Error Propagation Answer Key

Error Propagation Purpose: account for error in lab equipment and glassware Absolute Error (absolute uncertainty) Definition Notation Units? Relative Error (% error, inherent error, precision) Expresses the margin of uncertainty associated with the measurement; usually given by manufacturer (page 27 lab manual) ∆𝐶 (Represented with a delta sign and a variable) Ratio of the absolute error to the size of its associated measurement Has units (unless no units specified) IMPORTANT: Absolute error must have same units AND same number of decimal places as measured quantity NO units Most of the time relative error is expressed as a percentage ∆𝐶 𝐶 (absolute error divided by the measurement) Important Notes:   If an absolute error is not specified, assume the absolute error is ±1 in the last digit In general, absolute and relative errors are reported to 1 significant figure Practice: 1. In the 10.00mL graduated cylinder, the uncertainty in the reading, based on manufacturer specifications, is ±0.01𝑚𝐿  Graduated Cylinder Reading = about 6.63 mL  Absolute Error = ± 0.01 mL  Range of Measurement = (6.63 ± 0.01)𝑚𝐿 Range: 6.63 − 0.01 𝑚𝐿 𝑡𝑜 6.63 ± 0.01𝑚𝐿 = 6.62𝑚𝐿 𝑡𝑜 6.64 𝑚𝐿  % Relative Error = ∆𝐶 𝐶 = (± 0.01𝑚𝐿 6.63𝑚𝐿 ) × 100% = .15%0.2%(1𝑆𝐹) 2. Is the following expression correct? Explain why or why not. (1.8770 ± 0.002)𝑀 The expression is incorrect. Absolute error should have the same number of decimal places as the measured value 3. Determine the percentage error in the following quantities: a) (3.67 ± 0.05) 𝑚𝐿 ∆𝐶 C ∆𝐶 % error = × 100% 𝐶 0.05𝑚𝐿 = (± ) × 100% 3.67𝑚𝐿 = ±1% (𝑟𝑒𝑝𝑜𝑟𝑡 𝑡𝑜 1 𝑠𝑖𝑔 𝑓𝑖𝑔) b)0.2050𝑀 *Assume uncertainty of 1 in last digit* 0.2050𝑀 ± 0.0001) 𝑀 C ∆𝐶 ∆𝐶 % error = × 100% 𝐶 0.0001𝑀 = (± ) × 100% 0.2050𝑀 = ±0.05% (𝑟𝑒𝑝𝑜𝑟𝑡 𝑡𝑜 1 𝑠𝑖𝑔 𝑓𝑖𝑔) 4. Prepare a 0.5 M solution that has uncertainty of ±2%. How many significant figures are required in the original concentration? ∆𝐶 = 𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑒𝑟𝑟𝑜𝑟 = 2% = 0.02 𝐶 ∆𝐶 = 𝐶 × 𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑒𝑟𝑟𝑜𝑟 = 0.5𝑀 × 0.02 = ±0.01𝑀 0.50𝑀 ± 0.01𝑀 − 0.49 𝑡𝑜 0.51𝑀 (2 𝑠𝑖𝑔 𝑓𝑖𝑔𝑠) What happens if multiple measurements are made to obtain a result? The result is affected by errors in each of the measurements. NOTE: there is a difference in calculating error when you add/subtract or multiply/divide Addition/Subtraction Equation Relative Error = ∆𝐶 𝐶 ∆𝐶 ∆𝐴 ∆𝐵 = + 𝐶 𝐶 𝐶 𝐶 =𝐴+𝐵 𝐶 =𝐴−𝐵 Multiplication/Division Equation Relative Error = Absolute Error= ∆𝐶 ( ∆𝐶 = ∆𝐴 + ∆𝐵 The absolute error of a result is the sum of all the absolute errors (of the measurements) ∆𝐶 𝐶 ∆𝐶 ∆𝐴 ∆𝐵 + = 𝐵 𝐶 𝐴 The relative error of a result is the sum of the relative errors (of the measurements) 𝐶 =𝐴×𝐵 𝐶 = 𝐴/𝐵 ∆𝐶 ∆𝐴 ∆𝐵 = + )×𝐶 𝐶 