Atomic theory
Odds and ends
AMU
 1.6 x 10 -27 Kg
 Mass of 1/12 of carbon – 12 isotope
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
Atomic masses
 Average mass of various isotopes
 Found by multiplying isotope mass by its percentage in
nature, then adding them all up.
examples
 Oxygen-16
x 99.762%
 Oxygen-17
x
 Oxygen – 18
x 0.200% =

=
0.038% =
Then add them up__________
example
 Copper 63
x 69.17%
=
 Copper 65
x 30.83%
=

Then add
_________________
example
 Argon - 40
99.600%
 Argon - 38 .063%
 Argon - 36

.337%
then add _________
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?
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!
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 1023 particles
calculation
 Average Amu of carbon = 12.01
 12.01amu x 1.661x10-27 Kg x 1000 g = 1.99486 x 10-23
Amu
 1.99486 x 10 -23 g

Kg
x 6.023 x 1023 particles
mole
= 12.01 g
mole
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!!!!!
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
 MULTIPLE PROPORTIONS
 WHEN DIFFERENT COMPOUNDS ARE FORMED
FROM THE SME TWO ELEMENTS, THE RATIOS ARE
ALWAYS SMALL WHOLE NUMBERS.
 WATER
H20 vs peroxide
H 20 2
 CARBON DIOXIDE CO2 vs CARBON MONOXIDE CO