Error Analysis. Table 1. Tolerances of Class A Pipets and Volumetric Flasks

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1 Error Analysis Significant Figures in Calculations Most lab report must have an error analysis. For many experiments, significant figure rules are sufficient. Remember to carry at least one extra significant figure to avoid round- off error through intermediate calculations. Non-significant figures are often written smaller or like "subscripts" to avoid confusion. For example, ± would be written as in keeping with the four significant figures. On the other hand, ± would be written Retain at least one insignificant figure to avoid introducing round-off errors into subsequent calculations using Of course the final result is reported to the proper number of significant figures. Keeping track of significant figures in long calculations is easy. Just underline the insignificant digit for all the intermediate and final results using a pen or pencil. To find the uncertainties and approximate number of significant figures when using volumetric glassware use Table 1. Table 1. Tolerances of Class A Pipets and Volumetric Flasks Pipets Flasks 1 ml ± ml ± The significant figure rules are: SF Rule 1: In multiplication and division the number of significant figures in the result is the same as the smallest number of significant figures in the data. SF Rule : In addition and subtraction the number of decimal places in the result is the same as the smallest number of decimal places in the data. SF Rule 3: The number of significant figures in the mantissa of log x is the same as the number of significant figures in x. Use the same rule for ln x. (In log = -.374, the mantissa is the.374 part and the characteristic is.) SF Rule 4: The number of significant figures in 10 x is the number of significant figures in the mantissa of x. Use the same rule for e x.

2 Example 1: Concentration Calculations: A solution is made by transferring 1 ml of a M solution, using a volumetric pipet, into a 00-mL volumetric flask. Calculate the final concentration. Solution: The volume of the flask has 5 significant figures and all other values have 4. The calculations all involve multiplication and division, so the final answer should be expressed with 4 significant figures M / = M = M Example : Logs: The equilibrium constant for a reaction at two different temperatures is 0.03 at 98. and at 353. K. Calculate ln(k /k 1 ). Solution: Both k s have significant figures, so k /k 1 should also have significant figures: k /k 1 = Then using SF Rule 3 shows that ln k /k 1 should have significant figures in the mantissa: ln k /k 1 = ln =.68 7 =>.69 as the final result Example 3: Antilogs: The rate of a reaction depends on temperature as ln k = ln A E a/ RT. Using curve fitting it was found that ln A = Calculate A Solution: The result is e. = The mantissa has only significant figures, so the result should only have two significant figures (SF Rule 4) , or just Example 4: Antilogs: The rate of a reaction depends on temperature as ln k = ln A E a /RT. Using curve fitting it was found that ln A = and E a = 8. 6 kj/mol. Calculate k at T = K. Solution: The result for ln k has significant figures: both ln A and E a have 3 significant figures (SF Rule 1), but the difference has only one past the decimal point: ln k = /[(8.314)(98.15)] = = The mantissa has only 1 significant figure. Then solving for k and using SF Rule 4 gives k = e = or finally just 0.

3 Propagation of Errors Significant figure rules are sufficient when you do not have good estimates for the measurement errors. If you do have good estimates for the measurement errors then a more careful error analysis based on propagation of error rules is appropriate. Least squares curve fitting provides very good estimates for uncertainties. For error analysis with the slope or intercept from least squares curve fitting, a little more care is justified than is provided by significant figure rules. Use propagation of error rules to find the error in final results derived from curve fitting. The propagation of error rules listed below come from Table 3-1 of our text. The text uses e to denote the error or uncertainty in Chapter 3 and then σ and s for the standard deviation and its estimate in Chapter 4. For now, all of these symbols are interchangeable. The variance of x is the square of its standard deviation, σ x. (or error e or uncertainty). Rule 1: Variances add on addition or subtraction. For z = x ± y, e z = e x + e y Rule : Relative variances add on multiplication or division. For z = x y or z = x/y, e e z ex y = + z x y Rule 3: The variance in ln x is equal to the relative variance in x and the variance in log x is the (relative variance in x)/(ln 10) = (relative variance in x)/(.303). e For z = ln x, x 1 ex ez = and for z = log x, where ln 10 =.303, ez = x ln10 x ( ) Rule 4: The relative variance in e x is equal to the variance in x and the relative variance in 10 x is equal to the (variance in x)(.303). For z = e x e, z = ex and for z = 10 x e, z = ( ln10) ex z z Rule 5: In calculations with only one error term, you can work with the error itself rather than the variance. Rule 6: The variance of an average of N numbers, each with variance e, is e /N. (The standard deviation in the average improves as (1/N)). These rules all follow from the general formula from calculus for error propagation F F F ef = ex + e y + ez x y z F F F or σ F = σ x + σ y + σ z x y z

