Weighted Averages and Relative Atomic Mass PowerPoint

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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.
Page 1 and 2: Weighted AveragesHow relative atomi
Page 3 and 4: An Example Let’s s imagine a clas
Page 5 and 6: Weighted Average The Greek letter s
Page 7 and 8: Weighted Average There was only 1 s
Page 9 and 10: Weighted Average Because the data c
Page 11: Why bother? At this point you may b
Page 14 and 15: Relative Atomic Mass Actually, we h
Page 18: Conclusion By rounding an element

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