1. Atomic theory Odds and ends

2. AMU 1.6 x 10 -27 Kg Mass of 1/12 of carbon – 12 isotope

3. Masses of particles Proton plus charge 1.6 x 10 -27 Kg = 1 amu Neutron no charge 1.6 x 10 -27 Kg = 1 amu Electron negative charge 9.11 x 10 -31 Kg = 0 amu

4. Atomic masses Average mass of various isotopes Found by multiplying isotope mass by its percentage in nature, then adding them all up.

5. examples Oxygen-16 x 99.762% = Oxygen-17 x 0.038% = Oxygen – 18 x 0.200% = Then add them up__________

6. example Copper 63 x 69.17% = Copper 65 x 30.83% = Then add _________________

7. example Argon - 40 99.600% Argon - 38.063% Argon - 36.337% then add _________

8. FINAL EXAMPLE Naturally occuring Boron is 80.20% boron-11 and and 19.80 % of some other isotope. The average mass is 10.81. What is the mass of the other isotope?

9. MORE ATOM STUFF NOW THAT WE KNOW AN AMU IS 1.6 X 10 -27 Kg WHAT CAN WE DO WITH THAT INFO? IF WE KNOW THE AVERAGE AMU FOR AN ATOM, WE CAN TRANSLATE THAT INTO A QUANTITY WE CAN WORK WITH IN THE LAB!

10. THE MOLE The amount of a substance that contain as many particles as there are atoms in exactly 12g of carbon - 12. By experiment that amount is 6.02 x 10 23 particles

11. calculation Average Amu of carbon = 12.01 12.01amu x 1.661x10 -27 Kg x 1000 g = 1.99486 x 10 -23 Amu Kg 1.99486 x 10 -23 g x 6.023 x 10 23 particles = 12.01 g mole mole

12. RESULT WE CAN NOW HANDLE ATOMS IN THE LABORATORY AND KNOW HOW MANY WE HAVE! IF I HAVE 12.01 GRAMS OF CARBON, I HAVE 1 MOLE OF CARBON, I HAVE 6.023 X 10 23 ATOMS SAME WITH ANY OTHER MASS ON THE PERIODIC TABLE!!!!!

13. 3 MAJOR LAWS OF CHEM CONSERVATION OF MATTER IN ANY ORDINARY CHEMICAL REACTION, MATT IS NOT CREATED OR DESTROYED DEFINITE PROPORTIONS IN A GIVEN COMPOUND, THERE ARE ALWAYS THE SAME ELEMNTS IN EXACTLY THE SAME MASS RATIO

14. MULTIPLE PROPORTIONS WHEN DIFFERENT COMPOUNDS ARE FORMED FROM THE SME TWO ELEMENTS, THE RATIOS ARE ALWAYS SMALL WHOLE NUMBERS. WATER H 2 0 vs peroxide H 2 0 2 CARBON DIOXIDE CO 2 vs CARBON MONOXIDE CO