Significant Figures in Calculations Error Analysis Every lab report must have an error analysis. For many experiments, significant figure rules are sufficient. For a brush up on significant figure rules, see your General Chemistry or Analytical text. Remember to carry at least one extra significant figure to avoid round off error through intermediate calculations. Non-significant figures are written smaller or like "subscripts" to avoid confusion. Insignificant figures can also be underlined. For example, 0.1234 0.001 or 0.002 would be written 0.1234 or 0.1234. On the other hand, 0.1234 0.003 would be written 0.123 or 0.123. Keeping track of significant figures in long calculations is easy. Just underline the insignificant digit in your Excel spreadsheets 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. Capacity Tolerances for Class A Volumetric Glassware. Pipets Sig. Figs Flasks Sig. Figs 1 ml. ±0.006 3 10 ml. ±0.02 4 2 0.006 3 25 0.03 3 5 0.01 3 50 0.05 3 10 0.02 4 100 0.08 4 15 0.03 4 200 0.10 4 20 0.03 4 250 0.12 4 25 0.03 4 1000 0.30 4 A 10-ml pipet is listed as 10.00 0.02, which is close enough to 4 significant figures, 10.00 ml. But a 1-ml pipet is listed as 1.000 0.006, which is really only 3 significant figures, 1.00 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 2: 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 4.23x10-3 = -2.374, the mantissa is the.374 part.) 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.
Example 1: Concentration Calculations: A solution is made by transferring 1 ml of a 0.12453 M solution, using a volumetric pipet, into a 200-ml volumetric flask. Calculate the final concentration. Solution: The 1-mL volumetric pipet has 3 significant figures; all the other values have 4. The calculations all involve multiplication and division, so the final answer should be expressed with 3 significant figures. 1.00 x 0.12453 M / 200.0 = 0.0006226 M = 6.22x10-4 M Example 2: Logs: The equilibrium constant for a reaction at two different temperatures is 0.0322 at 298.2 and 0.473 at 353.2 K. Calculate ln(k 2 /k 1 ). Solution: Both k s have 2 significant figures, so k 2 /k 1 should also have 2 significant figures: k 2 /k 1 = 13.89. Then using SF Rule 3 shows that ln k 2 /k 1 should have 2 significant figures in the mantissa: ln k 2 /k 1 = ln 13.89 = 2.631 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 = 9.874. Calculate A. Solution: The result is e 9.874 = 1.9427x10 4. The mantissa has only 2 significant figures, so the result should only have two significant figures (SF Rule 4): 1.94x10 4, or just 1.9x10 4. 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 = 9.874 and E a = 28.26 kj/mol. Calculate k at T = 298.15 K. Solution: The result for ln k has 2 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: limiting term on subtraction ln k = 9.874 28.26x10 3 /8.314/298.15 = 9.874 11.40 = -1.5266 The mantissa has only 1 significant figure. Then solving for k and using SF Rule 4 gives k= e -1.5266 = 0.217 or finally just 0.2. 2
Propagation of Errors Significant figure rules are sufficient when you don't have god 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 are listed below. The variance of x, s(x) 2, is the square of the standard deviation. Rule 1: Variances add on addition or subtraction. s(z) 2 = s(x) 2 +s(y) 2 Rule 2: Relative variances add on multiplication or division. s(z)2 z 2 = s(x)2 x 2 +s(y)2 y 2 Rule 3: The variance in ln(x) is equal to the relative variance in x. s(z) 2 = s(x)2 x 2. The variance in log(x) is the (relative variance in x)/(2.303) 2. Rule 4: The relative variance in e x is equal to the variance in x. s(z)2 z 2 = s(x)2. The relative variance in 10 x is equal to the (variance in x)(2.303) 2. Rule 5: In calculations with only one error term, you can work with standard deviation instead of variance. Rule 6: The variance of an average of N numbers, each with variance s 2, is s 2 /N. (The standard deviation in the average improves as 1/ N ). Example 5: Subtraction: If Z = A B with A = 163.455±0.002 and B = 1.34±0.02 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): variance in A = (0.002) 2 + variance in B = (0.02) 2 Variance in result (0.02) 2 Standard deviation = 0.02 Result (163.455±0.002) (1.34±0.02) = 162.12±0.02 3
