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Understanding Atomic Mass and Mole Concepts in Chemistry

This guide explores atomic theory, focusing on atomic mass calculations and the concept of the mole. It details how to determine the average mass of isotopes by multiplying isotope masses by their natural abundances. Examples are provided for elements such as oxygen, copper, and argon. The mole is defined as the amount of substance containing 6.02 x 10²³ particles, derived from the atomic mass unit (amu). This understanding allows for precise laboratory work with elemental and compound substances while adhering to key chemical laws like conservation of matter and definite proportions.

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Understanding Atomic Mass and Mole Concepts in Chemistry

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  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 1023 particles

  11. calculation • Average Amu of carbon = 12.01 • 12.01amux 1.661x10-27 Kgx 1000 g = 1.99486 x 10-23 Amu Kg • 1.99486 x 10 -23 gx 6.023 x 1023 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 H20 vs peroxide H20 2 • CARBON DIOXIDE CO2vs CARBON MONOXIDE CO

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