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First SAT given (1926). Reading: Chapter 5 sections 1-5 (quiz) Electron Configuration HW Due Tomorrow HW 11: DUE Tomorrow Chap. 11#’s 36, 45, 47, 49-52 (all), 59, 60, 63, 64, 65, 67, 99 (no extra credit) HW #5.1: DUE Wednesday
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First SAT given (1926) • Reading: Chapter 5 sections 1-5 (quiz) • Electron Configuration HW Due Tomorrow • HW 11: DUE Tomorrow • Chap. 11#’s 36, 45, 47, 49-52 (all), 59, 60, 63, 64, 65, 67, 99 (no extra credit) • HW #5.1: DUE Wednesday • 3, 4, 5, 8, 9, 11, 13, 25-35 (odd), 41, 45, 49, 57, 80, 81, 85, 89 • HW #5.2: DUE Thursday • 17, 19, 21, 39, 43, 47, 53, 59, 71-77 (all), 83 • For lab tomorrow (dry lab)
Atomic Masses • Measured in the unit amu • Stands for atomic mass unit • Defined as 1/12th the mass of a 12C atom • 35Cl = 34.9689 amu • 50V = 49.9472 amu • 79Br = 78.9183 amu • Are these what the P.T. say?
All Isotopes are NOT Created Equal • Weighted averages • An equation (oh crap) • 35Cl = 34.9689 amu • Abundance = 75.77% • 37Cl = 36.9659 amu • Abundance = 24.23% • These must be the only 2 isotopes... Why?
All Isotopes are NOT Created Equal • 35Cl = 34.9689 amu • Abundance = 75.77% • 37Cl = 36.9659 amu • Abundance = 24.23% =0.7577 =0.2423
All Isotopes are NOT Created Equal • 35Cl = 34.9689 amu • Abundance = 75.77% • 37Cl = 36.9659 amu • Abundance = 24.23% =0.7577 Where have I seen this before? =0.2423 x 0.2423 = 35.45amu x 0.7577 34.9689 + 36.9659 Periodic Table gives AVERAGE ATOMIC MASS in amu!!
Some Algebra and a New Equation decimal, not % (i.e. 0.1234 not 12.34%) For solving for % abundance multiply by 100 to get percentages
Isotopes Calculation Summary • DO NOT use “normal” averages • If asked for two percent-abundances, use: • If asked for ANYTHING else:
A Problem! What is the average atomic mass of magnesium? 78.90% of all magnesium is 24Mg 23.985 amu 10.00% of all magnesium is 25Mg 24.986 amu 11.10% of all magnesium is 26Mg 25.983 amu 11.10% 100.00% = 24.30 amu
Another Problem! Lead has 4 isotopes, the masses and abundances of which can be found in the following table. Your job? Calculate the isotope mass of lead-207. 22.10% ???? X = Lead-207 = 206.9 amu
A Problem! Bromine has 2 isotopes, and . has a mass of 78.9193 amu and has a mass of 80.9163 amu. What are the percent abundances of each of the isotopes? Abundance 1 = 50.44% Abundance 2 = 49.56%
A Problem! Thallium has 2 isotopes, and . has a mass of 202.973 amu and has a mass of 204.974 amu. What are the percent abundances of each of the isotopes? Abundance 1 = 30.2% Abundance 2 = 69.8%
The Modern Periodic Table • Mendeleev • Set up the periodic table based on atomic mass • Moseley • Set it up by atomic number rather than atomic mass • Periodic Law – properties occur in a repeating pattern when set up by atomic number.
The Modern Periodic Table • Periods (series) • Labeled 1 - 7
The Modern Periodic Table • Groups (families) • Columns • 2 different labeling systems • 1-18 • A-B
A few definitions… • Main Group Elements: • Groups 1 and 2 and 13 – 18 • Groups A
A few definitions… • Main Group Elements: • Groups 1 and 2 and 13 – 18 • Groups A • Transition Elements: • Groups 3 – 12 • Groups B
Groups of Elements • Vertical columns are called groups • Group 1 = Alkali Metals
Groups of Elements • Vertical columns are called groups • Group 1 = Alkali Metals • Group 2 = Alkaline Earth Metals
Groups of Elements • Vertical columns are called groups • Group 1 = Alkali Metals • Group 2 = Alkaline Earth Metals • Group 17 = Halogens
Groups of Elements • Vertical columns are called groups • Group 1 = Alkali Metals • Group 2 = Alkaline Earth Metals • Group 17 = Halogens • Group 18 = Noble Gases
The Modern Periodic Table • Representative elements • Also known as main group elements • Transition metals • Inner transition metals • Lanthanide series
The Modern Periodic Table • Representative elements • Also known as main group elements • Transition metals • Inner transition metals • Lanthanide series • Actinide series
Diatomic Elements • Elements that exist as diatomic molecules when UNCOMBINED (meaning not in a compound with other elements). H2, N2, O2, F2, Cl2, Br2, I2
Diatomic Elements • Elements the exist as diatomic molecules when UNCOMBINED (meaning not in a compound with other elements). H2, N2, O2, F2, Cl2, Br2, I2 NaCl BiF5 K3N Examples of when these elements are NOT diatomic AlI3 TiBr4
Atoms vs. Ions • Atoms are NEUTRAL!!!!! • This means that they have zero charge • #p+ = #e-
Charge!!! e- = -1 p+ = 1 + 0 47 e- = -47 47 p+ = 47 + 0 When #p+ = #e-, the atom has no charge and is neutral
Atoms vs. Ions • Atoms can gain or lose e- to form IONS • ANY charged particle is called an ion • Losing e- gives POSITIVELY charge
Charge!!! When an atom LOSES electrons 11 p+ = 11 -10 10 e- = 11 e- = +1
Atoms vs. Ions • Atoms can gain or lose e- to form IONS • ANY charged particle is called an ion • Losing e- gives POSITIVELY charge • Called cations • Usually formed from metals • Gaining e- gives a NEGATIVELY charged ion
Atoms vs. Ions • Atoms can gain or lose e- to form IONS • ANY charged particle is called an ion • Losing e- gives POSITIVELY charge • Called cations • Usually formed from metals • Gaining e- gives a NEGATIVELY charged ion • Called anions • Usually formed from non-metals
Charge!!! When an atom LOSES electrons 11 p+ = 11 10 e- = -10 +1 16 p+ = 16 When an atom GAINS electrons -18 16 e- = 18 e- = -2
Atoms vs. Ions • Atoms can gain or lose e- to form IONS • ANY charged particle is called an ion • Losing e- gives POSITIVELY charge • Called cations • Usually formed from metals • Gaining e- gives a NEGATIVELY charged ion • Called anions • Usually formed from non-metals
How many electrons are needed to be neutral? 15 protons
15 protons 2 electrons
15 protons 10 electrons
Total charge = 0 Stable!!! 15 protons 15 electrons
Total charge = p+ - e- = 15 – 18 = -3 Stable!! 15 protons 18 electrons
How many electrons are needed to be neutral? 12 protons
12 protons 2 electrons