大学化学(梁永晔)03-lecture-prev.ppt

9/8/2023,Chapter 3 Stoichiometry:Calculations with Chemical Formulas and Equations,John D.BookstaverSt.Charles Community CollegeCottleville,MO,Lecture Presentation,2012 Pearson Education,Inc.,Law of Conservation of Mass,“We may lay it down as an incontestable axiom that,in all the operations of art and nature,nothing is created;an equal amount of matter exists both before and after the experiment.Upon this principle,the whole art of performing chemical experiments depends.”-Antoine Lavoisier,1789,2012 Pearson Education,Inc.,Chemical Equations,Chemical equations are concise representations of chemical reactions.,2012 Pearson Education,Inc.,Anatomy of a Chemical Equation,CH4(g)+2O2(g)CO2(g)+2H2O(g),2012 Pearson Education,Inc.,Anatomy of a Chemical Equation,CH4(g)+2O2(g)CO2(g)+2H2O(g),Reactants appear on the left side of the equation.,2012 Pearson Education,Inc.,Anatomy of a Chemical Equation,CH4(g)+2O2(g)CO2(g)+2H2O(g),Products appear on the right side of the equation.,2012 Pearson Education,Inc.,Anatomy of a Chemical Equation,CH4(g)+2O2(g)CO2(g)+2H2O(g),The states of the reactants and products are written in parentheses to the right of each compound.,2012 Pearson Education,Inc.,Anatomy of a Chemical Equation,CH4(g)+2O2(g)CO2(g)+2H2O(g),Coefficients are inserted to balance the equation.,2012 Pearson Education,Inc.,Subscripts and Coefficients Give Different Information,Subscripts tell the number of atoms of each element in a molecule.Coefficients tell the number of molecules.,2012 Pearson Education,Inc.,Reaction Types,2012 Pearson Education,Inc.,Combination Reactions,Examples:2Mg(s)+O2(g)2MgO(s)N2(g)+3H2(g)2NH3(g)C3H6(g)+Br2(l)C3H6Br2(l),In combination reactions two or more substances react to form one product.,2012 Pearson Education,Inc.,In a decomposition reaction one substance breaks down into two or more substances.,Decomposition Reactions,Examples:CaCO3(s)CaO(s)+CO2(g)2KClO3(s)2KCl(s)+O2(g)2NaN3(s)2Na(s)+3N2(g),2012 Pearson Education,Inc.,Combustion Reactions,Examples:CH4(g)+2O2(g)CO2(g)+2H2O(g)C3H8(g)+5O2(g)3CO2(g)+4H2O(g),Combustion reactions are generally rapid reactions that produce a flame.Combustion reactions most often involve hydrocarbons reacting with oxygen in the air.,2012 Pearson Education,Inc.,Formula Weights,2012 Pearson Education,Inc.,Formula Weight(FW),A formula weight is the sum of the atomic weights for the atoms in a chemical formula.So,the formula weight of calcium chloride,CaCl2,would be Ca:1(40.08 amu)+Cl:2(35.453 amu)110.99 amuFormula weights are generally reported for ionic compounds.,2012 Pearson Education,Inc.,Molecular Weight(MW),A molecular weight is the sum of the atomic weights of the atoms in a molecule.For the molecule ethane,C2H6,the molecular weight would be,C:2(12.011 amu),30.070 amu,+H:6(1.00794 amu),2012 Pearson Education,Inc.,Percent Composition,One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation:,2012 Pearson Education,Inc.,Percent Composition,So the percentage of carbon in ethane is,2012 Pearson Education,Inc.,Moles,2012 Pearson Education,Inc.,Avogadros Number,6.02 x 10231 mole of 12C has a mass of 12.000 g.,2012 Pearson Education,Inc.,Molar Mass,By definition,a molar mass is the mass of 1 mol of a substance(i.e.,g/mol).The molar mass of an element is the mass number for the element that we find on the periodic table.The formula weight(in amus)will be the same number as the molar mass(in g/mol).,2012 Pearson Education,Inc.,Using Moles,Moles provide a bridge from the molecular scale to the real-world scale.,2012 Pearson Education,Inc.,Mole Relationships,One mole of atoms,ions,or molecules contains Avogadros number of those particles.One mole of molecules or formula units contains Avogadros number times the number of atoms or ions of each element in the compound.,2012 Pearson Education,Inc.,Finding Empirical Formulas,2012 Pearson Education,Inc.,Calculating Empirical Formulas,One can calculate the empirical formula from the percent composition.,2012 Pearson Education,Inc.,Calculating Empirical Formulas,The compound para-aminobenzoic acid(you may have seen it listed as PABA on your bottle of sunscreen)is composed of carbon(61.31%),hydrogen(5.14%),nitrogen(10.21%),and oxygen(23.33%).Find the empirical formula of PABA.,2012 Pearson Education,Inc.,Calculating Empirical Formulas,2012 Pearson Education,Inc.,Calculating Empirical Formulas,Calculate the mole ratio by dividing by the smallest number of moles:,2012 Pearson Education,Inc.,Calculating Empirical Formulas,These are the subscripts for the empirical formula:C7H7NO2,2012 Pearson Education,Inc.,Combustion Analysis,Compounds containing C,H,and O are routinely analyzed through combustion in a chamber like the one shown in Figure 3.14.C is determined from the mass of CO2 produced.H is determined from the mass of H2O produced.O is determined by difference after the C and H have been determined.,2012 Pearson Education,Inc.,Stoichiometric Calculations,The coefficients in the balanced equation give the ratio of moles of reactants and products.,2012 Pearson Education,Inc.,Stoichiometric Calculations,Starting with the mass of Substance A,you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed(if its a product)or used(if its a reactant).,2012 Pearson Education,Inc.,Stoichiometric Calculations,Starting with 1.00 g of C6H12O6 we calculate the moles of C6H12O6use the coefficients to find the moles of H2Oand then turn the moles of water to grams.,C6H12O6+6 O2 6 CO2+6 H2O,2012 Pearson Education,Inc.,Limiting Reactants,2012 Pearson Education,Inc.,Limiting Reactants,The limiting reactant is the reactant present in the smallest stoichiometric amount.In other words,its the reactant youll run out of first(in this case,the H2).,2012 Pearson Education,Inc.,Limiting Reactants,In the example below,the O2 would be the excess reagent.,2012 Pearson Education,Inc.,Theoretical Yield,The theoretical yield is the maximum amount of product that can be made.In other words,its the amount of product possible as calculated through the stoichiometry problem.This is different from the actual yield,which is the amount one actually produces and measures.,2012 Pearson Education,Inc.,Percent Yield,One finds the percent yield by comparing the amount actually obtained(actual yield)to the amount it was possible to make(theoretical yield):,