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## Chemical Equations and Stoichiometry

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**3**Chemical Equations and Stoichiometry 3.1 Formulae of Compounds 3.2 Derivation of Empirical Formulae 3.3 Derivation of Molecular Formulae 3.4 Chemical Equations 3.5 Calculations Based on Chemical Equations 3.6 Simple Titrations**Stoichiometry (化學計量學)p.19**Deals with quantitative relationships (a) among atoms, molecules and ions RAM / RMM / Formula Mass**Stoichiometry (化學計量學)p.19**Deals with quantitative relationships (b) among the constituent elements of a compound Empirical / Molecular Formulae**Stoichiometry (化學計量學)p.19**Deals with quantitative relationships (c) among the substances participating in chemical reaction Calculations involving chemical equation**3.1**Formulae of Compounds**3.1 Formulae of compounds (SB p.43)**Empirical formula Shows the simplest whole number ratio of the atoms or ions present E.g. Methane, CH4 Sodium chloride, NaCl**3.1 Formulae of compounds (SB p.43)**Molecular formula Shows the actual number of each kind of atoms present in one molecule E.g. CH4 methane Ionic compounds do not have molecular formulae**3.1 Formulae of compounds (SB p.43)**E.g. CH4 methane Structural formula Shows the bonding order of atoms in one molecule**3.1 Formulae of compounds (SB p.44)**Different types of formulae of some compounds**3.2**Derivation of Empirical Formulae**3.2 Derivation of empirical formulae (SB p.45)**Example 1 Mg + N2 MgxNy 0.450 g excess 0.623 g From composition by mass Mass of N used = (0.623-0.450)g = 0.173 g**3.2 Derivation of empirical formulae (SB p.45)**Example 1 Mg + N2 MgxNy 0.450 g excess 0.623 g From composition by mass Mg3N2**3.2 Derivation of empirical formulae (SB p.45)**Example 3-3C Check Point 3-3A Water of Crystallization Derived from Composition by Mass Example 2 Q.14**CuSO4 : H2O =**Q.14 heat CuSO4.xH2O CuSO4 + xH2O 10.0 g 6.4 g (10.0–6.4) g = 3.6 g x = 5**CxHyOz**heat From combustion data Vitamin C CxHyOz + O2 CO2(g) + H2O(g) excess 0.2000 g 0.2998 g 0.0819 g**CxHyOz + O2 CO2(g) + H2O(g)**Mass of C in sample = mass of C in CO2 formed Mass of H in sample = mass of H in H2O formed Mass of O in sample = total mass of sample – mass of C – mass of H**C**H O Mass (g) 0.0818 0.0091 0.109 Number of moles (mol) Simplest whole no. ratio 3 4 3 C3H4O3**3.3**Derivation of Molecular Formulae**3.3 Derivation of molecular formulae (SB p.49)**What is molecular formulae? Molecular formula = (Empirical formula)n**3.3 Derivation of Molecular Formulae (SB p.49)**Example 3-3A Example 3-3B Molecular formula From empirical formula and known relative molecular mass Empirical formula Molecular mass**473 K, 1.00 atm**CnH2n+2(l) CnH2n+2(g) 3.28 dm3 12.0 g Q.15 = 142 g mol1 Relative molecular mass = 142**n = 10**Q.15 473 K, 1.00 atm CnH2n+2(l) CnH2n+2(g) 3.28 dm3 12.0 g RMM = 12n + 2n+2 = 142 The molecular formula is C10H22**C, H, O are present**Qualitative Analysis Structural formula CH3COOH Quantitative Analysis IR, NMR, MS Empirical formula CH2O Molecular formula C2H4O2 RMM = 60.0 Determination of Chemical Formulae Ethanoic acid**Na :**Al : S : Calculate the % by mass of the constituent elements