Analytical Chem istry - DePauw University

Chapter 4 Evaluating Analytical Data774C.1 A Few SymbolsA propagation of uncertainty allows us to estimate the uncertainty ina result from the uncertainties in the measurements used to calculate theresult. For the equations in this section we represent the result with thesymbol R, and the measurements with the symbols A, B, and C. The correspondinguncertainties are u R , u A , u B , and u C . We can define the uncertaintiesfor A, B, and C using standard deviations, ranges, or tolerances (orany other measure of uncertainty), as long as we use the same form for allmeasurements.The requirement that we express each uncertaintyin the same way is a critically importantpoint. Suppose you have a rangefor one measurement, such as a pipet’stolerance, and standard deviations for theother measurements. All is not lost. Thereare ways to convert a range to an estimateof the standard deviation. See Appendix 2for more details.4C.2 Uncertainty When Adding or SubtractingWhen adding or subtracting measurements we use their absolute uncertaintiesfor a propagation of uncertainty. For example, if the result is given bythe equationthen the absolute uncertainty in R isExample 4.5R = A + B - Cu = u + u + u2 2 24.6R A B CWhen dispensing 20 mL using a 10-mL Class A pipet, what is the total volumedispensed and what is the uncertainty in this volume? First, completethe calculation using the manufacturer’s tolerance of 10.00 mL ± 0.02 mL,and then using the calibration data from Table 4.9.So l u t i o nTo calculate the total volume we simply add the volumes for each use of thepipet. When using the manufacturer’s values, the total volume isV = 10. 00 mL + 10. 00 mL = 20.00 mLand when using the calibration data, the total volume isV = 9. 992 mL + 9. 992 mL = 19.984 mLUsing the pipet’s tolerance value as an estimate of its uncertainty gives theuncertainty in the total volume as2 2u R= ( 002 . ) + ( 0. 02) = 0.028 mLand using the standard deviation for the data in Table 4.9 gives an uncertaintyof2 2u R= ( 0. 006) + ( 0. 006) = 0.0085 mL