4 where one is calculating some function F which depends upon experimental quantities x, y, z, which have random errors e x, e y, e z, associated with them (see Appendix B). KNOW rule 1 rule 1 rule 3 rule rule 3 rule rule 4 rule ` y = constant x e y = constant e x rule 4 Note: if one requires an absolute error e x it is easier to work with the relative error e x /x rather than percent error since % e = 100 e x /x Example 5: Subtraction: If Z = A B with A = ± 0.00 and B = 1.34 ± 0.0 calculate Z. Find the uncertainty in the result. Solution: Work with absolute variances (Rule 1) and note that variances always add (i.e. the error always builds up) [calculus note: ( Z/ A) = ( Z/ B) = 1 => just add variances, take ]: [variance in A = (0.00) ] + [variance in B = (0.0) ] variance in result = => uncertainty = ( ) = Result: ( ± 0.00) (1.34 ± 0.0) = ± = 16.1 ± 0.0 Example 6: Multiplication: The result of an experiment is given by Z = slope Cp. Let slope = m = 13. ±.4 and Cp = ± Find the uncertainty in the result. Solution: The relative variance of the result is the sum of the relative variances of the data (Rule ). [calculus note: (e Z /Z) = [( Z/ m) e m + ( Z/ Cp) e Cp ]/Z where Z/ m = Cp and Z/ Cp = m so (e Z /Z) = [(Cp) e m + m e Cp ]/(mcp) = (e m /m) + (e Cp/Cp) and the error is just e Z = Z { (e m /m) + (e Cp/Cp) } never requiring a % error calculation] [relative variance in m = (.4/13.) = ( ) = ] + [relative variance in Cp = (0.031/4.184) = ( ) = ] relative variance in product = relative uncertainty in product = ( ) = or.0 84 % from above calculus: e Z = (13.)(4.184) {(.4/13.) + (0.031/4.184) } = Result: (13. ±.4) (4.184 ± 0.031) = ±.0 84 % = 515 ± 11

5 Example 7: Logs: The equilibrium constant for a reaction at two different temperatures is K 1 = 0.03 ± at 98. and K =0.473 ± at 353. K. Calculate Z = ln(k /K 1 ) and find the uncertainty in the result. Solution: Just like Example 6 the relative variance in K /K 1 is the sum of the relative variances [calculus note: e Z = [( Z/ K 1 ) e K1 + ( Z/ K ) e K ] where Z/ K 1 = -1/K 1 and Z/ K = 1/K so e Z = (-e K1 /K 1 ) + (e K /K ) and e Z = ]: [relative variance in K = (0.0007/0.03) = ( ) = ] + [relative variance in K 1 = (0.006/0.473) = ( ) = ] Using Rule 3 shows that absolute variance in ln K /K 1 is the relative variance in K /K 1. variance in ln K /K 1 = relative variance in K /K 1 = uncertainty in result = ( ) = from above calculus: e Z = {(0.0007/0.03) + (0.006/0.473) } = Result: ln(k /K 1 ) = ln => ± =.69 ± 0.03 See Example for the significant figure version of this error analysis. Example 8: Antilogs: The rate of a reaction depends on temperature as ln k = ln A E a /RT. Using curve fitting it was found that ln A = ± Calculate A and find the uncertainty. Solution: ln A = x => A = e x = e = The relative variance of the result is the absolute variance of A (Rule 4) [calculus note: e A = (da/dx) e x where da/dx= e x = A => e A = A e x ]: relative variance of e = variance of A = (0.041) = relative uncertainty in result = ( ) = or % from above calculus: e A = e = Result: ± % = ( ± ) 10 4 = (1.94 ± 0.08) 10 4 See Example 3 for the significant figure version of this error analysis. Example 9: Multiplication with 'certain' numbers: The result of a calculation is Z = (slope R), where R is the gas constant in J K -1 mol -1. Let slope = m = 1.3 ± 0.0. Find the uncertainty in the result. Solution: Since R is known to several more significant figures than the slope, the uncertainty in R will add very little to the error in the final result. Therefore, R is 'certain' for this calculation. Rules 5 and 1` apply since only the slope is in error. Therefore, the error in the final result is then just the standard deviation in the slope multiplied by R [calculus note: e Z = (dz/dm) e m where = dz/dm = R => e Z = R e m ]: Result: slope R = 1.3 ± = ± = 10. ± 0.

