HL11-2

1. HL11-2.ppt Objectives 11.2.1 & 11.2.2 Uncertainties in Calculated Results

2. Objectives 11.2.1 State uncertainties as absolute and percentage uncertainties. 11.2.2 Determine the uncertainties in results.

3. Uncertainties in Calculated Results • Like Sig Figs, the uncertainties in individual measurements can be combined to determine the uncertainty in the final value being calculated. • All you’ve got to do is follow some simple rules…

4. The “Simple” Rules • Add absolute uncertainties when adding or subtracting numbers • Q - What are absolute uncertainties? • Add % uncertainties when multiplying or dividing numbers • Multiply % uncertainties by the exponent when raising to a power • Q – How would you change an absolute uncertainty to a % uncertainty?

5. Changing Absolute to % Uncertainty • % = 100 x (abs/actual measurement) • Q – How would you go from % to absolute? • Abs = (%/100) x actual measurement

6. Practice Problem # 1 • 1. X = A2(B-C) • A = 123 ±0.5 B = 12.7 ±0.2 C = 4.3 ±0.1 • B-C = 12.7-4.3 = 8.4 Unc: add absolute uncertainties • 0.2% + 0.1% = ±0.3 % • Answer so far B-C = 8.4 ± 0.3 • A2 = 1232 = 15,129 Unc – switch to %, then multiply by 2 • % = (.5/123) x 100 = ±0.4% • 2x0.4% = ±0.8% • Answer so far A2= 15,100 ± 0.8% • [Note – the value was rounded, taking the uncertainty into account.]

7. Practice Problem cont. X = A2(B-C) • A2 = 15,100 ±0.8% • (B-C) =8.4 ±0.3 Unc : change to % % = (.3/8.4) x 100 = ±4% X = 15,100 x 8.4 = 126,840 Unc: add % % = 4 + 0.8= ± 4.8% Answer = 126,840 ± 4.8% Unc: convert back to abs Abs = (4.8/100) 126,840 = 6088 = ± 6100 X = 127,000 ± 6000 Note: 3 sig figs in answer Notice that the uncertainty has been rounded to 1 sig fig – a typical practice. Then the value has been rounded, taking the uncertainty into account

8. Practice Problem # 2 • X = A (B-C) • A = 123 ±0.5 B = 12.7 ±0.2 C = 4.3 ±0.1 • Answer = 1030 ± 40 • See the answer worked out on p.292-293 of your book.

9. Practice Problem # 3 • Q = m c ΔT ΔH = Q / n • m = 200 ± 0.5 g (± .3%) • c = 4.183 ± 0.0005 J g-1K-1 (± .01%) • Tinitial = 21.6 ± 0.1⁰C (± .5%) • Tfinal = 24.2 ± 0.1⁰C (± .4%) • n = 0.0500 ± 0.00005 mol (± .1%) • Notice that the% uncertainty for the temperature is much greater than the other % uncertainties…so you CAN ignore the minor uncertainties in the other data – greatly simplifying your calculations! However you MUST make an explicit statement to address this assumption.

10. Practice Problem # 3 cont • Q = m c ΔT ΔH = Q / n • m = 200. ± 0.5 g (± .3%) • c = 4.183 ± 0.0005 J g-1K-1 (± .01%) • Tinitial = 21.6 ± 0.1⁰C (± .5%) • Tfinal = 24.2 ± 0.1⁰C (± .4%) • n = 0.0500 ± 0.00005 mol (± .1%) • Q = (200.)(4.18)(2.6) = 2173 J • ΔH = 2173 / 0.0500 = 43,472 J/mol = 43 kJ/mol • unc focused on temp: (0.2/2.6)100= 7.7% • 7.7% = 3.3 kJ/mol • Answer = −43 ± 3 kJ/mol • [The negative is because the temperature went up – revealing the reaction is exothermic.]

11. Practice Problem # 3 cont. • Experimental Results = −43 ± 3 kJ/mol (± 7.7 %) • Lit Value = −45.6 kJ/mol • Percent Error = (43 – 45.6)/45.6 = − 5.7 % • Since the % error < experimental uncertainty, the experimental result is in agreement with the literature value. This means random error is the biggest factor here… and that the largest source of this uncertainty is in the measurement of temperature, so any suggestions for improvement would need to deal with this aspect.

12. Practice Problem # 3 cont. • Experimental Results = −43 ± 3 kJ/mol (± 7.7 %) • Lit Value = −55.6 kJ/mol • Percent Error = (43–55.6)/55.6 = −23 % • Since the % error > experimental uncertainty, the experimental result is NOT in agreement with the literature value. This means some type of systematic error is the biggest factor here… you would need to identify what may have caused you to get results that were 23 percent lower than the literature value. What could cause this?

13. Practice Problem # 3 cont. • What could cause this? • heat loss to the surroundings • cover the calorimeter to reduce heat loss • use a Styrofoam calorimeter • Not addressing the heat capacity of the calorimeter • Measure the heat capacity of the calorimeter • Assuming the solution’s heat capacity was equal to that of water • measure the heat capacity of the solution • NOTE: You could address random errors, too… but they should not be the main focus.

14. Homework • 11.2 Text Assignment • Read pp 292-294 • Do Ex 11.2 pp 298 #1-4 • Due Tuesday