowest value regardless of decimal point.

41. 41 Note that: When the uncertainty of a number is not known, the uncertainty is assumed to be +1 of the last digit to the right

42. 42 2.5 x 3.42 = ? The answer to this problem would be 8.6 (which was rounded from the calculator reading of 8.55). Why? 2.5 is the key number which has two significant figures while 3.42 has three. Two significant figures is less precise than three, so the answer has two significant figures.

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44. 44 Relative uncertainty in key number = ( +1/25) = +0.04 Now find the absolute uncertainty in answer: (8.55/25) = 0.342 Therefore, the uncertainty in the answer should be known to one decimal point. The answer can be written as 8.6 + 0.3. The relative uncertainty in answer can now be calculated: S rel = (+0.342/8.6) = +0.04 The relative uncertainty in answer is the same as that of the key number, which should be the case.

45. 45 2.33 x 6.085 x 2.1= ? How many significant figures should be in the answer? Answer - two. Which is the key number? Answer - the 2.1 Why? It has the least number of significant figures in the problem. It is, therefore, the least precise measurement. Answer = 30.

46. 46 Relative uncertainty in key number = ( +1/21) = +0.048 Now find the absolute uncertainty in answer: (29.77/21) = 1.4 Therefore, the absolute uncertainty in the answer should be known to integers. The answer can be written as 30. + 1. The relative uncertainty in answer can now be calculated: S rel = (+1/30.) = +0.033 The relative uncertainty in answer is close to that of the key number, which should be the case.

47. 47 How many significant figures will the answer to 3.10 x 4.520 = (Calculator gives 14.012) have? 3.10 is the key number which has three significant figures. Three is supposed to be the correct answer. 14.0 has three significant figures. Note that the zero in the tenth's place is considered significant. All trailing zeros in the decimal portion are considered significant.

48. 48 Another common error is for the student to think that 14 and 14.0 are the same thing. THEY ARE NOT. 14.0 is ten times more precise than 14. The two numbers have the same value, but they convey different meanings about how trustworthy they are. However, the correct answer should be reported as 14.0 1. Note that an additional significant figure is included in the answer. This is because the answer is less than the key number.

49. 49 Relative uncertainty in key number = ( +1/310) = +3.2*10 -3 Now find the absolute uncertainty in answer: (14.012/310) = 0.0452 Therefore, the absolute uncertainty in the answer should be known to one hundred th. The answer can be written as 14.0 1 + 0.05. The relative uncertainty in answer can now be calculated: S rel = (+0.05/14.0 1 ) = +3.5*10 -3 The relative uncertainty in answer is very close to that of the key number, which should be the case.

50. 50 Why do we add an additional significant figure in the answer when the answer is less than the key number? The answer to this question simply is to reduce the uncertainty associated with the answer. When the answer is less than the key number, the uncertainty associated with the answer is unjustifiably large.

51. 51 (4.52 x 10 -4 ) ÷ (3.980 x 10 -6 ). How many significant figures in the answer? Answer - three. Which is the key number? Answer - the 4.52 x 10 -4. Why? It has the least number of significant figures in the problem. It is, therefore, the least precise measurement. Notice it is the 4.52 portion that plays the role of determining significant figures; the exponential portion plays no role. However, since the answer is less than the key number, an additional significant figure is used and written as a subscript. Answer = 113. 6

52. 52 Relative uncertainty in key number = ( +1/452) = +2.2*10 -3 Now find the absolute uncertainty in answer: (113.568/452) = 0.251 Therefore, the uncertainty in the answer should be known to one decimal point. The absolute uncertainty in the answer is therefore +0.3. The answer can be written as 113.6 + 0.3. The relative uncertainty in answer can now be calculated: S rel = (+0.3/113. 6 ) = +2.6*10 -3 The relative uncertainty in answer is very close to that of the key number, which should be the case.

53. 53 4.20x3.52 = Which is the key number? Both have 3 significant figures. In this case, the number with smaller value, regardless of the decimal point, is the key number (3.52). The correct answer should be reported as 14.7 8 Once again, we have added an additional significant figure as a subscript since the answer is less than the key number

54. 54 Uncertainty in key number = ( +1/352) = +2.84*10 -3 Now find the absolute uncertainty in answer : (14.784/352) = 0.042 Therefore, the absolute uncertainty in the answer should be known to one hundred th. The absolute uncertainty in the answer is therefore +0.042. The answer can be written as 14.7 8 + 0.04. The relative uncertainty in answer can now be calculated: S rel = (+0.042/14.7 8 ) = +2.84*10 -3 The relative uncertainty in answer is the same as that of the key number, which should be the case.

