Unit 11 Measurement and data Processing

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Unit 11 Measurement and data Processing Learning outcomes: 11.1 Uncertainty and data error in measurement - Describe random uncertainties and systematic errors - Distinguish between precision and accuracy - Describe how effects of Random uncertainties can be reduced - State random uncertainties as an uncertainty range 11.2 Uncertainty in Calculate results - State uncertainties as an absolute and a percentage uncertainty - Determine the uncertainties in results

11.3 Graphical Techniques - Sketch graphs to represent dependences and interpret graph behavior - Construct graphs from experimental data - Draw best fit lines through data points on a graph - Determine the values of physical quantities from graphs

11.1 Uncertainty and error in measurements * Uncertainty in analogue instruments : Depends on the scale of the apparatus: Uncertainty is half the smallest division Example 1 : Measuring Length with meter: smallest scale is 1 mm so uncertainty is ± 0.5 mm length of a marker is 12.6 cm uncertainty is 0.5mm Indicating the length of the marker could be 12.65 cm or 12.55 Example 2 : Measuring volume liquid in a cylinder: smallest scale is 2 ml so uncertainty is ±1 ml. Measured volume is 34.4 ml uncertainty is 1 ml so value can be 34.5 ml or 34.3 ml

Example 3 : Measuring temperature with thermometer: smallest scale is 0.1 °c So uncertainty is ± 0.05 °c Measured temperature of a piece of metal is 83.3°c Value temperature can be 83.35 °c or 83.25°c * Uncertainty in digital instruments: Uncertainty is ± the smallest scale division (smallest digit) Example 1 : Measuring a mass with a digital weigh scale smallest measurement is 0.1 g so uncertainty is ± 0.1 g A measurement of a sample is 23.3 g so the value can be 23.2or 23.4g g

Question A sample is measured on an analogue weigh scale and a digital weigh scale the value is 35.5 g , what will be the uncertainty and the expected values ? Answer: Analogue weigh scale uncertainty is ÷ 0.05 g Digital weigh scale uncertainty is ÷ 0.1 g Analogue expected values between 35.55 and 35.45 g Digital expected values between 35.6 and 35.4 g The analogue weigh scale has a lower uncertainty !! Other sources of uncertainty: Time measurement of a chemical process (reaction time ) Voltage electrochemical cell

* Experimental errors (two types) : 1 Random errors (approximating a reading) - Readability of measurement instrument - Effects of changing surrounding (air currents change of temperature) - Insufficient data - Observer misinterpreting data Random errors can be reduced by repeated measurement !! 2 systematic errors (poor design or procedure, human error) - Reading wrong meniscus height volume liquid in cylinder - Overshooting volume in titration - Heat losses in exothermic reactions systematic errors cannot be reduced by repeated measurement !!

X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X Accuracy and Precision of Measurements X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X Not Accurate Not Precise Not Accurate Precise Accurate Precise

Examples of readings (accuracy and precision) Measurement of Data Accurate Precise Value volume liquid = 67.5 ml 55.3ml, 69.4 ml, 56.8 ml, 63.7 ml, 66.5 ml Mass of a sample of NaOH = 7.67 g 8.32 g, 8.34 g, 8.35 g, 8.31 g, 8.33g, 8.36 g Value temperature of water = 24.5 °c 24.8 °c, 23.8 °c, 25.1°c, 24.0 °c, 24.9 °c, 24.1 °c PH value of a liquid = 7.83 7.82, 7.85, 7.84, 7.81, 7.83, 7.82, 7.84 No data far away from real value No repeated data far away from each other No data far away from real value Yes repeated data close to each other Yes data close to real value No repeated data far away from each other Yes data close to real value Yes repeated data close to each other

The accuracy of a measurement is given by the percent error 11.2 Uncertainty in Calculated results * Percentage uncertainties and errors The accuracy of a measurement is given by the percent error Measured Value Accepted Value - % Error = x 100 Accepted Value Example An object is known to weigh 25.0 grams. You weight the object as 26.2 grams. What is the accuracy of your measurement? 26.2 g – 25.0 g % Error = x 100 = 4.8 % 25.0 g