𝐶 𝐶 Absolute Error= ∆𝐶 ( ∆𝐶 ∆𝐴 ∆𝐵 )×𝐶 = + 𝐵 𝐶 𝐴 ∆𝐶 = ( Example: ∆𝐴 ∆𝐵 + )×𝐶 𝐴 𝐵 5. Find the percent error and absolute error of the following: a) (18.56 ± 0.02)𝑚𝐿 + (2.85 ± 0.04)𝑚𝐿 𝐶 = 𝐴 + 𝐵 = 18.56 𝑚𝐿 ÷ 2.85 𝑚𝐿 = 21.41 𝑚𝐿 Absolute error = ∆𝐶 = ∆𝐴 + ∆𝐵 = ±0.02 𝑚𝐿 + ±0.04 𝑚𝐿 = ±0.06 𝑚𝐿 ∆𝐶 Percent error = 𝐶 ∆𝐴 =(𝐶 + ∆𝐵 )× 𝐶 100% = [(±0.02 mL / 21.41 mL) + (± 0.04 mL / 21.41 mL)] × 100% = ±0.3% b) (11.56 ± 0.02)𝑐𝑚 − (7.43 ± 0.01)𝑐𝑚 C = A – B = 11.56 cm – 7.43 cm = 4.13 cm Absolute error = ΔC = ΔA + ΔB = ± 0.02 cm + ± 0.01 cm = ±0.03 cm Percent error = ΔC/C = [ΔA/C + ΔB/C]× 100% = [(± 0.02 cm/4.13 cm) + (± 0.01 cm/4.13 cm)] × 100% = ±0.7% 6. Find the percent error and absolute error of the following: a) 10.42𝑐𝑚 × 87.5𝑐𝑚 Percent error = ΔC/C = [ΔA/A + ΔB/B]*100% = [(± 0.01 cm/10.42 cm) + (± 0.1 cm/87.5 cm)] × 100% = 0.00210255×100% = ±0.2% Absolute error = ΔC = (ΔC/C) × C = Relative error × C = (0.00210255)× 911.75 cm2 = ±1.917 cm = ± 2 cm2 C = A × B = 10.42 cm × 87.5 cm = 911.75 cm2 b) 10.42𝑚𝑜𝑙 ÷ 87.50𝐿 Percent error = ΔC/C = [ΔA/A + ΔB/B]× 100% = [(± 0.01 mol/10.42 mol) + (± 0.01 L/87.50 L)] × 100% = 0.001073978613× 100% = ±0.1% Absolute error = ΔC = (ΔC/C) × C = Relative error × C = (0.001073978613)× 0.1190 mol/L = ±0.0001278955102 mol/L = ± 0.0001 mol/L C = A ÷ B = 10.42 mol ÷ 87.50 L = 0.1190 mol/L 7. Calculate the percent error of the following: (Hint PEMDAS) 12.635 ± 0.005 12.635 ± 0.005 = [(5.967 ± 0.003) + (0.478 ± 0.004 )] 6.445 ± 0.007 0.005 0.007 % 𝑒𝑟𝑟𝑜𝑟 = ( + ) × 100% = 0.148%~0.1 12.635 6.445 Post-Lab Considerations    Note: You will be using these absolute errors in your preAbsolute Errors in Lab Equipment labs and post-labs. This is located on page 27 of your lab 5.00 mL pipet ± 0.01 mL manual 10.00 mL pipet ± 0.02 mL Systematic errors result from fundamental flaws in the Digital balance: ± 0.2 mg = ± 0.0002 g design, construction, calibration, operation, or 50.00 mL volumetric flask ± 0.04 mL 100.00 mL volumetric flask ±0.08 mL interpretation of the experiment. Therefore, this affects 25.00 mL buret ± 0.03 mL the _______accuracy_____ of a measurement and may also influence its precision. o Ex. When measuring water, the meniscus of water was below the line of the pipet (Should be at the line & need to be placed on flat surface) Random errors vary arbitrarily in both _____sign___ and _________magnitude_______ between successive measurements. These errors may result from the inability to control the constancy of an experimental variable or may arise from the limited _______precision________ with which the scale of an instrument can be read. o Ex. Limited precision of the 10.00 mL pipet; the absolute error (based on the manufacturer) in a 10.00mL pipet is ±0.02 mL Relationship between %RAD and % inherent