Example 6: Multiplication: The result of an experiment is given by (slope x C p ). Let slope = 123.2±2.4 and C p =4.184±0.031. Find the uncertainty in the result....solution: The relative variance of the result is the sum of the relative variances of the data (Rule 2). Relative variance in slope = (2.4/123.2) 2 = (0.019) 2 =3.6x10-4 + Relative variance in C p = (0.031/4.184) 2 = (0.0074) 2 =5.5x10-5 Relative variance in product =4.2x10-4 Relative st.dev in product= 4.2x10-4 = 0.020 or 2.0% Result =( 123.2±2.4 )x( 4.184±0.031)= 515.47 ± 2.0% = 515. ± 10. Example 7: Logs: The equilibrium constant for a reaction at two different temperatures is K 1 = 0.0322±0.0007 at 298.2 and K 2 =0.473±0.006 at 353.2 K. Calculate ln(k 2 /K 1 ) and find the uncertainty in the result. Solution: Just like Example 6 the relative variance in k 2 /k 1 is the sum of the relative variances: Relative variance in K 2 = (0.0007/0.0322) 2 = (0.022) 2 = 4.7x10-4 + Relative variance in K 1 = (0.006/0.473) 2 = (0.013) 2 = 1.6x10-4 Relative variance in K 2 /K 1 = 6.3x10-4 Then using Rule 3 shows that absolute variance in ln K 2 /K 1 is the relative variance in K 2 /K 1. Variance in ln K 2 /K 1 = Relative variance in K 2 /K 1 = 6.3x10-4 Standard deviation in result = 6.3x10-4 = 0.025 ln K 2 /K 1 = ln 14.689 = 2.687± 0.025 = 2.69±0.03 See example 2 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 = 9.874±0.0041. Calculate A, and find the uncertainty. Solution: The result is e 9.874 = 1.9427x10 4. The relative variance of the result is the absolute variance of A (Rule 4): Relative variance of e 9.874 = variance of A = (0.0041) 2 = 1.68x10-5 Relative st. dev in result = 1.68x10-5 = 4.1x10-3 or 0.41% Result = 1.9427x10 4 ± 0.41% = 1.943x10 4 ± 0.080x10 4. See example 3 for the significant figure version of this error analysis. 4
Example 9: Multiplication with 'Certain' Numbers: The result of a calculation is (slope x R), where R is the gas constant in J K -1 mol -1. Let slope = 1.23±0.02. 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. Rule 5 applies 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: Result = slope x R = 1.23 ±0.02 x 8.3144126 = 10.23 ± 0.17 Example 10: The Inverse of the Slope or Intercept: The result of an experiment is the inverse of the intercept from a graph, 1/b. Let b=0.523 ± 0.043. Find the uncertainty in the result....solution: The relative variance in the result is equal to the relative variance in the intercept (Rule 2). We can also work directly in terms of standard deviation (Rule 5): Relative st.dev in b = 0.043/0.523 = 0.082 or 8.2% Relative st. dev of Result = Relative st. dev in b = 8.2% Result = 1/b = 1/0.523 = 1.912 ± 8.2% = 1.91 ± 0.16 Example 11: Division and Subtraction: The term (1/T 2-1/T 1 ) is a very common factor in many equations. For T 1 = 298.2 ±0.2 K and T 2 = 353.2 ±0.2 K calculate (1/T 2-1/T 1 ). Find the uncertainty in the result. Solution: Don t be put off by multi-step problems, just work one step at a time. First, get the uncertainty in 1/T 2 and 1/T 1. Since both of these are divisions the relative variance of 1/T is just the relative variance of T (Rule 2). Then convert to absolute variance to calculate the error in (1/T 2-1/T 1 ) using Rule 1: Relative variance in 1/T 2 = (0.2/353.2) 2 = 3.2x10-7 Relative variance in 1/T 1 = (0.2/298.2) 2 = 4.5x10-7 Variance in 1/T 2 = 3.2x10-7 (1/353.2) 2 = 2.6x10-12 + Variance in 1/T 1 = 4.5x10-7 (1/298.2) 2 = 5.1x10-12 Variance in (1/T 2-1/T 1 ) = 7.7x10-12 Standard deviation in result = 7.7x10-12 = 2.8x10-6 Result = -5.22x10-4 ± 3x10-6 5