55. 55 Look at the following multiplication problem: The key number is 35.63 which has 4 significant figures. Therefore, the answer should be 88.55% The answer is larger than the key number, therefore no additional figure is added.

56. 56 Why not keep additional significant figures as we did in the previous example? Uncertainty in key number = ( +1/3563) = +2.8*10 -4 Now find the absolute uncertainty in answer : (88.547/3563) = 0.025 Therefore, the uncertainty in the answer should be known to one hundred th. The absolute uncertainty in the answer is therefore +0.02 and can be written as 88.55 + 0.02. The relative uncertainty in answer can now be calculated: S rel = (+0.025/88.55) = +2.8*10 -4 The relative uncertainty in answer is the same as that of the key number, which should be the case.

57. 57 Find the answer to the following calculation: It is clear that the key number is 891 and the answer should have 3 significant figures. However, in cases where the answer is less than the key number as is our case, the answer retains an extra digit as a subscript. The answer is 546. 6 as we have seen before.

58. 58 Uncertainty in key number = ( +1/891) = +1.1*10 -3 Now find the absolute uncertainty in answer : (546.57/891) = 0.6 Therefore, the uncertainty in the answer should be known to one decimal point. The uncertainty in the answer is therefore +0.6. The answer can be written as 546. 6 + 0.6. The uncertainty in answer can now be calculated: S rel = (+0.6/546. 6 ) = +1.1*10 -3 The relative uncertainty in answer is the same as that of the key number, which should be the case.

59. 59 Find the answer to the following calculation: When multiple operations are involved, do it in a step by step procedure. The parenthesis above has 97.7 as the key number.

60. 60 For the initial calculation : ( 97.7 x 100.0 ) 32.42 Uncertainty in key number = ( +1/977) = +1.0*10 -3 Now find the absolute uncertainty in answer : (301.36/977) = 0.308 Therefore, the uncertainty in the answer should be known to one decimal point. The uncertainty in the answer is therefore +0.31. The answer can be written as 301. 36 + 0.3. The relative uncertainty in answer can now be calculated: S rel = (+0.31/301. 36 ) = +1.0*10 -3 The relative uncertainty in answer is the same as that of the key number, which should be the case.

61. 61 The second process has 687 as the key number. Finally as the answer is less than the key number an additional digit was added as a subscript. It is noteworthy to observe that extra digits were retained temporarily in all steps and rounding off to the correct number of significant figures was done in the final answer. This is simply to reduce the error resulting from multiple rounding-off.

62. 62 The second part of the problem (337. 4 /687) = 0.491 1 Uncertainty in key number = ( +1/687) = +1.46*10 -3 Now find the absolute uncertainty in answer : (0.4911/687) = 0.0007 Therefore, the uncertainty in the answer should be known to one in 10,000 th. The uncertainty in the answer is therefore +0.0007. The answer can be written as 0.491 1 + 0.0007. The relative uncertainty in answer can now be calculated: S rel = (+0.0007/0.491 1 ) = +1.46*10 -3 The relative uncertainty in answer is the same as that of the key number, which should be the case.

63. 63 Lecture 4 Data Handling Significant Figures … Continued Error and Standard Deviation

64. 64 Logarithms The digits to the left of the decimal point in a logarithmic value are not counted since they merely reflect the log 10 x and they are not considered significant. The zeros to the right of the decimal point are all significant. Examples Log 2.0x10 3 = 3.30 (two significant figures in both terms). The blue digit in the answer is not significant as it comes from the 10 3 portion which has nothing to do with expressing the number of significant figures

65. 65 Log 1.18 = 0.072 (three significant figures in both terms, the blue zero in the answer is significant) Antilog of 0.083 = 1.21 (three significant figures in both terms) Log 12.1 = 1.083 (three significant figures in both terms, the blue digit in the answer is not significant)

66. 66 Errors Errors can be classified according to nature of the error into two types, determinate and indeterminate errors. A determinate error (sometimes called a systematic error) is an error which has a direction either positive or negative. An example of such an error is performing a weight measurement on an uncalibrated balance (for instance it always add a fixed amount to the weight).

67. 67 Another important example is measuring volume using a burette that has extra or less volume than indicated on its surface. When using the abovementioned balance or burette, our results will always be higher or lower depending on whether these tools have positive or negative bias. This means that determinate error is unidirectional. Sometimes a determinate error can be significant if the analyst is careless or inexperienced neglecting enough drying times in a gravimetric procedure, using a too high indicator concentration in a volumetric procedure, etc.

