Accuracy and Precision Chapter 1 Exercises Key

Accuracy and Precision Accuracy and precision are terms used to describe data. Accuracy refers to how close a value or set of values is to the true value. Example 1: Determine which set of data is more accurate for three measurements of the same object, which has an accepted length of 3.651 cm? Set A: 3.84 cm, 3.58 cm, 3.62cm. Set B: 3.25 cm, 3.58 cm, 3.49 cm Plan – Determine the average of each set and determine which average is closest to the accepted value. Solution – Set A average = 3.68 cm Set B Average = 3.44 cm Set A differs from the accepted value by a smaller amount than set B, therefore, set A is more accurate. 1. The density of water at 50oC is 0.9881 g/mL. One student performs three trials and finds that the density is 0.987 g/mL, 0.974 g/mL and 0.991 g/mL. A second student finds the density is 0.968 g/mL, 0.980 g/mL and 0.989 g/mL. Which student is more accurate? Student 1 Average = 0.984 g/mL Student 2 Average = 0.979 g/mL Student 1 difference from accepted = 0.984 – 0.9881 = -0.0041 Student 2 difference from accepted = 0.979 – 0.9881 = -0.0091 Student 1’s average is closer, therefore, student 1 is more accurate. 2. The mass of an electron is 9.1093897 x 10-34 g. In trying to find the mass of an electron on one day you measure 9.105 x 10-34 g, 9.086 x 10-34g and 9.163 x 10-34 g. You decide to repeat your experiment on the next day and get 9.085 x 10-34 g, 9.113 x 10-34g and 9.078 x 10-34g. Were you more accurate on the first or second day? Day 1 average = 9.118 x 10-34g Day 2 average = 9.092 x 10-34g Day 1 difference from accepted 9.118 x 10-34g - 9.1093897 x 10-34 g = 9 x 10-37g Day 2 difference from accepted 9.092 x 10-34 g - 9.1093897 x 10-34 g = - 1.7 x 10-36g Day 1’s average is closer to the accepted and more accurate. Precision refers to how close a set of measured values are to one another. It does not reference the true value. Example 1: Determine which set of data is more precise for three measurements of the same object, which has an accepted length of 3.651 cm? Set A: 3.84 cm, 3.58 cm, 3.62cm. Set B: 3.25 cm, 3.58 cm, 3.49 cm Plan – Determine the average deviation of the values from the average of the set. Solution – Set A average = 3.68 cm Data point 1 deviation from average = |3.84 cm – 3.68 cm| = 0.16 cm Data point 2 deviation from average = |3.58 cm – 3.68 cm| = 0.10 cm Data point 3 deviation from average = |3.62 cm – 3.68 cm| = 0.06 cm Average deviation = (0.16 + 0.10 + 0.06)/3 = 0.11 Set B Average = 3.44 cm Data point 1 deviation from average = |3.25 cm – 3.44 cm| = 0.19 cm Data point 2 deviation from average = |3.58 cm – 3.44 cm| = 0.14 cm Data point 3 deviation from average = |3.49 cm – 3.44 cm| = 0.05 cm Average deviation = (0.19 + 0.14 + 0.05)/3 = 0.13 Set A has a smaller average deviation than set B. Set A is, therefore, more precise. 3. The density of water at 50oC is 0.9881 g/mL. One student performs three trials and finds that the density is 0.987 g/mL, 0.974 g/mL and 0.991 g/mL. A second student finds the density is 0.968 g/mL, 0.980 g/mL and 0.989 g/mL. Which student is more precise? Student 1 Average = 0.984 g/mL Student 2 Average = 0.979 g/mL Student 1 Deviations from Average = 0.003, -0.010, 0.007 Student 2 Deviations from Average = -0.011, 0.001, 0.010 Student 1 Average Deviation = 0.00667 Student 2 Average Deviation = 0.00733 Student 1 has a smaller average deviation of their values and is more precise 4. The mass of an electron is 9.1093897 x 10-34 g. In trying to find the mass of an