1. Isotopes atoms of a given element that differ in the number of neutrons …and consequently in mass.

2. Working with Atomic Notation AzXAzX Atomic Number Chemical Symbol Atomic Mass (of Isotope)

3. Example #1 12 6 C Atomic Number Chemical Symbol Atomic Mass (of Isotope)

4. Example #2 13 6 C Atomic Number Chemical Symbol Atomic Mass (of Isotope)

5. Example #3 14 6 C Atomic Number Chemical Symbol Atomic Mass (of Isotope)

6. Some isotopes of carbon 12 6 C 13 6 C 14 6 C

7. 10Ne Neon 20.1797 3Li Lithium 6.941 47Ag Silver 107.8682 11Na Sodium 22.98977

8. Various ways of identifying Isotopes Using atomic notation, Example: 12 6 C or simply 12 C Using the mass notation. Example: Carbon-12 or C-12 (read “carbon twelve” or “C twelve”)

9. Isotopes Some Isotopes of Carbon mass notation atomic notation # of p + # of e - # of n o C-11 11 C66 C-12 12 C66 C-13 13 C66 C-14 14 C66

10. Another example of isotopes 1 1 H 2 1 H 3 1 H or H-1H-2H-3

11. Isotopes The Isotopes of Hydrogen mass notation atomic notation # of p + # of e - # of n o Hydrogen-1 1H1H11 Hydrogen-2 2H2H11 Hydrogen-3 3H3H11

12. p + = n o = e - = p + = n o = e - = p + = n o = e - = Figure #1 Figure #2 Figure #3

13. Isotopes atoms of a given element that differ in the number of neutrons …and consequently in mass.

14. Why are masses on the periodic table usually expressed as decimal numbers? masses on the table are weighted averages of all known isotopes of the element of interestmasses on the table are weighted averages of all known isotopes of the element of interest

15. Keep in mind:  It is not possible to determine how many different isotopes exist by looking at the periodic table.  It is not possible to determine the frequency of various nuclides by looking at the periodic table.

16. The following does not occur in nature! 1 1 Hoccurrence 33.3% 2 1 Hoccurrence 33.3% 3 1 Hoccurrence 33.3%

17. The following does occur in nature! 1 1 Hoccurrence 99.98% 2 1 Hoccurrence 0.0156% 3 1 Hoccurrence 0.0044%

18. another way of looking at it: 1 1 Hoccurrence 9,998 2 1 Hoccurrence 1.56 3 1 Hoccurrence 0.44 Imagine having 10,000 H atoms

19. That means the weighted average is: 1 1 H 1 x 0.9998 = 0.9998 2 1 H 2 x 0.00156 = 0.00312 3 1 H 3 x 0.00004 = 0.00012 Weighted Average (0.9998 + 0.00312 + 0.00012)  1.01

20. Zn has 5 naturally occurring isotopes 64 30 Znoccurrence 66 30 Znoccurrence 67 30 Znoccurrence 68 30 Znoccurrence 70 30 Znoccurrence

21. Zn has 5 naturally occurring isotopes 64 30 Znoccurrence  49 % 66 30 Znoccurrence ~28 % 67 30 Znoccurrence ~4 % 68 30 Znoccurrence ~18 % 70 30 Znoccurrence ~1 %

22. Zn has 5 naturally occurring isotopes 64 30 Zn  49 % 66 30 Zn~28 % 67 30 Zn~4 % 68 30 Zn~18 % 70 30 Zn~1 % 65.39

23. (parenthesis) on the Periodic Table indicate the most stable isotope stable means “longest living” Parenthesis also suggest the element of interest is radioactive.

24. Review problem #1 Represent the following using atomic notation. 92U Uranium 238.0289

25. Review problem #2 Represent the following using mass notation. 10Ne Neon 20.1797

26. Review problem #3 If the atom described below had 2 naturally occurring isotopes, which of the 2 would have a greater frequency of occurrence? Express your answer in atomic and mass notation. 3Li Lithium 6.941

27. Review problem #4 How many total subatomic particles are in the following “neutral” atoms of Fe-55 and Fe-57?