Significant Figures The Jacobs Way

Transcription

1 Significant Figures The Jacobs Way Measured values are limited by the measuring device used. This limitation is described in the precision of the value. This precision is expressed by the number of digits found in the value, or even better, by the place of the final digit found in the value. Understanding measurement is important (and more difficult than you may think), so much so that the class will complete a short, in class lab experience early in the year dealing with measurement. Regardless, the basics of significant figures are review for you. Students should be able to identify both the number of significant digits found in a value as well as the point of precision. The difference being described in this presentation is how you propagate measured values. Propagation is the act of performing the math associated with known physical relationships. The scientist must have a methodology that reduces the chances for rounding errors. During Chemistry I you may have used the odd-even rule, a method which is valid statistically, but less so when a small number of significant digits are present. In this course we will be evaluating quantitative error, which necessitates the use of a new methodology for handling our calculations because this error value only contains a single significant digit.

2 Imagine a cylinder of copper with the following dimensions: Height = cm diameter = 9.10 cm and a mass of mass = 13.1 kg Cu Problem: Calculate the density of copper in units of grams/cm 3. Step 1: Ask yourself how do I calculate density? The formula for density is Density = mass volume Step 2: Ask yourself what is the mass of the object? The mass is a given, so no real calculations need to be done. You should notice that the units associated with the mass do not coincide with those that the problem are asking for. In this particular case a metric conversion must be done kg 1000 x 1 1kgg = 13,100 g

3 This particular operation was accomplished by using a defined relationship (1 kg = 1000 g); defined values do not limit the number of significant figures found in the answer of a calculation. Another way to look at this is to think of defined values as being infinitely significant. The 1.31 kg is measured, the relationship defined. Step 3: Determine the volume of the cylinder. This can be found by using a known geometric relationship. Density cylinder = π r 2 h Part (a) π is an irrational constant. You could choose to use an infinite number of figures, but this would be statistical overkill. A common sense approach would be to use one digit beyond what your measured values are limiting you to, and in this case that would be 5 digits. Since every value in this constant is accurate, rounding this fourth digit is allowed (confused by this statement? More on this later!). So, rounds to five significant figures so that π = Part (b): Calculate the radius. This is a simple calculation with a sophisticated explanation. Please jump to the next slide.

4 Radius = Diameter 2. The value 2 is a defined mathematical concept, and therefore does not limit. Additionally, the process is a repetitive subtraction. In other words, you are taking half of the diameter and subtracting that value from the diameter cm 4.55 cm = 4.55 cm. This is important to note because when we add or subtract the answer arrived upon can only be as precise as the least precise measured value. In repetitive operations all of the values should be equally precise, and no precision should be lost. This will be an important point during molar mass calculations. Part (c): Complete the squaring of the radius (4.55 cm) 2 = cm 2 Two unique operations have been performed during this calculation. Notice how both the numbers and the units have been squared; mathematical operations are also performed on the units associated with the values. The next concern centers on what do we do with all of the numbers we find on the calculator screen? I m sure the lazy part in all of us would like to leave the numbers in the calculator and move on to the next step, but technically that isn t allowed

5 (while leaving the entire value in the calculator may not effect the final answer, there is a chance that it might.) The value contains three significant digits, one insignificant digit and two ludicrous digits. Let me explain these terms to you, the last of which is one that you should only find used in my classroom. The radius value (4.55) contains three significant figures; during multiplication and division our rule is that value with the fewest SF s will limit how many the answer can contain. Squaring the value 4.55 means that we ve used it twice in the calculation and, logically, the answer will be limited to three significant figures. The fourth value does contain some statistical value, and can be retained to help reduce the errors that can occur due to rounding. Any values occurring thereafter (the second two and the five) have no statistical value whatsoever; they amount to numbers spinning on a slot machine where any arbitrary value could appear. They are ludicrous, and have no mathematical value. In fact, those values could themselves cause a rounding mistake. Ludicrous values should be truncated, or dropped from the value, without having any impact on rounding. In other words, we always leave the insignificant number alone; it is never rounded through the use of ludicrous values. The

6 insignificant figure must be denoted in a way that it can be identified. This figure is written as a subscript, so on a hard copy of calculations the squared radius should be written as cm 2. While the zero will eventually have no impact on our final answer, it is good form to apply this practice to all insignificant figures notations. Part (d): Since h is known, the volume can now be calculated. V = (3.1416)( cm 2 )(22.50 cm) = 1,46 3 cm 3 Notice how the volume contains three significant figures, one insignificant figure as well as a cubic unit. Part (e): Calculation of the density of copper. 13,100 g Density = cm g Density = or cm since the 4 doesn t round cm the 5. Final answers NEVER contain an insigfig!! Just for fun s and haha s check that fancy thing we call the internet for the accepted value for the density of copper. Did the measurements associated with this problem yield an accurate value? A synonym for accurate is correct. Is correct synonym for precise? Please think about this point!! g

7 Try one for yourself! Problem: What is the kinetic energy associated with a large, ton boulder rolling down a mountain side traveling at a velocity of 47.3 miles per hour? What will you need to answer this problem? You need to know the relationship for translational kinetic energy. You know, the one you learned in 8 th grade?!? You will soon find that I expect you to learn for life, not for the test. This class will build (not isolate) concepts. You should know the definition for the Imperial ton. While I don t expect you to know some of the more esoteric units found in the Imperial system (such as a Furlong; few of you have fathers who play the horses, as I did) there are many that I will. It is our measurement system, you know. The conversion factor from lbs to kg. Is this defined or measured? You should figure this out! The conversion factor from inches to cm. Is this defined or measured? Again, an important distinction! A knowledge of SI units. I want this answer in SI units.

8 The Answer Don t cheat, do the problem before you check! (1.384 tons)(2000 lbs/ton) = 2, lbs (2, lbs) (1 kg/ lbs) = 1, kg (47.3 miles/hour)(5280 ft/mile)(12 in/ft)(2.54 cm/in) (1 m/100 cm)(1 hour/3600 sec) = m/sec KE = ½ mν 2 KE = ½ (1, kg)( m/sec) 2 = 13,2 7 0 kg m 2 /sec 2 Final Answer is 13,300 Joules What is the actual point to this problem? Do not.. get trapped under a boulder in this class! Keep up with your work. And work smart do not live the legend of Sisyphus!

9 A little extra help Stoichiometry requires the use of molar masses. In any class I teach we round all the values on the periodic table to the hundreths place (that would be two places past the decimal point) unless the table doesn t provide you that degree of precision (very often the case with Pb, or lead.) Calculating molar masses is often an example of repetitive addition. For practice let s calculate the molar mass of Aluminum Sulfate, or Al 2 (SO 4 ) 3. 2 x (26.98 g/mol) = g/mol 3 x (32.06 g/mol) = g/mol 12 x (16.00 g/mol) = huh? There is often confusion here but I want you to remember our concept of repetitive addition or subtraction. Students understand that 12 is counted and therefore doesn t impact the number of significant figures. BUT, with containing 4 SF s they will often leave this value as g/mol. NO!! Multiplying a measured value by a counted value isn t really multiplication, it is repetitive addition, and you shouldn t lose any precision. Therefore, the answer is g/mol. Final answer = = g/mol