electron on one day you measure 9.105 x 10-34 g, 9.086 x 10-34g and 9.163 x 10-34 g. You decide to repeat your experiment on the next day and get 9.085 x 10-34 g, 9.113 x 10-34g and 9.078 x 10-34g. Were you more precise on the first or second day? Day 1 average = 9.118 x 10-34g Day 2 average = 9.092 x 10-34g Day 1 Deviations from Average = -1.3 x 10-36, -3.2 x 10-36, 4.5 x 10-36 Day 2 Deviations from Average = -7 x 10-37, 2.1 x 10-36, -1.4 x 10-36 Day 1 Average Deviation = 3.0 x 10-36 Day 2 Average Deviation = 1.4 x 10-36 Day 2 has a smaller average deviation and is more precise Systematic and Random Error There are two types of error that can occur when making measurements: systematic and random error. How you correct the error depends on the type(s) of error present. Systematic error is present when all (or most) of the data points collected are higher or lower than the true value for measurements. Random error is present when the data points lie above and below the true value in nearly equal numbers. If the random error is free from systematic error, the average of the data points will be close to the true value for what is being measured. If the random error is large, then many measurements may need to be made to obtain the true value. Example 3: A student takes the temperature of an ice water bath 5 times and records the following data: 5.01 oC, 5.02 oC, 4.99 oC, 5.00 oC, and 5.01 oC. Does the data show systematic or random error? The average temperature for the bath is: 5.01 oC + 5.02 oC + 4.99 oC + 5.00 oC + 5.01 oC = 5.01 oC 5 While this data appears to be very precise, it shows systematic error. The expected value for an ice water bath should be close to 0.00 oC! We can correct this type of error in a piece of equipment (in this case, the thermometer) by calibrating it. When we calibrate a thermometer, we make multiple measurements at known temperatures (ice water bath and boiling water) and then determine the difference between the average of the thermometer readings and the true (or expected) temperature. We then add (if true value is higher) this average (or subtract it if true value is lower) from any measurements made using the thermometer. 5. The density of lead is 11.34 g/mL. In their CHEM 122N General Chemistry Lab, three students measure the density of lead in 5 different trials and obtain the following data: Student 1: 11.56 g/mL, 11.89 g/mL, 11.72 g/mL, 11.33 g/mL, 11.44 g/mL Avg = 11.59 Avg. Deviation = 0.17 Student 2: 11.39 g/mL, 11.30 g/mL, 11.31 g/mL, 11.36 g/mL, 11.33 g/mL Avg = 11.34 Avg. Deviation = 0.03 Student 3: 11.28 g/mL, 11.14 g/mL, 11.56 g/mL, 11.39 g/mL, 11.34 g/mL Avg = 11.34 Avg. Deviation = 0.11 a. Which student(s) has (have) the most accurate data? __Students 2 and 3____________ b. Which student(s) has (have) the smallest random error in their data?___Student 2 _________ c. Which student(s) has (have)data which appears to have systematic error? __Student 1 – Except for 11.33g/mL, all values are high______ Exercises 6. The following are two sets of data measuring the speed of an object with an accepted speed of 5.48 m/s. Data Set 1 Data set 2 5.85 m/s 5.35 m/s 5.47 m/s 5.29 m/s 5.25 m/s 5.64 m/s a. Which data set is more accurate? Data set 1 average = 5.52, Difference from accepted = 0.04 (More accurate) Data set 2 average = 5.43. Difference from accepted = 0.05 b. Which data set is more precise? Data set 1 deviations from average = 0.33, -0.05, -0.27, Average deviation = 0.22 Data set 2 deviations from average = -0.08, -0.14, 0.21. Average