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ATOMIC STRUCTURE P + = PROTON N o = NEUTRON E --- = ELECTRON DIFFERENT WAYS TO WRITE ELEMENTS & THEIR SYMBOLS Phosphorus—31 AND 31 15 P ***NEUTRONS = mass # -- atomic # ***isotopes--- atoms of the same element but with different numbers of neutrons (different mass #s) Ex: P—32 and P—30

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Ex: phosphorus—31 Protons = 15 Electrons = 15 Neutrons = 16

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NUMBER THESE COLUMNS ON YOUR PERIODIC TABLE: 12131415 161718 (1A)(2A)(3A)(4A)(5A)(6A) (7A)(8A) Vertical columns—groups or families --- there are 18 total (8A groups and 10B groups) Horizontal rows—periods or series ---there are 7 rows total

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VALENCE ELECTRONS-- the outermost electrons having the highest amount of energy and are farthest from the nucleus and the ones involved in chemical reactions ** # of valence electrons = “A” column # on the periodic table (“A” group #) ** # of energy levels = row # on the periodic table Ex: Beryllium (Be): column 2A= 2 valence electrons Row 2 = 2 energy levels Ex: Phosphorus (P): column 5A = 5 valence electrons Row 3 = 3 energy levels

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ATOMIC MATH PROBLEM 1. Take the # of protons in oxygen X the # of electrons in fluorine = _______________ 2.Take this number --- # of neutrons in chlorine = _________________ 3. Take this number + silicon’s atomic # = ____________ 4. Take this number / # of neutrons in helium = _____________ 5.Take this number --- aluminum’s mass # = _____________

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Calculating average atomic mass using % abundance information atomic mass-the average mass of all of an element's isotopes; you round this to actually get the mass number ***every element has at least 2 isotopes * **the most abundant isotope matches the mass # from the periodic table!! Ex: Which is most common? O-16 O-17 O-18

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Hydrogen has 3 known isotopes: H-1 (protium) occurs 99.9885% of the time with an atomic mass of 1.0078 amu H-2 (deuterium) occurs 0.0115% of the time with an atomic mass of 2.0141 amu H-3 (tritium) is synthetically made MASS OF EACH ISOTOPE X DECIMAL FORM OF THE % OF THE TIME IT OCCURS

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(1.0078 X 0.999885 ) + (2.0141 X 0.000115) 1.007684103 + 0.000231622 1.007915725 amu

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Calculate avg. atomic mass for “element Z” Z—10 occurs 75.5% of the time Z—11 occurs 20.5 % of the time Z—12 occurs 4.0 % of the time (10 x 0.755) + (11 x 0.205) + (12 x 0.040) 10.285 amu

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