Uncertainty in Measurement
When recording measurements it’s very important to have the correctnumber of significant digits.
This is determined by the increments onthe instrument
The significant digits are all of the numbersthat you know with certainty plus one more that you estimate.
4.8 cm
4.83 cm
Whatever the increments are, estimate onemore place, but no more!
For digital instruments, record exactly what appears on the display.
4.830 cm
Rules for Significant Figures
1. All nonzero digits are significant figures
2. Leading zeros are never significant
Ex: 0.00025 has 2 sig figs
3. Trailing zeros are significant figures if a decimal point is present
Ex: 0.0002500 has 4 sig figs
If no decimal point is present, then the zerosare not sig figs
Ex: 25000 has 2 sig figs
Significant Figures in Calculations
multiplication & division
The number of significant figures in the answeris determined by the measurement having thefewest significant figures
Ex: (2.500 x 10-4)(4.3 x 107) = 10.75 x 103
10.75 x 103 =11 x 103=1.1 x 104
addition & subtraction
the measurement with the least accurate placevalue determines the place value in the answer.
Ex: 19.5 + 200.060 + 0.25 = 219.81 =219.8
Ex: 250 – 12 = 238 = 240
Combination Problems:
Ex: (80.75) (4.18) (32.8- 24.5)
=2801.5405 =2800
8.3
UncertaintyWhenever you record a measurement you muststate the uncertainty in the value; i.e. the rangeof possible values (above or below your estimate)
centigram balance
Ex: 123.08 g +/- 0.01 g
analytical balance
Ex: 123.0835 g
+/- 0.0001 gDigital: +/- 1 of last digit !
Non-electronic instruments
87.4◦ C
+/- 0.3 ◦
52.9 mL
+/- 0.2 mL
+/- 0.1- 0.5
buret
21.30 mL
+/- 0.02 mL
+/- 0.02 – 0.05 mL
Acceptable Uncertainties?
103.25 mL +/- 0.005
24.7762 g +/- 0.001
32.95 ◦ +/- 0.01
15.38 g +/- 0.1
3.61 mL +/- 0.05
no
no
yes
no
yes
Propagating Uncertainty
Calculated answers should have the right number of significant figures and theuncertainty
Not the same as percent error:
% error = |Δaccepted and experimental value| accepted value
100
Ex: Actual melting pt. = 84.8 ◦ C Experimental m.p. = 86.5◦C % error = |86.5 – 84.8 |
84.8 100 = 2.0 %
Just add the uncertainties in the measurementsto get the uncertainty in the answer
answers derived from addition or subtraction:
Ex: mass of crucible and compound 24.31 g +/-0.01mass of empty crucible 19.94 g +/- 0.01
mass of compound 24.31 – 19.94 = 4.37 g +/- 0.02
If getting an average of several values, theuncertainty remains the same
Ex: 4.5 mL +/- 0.2 + 4.7 mL +/- 0.2
avg= 4.6 mL +/- 0.2 not +/- 0.4
multi-step problems involving mult. & division:1. Convert the absolute uncertainties in all the measurements into a percentage
2. Add all the percentages3. Convert the percentage uncertainty back into an absolute uncertainty in the final answer
Ex: An experiment is performed to determine the enthalpy change, ΔH, for the reaction between HCl and magnesium metal.If 2.00 g of Mg are added to 50.08 g of HCl and the temperature ofthe solution rises from 23.8◦ C to 38.2◦ C, calculate ∆H and specifythe uncertainty. Specific heat of the solution = 4.18 J/g ◦ C
ΔH = m C ∆T
ΔH = (50.08 +2.00 g) (4.18 J/g◦C)(38.2 – 23.8 ◦ C)
ΔH = 3,134.7993 J 3.1347993 KJ
3.13 KJ
balance uncertainty = +/- 0.01 gthermometer uncertainty = +/- 0.2
mass of solution: 0.01 + 0.01 = 0.02uncertainty in the answer:
% = 0.02 x 100 52.08
= 0.0384 %
temp. of solution: 0.2 + 0.2 = 0.4
% = 0.4 x 100 14.4
= 2.777 %
Total = 0.0384 + 2.777 = 2.815 %
3.13 KJ (3.13) (0.02815) =0.0881
3.13 KJ +/- 0.09
Ex: 25.00 mL of an acid measured from a pipet (+/- 0.05) istitrated with sodium hydroxide solution measured from a buret. The initial buret reading at the beginning of the titration is 0.82 mL (+/-0.02). After the titration the final buret reading is 33.87 mL (0.02 ). If the molarity of the NaOH is 0.25 M (+/- 0.01 ), what is the molarity of the acid with the uncertainty?
M1V1 = M2V2
acid base
M1 (25.00 mL) = (0.25 M )(33.87 – 0.82 mL)25.00 mL 25.00 mL
M1= 0.3305 = 0.33 M
uncertainty in the answer:volume of NaOH: 0.02 + 0.02 = 0.04
% = 0.04 x 100 33.05
= 0.121 %
volume of acid: 0.05
% = 0.05 x 100 25.00
= 0.2 %
molarity of NaOH: 0.05
% = 0.01 x 100 0.25
= 4 %
Total= 0.121 + 0.2 + 4 = 4.321 %
Total= 0.121 + 0.2 + 4 = 4.321 %
(0.33) (0.04321) = 0.01426
= 0.33 M +/- 0.01