68. 68 An indeterminate error is a random error and has no direction where sometimes higher or lower estimates than should be observed are obtained. In many cases, indeterminate errors are encountered by lack of analyst experience and attention. Indeterminate errors are always present but can be minimized to very low levels by good analysts and procedures.

69. 69 Absolute Error The difference between the measured value and the true value is referred to as the absolute error. Assume that analysis of an iron ore by some method gave 11.1% while the true value was 12.1%, the absolute error is: AE = 11.1% - 12.1% = -1.0% The negative sign indicates a negative error

70. 70 Relative Error The relative error is the percentage of the absolute error to the true value. For the argument above we can calculate the relative error as: Relative error = (absolute error/true value)x100% RE = (-1.0/12.1)x100% = -8.3%

71. 71 Relative Accuracy The percentage of the quotient of observed result to the true value is called relative accuracy. Relative accuracy = (observed value/true value)x100% For the abovementioned example: Relative accuracy = (11.1/12.1)x100% = 91.7%

72. 72 Standard Deviation The standard deviation for a set of data provides information on the spread of the values.

73. 73 What we mean by spread is clear from the graphs blue and red. Values used to draw the red graph are not as close to each others as values in the blue graph. Therefore, values in the red graph have higher spread from the mean and have higher standard deviation. On the contrary, values in the blue graph are closer together and have a lower spread from their mean, and thus have a lower standard deviation.

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75. 75 For an infinite or large number of data points (more than 20) or when the true mean is known, the population standard deviation is defined as:  = (  (x i -  ) 2 / N ) 1/2 Where  is the population standard deviation,  is the population mean, x i is the individual data point, and N is the number of data points

76. 76 However, in real chemical laboratories where a sample is analyzed, an experiment is repeated three to five times and thus a very limited number of data points (3-5) is collected. The sample standard deviation (s) is defined as: s = (  (x i - x) 2 / (N-1) ) 1/2 x is the average (mean) of the data points. The sample standard deviation is also called estimated standard deviation since it is only an estimate of .

77. 77 Lecture 5 Data Handling Standard Deviation,.. Continued Propagation of Error

78. 78 Standard deviation of the mean (s(mean)) S(mean) = s / N 1/2 Another important expression of deviation is the relative standard deviation (RSD) or sometimes called coefficient of variation (CV) where RSD = ( s / x ) X100% RSD (mean) = ( s(mean) / x ) X100%

79. 79 Example The following replicate weights were obtained for a sample: 29.8, 30.2, 28.6, and 29.7 mg. Calculate s, s(mean), RSD, and RSD(mean) Solution First, we find x X = (29.8+30.2+28.6+29.7)/4 = 29.6

80. 80 x i x i – x (x i – x) 2 29.8 0.20.04 30.2 0.60.36 28.6 1.01.00 29.7 0.10.01  = 1.41 s = (  (x i - x) 2 / (N-1) ) 1/2

81. 81 s = (1.41/3) 1/2 s = 0.69 mg S(mean) = s / N 1/2 S(mean) = 0.69/(4) 1/2 S(mean) = 0.34 mg RSD or CV = (0.69/29.6)x100% = 2.3% RSD(mean) = (0.345/29.6)x100% = 1.1%

82. 82 It should be recognized that as the number of experiments is increased, the precision of the measurement is increased as well. This is because s  1/N 1/2 which means that decrease in s as N increases is not linear which implies that, after some number of experiments, further increase in the number of experiments will result in very little decrease in s, which does not justify extra time and effort.

83. 83 Pooled Standard Deviation (s p ) When replicate samples are done using two different methods, the standard deviation can be pooled in order to determine the reliability of the analytical method (proposed or new). S p = {  (x i1 – x 1 ) 2 +  (x i2 – x 2 ) 2 )/(N 1 + N 2 –2)} 1/2 S p is the pooled standard deviation, x 1, x 2 are average values for data sets 1 and 2, respectively, N 1 and N 2 are the number of data points of data set 1 and 2, respectively.

