Presentation is loading. Please wait.

Presentation is loading. Please wait.

1/48 Chapter 6: Random Errors in Chemical Analysis.

Similar presentations


Presentation on theme: "1/48 Chapter 6: Random Errors in Chemical Analysis."— Presentation transcript:

1 TMHsiung@2014 1/48 Chapter 6: Random Errors in Chemical Analysis

2 TMHsiung@2014 2/48 Contents in Chapter 06 1.The Nature of Random Errors 1)Random Errors and Their Mathematical Equations 2)Central Limit Theorem 2.Gausian (Normal) Distribution 1)Define Gausian Distribution 2)Z Value Transformation 3)Z value Applications 4)Constructing Gaussian Curve From Experimental Data 3.Standard Deviation of Calculated Results 4.Significant Figures

3 TMHsiung@2014 3/48 1.The Nature of Random Errors 1)Random Errors and Their Mathematical Equations Random (Indeterminate, Statistic) error: Errors affecting the precision (uncertainty), caused by uncontrollable variables. Positive and negative (  ) fluctuation occur with approximate equal frequency. Population: The set of infinite objects in the system being investigated. Sample: The finite members of a population that we actually collect and analyze.

4 TMHsiung@2014 4/48 Mean (Average) Standard deviation - For population: - For sample: ● Spread (range, w): The difference between extreme values in a set of d

5 TMHsiung@2014 5/48 Pooled standard deviation: For several subsets of data, estimating standard deviation by pooling (combining) the data. Variance = s 2 * variance are additive Relative standard deviation (RSD) by 100% (also called coefficient of variation, CV):

6 TMHsiung@2014 6/48 Degree of freedom (dof) In statistics, the number of independent observations on which a result is based. -For standard deviation: dof = n-1, n is munber of measurement. -For pooled standard deviation: dof = N – M, N:Total number of measurements M:Number of the subset.

7 TMHsiung@2014 7/48 2)Central Limit Theorem i)Definition For the probability distribution plot for the frequency of individual values. The measurements subject to indeterminate errors arise a normal (Gaussian) distribution. The sources of individual error must be independent. The individual error must have similar magnitude (no one source of error dominates the final distribution).

8 TMHsiung@2014 8/48 Combination of uncertaintyMagnitude of Combination Frequency of combinations Relative frequency +U 1 +U 2 +U 3 +U 4 +4U11/16=0.0625 -U 1 +U 2 +U 3 +U 4 +U 1 -U 2 +U 3 +U 4 +U 1 +U 2 -U 3 +U 4 +U 1 +U 2 +U 3 -U 4 +2U44/16=0.250 -U 1 -U 2 +U 3 +U 4 +U 1 +U 2 -U 3 -U 4 +U 1 -U 2 +U 3 -U 4 -U 1 +U 2 -U 3 +U 4 -U 1 +U 2 +U 3 -U 4 +U 1 -U 2 -U 3 +U 4 066/16=0.375 +U 1 -U 2 -U 3 -U 4 -U 1 +U 2 -U 3 -U 4 -U 1 -U 2 +U 3 -U 4 -U 1 -U 2 -U 3 +U 4 -2U44/16=0.250 -U 1 -U 2 -U 3 -U 4 -4U11/16=0.0625 (Frequency of Combinations of four equal-sized uncertainty, u) ii)Simulated Central Limit Theorem

9 TMHsiung@2014 9/48 4 random uncertainties

10 TMHsiung@2014 10/48 10 random uncertainties Infinite number of random uncertainties -A Gaussian (normal) distribution curve. -Symmetrical about the mean.

11 TMHsiung@2014 11/48 2.Gausian (Normal) Distribution 1)Define Gausian Distribution A “bell-shaped” probability distribution curve for measurements showing the effect of random error, which encountered continuous distribution. The equation for Gaussian distribution:

12 TMHsiung@2014 12/48 IF: Same Mean, Different SD 982_Ch04 _Treat_Ra ndDistr_D ata.xls

13 TMHsiung@2014 13/48 IF: Different Mean, Same SD 982_Ch04 _Treat_Ra ndDistr_D ata.xls

14 TMHsiung@2014 14/48 IF: Different Mean, Different SD 982_Ch04 _Treat_Ra ndDistr_D ata.xls

15 TMHsiung@2014 15/48 2)Z value Transformation For x-axis: transform x to z by: For y-axis: transform to f(z) by :

16 TMHsiung@2014 16/48 Now, the Gaussian Curve Always Consistent 982_Ch04 _Treat_Ra ndDistr_D ata.xls

17 TMHsiung@2014 17/48 教戰守則 1. 經過 z transform, Gaussian curve 長得一模一樣, 此時 x-axis 的單位為 σ 。 2. 看到標示為 x ,單位為 σ 時,代表已經經過 z transform 。 3.Relative frequency (y) at mean (x), 約為 0.4 。 4.The entire area of the Gaussian curve is 1.