deviation = 0.14 (More precise) 7. A student tries to determine the mass of an object based on a new method he is developing. On an accurate balance, the student finds the mass of the object to be 3.128g. His first alternate method gives him 3.423g, 3.245g, and 3.123g. He tries a second method and gets 2.984g, 3.122 and 3.281g. Which method is more accurate and which is more precise? First method average = 3.264g, Difference from accepted = -0.136g Second method average = 3.129g, Difference from accepted = 0.001g (More accurate) First method deviations from average = 0.159, -0.019, -0.141. Average deviation = 0.106 Second method deviations from average = -0.145, -0.007, 0.152. Average deviation = 0.101 (More precise) Significant Digits Significant digits keep track of the accuracy of a measurement and any calculation performed with that measurement. When making calculations with a measurement, one cannot give a result that is more accurate than the least accurate measurement used in the calculation. In a simple sense, significant digits clue you in on how much to round a number from a calculation. Counting Significant digits For any measurement, all digits are considered significant except zeroes at the beginning of a number and zeroes at the end of a number that does not contain a decimal. For numbers in scientific notation, the coefficient indicates the significant digits. Example 1: Underline and count the significant digits in the following numbers. a. 0.23 cm (two significant digits) b. 14.00 g (four significant digits) c. 1200 km (two significant digits) d. 3.040 x 1023 atom (four significant digits) Calculations with Significant Digits In calculations, you must round your result to the correct accuracy indicated by your original measurements. There are two general rules associated with the common operations of addition, subtraction, multiplication and division. Example 1: What is the result of the following calculation with the correct significant digits? 5.2g + 0.0236g Plan – For addition we consider the place of the final significant digit in each value used. We round our answer to the one with the highest place. In most cases this can be considered rounding to the least number of decimal places. 5.2 has 2 as its final significant digit, which is in the tenths spot (one decimal place). 0.0236 has 6 as its last significant digit in the ten thousandths spot (four decimal places). We will round our result to the tenths spot (one decimal place). Calculator: 5.2g + 0.0236g = 5.2236g Answer : 5.2 g with one decimal places Give the answers to the following calculations ensuring you round the answer to the correct number of significant digits 8. 18.04 cm + 2.5 cm = 20.5 cm 9. 105 g – 2.6 g = 102 g 10. 16.2050 m + 5.102 m = 21.307 m 11. 17.48 g – 0.483 g = 17.00 g Example 2: What is the result of the following calculation with the correct significant digits? 5.2 cm x 0.0236 cm Plan – For multiplication and division, we consider the number of significant digits. The result of the calculation will have the smaller number of significant digits of the measured values. 5.2 has two significant digits and 0.0236 has three significant digits. We will round our result to two significant digits. Calculator: 5.2 cm * 0.0236 cm = 0.12272 cm2 Answer : 0.12 cm2 with two significant digits Give the answers to the following calculations ensuring you round the answer to the correct number of significant digits 12. 13. 14. 15. 