Error   Purpose of Calibration: to check experimenter’s precision/accuracy of lab equipment A comparison between %RAD and % inherent error is used to check if the experiment was performed within the precision warranted by the equipment %𝑅𝐴𝐷 = 𝑚𝑒𝑎𝑠𝑢𝑟𝑒 𝑜𝑓 𝑝𝑟𝑒𝑐𝑖𝑠𝑖𝑜𝑛 𝑜𝑓 𝑑𝑎𝑡𝑎 (𝑜𝑓 𝑡ℎ𝑒 𝑡𝑟𝑖𝑎𝑙𝑠 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑒𝑑 𝑖𝑛 𝑙𝑎𝑏) % 𝑖𝑛ℎ𝑒𝑟𝑒𝑛𝑡 𝑒𝑟𝑟𝑜𝑟 = 𝑚𝑒𝑎𝑠𝑢𝑟𝑒 𝑜𝑓 𝑝𝑟𝑒𝑐𝑖𝑠𝑖𝑜𝑛 𝑜𝑓 𝑙𝑎𝑏 𝑖𝑛𝑠𝑡𝑟𝑢𝑚𝑒𝑛𝑡 𝑢𝑠𝑒𝑑 The lower the %RAD, the higher the precision of the data The higher the %RAD, the lower the precision of the data Comparison %𝑅𝐴𝐷 ≤ 𝐼𝑛ℎ𝑒𝑟𝑒𝑛𝑡 𝑒𝑟𝑟𝑜𝑟 Meaning Experiment was performed within the precision warranted by the equipment; no need to repeat the experiment Experiment was NOT performed within the precision warranted by the equipment; experiment should be repeated %𝑅𝐴𝐷 > % 𝐼𝑛ℎ𝑒𝑟𝑒𝑛𝑡 𝑒𝑟𝑟𝑜𝑟 Examples: (Application of %RAD and Inherent Error) 8. Using a 10-mL volumetric pipet, a student obtained the following amounts of water on three trials: 10.10 mL, 10.02mL and 10.50 mL. The absolute error associated with a 10-mL volumetric pipet is ±0.02 mL. What can you conclude about the experimenter’s results? Should the experimenter repeat the experiment? % inherent error = (absolute error in 10-mL volumetric pipet / 10.00 mL) × 100% = (0.02 mL/ 10.00 mL) × 100% = ±0.2 % Mean of data = (10.10 mL + 10.02 mL + 10.50 mL) / 3 = 10.21 mL Average deviation = [ǀ10.10 – 10.21ǀ + ǀ10.02 – 10.21ǀ + ǀ10.50 – 10.21ǀ] / 3 = .20 mL % RAD (1.9%) > % inherent error (0.2 %); The experimenter is not within the precision warranted by the equipment. Therefore, the experimenter should repeat the experiment. 9. Calculate the percent inherent error and absolute error of the following: a) (20.54 𝑚𝐿 ± 0.02𝑚𝐿 × 0.003𝑔 3.21𝑔 0.05𝑔 0.254𝑔 ± 𝑚𝐿 ) ÷ ( 𝑚𝑜𝑙 ± ) 𝑚𝐿 𝑚𝑜𝑙 Notice that the problem involves multiplication AND division. Make sure to use the correct equation! Let A, B, and C refer to quantities. Let ΔA, ΔB, and ΔC refer to the absolute error of the respective quantities. (20.54 ± 0.02 x 0.254 ± 0.003) / (3.21 ± 0.05) ∆A A B ∆B ∆C C Percent Inherent Error (Another name for percent relative error) The problem has three measurements with their respective absolute errors. The equation can be written as: ∆𝐷 ∆𝐶 ∆𝐴 ∆𝐵 = + + 𝐵 𝐷 𝐶 𝐴 Therefore, the relative error is: ΔD/D = (± 0.05 g/mol / 3.21 g/mol) + (± 0.003 g/mL / 0.254 g/mL) + (± 0.02 mL/ 20.54 mL) = .03 Multiply by 100% to find the percent relative error: 100% × .03 = ±3% (1 SF) Absolute Error: Recall the relative error equation: ∆𝐷 𝐷 = ∆𝐶 𝐶 + ∆𝐴 ∆𝐵 + 𝐵 𝐴 ∆𝐶 Remember the absolute error is ΔD. Therefore, the equation is: ∆𝐷 = ( 𝐶 + For the problem, D = (A * B) / (C) ∆𝐴 𝐴 + ∆𝐵 𝐵 )×𝐷 D = (20.54 mL x 0.254 g/mL) / (3.21 g/mol) = 1.625 mol = 1.63 mol The relative error was already found in the previous