6 Example 10: The Inverse of the Slope or Intercept: The result of an experiment is the inverse of the intercept from a graph, Z = 1/b. Let b = 0.53 ± Find the uncertainty in the result. Solution: The relative variance in the result is equal to the relative variance in the intercept (Rule ). We can also work directly in terms of the error (Rule 5) [calculus note: e Z = (dz/db) e b where dz/db = -1/b => e Z = e b /b 4 or, since Z = 1/b, (e Z /Z) = (e b /b) and e Z = Z (e b /b) = e b /b ] : relative error in b = 0.043/0.53 = or 8. 1 % relative uncertainty of result = relative error in b = 8. 1 % from above calculus: e Z = 0.043/(0.53) = Result: 1/b = 1/0.53 = ± 8. 1 % = ± = 1.91 ± 0.16 Example 11: Division and Subtraction: The term Z = (1/T 1/T 1 ) is a very common factor in many equations. For T 1 = 98. ± 0. K and T = 353. ± 0. K calculate (1/T 1/T 1 ). Find the uncertainty in the result. Solution: Do not fret about a multi-step problem, just do one step at a time. First, get the uncertainty in 1/T and 1/T 1. Since both of these are divisions the relative variance of 1/T is just the relative variance of T (Rule ). Then convert to absolute variance to calculate the error in (1/T 1/T 1 ) using Rule 1 [calculus note: e Z = ( Z/ T 1 ) e T1 + ( Z/ T ) e T where ( Z/ T) e T = e T /T 4 since Z/ T = 1/T ]: relative variance in 1/T = relative variance in T = (0./353.) = relative variance in 1T 1 = relative variance in T 1 = (0./98.) = [variance in 1/T = (1/353.) = ] + [variance in 1/T 1 = (1/98.) = ] variance in (1/T 1/T 1 ) = from above calculus: variance in (1/T 1/T 1 ) = (0.) /(353.) 4 + (0.) /(98.) 4 = uncertainty in result = ( ) = Result: ± = (-5. ± 0.03) 10-4 Note that there are only 3 significant figures due to the rules for decimals in (1/T 1/T 1 ). Example 1: A Multi-Step Problem: An example of a more realistic problem is the temperature dependence of the equilibrium constant. We wish to evaluate ΔH, knowing K 1, K, R, T 1, and T in the equation: ln(k /K 1 ) = ΔH/R (1/T 1/T 1 ) where K 1 = 0.03 ± at 98. and K = ± at 353. K with T = ± 0. K (The same data as Examples 7 and 11). Find the uncertainty in the result. Solution: Solving for ΔH: ΔH = 8.314[ln(K /K 1 )]/(1/T 1/T 1 ) = kj/mol.

7 We already know the uncertainty in ln(k /K 1 ) from Example 7, ± , and the uncertainty in (1/T 1/T 1 ) from Example 11, ( ± ) R is a certain number. The relative variance in ΔH is then just the sum of the relative variances: relative variance ΔH = ( / ) + ( / ) = relative uncertainty of ΔH = ( ) = or % uncertainty of ΔH = ( )(4.7 8 ) = kj/mol Result: ΔH = 4.8 ± 0.46 => 4.8 ± 0.5 kj/mol Note that in the first line that the relative variance of ΔH has significant figures even though the two relative variances being added both have only 1. Again following the rules for decimals when adding, the terms are ( ) 10-4 which exceeds one and gives significant figures in the result.