18 TMHsiung@2014 18/48 3)Z value Applications Area (probability, percentage) in defined z interval

19 TMHsiung@2014 19/48 Step 1: Raw data collection Step 2:Arrange the data in order from lowest to highest Step 3:Condense the data by grouping them into cells Step 4:Pictorial representation of the frequency distributions Step 5:Estimating the σ from s Step 6:Plotting relative frequency versus x or z 4)Constructing Gaussian Curve From Experimental Data i)General procedure

20 TMHsiung@2014 20/48 Example: Replicate Data for the Calibration of a 10 mL Pipet 982_Ch04 _Treat_Ra ndDistr_D ata.xls

21 TMHsiung@2014 21/48 Cont’d 982_Ch04 _Treat_Ra ndDistr_D ata.xls

22 TMHsiung@2014 22/48 982_Ch04 _Treat_Ra ndDistr_D ata.xls

23 TMHsiung@2014 23/48 Number in rangePercentage in range Relative frequency versus x 962_Ch04_Trea t_RandDistr_D ata.xls X-axis 刻度為體積之間隔

24 TMHsiung@2014 24/48 Matching Histogram with Gaussian curve

25 TMHsiung@2014 25/48 Z Transformed Gaussian Curve 982_Ch04 _Treat_Ra ndDistr_D ata.xls

26 TMHsiung@2014 26/48 4.Standard Deviation of Calculated Results (for Random Errors)

27 TMHsiung@2014 27/48 1)Sum or Difference Example 1: The calibration result of class A 10 mL pipet showed that the marker reading is 9.992  0.006 mL. When it is used to deliver two successive volumes. What is the absolute and percent relative uncertainties for the total delivered volume. Solution: Total volume = 9.992 mL + 9.992 mL = 19.984 mL Ans:19.984(  0.008) mL

28 TMHsiung@2014 28/48 Example 2: For a titration experiment, the initial reading is 0.05(  0.02) mL and final reading is 17.88 (  0.02) mL. What is the volume delivered? Solution: Delivered volume = 17.88 mL – 0.05 mL = 17.83 mL Ans:17.83(  0.03) mL

29 TMHsiung@2014 29/48 Example 1: The quantity of charge Q = I x t. When a current of 0.15  0.01 A passes through the circuit for 120  1 s, what is the total charge? Solution: Total charge 0.15 A x 120 s = 18 C Ans:18(  1) C 2)Product or Quotient

30 TMHsiung@2014 30/48 3)Mixed operations

31 TMHsiung@2014 31/48 Ans:1.9 (  0.1)x10 –4 M 4)Exponents and logarithms Example: The pH of a solution is 3.72  0.03, what is the [H 3 O + ] of this solution? Solution: [H 3 O + ] = 10 –pH y = 10 –3.72 =1.9 1 x10 –4 y(  s y ) = 1.9 1 (  0.1 3 )x10 –4

32 TMHsiung@2014 32/48 4.Significant Figures 1)Statement of significant figures The digits in a measured quantity, including all digits known exactly and one digit (the last figure) whose quantity is uncertain. The more significant digits obtained, the better the precision of a measurement. The concept of significant figures applies only to measurements. The Exact values (e.g., 1 km = 1000 m) have an unlimited number of significant figures.

33 TMHsiung@2014 33/48 012 3 Recording: 1.51 or 1.52 or 1.53 012 3 Recording: 1.4 or 1.5 or 1.6 2)Recording with significant figures 2 significant figure, 1 certain, 1 uncertain 3 significant figure, 2 certain, 1 uncertain

34 TMHsiung@2014 34/48 3)Rules for Zeros in Significant Figures Zeros between two other significant digits are significant e.g. 10023no. of sig. fig.: 5 A zero preceding a decimal point is not significant e.g., 0.10023no. of sig. fig.: 5 Zeros preceding the first nonzero digit are not significant e.g. 0.0010023no. of sig. fig.: 5 Zeros at the end of a number are significant if they are to the right of the decimal point e.g. 0.1002300no. of sig. fig.: 7

35 TMHsiung@2014 35/48 Zeros at the end of a number may or may not be significant if the number is written without a decimal point e.g. 92500no. of sig. fig.: N/A Scientific notation is required: e.g. 9.25x10 4 no. of sig. fig.: 3 e.g. 9.250x10 4 no. of sig. fig.: 4 e.g. 9.2500x10 4 no. of sig. fig.: 5

36 TMHsiung@2014 36/48 2.Significant Figures in Arithmetic 1)Rules for rounding off numbers i)If the digit immediately to the right of the last sig. fig. is more than 5, round up. ii)If the digit immediately to the right of the last sig. fig. is less than 5, round down. iii)If the digit immediately to the right of the last sig. fig. is 5 followed by nonzero digits, round up. iv)If the digit immediately to the right of the last sig. fig. is 5 round up if the last sig. fig. is odd. round down if the last sig. fig. is even. v)If the resulting number has ambiguous zeroes, it should be recorded in scientific notation to avoid ambiguity.

