1. SECTION 4.3 HOW ATOMS DIFFER

2. PENNIES AS ATOMS MASS OF SAMPLE 1: MASS OF SAMPLE 2: IF THEY HAD THE SAME NUMBER OF PENNIES, WHY WERE THE MASSES DIFFERENT? THERE ARE TWO DIFFERENT TYPES OF PENNIES (BEFORE AND AFTER 1983) THAT HAVE DIFFERENT MASSES

3. THE PERIODIC TABLE Hydrogen 1 H 1.008 What information does it give us? Chemical Name Atomic Number Chemical Symbol Average Atomic Mass

4. ATOMIC NUMBER ATOMIC NUMBER: THE NUMBER OF PROTONS IN AN ATOM IDENTIFIES AN ATOM AS A PARTICULAR ELEMENT ALSO EQUAL TO THE NUMBER OF ELECTRONS WHY? POSITIVE PROTONS AND NEGATIVE ELECTRONS CANCEL EACH OTHER OUT FOR A NEUTRAL ATOM Hydrogen 1 H 1.008

5. COMPLETE THE FOLLOWING TABLE ElementSymbolAtomic Number ProtonsElectrons LeadPb82 OxygenO888 ZincZn30

6. ISOTOPES AND MASS NUMBER WHAT ARE THE RELATIVE MASSES OF PROTONS AND NEUTRONS? MASS NUMBER: THE SUM OF NUMBER OF PROTONS AND NEUTRONS IN THE NUCLEUS MASS NUMBER = 4

7. ISOTOPES AND MASS NUMBER ISOTOPES: ATOMS WITH THE SAME NUMBER OF PROTONS AND DIFFERENT NUMBER OF NEUTRONS Isotopes of Hydrogen Protium DeuteriumTritium

8. NOTATION # Protons12 # Electrons12 # Neutrons12

9. NOTATION Chemical Name – Mass Number Potassium - 40 # Protons19 # Electrons19 # Neutrons21

10. ElementAtomic Number Mass Number Protons & Electrons NeutronsIsotopeSymbol Neon22 Calcium46 Oxygen17 Iron57 Zinc64 Mercury204

11. REVIEW: ISOTOPE SYMBOLS Cu 63 29 # Protons # Neutrons # Electrons

12. MASS OF ATOMS Atomic mass unit (AMU) – 1/12 the mass of a carbon-12 atom ParticleMass (amu)Relative Mass Electron0.0005491/1840 Proton1.0072761 Neutron1.0086651

13. ATOMIC MASS -Most elements exist as a mixture of isotopes, so not every atom is exactly the same -Atomic mass- the weighted average mass of the isotopes of an element

14. Given the data in the table, calculate the atomic mass of unknown Element X. Then, identify the unknown element. SOLVE: 6 X: mass contribution = (mass)(percent abundance) 7 X: mass contribution = (mass)(percent abundance) mass contribution = (6.0115 amu)(0.0759) mass contribution = (7.016 amu)(.9241) = 0.4565 amu = 6.483 amu Isotope Abundance for Element X IsotopeMass (amu)Percent Abundance 6X6X 6.0157.59% 7X7X 7.01692.41%

15. SOLVE: 6 X: mass contribution = (mass)(percent abundance) 7 X: mass contribution = (mass)(percent abundance) mass contribution = (6.0115 amu)(0.0759) mass contribution = (7.016 amu)(.9241) = 0.4565 amu = 6.483 amu Atomic mass = 6.939 amu What is the element?