Slide 1Conceptual Problem 2.26: model of water, KCI, methanol, ammonia.Problem 2.64: Models of sulfuric acid, benzene, and 2-propane.Slide 4Figure 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 10Slide 11Slide 12Chapter 3 Calculations with Chemical Formulas and EquationsSlide 14Slide 15The MoleSlide 17Slide 18Figure 3.2: One mole each of various substances. Photo courtesy of American Color.Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29Slide 30Slide 31Slide 32Slide 33Slide 34Slide 35Conceptual Problem 2.26: model of water, KCI, methanol, ammonia.waterpotassium chlorideethanolammoniaProblem 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) + CO2(g)Write this reaction in words:Methane gas plus 2 oxygens make 2 waters and one carbon dioxideFigure 2.26: The burning of propane gas. Photo courtesy of American Color. Write a balanced equation for this combustionreaction. The products are carbon dioxide andwater. First write the reaction.C3H8 + O2  H2O + CO2 Then balance it.Then add physical states.C3H8 + 5 O2  4 H2 O + 3 CO2 C3H8 (g) + 5 O2 (g)  4 H2 O (l) + 3 CO2 (g)Figure 3.3: Reaction of zinc and iodine causing iodine to vaporize. Photo courtesy of James Scherer.Chemical EquationsReactants ProductsQualitative Information: States of Matter: (s) solid(l) liquid(g) gaseous(aq) aqueous2 H2 (g) + O2 (g) 2 H2O (g)Balanced Equations1 CH4 (g) + O2 (g) 1 CO2 (g) + H2O (g)• mass balance (atom balance)- same number of each element(1) start with simplest element(2) progress to other elements(3) make all whole numbers(4) re-check atom balance•charge balance (no “spectator” ions)1 CH4 (g) + 2 O2 (g) 1 CO2 (g) + 2 H2O (g)1 CH4 (g) + O2 (g) 1 CO2 (g) + 2 H2O (g)Ca2+ (aq) + 2 OH- (aq) Ca(OH)2 (s)+ Na++ Na+Balancing Chemical Equations - IProblem: The hydrocarbon hexane is a component of Gasoline that burns in an automobile engine to produce carbon dioxide and water as well as energy. Write the balanced chemical equation for the combustion of hexane (C6H14).Plan: Write the skeleton equation from the words into chemical compounds with blanks before each compound. begin the balance with the most complex compound first, and save oxygen until last!Solution:C6H14 (l) + O2 (g) CO2 (g) + H2O(g) + EnergyC6H14 (l) + O2 (g) CO2 (g) + H2O(g) + EnergyBegin with one Hexane molecule which says that we will get 6 CO2’s!1 6C6H14 (l) + O2 (g) CO2 (g) + H2O(g) + EnergyThe H atoms in the hexane will end up as H2O, and we have 14 H atoms, and since each water molecule has two H atoms, we will geta total of 7 water molecules.Balancing Chemical Equations - II1 6 7Since oxygen atoms only come as diatomic molecules (two O atoms, O2),we must have even numbers of oxygen atoms on the product side. We do not since we have 7 water molecules! Therefore multiply the hexane by 2, giving a total of 12 CO2 molecules, and 14 H2O molecules.C6H14 (l) + O2 (g) CO2 (g) + H2O(g) + Energy2 12 14This now gives 12 O2 from the carbon dioxide, and 14 O atoms from thewater, which will be another 7 O2 molecules for a total of 19 O2 !C6H14 (l) + O2 (g) CO2 (g) + H2O(g) + Energy2 12 1419Predicting the Ion an Element Will Form in Chemical ReactionsProblem: What monoatomic ions will each of the elements form?(a) Barium(z=56) (b) Sulfur(z=16) (c) Titanium(z =22) (d) Fluorine(z=9)Plan: We use the “z” value to find the element in the periodic table and which is the nearest noble gas. Elements that lie after a noble gas will loose electrons, and those before a noble gas will gain electrons.Solution: (a) Ba+2, Barium is an alkaline earth element, Group 2A, and is expected to loose two electrons to attain the same number of electrons as the noble gas Xenon! (b) S -2, Sulfur is in the Oxygen family, Group 6A, and is expected to gain two electrons to attain the same number of electrons as the noble gas Argon! (c) Ti+4, Titanium is in Group 4B, and is expected to loose 4 electrons to attain the same number of electrons as the noble gas Argon! (d) F -, Fluorine is in a halogen, Group 7A, and is expected to gain one electron, to attain the same number of electrons as the noble gas Neon!Chapter 3 Chapter 3 Calculations with Chemical Formulas and Calculations with Chemical Formulas and EquationsEquationsHomework Chapter 3: 6, 7, 13, 15, 18, 19, 23, 25, 29, 35, 37, 41, 45, 47, 49, 51, 53, 55, 59, 63, 67, 69, 71, 77, 81, 83, 85, 91, 95, 99, 105,109, 111Counting Objects of Fixed Relative Mass12 red marbles @ 7g each = 84g12 yellow marbles @4e each=48g55.85g Fe = 6.022 x 1023 atoms Fe32.07g S = 6.022 x 1023 atoms S•The Mole is based upon the definition:•The amount of substance that contains as many elementary parts (atoms, molecules, or other?) as there are atoms in exactly •12 grams of carbon -12.•1 Mole = 6.022045 x 1023 particlesThe MoleMole - Mass Relationships of ElementsElement Atom/Molecule Mass Mole Mass Number of Atoms1 atom of H = 1.008 amu 1 mole of H = 1.008 g = 6.022 x 1023 atoms1 atom of Fe = 55.85 amu 1 mole of Fe = 55.85 g = 6.022 x 1023 atoms1 atom of S = 32.07 amu 1 mole of S = 32.07 g = 6.022 x 1023 atoms1 atom of O = 16.00 amu 1 mole of O = 16.00 g = 6.022 x 1023 atoms1 molecule of O2 = 32.00 amu 1 mole of O2 = 32.00 g = 6.022 x 1023 molecule1 molecule of S8 = 2059.52 amu 1 mole of S8 = 2059.52 g = 6.022 x 1023 moleculesCaCO3 100.09 gOxygen 32.00 gCopper 63.55 gWater 18.02 gOne Mole of Common SubstancesFigure 3.2: One mole each of various substances.Photo courtesy of American Color.Information Contained in a Balanced EquationViewed in Reactants Productsterms of: 2 C2H6 (g) + 7 O2 (g) = 4 CO2 (g) + 6 H2O(g) +