84. 84 Example Mercury in a sample was determined using a standard method and a new suggested method. five replicate experiments were conducted using the two procedures giving the following results in ppm New Method Standard method 10.5 10.1 9.9 10.3 10.4 10.2 11.2 10.3 10.5 10.4 Find the pooled standard deviation

85. 85 First find x 1 and x 2 X 1 = (10.5+9.9+10.4+11.2+10.5)/5 = 10.5 X 2 = (10.1+10.3+10.2+10.3+10.4)/5 = 10.3  (X i1 -X 1 ) 2 = { (10.5-10.5) 2 + (9.9-10.5) 2 +(10.4-10.5) 2 + (11.2-10.5) 2 + (10.5-10.5) 2 }  (X i1 -X 1 ) 2 = 0.86  (X i2 -X 2 ) 2 = { (10.1-10.3) 2 + (10.3-10.3) 2 + (10.2-10.3) 2 + (10.3-10.3) 2 +(10.4-10.3) 2 }  (X i2 -X 2 ) 2 = 0.06 S p = {(  (X i1 -X 1 ) 2 +  (X i2 -X 2 ) 2 )/ (N 1 + N 2 –2)} 1/2 S P = {(0.86+0.06)/(5+5-2)} 1/2 = 0.34

86. 86 Standard Deviation of the Difference When multiple samples are analyzed by a proposed and standard methods, S d is the calculated standard deviation for the difference. S d = (  ( D i – D ) 2 / (N-1)) 1/2 S d is the standard deviation of the difference, D i is the difference between a result obtained by the standard method from that obtained by the proposed method for the same sample. D is the average of all differences.

87. 87 Example Mercury in multiple samples was determined using a standard method and a new suggested method. Six different samples were analyzed using the two procedures giving the following results in ppm Sample New Method Standard method 110.2 10.5 212.7 11.9 38.6 8.7 417.5 16.9 511.2 10.9 611.5 11.1 Find the standard deviation of the difference.

88. 88 It is wise to construct a table as below New Method Standard method D i 10.3 10.5 -0.2 12.7 11.9 +0.8 8.6 8.7 -0.1 17.5 16.9 +0.6 11.2 10.9 +0.3 11.5 11.1 +0.4 _____________________________ ________  D i = 1.8 D = 1.8/6 = 0.30

89. 89  (D i – D) 2 = { (-0.2-0.3) 2 + (+0.8-0.3) 2 + (- 0.1-0.3) 2 + (+0.6-0.3) 2 + (+0.3-0.3) 2 + (+0.4-0.3) 2 } = {0.25+0.25+0.16+0.09+0+0.01}  (D i – D) 2 = 0.76 S d = (  ( D i – D ) 2 / (N-1) ) 1/2 S d = (0.76/5) 1/2 = 0.39

90. 90 Propagation of Errors As seen earlier, each measurement has some uncertainty associated with it. During a process of calculations the uncertainty in the answer can be calculated from uncertainties in individual measurements.

91. 91 Calculation of error in the answer depends on whether the mathematical operation is a summation /subtraction or multiplication/division. It should be clear that in a process of calculating a final answer, as the number of mathematical operations increase, error will propagate.

92. 92 Addition and subtraction The absolute uncertainty in the answer S a can be evaluated from absolute uncertainties in individual numbers (b, c, d,.. ) as below: S a 2 = S b 2 + S c 2 + S d 2 + … Where, S a, S b, S c, and S d are absolute uncertainties in answer, b, c, and d (estimated standard deviation in answer, b, c, and d).

93. 93 Example Three samples were analyzed for iron content. The average percentage of iron in the first sample was 65.06, the second sample contained 56.13, and the third contained 62.68%. The estimated standard deviation of each of the three samples were + 0.07, + 0.01, and + 0.02%, respectively. What is the average iron content of the samples depending on these results?

94. 94 % Iron = {(65.06 + 0.07%) + (56.13 + 0.01%) + (62.68 + 0.02%)}/3 = (183.87/3) + S a % % iron = 61.29 + S a % S a 2 = (+ 0.07) 2 + ( +0.01) 2 + (+0.02) 2 = + 5.4x10 -3 S a = 7.3x10 -2 % Iron = 61.29 + 0.073% It is clear that we should retain two digits in the uncertainty as the answer is known to the nearest one hundredth (to get accurate number of significant figures). Therefore, the answer should be reported as: % Iron = 61.29 + 0.07%

95. 95 Lecture 6 Propagation of Error Tests of Significance

96. 96 Multiplication and Division The absolute uncertainty in calculations involving multiplication and division can not be estimated directly. The first step in such operations is to find the relative uncertainty in the answer from relative uncertainties in individual measurements as follows (S a 2 ) rel = (S b 2 ) rel + (S c 2 ) rel + (S d 2 ) rel + … Where, (S b ) rel = Estimated standard deviation in b (i.e. uncertainty in b)/ absolute value of b S a = Answer x (S a ) rel

