1. Isotopes and Ions Variations on the Atom Dr. M. Hazlett Mandeville High School

2. Isotopes All atoms of an element have the SAME number of protons (p + ) The p + number is the atomic number (Z) – This is a constant – For example: All Sodium (Na) atoms have 11 p + – If an atom loses a proton, it becomes a different element If Na loses 1 p +, then it has become Neon (Ne)

3. Z = atomic number = p + The number of protons identifies the atom and which element it is In a stable atom: – # p + = # n 0 = # e - – Thus, Na in its stable form has 11 p + ; 11 n 0 ; and 11 e - – If it has an unequal number of p + and n 0, then it is called an ISOTOPE

4. Theoretically – an element can have as many isotopes of itself as it has neutrons, or it can add an unlimited number of n 0 For example: H has 3; C has 16; Al has 25 – These can be looked up in the CRC (the Chemistry/Physics Data Bible) or on the internet – Remember – a change in the number of n 0 does not change the element’s atom – only a change in the number of protons can do that!

5. The Carbon Isotope

6. Ions Ions are when an atom has an unequal number of p + and e - Remember – a stable atom has a neutral overall charge due its equal number of p + and e - When an atom loses or gains an e -, its charge changes accordingly – Loss of e - means a + charge; gaining an e - means a – charge for the atom

7. Losing or Gaining e -..... If an atom loses an e -, then it has more p + than e - and it will have an overall positive charge Different elements’ atoms can lose 1, 2, 3, or even 4 electrons depending on various factors If an atom has LOST e -, then it is called a CATION or a positive ion – A Cation would be written as Al + (the one being understood) or Al +3

8. Atoms can also gain electrons If an atom gains electrons (from 1 up to 4), then it will have more e - than p + and will end up having an overall negative charge A negatively charged ion is called an ANION – The element is shown this like: Na - (the 1 is understood) or Na -2 The losing or gaining of electrons determines what type of bonds the atoms will form, and which atoms will bond to others

11. Ions in Water Solution

12. Using the Periodic Table Elements in the Main Groups (A), form fairly consistent ions – LEARN TO USE THE CHART Group IA will form +1 ions; Group 2A form up to +2; Group 3A form up to +3 ions Group 4A will form either up to -4 or +4 ions Group 5A will form up to -3 ions; Group 6A up to -2; Group 7A form -1; and Group 8A will not form ions at all Those elements in the B groups vary and we’ll learn those later

14. Ions and Isotopes in Review Stable atom: #p + = #n 0 = #e - Atomic Mass - #n 0 = # p + Atomic Mass - #p + = #n 0 If charge is 0, then #p + = #e - If charge is positive, then #p + > #e - Cation If charge is negative, then #p + < #e - Anion

15. Examples: Li -1 has gained an electron, meaning there is one more negative charge than positive ones – It has 3 p + and 4 e - Li +1 has lost an electron, meaning there is one more positive charge than negative ones – It has 3 p + and 2 e - REMEMBER: The # of p + DO NOT CHANGE Only the number of n 0 (isotope) and e - (ion) change

16. Cf -3 has an atomic number of 98 – This means it has 98 p + – Its atomic mass is 216 – It has 118 n 0, (216 – 98), making it an ion and an isotope! – Since it has a -3 charge, the number of e - will be 101; (98 + 3) – Zn +1 has 30 p + and n 0 ; but due to the +1 charge, it has only 29 e -

18. Mass Number and Atomic Mass An atom’s mass number = # p + + # n 0 The atomic mass unit (amu or u) is a little more complex – It is an average of all of an atom’s isotopes and what percent abundance that isotope is in nature Abundances will add up close to 100% The closer to a whole number the amu is, the fewer the isotopes that exist

19. Determining the average atomic mass: Average Atomic Mass = (Mass of Isotope 1)(% Abundance of Isotope 1) + (Mass of Isotope 2)(% Abundance of Isotope 2) + (Mass of Isotope 3)(% Abundance of Isotope 3) + (Mass of Isotope ∞)(% Abundance of Isotope ∞) AMU is a little different.......

20. AMU (sometimes just an ‘u’) Average Mass Unit – It uses C-12 as a reference point C-12 has 6 protons and 6 neutrons 1 amu is the equivalent of 1/12 of a Carbon’s mass Mass amu n 0 1.675 x 10 -24 g 1.008665 p + 1.673 x 10 -24 g 1.007276 e - 9.1 x 10 -28 g 0.000549

21. Average Atomic Weight example: For an unknown element we know that: the mass of Isotope 1 is 6.015 amu and its abundance is 7.5% The mass of Isotope 2 is 7.016 amu with a 92.5% abundance Therefore – – (6.015)(.075) + (7.016)(.925) = 6.941 amu – Looking on the Periodic Chart we can see the element is Lithium (Li)

22. Another example: N 14 and N 15 have a total amu of 14.007. What are the percentages of abundance? Make the abundances equal to x and (x-1). Thus: 14(x) + 15(1 - x) = 14.007 14x + (15 – 15x) = 14.007 - x = 14.007 - 15 so, x = 99.3 % for N 14 and, 1 – x = 0.7% for N 15

23. On the Periodic Table: The top number is Z, the Atomic Number or number of p + The Element’s Symbol The element average atomic weight set by isotopes and abundances

24. If the Atomic Weight is in (parentheses), then it is a synthetically made element and it has no known isotopes The closer to a whole number the atomic weight is, the fewer isotopes the element has To discover known isotopes and abundances – use the CRC Handbook

25. Conservation of Mass Conservation of Mass means that the mass of the reactants will equal the mass of the products after the reaction – This is true no matter how many reactants or products exist in the reaction – Example: Fe with a mass of 15.72 g; placed in a solution of 21.2 g Cu(II)Sulfate. Cu separates. How much Fe (II) Sulfate created?

26. – The final masses of the reaction (rxn) are Fe = 8.33 g; and Cu = 8.41 g – Thus – 15.72 g – 8.33 g = 7.39 g – m reactant 1 + m reactant 2 = m product 1 + m product 2 m Fe + m Cu = m CuS + m FeS m FeS = m Fe + m CuS - m Cu m FeS = 7.39 g + 21.12 g – 8.41 g = 20.10 g

27. Law of Definite Proportions In a compound, the same elements will be in the same proportion by mass Example: – 100 g H 2 O contains 11.19 g of H 2 and 88.81 g O – % Composition = mass element x 100 mass compound Well, what does it equal???????

28. OK – try another one.... 25 g of a compound with 6.77 g tin and 18.23 g bromine. What percent is tin by mass? – mass tin x 100 = 6.77 x 100 = mass compound 25 Did you get the answer?

29. The End Now, onto the Periodic Table!