Example 12: A Multi-Step Problem An example of a more realistic problem is the temperature dependence of the equilibrium constant. Let s assume we wish to evaluate rh, knowing K 2, K 1, R, T 2, and T 1 in the equation: ln( K 2 rh K 1 ) = R ( 1 1 T - 2 T 1 ) where K 1 =0.0322±0.0007 at 298.2 and K 2 =0.473±0.006 at 353.2 K with T = ±0.2 K (The same data as Examples 7 and 11!). Find the uncertainty in the result. Solution: Solving for rh: r H = - 8.314 ln( K 2 K 1 ) ( 1 T 2-1 T 1 ) = 42.785 kj/mol We already know the uncertainty in ln(k 2 /K 1 ) from Example 7, 2.687± 0.025, and the uncertainty in (1/T 2-1/T 1 ) from Example 11, -5.22x10-4 ± 2.8x10-6. R is a certain number. The relative variance in r H is then just the sum of the relative variances: Relative variance r H = (0.025/2.687) 2 + (2.8x10-6 / 5.222x10-4 ) 2 = 1.19x10-4 Relative standard deviation of r H = 1.19x10-4 = 0.011 or 1.1% Standard deviation of r H = (0.011)(42.785) = 0.47 kj/mol Result r H = 42.78 ± 0.47 kj/mol As the calculations get longer, the error analysis can get to be rather tedious. However, spreadsheets can come to the rescue. When you set up your spreadsheets just include the error analysis. For this problem you might set up the following: B C D E F G H I 4 relative relative 5 K ± variance T (K) ± variance 6 K2 0.473 0.006 1.609E-04 T2 353.2 0.2 3.206E-07 7 K1 0.0322 0.0007 4.726E-04 T1 298.2 0.2 4.498E-07 8 9 rel variance variance ± 10 1/T2-1/T1= -5.222E-04 2.798E-05 7.629E-12 2.762E-06 K-1 11 12 ln K2/K1= 2.6871289 8.773E-05 6.335E-04 0.0251694 13 14 rh= 4.278E+04 1.157E-04 2.118E+05 460.22886 J/mol 15 rh= 42.785 ± 0.460 kj/mol 16 17 R= 8.314447 J/mol/K The important cells are Variance in (1/T 2-1/T 1 ) E10: =(I6/G6^2+I7/G7^2) 6
Variance in ln K 2 /K 1 Relative variance r H E12: =E6+E7 D14: =D10+D12. It is a good habit to use spreadsheets for all your lab calculations and to include the error analysis, if the lab requires a full propagation of errors treatment (see below). Summary: Keep this tutorial handy, you will need it all year long. Every lab requires an error analysis for the final results, even calculations-only lab reports. All lab reports should have an error analysis If the calculations don t use least squares curve fitting or the curve fitting is used at a very early stage of the calculation with many subsequent calculations, just use significant figure rules (e.g. for all the calorimetry experiments just use significant figure rules). From least squares curve fitting use propagation of error rules 7
HOMEWORK Name 1. 5 ml of a 0.2134 M HCl is diluted to 100 ml using volumetric glassware. Calculate the ph of the final concentration. Use significant figure rules. Present the results using the proper number of significant figures. 2. The binding constant for a guest-host complex is given by slope/intercept of a graph. Using curve fitting it was found that the slope=0.2265±0.0379 and intercept=1.601x10-4 ± 1.99x10-5. Calculate the binding constant and the uncertainty using propagation of errors rules. (This is real Colby research data, by the way.) 3. The number of ligands in a complex is determined by the inverse of the intercept of a graph. The intercept is b=0.341 ± 0.023. Calculate the number of ligands and the uncertainty of the result. Is it safe to assume that the answer is = 3.00 to within experimental error? 4. 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 = 9.874±0.023 and E a = 28.26±0.03 kj/mol. Calculate k at T = 298.15 K. Use propagation of errors. 8
Y (units?) 5. Write a spreadsheet to fit the following data. The data and the final result are listed below to use to check your spreadsheet. But you use the formulas with the explicit sums you don t use the Excel linest() function for your assignment! Excel Least Squares Curve Fitting x y fit y -1.5-1.35-0.76 1 2.67 1.871 3.5 4.68 4.499 6 6.74 7.127 slope 1.0512 0.8198 intercept slope ± 0.136972 0.4914789 intercept ± r2 0.967159 0.7656958 s(y) ± F 58.89904 2 degrees-of-freedom regression SS 34.53192 1.17258 residual SS Test Chart 8 7 6 5 4 3 2 y fit y 1 0-2 0 2 4 6-1 -2-3 X (units?) 9