97. 97 Find the result of the following calculation, using the correct number of significant figures: (2.23 + 0.01)*(3.508 + 0.007) = 7.82 + s a Since 2.23 is the key number (S a ) rel = {(+0.01/2.23) 2 + (+0.007/3.507) 2 } 1/2 (s a ) rel = 4.91*10 -3 S a = 7.823*4.91*10 -3 = 0.0384 Answer = 7.82 + 0.04

98. 98 Chloride in a 25 mL sample was determined by titration with a 0.1167 + 0.0002 M AgNO 3 solution. If the titration required an average AgNO 3 volume of 36.78 mL and the standard deviation in the volume was 0.04 mL, find the uncertainty in the number of mmol of chloride contained in 250 mL chloride sample.

99. 99 You should remember that the standard deviation is the absolute error in volume of AgNO 3 mmol chloride = mmol AgNO 3 Since Ag + reacts with Cl - in a 1:1 mole ratio mmol AgNO 3 = molarity AgNO 3 * Volume (mL) AgNO 3 mmol AgNO 3 = (0.1167 + 0.0002) ( 36.78 + 0.04) = 4.292 + ?

100. 100 Since this is a multiplication process we use the equation for relative uncertainty (S a 2 ) rel = ( +0.0002/0.1167) 2 + ( + 0.04/36.78) 2 (S a ) rel = +2.03x10 -3 S a = 4.292 x (+2.03x10 -3 ) = 8.71x10 -3 ( This is the uncertainty in 25 mL chloride) S a in 250 mL chloride = 10 x 8.71x10 -3 = +0.0871 mmol

101. 101 If we are to report the number of mmol chloride in 250 mL sample, the answer would be Answer = 42.92 + 0.0871 mmol The final answer should be 42.92 + 0.09 mmol since only two digits after the decimal points are allowed here to express actual uncertainty depending on the number of significant figures.

102. 102 Significant Figures and Propagation of Errors Unlike the argument above, we are now ready to indicate that the uncertainty (if properly calculated) defines the number of significant figures in an answer, regardless of the key number. For example, consider the mathematical value of the following calculation (73.1 + 0.2)(2.245+0.008) = 164.1 + S a

103. 103 The key number is 73.1 which has 3 significant figures, the answer is supposed to have 3 sig figures as well. This means that the answer is supposed to be 164+S a However, calculation of the absolute uncertainty gives S a = +0.7 Since the uncertainty is known to the tenth position, the answer should be reported to the tenth position too. The correct answer is 164.1+0.7 The number of significant figures is more than the key number !!

104. 104 Let us look at the same calculation above but change the uncertainty of the first term to 0.7, we get the following answer: (73.1 + 0.7)(2.245+0.008) = 164.1 + 1.677 It is clear that we have uncertainty associated with the integer itself, rather than the digits after the decimal point. If there is an uncertainty in an integer then the decimals are exceedingly uncertain that they can never be included. Therefore, the answer should be reported as: (73.1 + 0.2)(2.245+0.008) = 164 + 2

105. 105 Find the result of the following calculation, using the correct number of significant figures: (22.3 + 0.1)*(35.08 + 0.07) = 782.3 + s a (S a ) rel = {(+0.1/22.3) 2 + (+0.07/35.07) 2 } 1/2 (s a ) rel = 4.91*10 -3 S a = 782.3*4.91*10 -3 = 3.84 Answer = 782 + 4

106. 106 Look carefully at this example (101.1 + 0.9)(1.562 + 0.008) = 157.9 + 1.62 Since there is an uncertainty in an integer then the decimals are exceedingly uncertain that they can never be included. Therefore, the answer should be reported as Answer = 158 + 2 The number of significant figures is less than the key number !! Therefore, the uncertainty actually determines the number of significant figures

107. 107 The Confidence limit The standard deviation of a set of measurements provides an indication of the precision inherent in these measurements. However, no indication of how close the obtained result is from the accurate result can be deduced from the standard deviation. The confidence interval presents the range within which the accurate value might occur. This range is called the confidence interval. The probability that the true value occurs within this range is called the confidence level.