* Propagation of uncertainties 1 Adding and subtraction When adding or subtracting measurements, the uncertainty is the sum of the absolute uncertainties ! Example Consider two burette readings to find volume difference Initial reading 15.05 cm3 ± 0.05 cm3 (15.00-15.10 cm3) Final reading 37.20 cm3 ± 0.05 cm3 (37.15-37.25 cm3) V max = 37.25 – 15.00 = 22.25 cm3 V min = 37.15 – 15.10 = 22.05 cm3 Therefore Volume = 22.15±0.1cm3 Error % = 0.1/22.15 × 100% = 0.45%

Now you ! Consider two burette readings to find volume difference Initial reading 12.3 cm3 ± 0.5 cm3 Final reading 39.2 cm3 ± 0.5 cm3 Find the absolute error and error % of the reading Answer V max = 39.7 – 11.8 = 72.9 cm3 V min = 38.7 – 12.8 = 25.9 cm3 Therefore Volume = 26.9±1cm3 Error % = 1/26.9 × 100% = 3.72%

1 Multiplication and division (formulas) General formula: Wherein ΔX is the absolute uncertainty When Multiplying or dividing measurements, the total uncertainty % is the sum individual % uncertainties ! The absolute uncertainty can then be calculated from uncertainty % We can calculate the Absolute uncertainty and Uncertainty in two ways

Example The value uncertainty of density from mass and Volume Measured values: Mass : 24.0 ± 0.5g Volume: 2.0 ±0.1 cm3 Method 1 % uncertainty Mass = (0.5/24.0) × 100 % = 2% % uncertainty Volume = (0.1/2.0) × 100 % = 5% Density = Mass/Volume D = 12.00 gcm-3 % uncertainty D = 2% + 5 % = 7 % Method 2 Max value D = 24.5/1.9 = 12.89 Min Value D = 23.5/2.1 = 11.19 Absolute uncertainty = 12.89- 12.00 = 0.89 gcm-3 % D = (0.89/12) × 100% = 7.4 % (value in 2 s.f ) answer: 7%

Method 3 (with use of the formula) ΔD/D = ΔM/M + ΔV/V ΔD/12 = 0.5/24.0 + 0.1/2.0  ΔD = 0.85  % uncertainty D = 7.08 %  7% Now you ! Find the value of the volume, uncertainty and % uncertainty from following measurements : L = 3.40 ±0.05 m, W = 1.70 ± 0.05m, H = 0.75 ± 0.05m

V = 4.34 ± 0.49 Percentage Uncertainty = 11% Answer: Value Percentage uncertainty Methods calculation Volume uncertainty and % uncertainty L = 3.40 ±0.05 m W = 1.70 ± 0.05m H = 0.75 ± 0.05m (0.05/3.40) × 100 % =1.4% [1] : %uncertainty V = % L + % W + % H = 1.4% + 2.9% + 6.7 % = 11% [2]Max value V =4.83 Min Value = 3.59 : absolute uncertainty = 0.49, % UncertaintyV = 0.49/4.34 = 11.3%  11% (0.05/1.70) × 100 % =2.9% [3] ΔV/V = ΔL/L + ΔW/W + ΔH/H ΔV/4.34 = 0.05/3.40 +0.05/1.70 + 0.05/0.75 ΔV = 0.111  absolute % uncertqinty V = 11% (0.05/0.75) × 100 % =6.7% V = 4.34 ± 0.49 Percentage Uncertainty = 11%

11.3 Graphical techniques * Plotting graphs - Give the graph a title - Label the axis with both quantities and units - Use available space as effectively as possible (minimum 50% of graph paper) - use linear scale (no uneven jumps) - Plot all points correct, the line of best fit should be smoothly (not from point to point) Example graphs

C = intercept is 50 °c Finding gradient and intercept For a straight line y= Mx + C, x is the independent variable, m is the gradient. To find gradient use triangle method (has to cover min 50 % of graph) In this example x = 50/5 = 10 °cmin-1 C = intercept is 50 °c

Other use of graph techniques will be discussed in Unit 6 kinetics (rate of reaction etc…)