18.00 cm * 2.5 cm = 45 cm2 105 g / 2.56 g = 41.0 16.2050 m * 5.102 m = 82.68 m2 17.48 g / 0.483 g = 36.2 Multi-step problems When combining operations, keep track of significant digits after each step; however, only round at the end. Example 2: What is the result of the following calculation with the correct significant digits? ( 1.5 cm−0.806 cm ) 0.085 s Plan – We will perform our steps one at a time and highlight the digits that are significant after each step. Step 1 is subtraction, so our answer will have the least number of decimal places. Our result can only have one decimal place. We will not round, rather we will highlight this to keep track of significant digits 1.5cm – 0.806 cm = 0.694 cm (one decimal place, one significant digit) Step 2 is division. We divide our unrounded result from the first step by 0.085 s. With division we consider the total number of significant digits in the values. 0.694cm has only one significant digit and 0.085 has two. Our result, therefore, can only have one significant digit. 0.694 cm / 0.085 s = 8.1647… cm/s = 8 cm/s Perform the following calculations and give the answer with the correct number of significant digits and units: ( 2.057 cm+ 9.84 cm ) = 4.854 cm2/s 2.451 s 16. 17. ( 5.84 cm∗6.74 cm ) −6.45 cm2 = 32.9 cm2 18. {[ ] } (12.45−8.1) ∗(7.15 x 10−4 ) −0.52=¿ 1.0 2.03 x 10−3 Exercises – please give the answers to the following questions with the correct number of significant digits. 19. The mass of the moon is 7.346 x 1025 g. If the volume of the moon is 2.1968 x 1025 cm3, what is the density of the moon in g/cm3? D = m/v D = (7.346 x 1025g) / (2.1968 x 1025cm3) = 3.344 g/cm3 20. If a liquid has a density of 0.845 g/mL, what volume of liquid will have a mass of 15 g? D = m/v 0.845 g/mL = 15g / V V = 18 mL 21. A cube is found to have sides of 2.5 cm, 1.5 cm and 6.8 cm. If the cube has a mass of 0.857 g, what is the density of the cube with the correct significant digits in units of g/cm 3? V = LxWxH = 2.5cm x 1.5cm x 6.8cm = 25.5 cm3 (only the first two digits are significant) D = m/v D = 0.857g / 25.5 cm3 = 0.034 g/cm3 22. A sample of a pure metallic element has a mass of 12.45g. If the sample is placed into a graduated cylinder that contains 16.5 mL of water and the water rises to 19.8 mL, what is the density of the metal in g/mL? D = m/v V = 19.8mL – 16.5mL = 3.3 mL D = 12.45g / 3.3mL = 3.8 g/mL 23. A 1.045 g sample of a solid is dissolved in 7.57 mL of a liquid that has a density of 1.21 g/mL. If the total volume of the sample is 8.28 mL, what is the density of the solution? This is what we need to find and how to find it: Dsolution = msolution / vsolution We are given the volume but not the mass of the solution. We must find the mass of the solution: msolution = msolid + mliquid We are given the mass of the solid but not the liquid. We must find the mass of the liquid: mliquid = Dliquid * vliquid = 1.21g/mL * 7.57mL = 9.1597 g (only the first three digits are significant) Now we can determine the mass of the solution: msolution = 1.045g + 9.1597g = 10.2047 (only the first four digits are significant) and now the density: Dsolution = 10.2047g / 8.28mL = 1.23 g/mL Dimensional Analysis Dimensional Analysis is used to convert the units of a number. It does not involve changing the overall amount of a substance, it just changes the way it is presented. The process involves multiplying a value by an equivalency setup as a fraction also called a conversion factor. Example 1: Convert 12 μm to m Plan – To convert units we first must know the relationship between the units. 