108. 108 Confidence limit = x + ts/N 1/2 Where; t is a statistical factor which depends on the confidence level and the number of degrees of freedom (number of experiments – 1). Confidence interval (range) = (x + ts/N 1/2 ) to (x - ts/N 1/2 )

109. 109

110. 110 The standard deviation for the analysis of a carbonate sample was 0.075% for the results 93.50, 93.58, and 93.43% carbonate. Find the confidence limit and range at the 95% confidence level. t 95% = 4.303 (from table) Confidence limit = x + ts/N 1/2 X = (93.50+93.58+93.43)/3 = 93.50 Confidence limit = 93.50 + 4.303x0.075/3 1/2 Confidence limit = 93.50 + 0.19% Range = (93.50 – 0.19%) – (93.50 +0.19%) Range = 93.31-93.69%

111. 111 Tests of Significance In this section we deal with two tests used for comparing two analytical methods, one is a new or proposed method and the other is a standard method. The two methods are compared in terms of whether they provide comparable precision ( the F test ), based on their standard deviations or variances. The other test ( t test ) tells whether there is a statistical difference between results obtained by the two methods.

112. 112 The F Test The precision of two methods could be compared based on their standard deviations using the F test which can be defined as the ratio between the variances ( the variance is the standard deviation squared ) of the two methods. The ratio should always be larger than unity. That is, the larger variance of either method is placed in the nominator. F = S 1 2 /S 2 2 > 1

113. 113 Where, S 1 2 > S 2 2 Values of F ( a statistical factor ) at different confidence levels which can be obtained from statistical F tables. When F calculated < F tabulated this is an indication of no statistical difference between precision or variances of the two methods.

114. 114

115. 115 Example In the analysis of glucose using a new developed procedure and a standard procedure, the variances of the two procedures were 4.8 and 8.3. If the tabulated F value at 95% confidence level at the number of degrees of freedom used was 4.95. Determine whether the variance of the new procedure differs significantly from that of the standard method

116. 116 F = S 1 2 /S 2 2 F = 8.3/4.8 = 1.7 3 (the subscript is because the answer is less than the key number) Since F calculated < F tabulated, there is no significant statistical difference between the variances of the two methods (i.e. there is no significant statistical difference between the precision of the two methods).

117. 117  

118. 118 Lecture 7 Data Handling Student t Test Q Test

119. 119 The Student t Test To check whether there is a significant statistical difference between the results of a new or proposed procedure and a standard one, the t test is used. As we did above, we calculate t and compare it to the tabulated value at the required confidence level and at the used degrees of freedom. There is no significant statistical difference between the results of the two methods when t calculated < t tabulated. There are three situations where the t test is applied:

120. 120

121. 121 a. When an Accepted Value is Known The t calc is calculated from the relation below and compared to t tab  = x + ts/N 1/2 or more conveniently, + t = (x -  ) N 1/2 /s

122. 122 Example A new procedure for determining copper was used for the determination of copper in a sample. The procedure was repeated 5 times giving an average of 10.8 ppm and a standard deviation of +0.7 ppm. If the true value for this analysis was 11.7 ppm, does the new procedure give a statistically correct value at the 95% confidence level? t tab = 2.776

123. 123 Substitution into equation below, we get: + t = (x -  ) N 1/2 /s + t = (10.8-11.7) 5 1/2 /0.7 + t = 2. 9 the t calc is larger than the t tab. Therefore, there is a significant statistical difference between the two results which also means that it is NOT acceptable to use the new procedure for copper determination.

124. 124 b. Comparison between two means When an accepted value is not known and the sample is analyzed using the new procedure and a standard procedure. Here, we have two sets of data, a standard deviation for each set of data and a number of data points or results in each set. Under these conditions, we use the pooled standard deviation for the two sets. The same equation in a is used but with some modifications. The t value is calculated from the relation (N 1 N s ) 1/2 (N 1 + N s ) 1/2 ( x 1 – x s ) S p + t = *

125. 125 Where, x 1 and x s are means of measurements using the new and standard methods. N 1 and N s are number of replicates done using the new and standard methods, respectively. S p is the pooled standard deviation. In such calculations it is wise to apply the F test first, and if it passes the t test is then applied.

126. 126 Example Nickel in a sample was determined using a new procedure where six replicate samples resulted in a mean of 19.65% and a variance of 0.4524. Five replicate analyses where conducted using a standard procedure resulting in a mean of 19.24% and a variance of 0.105. If the pooled standard deviation was +0.546, is there a significant difference between the two methods?

127. 127 First, let us find whether there is a significant difference in precision between the two procedures, by applying the F test F = 0.4524/0.105 = 4.31 The tabulated F value is 6.26. Since F calculated < F tabulated, then there is no significant statistical difference between the precision of the two procedures. Therefore, we continue with calculation of t test.

