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WWU CHEM 121 - Calculating the “Average” Atomic Mass of an Element

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Slide 1Slide 2Slide 3Slide 4Slide 5DefinitionsSlide 7Slide 8Slide 9Slide 10Slide 11Figure 2.14: A modern form of the periodic table.Slide 13Figure 2.16: Molecular and structural formulas and molecular models.Slide 15Slide 16Figure 2.19: A model of a portion of NaCl.Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Figure 2.23: Molecular model of nitric acid.Slide 26Conceptual Problem 2.26: model of water, KCI, methanol, ammonia.Conceptual problem 2.63: Ball and stick model of hydrogen, oxygen, and carbon.Problem 2.64: Models of sulfuric acid, benzene, and 2-propane.Slide 30Figure 2.28: Representation of the reaction of methane with oxygen.Figure 2.26: The burning of propane gas. Photo courtesy of American Color.Figure 3.3: Reaction of zinc and iodine causing iodine to vaporize. Photo courtesy of James Scherer.Chemical EquationsBalanced EquationsSlide 36Slide 37Slide 38546863 786612633934 791450710305 794349751983 815836767614 836029778599 836353780518 854332Exam R 1-2 pmReview T 2 SL120. NO GRAPHING CALULATORS!!!!!!!!!Calculating the “Average” Atomic Mass of an Element24Mg (78.7%) 23.98504 amu x 0.787 = 18.876226 amu 25Mg (10.2%) 24.98584 amu x 0.102 = 2.548556 amu26Mg (11.1%) 25.98636 amu x 0.111 = 2.884486 amu 24.309268 amu With Significant Digits = 24.3 amuProblem: Calculate the average atomic mass of Magnesium! Magnesium Has three stable isotopes, 24Mg ( 78.7%); 25Mg (10.2%); 26Mg (11.1%).Problem: Calculate the abundance of the two Bromine isotopes: 79Br = 78.918336 g/mol and 81Br = 80.91629 g/mol , given that the average mass of Bromine is 79.904 g/mol.Plan: Let the abundance of 79Br = X and of 81Br = Y and X + Y = 1.0Solution: X(78.918336) + Y(80.91629) = 79.904 X + Y = 1.00 therefore X = 1.00 - Y (1.00 - Y)(78.918336) + Y(80.91629) = 79.904 78.918336 - 78.918336 Y + 80.91629 Y = 79.904 1.997954 Y = 0.985664 or Y = 0.4933 X = 1.00 - Y = 1.00 - 0.4933 = 0.5067 %X = % 79Br = 0.5067 x 100% = 50.67% = 79Br %Y = % 81Br = 0.4933 x 100% = 49.33% = 81BrModern Reassessment of the Atomic Theory1. All matter is composed of atoms. Although atoms are composed of smaller particles (electrons, protons, and neutrons), the atom is the smallest body that retains the unique identity of the element.2. Atoms of one element cannot be converted into atoms of another element in a chemical reaction. Elements can only be converted into other elements in Nuclear reactions in which protons are changed.3. All atoms of an element have the same number of protons and electrons, which determines the chemical behavior of the element. Isotopes of an element differ in the number of neutrons, and thus in mass number, but a sample of the element is treated as though its atoms have an average mass.4. Compounds are formed by the chemical combination of two or more elements in specific ratios, as originally stated by Dalton.Definitions•ELEMENT - A substance that cannot be separated into simpler substances by chemical means•COMPOUND - A substance composed of atoms of two or more elements chemically united in fixed proportions•PERIODIC TABLE - “MENDELEEV TABLE” - A tabular arrangement of the elements, vertical groups or families of elements based upon their chemical properties - actually combining ratios with oxygenExam F 1-2 pmReview W (SL 110)and R (SL 120)at 4 pm. NO GRAPHING CALULATORS!!!!!!!!!Fig. 2.16Fig. 2.17HeNeArKrXeRn The Periodic Table of the Elements Most Probable Oxidation StateCrMn Fe Co NiMoWTcReRuOsRhIrPdPt+1+2+3 +4+3 +_4 - 3 - 2 - 1 0HLiNaKRbCsFrScYBeMgCaSrBaRaLaAcBAlGaInTlTiRfHfZrCSiGeSnPbFClBrIAtOSSeTePoNPAsSbBiZnCdHg+ 2+1CuAgAu+5VNbTaCeThPr Nd PmSmEu Gd Tb Dy Ho Er TmYb LuPa U Np Pu AmCmBk Cf Es FmMd No Lr+3+3Du Sg Bo Ha MeGroups in the Periodic TableMain Group Elements (Vertical Groups) Group IA - Alkali Metals Group IIA - Alkaline Earth Metals Group IIIA - Boron Family Group IVA - Carbon Family Group VA - Nitrogen Family Group VIA - Oxygen Family (Calcogens) Group VIIA - Halogens Group VIIIA - Noble GasesOther Groups ( Vertical and Horizontal Groups)Group IB - 8B - Transition MetalsPeriod 6 Group - Lanthanides (Rare Earth Elements)Period 7 Group - ActinidesFigure 2.14: A modern form of the periodic table.Chemical Compounds and BondsChemical Bonds - The electrostatic forces that hold the atoms of elements together in the compound.Ionic Compounds - Electrons are transferred from one atom to another to form Ionic Cpds.Covalent Compounds - Electrons are shared between atoms of different elements to form Covalent Cpds.“Cations” - Metal atoms lose electrons to form “ + ” ions.“Anions” - Nonmetal atoms gain electrons to form “ - ” ions.Mono-atomic ions form binary ionic compoundsFigure 2.16: Molecular and structural formulas and molecular models.EthanolFig. 2.18Figure 2.19: A model of a portion of NaCl.HeNeArKrXeRn The Periodic Table of the Elements Most Probable Oxidation StateCrMn Fe Co NiMoWTcReRuOsRhIrPdPt+1+2+3 +4+3 +_4 - 3 - 2 - 1 0HLiNaKRbCsFrScYBeMgCaSrBaRaLaAcBAlGaInTlTiRfHfZrCSiGeSnPbFClBrIAtOSSeTePoNPAsSbBiZnCdHg+ 2+1CuAgAu+5VNbTaCeThPr Nd PmSmEu Gd Tb Dy Ho Er TmYb LuPa U Np Pu AmCmBk Cf Es FmMd No Lr+3+3Du Sg Bo Ha MeFig. 2.20Don’t use thesewhen naming ioniccompounds--they’reONLY for covalentcompounds!!Chemical FormulasEmpirical Formula - Shows the relative number of atoms of each element in the compound. It is the simplest formula, and is derived from masses of the elements.Molecular Formula - Shows the actual number of atoms of each element in the molecule of the compound.Structural Formula - Shows the actual number of atoms, and the bonds between them ; that is, the arrangement of atoms in the molecule.Figure 2.23: Molecular model of nitric acid. What is the formula for nitric acid?HNO3Conceptual Problem 2.26: model of water, KCI, methanol, ammonia.waterpotassium chlorideethanolammoniaConceptual problem 2.63: Ball and stick model of hydrogen, oxygen, and carbon.PCl5NO2C3H6O2Problem 2.64: Models of sulfuric acid, benzene, and 2-propane.Figure 2.28: Representation of the reaction of methane with oxygen. Write a balanced chemical reaction for this combustion reaction. CH4(g) + 2 O2 (g) 2 H2 O(l)


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WWU CHEM 121 - Calculating the “Average” Atomic Mass of an Element

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