Structure of Atoms. Atoms and Elements. Structure of Atoms. Structure of Atoms. Structure of Atoms. Structure of Atoms 10/9/13

Transcription

1 Atoms and Elements Through a series of experiments conducted by a number of different people (Thomson, Milliken, Rutherford, etc.), we know that atoms are composed of three different types of subatomic particles: Æ electrons Æ protons Æ neutrons Thomson s cathode ray tube Millikan s oil drop experiment Determined mass to charge ratio of electrons Determined charge of an electron n Electrons are negatively charged particles charge = x C (C=coulomb) mass = 9.11 x kg The Nucleus: Rutherford s gold foil experiment Proved that atoms have a nucleus containing most of atom s mass 1

2 Protons and neutrons combine to form the nucleus of the atom Protons: n Protons are positively charged particles with a charge that is equal in magnitude to an electron s, but opposite in sign charge = x C mass = x 10-7 kg n The mass of a proton is ~1800 times that of an electron Neutrons: n Neutrons are uncharged particles that are slightly heavier than protons charge = 0.00 C mass = x 10-7 kg n Each element in the Periodic Table is identified by its Atomic Number n Atomic number indicates the number of protons in the nucleus of an atom n The Mass Number indicates the total number of particles in the nucleus n Mass number = protons + neutrons n Mass number is measured in atomic mass units (amu: 1 amu mass of p/n) n The atomic number and mass number can be expressed with the elemental symbol: n Atomic number (Z) is indicated by a left subscript n Mass number (A) is indicated by a left superscript A ZX Examples: 1 C = carbon Z = 6 protons 6C A = 1 ð 6 neutrons 1H 1 H = hydrogen Z = 1 proton A = 1 ð 0 neutrons

3 Examples: 35 U = uranium Z = 9 protons 9U A = 35 ð 143 neutrons 38 U = uranium Z = 9 protons 9U A = 38 ð 146 neutrons This is an example of isotopes: isotopes have the same atomic number, but different mass numbers n Atomic Mass (or atomic weight) is the weighted average mass of the naturally occurring isotopes of an element isotope mass abundance 1 1 H % 1 H % n The atomic mass of hydrogen is given as n The units of atomic mass are grams per mole: g/mol n 1 mole = 6.0 x 10 3 particles n A mole is just a convenient unit of measure for very large quantities (like a dozen [=1] or a score [=0]) Relationship of these quantities to the Periodic Table: Elemental symbol Z X Elemental name AM Atomic number Atomic mass He Helium Protons = Neutrons = 8 O Oxygen Protons = 8 Neutrons = 8 3

4 Example Problem #1 17 Cl Chlorine Protons = 17 Neutrons = 18.45? Chlorine has two isotopes: Cl = 75.53% (18 neutrons) Cl = 4.47% (0 neutrons) The weighted average results in the listed atomic mass If a gold bar has a volume of 107 ml, how many moles of gold are in the bar? Step 1: Determine mass of the bar ρ Au = 19.3 g/ml (Table 1.1) g m Au = ( 107 ml) 19.3 = g ml Step : Convert from grams to moles g = mol g mol Example Problem # 1 carat is equal to 00 mg. Diamond is a form of pure carbon with the C atoms arranged in a tetrahedral structure. How many moles of C are in a.31 carat diamond? Step 1: Convert mass from carats to grams (.31 carat)(00 mg/carat)(1 g/1000 mg) = 0.46 g Example Problem # (con t.) 1 carat is equal to 00 mg. Diamond is a form of pure carbon with the C atoms arranged in a tetrahedral structure. How many moles of C are in a.31 carat diamond? Step : Convert from grams to moles (0.46 g)/( g/mol) = mol Example Problem #3 A piece of copper wire is 5 ft long and has a diameter of.0 mm. How many moles of copper and how many copper atoms are in the wire? ρ Cu = 8.9 g/ml Step 1: Calculate volume of wire V cylinder = (cross sectional area)(height) Area = πr = π(1.0 mm) = 3.1 mm (3.1 mm )(1 cm/10 mm) = 3.1 x 10 - cm Example Problem #3 (con t.) Step 1: Calculate volume of wire h = (5 ft)(1 in/ft)(.54 cm/in) = 760 cm V = (3.1 x 10 - cm )(760 cm) = 4 cm 3 Step : Calculate mass of wire m = (density)(volume) = (8.9 g/ml)(4 cm 3 ) = 10 g 1 ml = 1 cm 3 4

5 Example Problem #3 (con t.) Example Problem #4 Step 3: Calculate moles (10 g)/( g/mol) = 3.3 mol Step 4: Determine number of atoms (3.3 mol)(6.0 x 10 3 atoms/mol) =.0 x 10 4 atoms What volume will 84.9 mol of water occupy at 0 o C? Step 1: Calculate molecular weight of water The molecular weight of a molecule is determined by summing the atomic weights of all elements comprising the molecule. Example Problem #4 (con t.) Step 1 (con t.): Calculate molecular weight of water H O has hydrogens and 1 oxygen MW = ( g/mol) g/mol = g/mol Step : Calculate mass of H O (84.9 mol)( g/mol) = 5133 g Example Problem #4 (con t.) Step 3: Calculate volume of water (5133 g)/( g/ml) = 514 ml (514 ml)(1 L/1000 ml) = 5.14 L 5