128. 128 + t = 1.2 3 The tabulated t value is 2.262. Since t calculated < t tabulated for nine degrees of freedom at 95% confidence level, we conclude that there is no significant statistical difference between the results of the two methods. (N 1 N s ) 1/2 (N 1 + N s ) 1/2 ( x 1 – x s ) S p + t = * (6*5) 1/2 (6+ 5) 1/2 ( 19.65 – 19.24) 0.546 + t = *

129. 129 c. The t Test with Multiple Samples Till now we have considered replicate measurements of the same sample. When multiple samples are present, an average difference is calculated and individual deviation from a mean difference is calculated and used to calculate a difference standard deviation, S d which is used in a successive step to calculate t. + t = DN 1/2 /S d S d = [  ( D i – D ) 2 / (N-1)] 1/2

130. 130 S d is the standard deviation of the difference, D i is the difference between a result obtained by the standard method subtracted from that obtained by the proposed method for the same sample. D is the average of all differences. Example Mercury in multiple samples was determined using a standard method and a new suggested method. Six different samples were analyzed using the two procedures giving the following results in ppm:

131. 131 Sample No. New Method Standard method 1.10.3 10.5 2.12.7 11.9 3.8.6 8.7 4.17.5 16.9 5.11.2 10.9 6.11.5 11.1 Find the standard deviation of the difference. If the two methods have comparable precisions, find whether there is any significant difference between the results of the two methods at the 95% confidence level. The tabulated t value for five degrees of freedom at 95% confidence level is 2.571.

132. 132 Sample No. New Method Standard method D i 1.10.3 10.5 -0.2 2.12.7 11.9 +0.8 3.8.6 8.7 -0.1 4.17.5 16.9 +0.6 5.11.2 10.9 +0.3 6.11.5 11.1 +0.4 _______________________________________________  D i = 1.8 D = 1.8/6 = 0.30

133. 133  ( D i – D ) 2 = { (-0.2-0.3) 2 + (+0.8-0.3) 2 + (-0.1-0.3) 2 + (+0.6-0.3) 2 + (+0.3-0.3) 2 + (+0.4-0.3) 2 } = {0.25+0.25+0.16+0.09+0+0.01}  ( D i – D ) 2 = 0.76 S d = (  ( D i – D ) 2 / (N-1) ) 1/2 S d = (0.76/5) 1/2 = 0.39 + t = 0.30x6 1/2 /0.39 =1.8 8 The calculated t value is less than the tabulated t value which means that there is no significant difference between the results of the two methods. + t = DN 1/2 /s d

134. 134 The Q Test In several occasions, when replicate experiments are done one of the data point may look odd or faulty. The analyst is confused whether to keep it or reject it. The Q test provides a means to judge if it should be retained or rejected. This can be done by applying the Q test equation: Q = a/w Where a is the difference between the suspected result and the result nearest to it in value, w is the difference between highest and lowest results.

135. 135 Once again, if the calculated Q value is less than the tabulated value, then the suspected data point should be retained. In contrast to F and t tests the statistical value of Q depends on the number of data points rather than the number of degrees of freedom.

136. 136

137. 137 Example In the replicate determination of gold you got the following results: 96, 99, 97, 94, 100, 95, and 72%. Check whether any point should be excluded at the 95% confidence level. Tabulated Q 95% = 0.568 for 7 observations Arrange results: 72, 94, 95, 96, 97, 99, 100 Q = a/w Q calc = (94-72)/(100 - 72) = 0.79 Q calc > Q tab The point 72% should be rejected.

138. 138 Example In the replicate determination of gold you got the following results: 96, 99, 97, 94, 100, 95, and 88%. Check whether any point should be excluded at the 95% confidence level. Tabulated Q 95 % = 0.568 for 7 observations. Arrange results: 88, 94, 95, 96, 97, 99, 100 Solution Q = a/w Q calc = (94-88)/(100-88) = 0.50 Q calc < Q tab The point 88% should be retained.

139. 139 Lecture 8 Linear Least Squares Stoichiometric Calculations

140. 140 Linear Least Squares Frequently, an analyst constructs a calibration curve using several standards and draws a straight line among the data points in the graph. In many cases, the line does not cross all points and the analyst starts judging where the straight line should pass. Human judgment is not perfect and, unfortunately, may be biased. The method of linear least squares is a mathematical method that help us choose the best path of the straight line.