problem. Multiply the relative error by the measuredquantity to find the absolute error. (ΔD/D) × D = ΔD = .03 × 1.625 mol = ± .05 mol (1 sig fig) b) (30.078 ± 0.003 − 20.174 ± 0.001) + 9.813 ± 0.005 Notice that the problem involves addition AND subtraction. Make sure to use the correct equation! Let A, B, and C refer to quantities. Let ΔA, ΔB, and ΔC refer to the absolute error of the respective quantities. (30.078 ± 0.003 - 20.174 ± 0.001) + 9.813 ± 0.005 A Percent Inherent Error: ∆𝐷 𝐷 = ∆𝐶 𝐷 + ∆A B ∆B C ∆C ∆𝐴 ∆𝐵 + 𝐷 𝐷 In the problem, D = A – B + C = 30.078 - 20.174 + 9.813 = 19.717 ΔD/D = (± 0.003/19.717) + (± 0.001/19.717) + (± 0.005/19.717) = .0005 Multiply by 100% to find the percent relative error: 100% × 0.0005 = ±0.05% (1 SF) ∆𝐶 Absolute Error: ∆𝐷 = ( 𝐷 + ∆𝐴 𝐷 + ∆𝐵 𝐷 ) × 𝐷 = ∆𝐴 + ∆𝐵 + ∆𝐶 ΔD = (± 0.003) + (± 0.001) + (± 0.005) = ± 0.009 (1 SF) 10. Suppose you prepared a stock solution by dissolving 0.0558 g (weight obtained from digital balance) of Mg(OH)2 in a 50.00 mL volumetric flask. Assume the molar mass of Mg(OH)2 is 58.3258 g/mol. Assume no uncertainty in the molar mass. The absolute uncertainty in the digital balance is ± 0.2 mg and the 50.00 mL volumetric flask is ± 0.04 mL. a) Calculate the molarity of Mg(OH)2 𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 = 𝑚𝑜𝑙 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝐿 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 Molarity = [0.0558 g (1 mol/ 58.3258 g)] / [(50.00 mL)(1 L/ 1000 mL) = 0.0191 M (3 SF) b) Calculate the percent inherent error (for the molarity) in the Mg(OH)2 solution Look at the calculation of molarity. What does the calculation involve? The calculation involves division; therefore, use the equation that corresponds to multiplying or dividing to calculate inherent error. ∆𝐶 ∆𝐴 ∆𝐵 = + 𝐶 𝐴 𝐵 There is uncertainty in the measurement of the mass of the Mg(OH)2 and the volume of the solution. Let: Let A = 0.0558 g and B= 50.00 mL. Let ΔA = ±0.2 mg = ±0.0002 g and ΔB = ± 0.04 mL. ΔC/C = (±0.0002 g/ 0.0558 g) + (± 0.04 mL/50.00 mL) = 0.004 x 100% = ±0.4% c) Calculate the absolute error in (molarity for the Mg(OH)2 solution Let C be the molarity of the solution. (ΔC/C) × C = 0.004 × 0.0191 M = ± 0.00008 M Remember that absolute error should have the same number of decimal places as the measured quantity. Therefore, since the measured quantity (the molarity) has 4 decimal places, the absolute error must have 4 decimal places. Absolute error = ± .0001 M 11. If a standard hydrochloric acid solution had a concentration of 0.375 M, with a percent relative error of ± 2% what is the absolute error? ΔC/C = Relative error = 2% = 0.02 Let C be the measured quantity (the concentration) ΔC = C × relative error = 0.375 M × 0.02 = ± 0.008 M