of soda alum. Na2SO4·Al2(SO4)3·24H2O**H :**O : Calculate the % by mass of the constituent elements of soda alum. Na2SO4·Al2(SO4)3·24H2O**A certain compound was known to have a formula which could**be represented as [PdCxHyNz](ClO4)2. Analysis showed that the compound contained 30.15% carbon and 5.06% hydrogen. When converted to the corresponding thiocyanate, [PdCxHyNz](SCN)2, the analysis was 40.46% carbon and 5.94% hydrogen. Calculate the values of x, y and z. (Relative atomic masses : C = 12.0, H = 1.0, N = 14.0, O = 16.0, Cl = 35.5, S = 32.0, Pd = 106.0)**Let M be the formula mass of [PdCxHyNz](ClO4)2**Then, the formula mass of [PdCxHyNz](SCN)2 = M + 2(32.0+12.0+14.0) – 2(35.5+416.0) = M – 83.0 % by mass of C in [PdCxHyNz](ClO4)2 % by mass of C in [PdCxHyNz](SCN)2**Solving by simultaneous equations,**x = 14, M = 557 % by mass of H in [PdCxHyNz](ClO4)2 y = 28 557 = 106.0 + 12.014 + 1.028 + 14.0z + 2(35.5+416.0) z = 4**3.4**Chemical Equations**3.4 Chemical equations (SB p.53)**Chemical equations a A + b B c C + d D a, b, c, d are stoichiometric coefficients**3.5 Calculations Based on Equations (SB p.65)**Example 3-5A Example 3-5B Check Point 3-4 Calculations based on equations Calculations involving reacting masses**2.43 g**excess ? 3.4 Chemical equations (SB p.53) Example (a) Excess oxygen 2Mg(s) + O2(g) 2MgO(s)**3.4 Chemical equations (SB p.53)**Example (a) Excess oxygen 2Mg(s) + O2(g) 2MgO(s) 2.43 g excess ? = 4.03 g**3.4 Chemical equations (SB p.53)**Example(b) limiting reagent to be determined 2Mg(s) + O2(g) 2MgO(s) 2.43 g 1.28g ?**3.4 Chemical equations (SB p.53)**Example(b) limiting reagent to be determined 2Mg(s) + O2(g) 2MgO(s) 0.100 mol 0.040 mol ? Mg is in excess, O2 is the limiting reagent**Check Point 3-4**3.4 Chemical equations (SB p.53) Example(b) limiting reagent to be determined 2Mg(s) + O2(g) 2MgO(s) 0.040 mol 0.080 mol = 3.22 g Q.16, 17**P4 (s) + 5O2(g) 2P2O5(s)**Q.16 4.00 g 6.00 g**excess**limiting reactant P4(s) + 5O2(g) 2P2O5(s) 0.0323 mol 0.188 mol O2 is in excess**P4(s) + 5O2(g) 2P2O5(s)**Q.16 0.0323 mol 20.0323 mol = 9.17 g**Q.17**= 2 : 3 2Al(s) + 3Cu2+(aq) 2Al3+(aq) + 3Cu(s)**3.5 Calculations Based on Equations (SB p.66)**Calculations involving volumes of gases • Gases not at the same conditions • Gases at the same conditions • - Gay Lussac’s Law**CH4(g) + 2O2(g) CO2(g) + 2H2O(g)**2.8 dm3 25C 1.65 atm 35.0 dm3 31C 1.25 atm ? dm3 125C 2.50 atm O2 is in excess and CH4 is the limiting reactant**CH4(g) + 2O2(g) CO2(g) + 2H2O(g)**0.189 mol 0.189 mol ? dm3 125C 2.50 atm = 2.47 dm3**Gay Lussac’s law :**When gases reacts, they do so in volumes which bears a simple whole number ratio to one another, and to the volumes of gaseous products, all volumes being measured under the same conditions of temperature and pressure. Watch video**39.5 cm3**Fe as catalyst Ammonia nitrogen + hydrogen heat H2(g) + CuO Cu(s) + H2O(l) 20 cm3 10 cm3 29.5 cm3 2NH3(g) 1N2(g) + 3H2(g)**Gay Lussac’s law is an application of the Avogadro’s**law. a A(g) + b B(g) c C(g) + d D(g) At fixed T & P, n V**Gay Lussac’s law**a A(g) + b B(g) c C(g) + d D(g) At fixed T & P**CxHy(g) + O2(g) xCO2(g) + H2O(?)**Determination of Molecular Formula from Reacting Volumes of Gases For CxHy