37 TMHsiung@2014 37/48 Examples: 35.76 in 3 sig. fig. is35.8 35.74 in 3 sig. fig. is35.7 24.258 in 3 sig. fig. is24.3 24.35 in 3 sig. fig. is 24.4 (rounding to even digit) 24.25 in 3 sig. fig. is24.2 (rounding to even digit) 13,052 in 3 sig. fig. is1.31 x 10 4

38 TMHsiung@2014 38/48 2)Addition and Subtraction: The reported results should have the same number of decimal places as the number with the fewest decimal places Example 2: MW of KrF 2 18.9984032 (F) + 18.9984032 (F) + 83.798 (Kr) 121.7948064 Ans: 121.795 g/mol Ans: 112.2 m 49.146 m + 72.13 m – 9.1 m 112.176 m Example 1:

39 TMHsiung@2014 39/48 Note: To avoid accumulating “round-off ” errors in calculations: 1)Go through all calculation by calculator, then rounding on the final answer OR 2)Retaining one extra insignificant figure (a subscribe digit) for intermediate results, then rounding on the final answer

40 TMHsiung@2014 40/48 Example 3: Write the answer with correct number of digits: 2.432x10 6 + 6.512x10 4 – 1.227x10 5 = ? Ans: 2.374x10 6 2.432 x 10 6 + 0.06512 x 10 6 – 0.1227 x 10 6 2.37442 x 10 6

41 TMHsiung@2014 41/48 3)Multiplication and Division: The reported results should have no more significant figures than the factor with the fewest significant figures Example 1: 1.827 m × 0.762 m = ? 1.827 m x 0.762 m = 1.392174 m 2 = 1.39 m 2 Ans: 1.39 m 2 Example 2: (4.3179 x 10 12 )(3.6x10 –19 ) = 1.554444x10 –6 = 1.6x10 –6

42 TMHsiung@2014 42/48 Stoichiometric coefficients in a chemical formula, and unit conversion factors etc, have an infinite number of significant figures. Example: Results of four measurements: 36.4 g, 36.8 g, 36.0 g, 37.1 g. What is the average? Solution: (36.4+36.8+36.0+37.1)/4 = 36.6 Ans: 36.6 g

43 TMHsiung@2014 43/48 4)Logarithms and Antilogarithms i)Mathematic terms For exponential expression, e.g. 4.2 x 10 –2 : “4.2” is called the coefficient, “–2” is called the exponent. For logarithm operation, e.g. log (4.2 x 10 –2 ) = –1.38 “–1” is called the characteristic, the integer part “38” is called the mantissa, the decimal part

44 TMHsiung@2014 44/48 ii)Logarithms operation Example 1: log(56.7 x 10 6 ) = ? Solution: log(5.67 x 10 7 ) = 7 + log(5.67) = 7.754 3 sig. fig. Example 2: log(0.002735) = ? Solution: log(0.002735) = – 2.5630 4 sig. fig. Or log(0.002735) = 2.735x10 –3 = (– 3) + log(2.735) = (–3) + (0.4370) = – 2.5630 Mantissa remain equal number of sig. fig. as the original digit in coefficient

45 TMHsiung@2014 45/48 Example 3: Percent transmittance (%T) is related to the absorbance (A) by the equation: A = –log (%T/100). What is correct digits of A if %T is 72.9. Solution: A = –log (%T/100) = –log (0.729) = – (– 0.137) = 0.137 Or log (0.729) = – log(7.29x10 –1 ) = – [(–1) + log (7.29)] = – [(–1) + (0.863)] = 0.137 Example 4: The pH is defined as pH = –log [H 3 O + ]. If the [H 3 O + ] is 3.8 x 10 –2 M, what is the pH of the solution? Solution: pH = –log [H 3 O + ] = –log (3.8 x 10 –2 ) = –[( – 2) + log(3.8)] =– (–1.42) = 1.42

46 TMHsiung@2014 46/48 iii)Antilogarithms operation Example 1: antilog(2.671) = ? Solution: antilog(2.671) = 10 2.671 = 10 0.671 x10 2 = 4.69 x 10 2 Example 2: What is the correct digits of %T if the absorbance is 0.931? Solution: Antilog (–0.931) = 10 –0.931 = 0.117 %T = 0.117 x 100% = 11.7% Final digit has same number of sig. fig. as number of digits in mantissa

47 TMHsiung@2014 47/48 Example 3: If the pH is 10.3, what is the [H 3 O + ]? Solution: [H 3 O + ] = 10 –pH = 10 –10.3 = 5 x 10 –11 Or 10 –10.3 = 10 0.7 x 10 –11 = 5 x 10 –11 Example 4: If the pH is 2.52, what is the [H 3 O + ]? Solution: [H 3 O + ] = 10 –pH = 10 –2.52 = 3.0 x 10 –3 Or 10 –2.52 = 10 0.48 x 10 –3 = 3.0 x 10 –3

48 TMHsiung@2014 48/48 Homework (Due 2014/3/6) Skoog 9 th edition, Chapter 06 Questions and Problems 6-1 6-4 6-5 6-9 (a) (c) (e) 6-11 (a) (c) 6-15 End of Chapter 06


Download ppt "1/48 Chapter 6: Random Errors in Chemical Analysis."

Similar presentations


Ads by Google