1 x 10 6 μm = 1m. We then take our original value and multiply it by the equality setup as a fraction such that the units cancel out properly. 12 μm ( 1 x 101 m μm )=1.2 x 10 −5 6 m 24. Convert the following – remember to show your work: a. 2.5 ms into s 2.5 ms ( 10001 sms )=0.0025 s b. 3.7 x 106 m into km 6 3.7 x 1 0 m ( 1 x110km m )=3.7 x 1 0 km 3 3 c. 5.34 lb to kg (1kg = 2.205 lb) 5.34 lb 1 kg ( 2.205lb ) =2.43 kg d. 12.5 cm to inches (2.54cm = 1 inch) 12.5 cm¿ In some cases, two or more steps will be required to convert a unit. Example 2: Convert 1.5 x 104 nm to cm Plan – We will convert from nanometers to meters first since 1 x 10 9 nm = 1m and then convert meters to centimeters knowing 100cm = 1m. 1.5 x 10 4 nm ( 1 x 101mnm )( 1001 mcm )=1.5 x 10 −3 9 cm 25. Convert the following – remember to show your work: a. 3.6 x 10-5 km to nm m 1 x 10 nm =3.6 x 10 nm ( 1000 1 km )( 1 m ) 9 −5 3.6 x 1 0 km 7 b. 456 μm to cm ( 1 x 101m μm )( 1 001mcm )=4.56 x 10 456 μm −2 6 c. 7834 seconds to hours 7834 s ( 160mins )( 601hrmin )=2.176 hr d. 3424.7 inches to m (2.54cm = 1 inch) 3424.7∈¿ cm When converting units in a denominator ensure that your conversion factors are setup correctly to cancel units. Example 3: Convert 3.4 km/min to m/s Plan – We will convert km to m using 1000m = 1km and will convert min to s using 1 min = 60 s ( 3.4 km 1000 m min 1 km )( 160mins )=57 m/ s 26. Convert the following – remember to show your work: a. 3.6 m/s to cm/μs ( 3.6 m 100 cm s 1m ) ( 1 x 110s μs )=3.6 x 10 −4 6 cm / μs b. 45.6 g/mL to mg/L ( )( ) 45.6 g 1000 mg 1000 mL =4.56 x 107 mg / L mL 1g 1L c. 3.7 J/mol to cal/mmol (Energy units: 4.184 Joule = 1 calorie) ( 3.7 J 1 cal mol 4.184 J )( 10001molmmol) =8.8 x 10 −4 cal/ mmol With squared or cubed units, one must convert the unit multiple times Example 4: Convert 5.6 x 103 cm3 to μm3. Plan – 100 cm = 1 m and 1x106μm = 1m Since cm is cubed, we have to do each conversion three times. ( 1001 mcm )( 1001 mcm)( 1001 mcm ) ( 1 x 101 m μm )( 1 x 101 mμm )( 1 x 101m μm )=5.6 x 10 6 5.6 x 103 cm3 6 6 15 μm 3 27. Convert the following – remember to show your work: a. 6.4 mm2 into m2 ( 10001 mmm )( 10001 mmm )=6.4 x 10 −6 6.4 m m2 m 2 b. 34.5 cm2 to in2 (2.54 cm = 1 in) 2 34.5 c m ¿ c. 4.5 g/cm3 into mg/mm3 ( 4.5 g 1000 mg 1g c m3 )(1001 mcm )( 1001mcm )( 1001 mcm )( 10001 mmm )( 10001mmm )(10001 mmm )=4.5 mg/ m m 3 Exercises – Use Dimensional Analysis to answer the following questions. Show your work! 28. The speed of light in a vacuum is 2.998 x 108 m/s. What is this in miles per hour? (1km = 0.621317 miles) ( )( )( 0.621317 mi 60 s 2.998 x 108 m 1 km s 1000 m 1 km 1 min )( 601hrmin )=6.706 x 10 mi/hr 8 29. The acceleration on earth due to gravity is 9.8 m/s2. What is this in km/min2? ( )( )( ) 2 9.8 m 1 km 60 s 60 s =35 km/min s 2 1000 m 1mim 1 mim 30. The charge to mass ratio of an electron is 1.76 x 108 C/1 g. If the charge of an electron is 1.602 x 10-19 C, what is the mass of an electron? 1.602 x 1 0−19 C ( 1.76 1x g1 0 C )=9.10 x 1 0 −28 8 g 31. One mole of atoms is equivalent to 6.022 x 1023 atoms. For carbon, 1 mole of carbon atoms has a mass of 12.011 grams. Convert 35.32g of carbon into number of atoms of carbon. 6.022 x 10 atoms C 1 mol C ( 12.011 )=1.771 x 10 g C )( 1mol C 23 35.32 g C 24 atoms of C 32. The hope diamond is the largest deep blue diamond with a mass of 45.52 carats. If 1 carat is equivalent to 0.200g and the density of diamond is 3.51 grams per cubic centimeter, what is the volume of the hope diamond in cm3? g 1cm =2.59 cm ( 0.200 1 carat )( 3.51 g ) 3 45.52 carat 3