1. Isotope l The average atomic mass of each element uses the masses of the various isotopes of an element l An isotope of an element is the same element with the same # of p +, but a varying number of neutrons and thus a varying atomic mass l NOT all elements contain isotopes

2. Isotope Determination ex. H There are 3 isotopes of Hydrogen atomic mass atomic # #p + #n 1.00 1 H1.00 110 2.00 1 H2.00 111 3.00 1 H3.00 112 The average atomic mass of H = 1.0079

3. Isotope Determination There are three isotopes of Carbon 12 CCarbon 12 13 CCarbon 13 14 CCarbon 14

4. Groups - # valence e - Periods - Size of atom Noble gases Group VIII Halogens Group VII Alkalli Metals Group I Metals Nonmetals Metalloids Alkali Earth Metals Group II

5. Formula Weights H2H2 N2N2 O2O2 F2F2 Cl 2 Br 2 I2I2

6. Chemical Equations Sodium + Diatomic Chlorine ------> Sodium Chloride Reactant(s) ----- yields -----> Product(s) Na + Cl 2 ----------> NaCl Equations must be balanced due to the Law of Conservation of Matter

7. Chemical Equations Bookkeep the Atoms Count the individual atoms of reactants and products line up under each atom in separate rows: Na + Cl 2 NaCl 1 Na 2 Cl 1 Cl Use whole number multipliers to equalize the atoms of reactants and products. Place these whole number prefixes before the element or the compound

8. Chemical Equations Bookkeep the Atoms Na + Cl 2 NaCl 1 Na 1 Na 2 Cl 1 Cl

9. Chemical Equations 2Na + Cl 2 2NaCl 2 Sodium atoms react with 1 chlorine molecule to form 2 “Molecules” of Sodium Chloride

10. Chemical Equations 1. (NH 4 )SO 4 H 2 SO 4 + NH 3 5. Fe + H 2 O Fe 3 O 4 + H 2 TRY in groups #6 and #9

11. % of an Element in a Compound (Elemental Analysis) l Mass of Element x 100 FW of Compound l % H in H 2 O 2H x 100 =2.00 amu x 100 = 11.1% H 2 O18.00 amu l % O in H 2 O O x 100 = 16.00 amu x 100 = 88.89% H 2 O 18.00 amu

12. Group Work l Determine the Elemental Analysis of: »C 6 H 12 O 6 »NH 3 »CO 2

13. Definition l Mole (mol or n) »1 mole contain the same number of entities (atoms, molecules, ions, particles) as there are in exactly 12.01 grams of Carbon and a mole contains 6.022 x 10 23 number of entities. This number is called Avogadro’s Number.

14. 1.00 g H 12.01 g C 18.00g H 2 O 35.45 g Cl - 180.06g C 6 H 12 O 6 6.022 x10 23 atoms 6.022 x10 23 atoms 6.022 x10 23 molecules 6.022 x10 23 ions 6.022 x10 23 molecules

15. Analogy l 1 pair = 2 l 1 trio = 3 l 1 dozen = 12 l 1 mol = 6.022 x 10 23

16. Molar Mass (MM) To Determine the Molar Mass (MM): For an element - it’s the atomic weight in grams For a compound - it’s the formula weight in grams A mole contains Avogadro’s Number (6.022 x 10 23 ) particles (atoms, ions, molecules).

17. Working with Moles l Formula given grams of a compound or element, determine the number of moles l Moles = grams MM OR l Dimensional Analysis

18. Working with Moles l Formula given moles of a compound or element, determine the number of grams moles (MM) = grams OR l Dimensional Analysis

19. Multiply by Avogadro’s #, Divide by Molar Mass Multiply by Molar Mass, Divide by Avogadro’s # Number of Moles Number of atoms, ions, molecules Mass in grams