141. 141

142. 142 It is well known that the equation of a straight line is mathematically represented by y = mx + b Where m is the line and b is the line intercept, x and y are variables. The slope, m, can be calculated from the relationship m = {  x i y i – [(  x i  y i )/n]}/{  x i 2 – [(  x i ) 2 /n]} b = y – mx x, y are average values of x i and y i.

143. 143 The standard deviation of any of the y i points (S y ) is given by the relation S y = {([  y i 2 – ((  y i ) 2 /n)] – m 2 [  x i 2 – ((  x i ) 2 /n)])/(n-2)} 1/2 The uncertainty in slope can then be calculated from S y as follows S m = {S y 2 / [  x i 2 – ((  x i ) 2 /n)]} 1/2

144. 144 Example Using the following data and without plotting, if the fluorescence of a riboflavin sample was 15.4 find its concentration.

145. 145 (  x i ) 2 = 2.250 x = (  x i )/n = 1.500/5 = 0.300 y = (  y i )/n = 83.6/5 = 16.7 2 m = {  x i y i – [(  x i  y i )/n]}/{  x i 2 – [(  x i ) 2 /n]} Substitution in the equation above gives m = {46.6 – [(1.500*83.6)/5]}/ {0.850 –[(2.250/5]} m = 53.7 5

146. 146 This Excel plot gives the same results for slope and intercept as calculated in the example.

147. 147 To calculate b we use the equation b = y – mx b = 16.7 2 – 53.7 5 *0.300 = 0.6 0 Now we are ready to calculate the sample concentration y = mx + b 15.4 = 53.75*x + 0.60 x = 0.275 ng/L

148. 148 Correlation Coefficient (r) When the points that are supposed to be on a straight line are scattered around that line then one should estimate the correlation between the two variables. The correlation coefficient serves as a measure for the correlation of these two variables. This can be very important if correlation between results obtained by a new method and a standard method is required. r = {n  x i y i – (  x i  y i )}/ {[n  x i 2 – (  x i ) 2 ][n  y i 2 – (  y i ) 2 ]} 1/2

149. 149 Two graphs with the same straight line equation??

150. 150 Example Calculate the correlation coefficient of the data : Solution First we find  y i 2 and (  y i ) 2 from the table in previous example  y i 2 = 2554.66 (  y i ) 2 = 6988.96

151. 151 Substituting in the correlation coefficient equation above: r = {5*46.6-(1.500*83.6)} / {[5*0.850 – 2.250][5*2554.66-6988.96]} 1/2 r = 1.00 The correlation coefficient occurs between + 1. As the correlation coefficient approaches unity, correlation increases and exact correlation occurs when r = 1. An r value less than 0.90 is considered bad while that exceeding 0.99 is considered excellent.

152. 152 Currently, many scientists prefer to use the square of the correlation coefficient, r 2 rather than r, to express correlation. Evidently, the use of r 2 is a more strict criterion as a smaller value is always obtained when fractions are squared.

153. 153 Detection Limits All instrumental methods have a degree of noise associated with the measurement that limits the amount of analyte that can be detected. 1.Detection Limit is the lowest concentration level that can be determined to be statistically different from an analyte blank.

154. 154 When a graphical display of results is obtained, the detection limit of the instrument can be defined as the concentration of analyte resulting in a signal that is twice as the peak to peak noise (the distance between the two dashed lines in the schematic below).

155. 155 Peak-to-peak noise level as a basis for detection limit. A “detectable” analyte signal would be 12 divisions above a line drawn through the average of the baseline fluctuations

156. 156 2.Detection Limit is the concentration that gives a signal three times the standard deviation of the background signal. To calculate the detection limit: a.Find the average of the blank signal b.Find the standard deviation of the blank c.Find the net analyte signal analyte conc. * 3*s analyte signal DL =

157. 157 Example A blank solution in a colorimetric analysis resulted in absorbance readings of 0.000, 0.008, 0.006, and 0.003. A 1 ppm standard solution of the analyte resulted in a reading of 0.051. Calculate the detection limit. The standard deviation of the four data points of the blank can be calculated to be + 0.0032 and the mean of the blank is 0.004

158. 158 The net reading of the standard = 0.051 – 0.004 = 0.047 The detection limit is the concentration which results in three times the standard deviation (3 x 0.0032 = 0.0096). Detection limits = 1 ppm x 0.0096/0.047 = 0.2 ppm The absorbance reading of the least detectable concentration = 0.0096 + 0.004 = 0.014