Weighted Averages and Relative Atomic Mass PowerPoint

Weighted AveragesHow relative atomic mass(or atomic weight) iscalculated.

Simple Average You know from your math classesthat averaging numbers (finding themean) ) is fairly easy. One simply adds all the numbers inthe set together and divides by thenumber of entries.

An Example Let’s s imagine a classof 10 students took a10 point quiz. Here are the scores: Calculate the averagescore. 7.4 is correct. Did any of thestudents actually get7.4 points? 10 10 9 8 8 8 7 6 6 2

Weighted Average There is another way to solve such aproblem. Rather than calculating a simpleaverage, we will calculate a weightedaverage. A weighted average is calculated asfollows: Σ(relativeabundance) x (value of data class)

Weighted Average The Greek letter sigma (Σ)() means“sum of”. Let’s s define the data classes first. There were several different scoreson the quiz. Each type of score is adata class. There were 10s, 9s, 8s, 7s, 6s and a2.

Weighted Average Next we need the relative abundanceof each. Relative abundance is the number ofitems in the data class/total entries. There were 2 scores of 10 and 10total scores. 2/10 = 0.2 The relative abundanceof 10s is 0.2.

Weighted Average There was only 1 score of 9; relativeabundance = 1/10 = 0.1 There were 3 scores of 8. RA = 0.3. Only 1 score of 7. RA = 0.1, etc. A table might make this easier tovisualize.

Weighted AverageScoreRelative Abundance10 0.29 0.18 0.37 0.16 0.22 0.1Total 1.0The total of the relative abundances must be1.0 (or very close.)

Weighted Average Because the data classes occur withdifferent frequencies, , we cannotsimply add the values of the dataclass together. We must give each data class aweight. . (Thus the term ‘weighted’average.) Weighting each class eliminates theneed for the simple average we usedearlier.

Weighted Average Average quiz score = Σ(testscores) x (relative abundance) 10•(0.2) + 9•(0.1) 9+ 8•(0.3) 8+7•(0.1) + 6•(0.2) 6+ 2•(0.1) 2 2.0 + 0.9 + 2.4 + 0.7 + 1.2 + 0.2 =7.4 This is the same result we obtainedearlier.

Why bother? At this point you may be saying,“That method looks like a lot ofwork. Why would I want to use it?” We probably wouldn’t t use it forcalculating the average of 10 quizscores, but what if you had 100scores, or 1000 to do?

Relative Atomic Mass This data is obtained using a devicecalled a ‘mass spectrometer.’

Relative Atomic Mass Actually, we have no choice but tocalculate average atomic mass in thisway. Atoms are to small to even see, let alonecount individually. So we can’t t calculate asimple average (like you did with theBeanium Lab) even if we wanted to. And, a small amount of matter contains ahuge number of atoms, so relativeabundances or % frequencies are the onlypractical way to go about it.

An Example Problem Chlorine has 2 naturally occurringisotopes: Cl-35 (accurate mass34.968amu) and Cl-37 (accuratemass 36.965amu). Cl-35 abundance is 75.77%(RA = 0.7577). Cl-37 abundance is 24.23%(RA = 0.2423). We cannot simply add 35 and 37 toget an average of 36.

An Example Problem To get chlorine’s s relative atomic mass, wecalculate the weighted average: 34.968amu•(0.7577) + 36.965amu•(0.2423)= 35.45 amu. This is the value reported on the periodictable of elements as the atomic mass ofchlorine. Is there any actual chlorine atom that hasa mass of 35.45amu? No! One cannot have ½ an atom.

Conclusion So, even though the mass number ofan isotope is given as a wholenumber (p + + n 0 ), we see that therelative atomic mass on the periodictable is really a weighted average ofall of an element’s s naturallyoccurring isotopes.

Conclusion By rounding an element’s s atomicmass to the nearest whole number,we can make a reasonable guessabout what that element’s s mostcommon isotope is.Rounding uranium’s atomicmass to 238 gives us themost common isotope, U-238. This doesn’t alwayswork, but it does most ofthe time.