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Chemical Equations and Stoichiometry

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.

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Chemical Equations and Stoichiometry

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  1. 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

  2. Stoichiometry (化學計量學)p.19 Deals with quantitative relationships (a) among atoms, molecules and ions RAM / RMM / Formula Mass

  3. Stoichiometry (化學計量學)p.19 Deals with quantitative relationships (b) among the constituent elements of a compound Empirical / Molecular Formulae

  4. Stoichiometry (化學計量學)p.19 Deals with quantitative relationships (c) among the substances participating in chemical reaction Calculations involving chemical equation

  5. 3.1 Formulae of Compounds

  6. 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

  7. 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

  8. 3.1 Formulae of compounds (SB p.43) E.g. CH4 methane Structural formula Shows the bonding order of atoms in one molecule

  9. 3.1 Formulae of compounds (SB p.44) Different types of formulae of some compounds

  10. 3.2 Derivation of Empirical Formulae

  11. 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

  12. 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

  13. 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

  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

  15. CxHyOz heat From combustion data Vitamin C CxHyOz + O2 CO2(g) + H2O(g) excess 0.2000 g 0.2998 g 0.0819 g

  16. 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

  17. Mass of O in sample = (0.2000-0.0818-0.0091) g = 0.109 g

  18. C H O Mass (g) 0.0818 0.0091 0.109 Number of moles (mol) Simplest whole no. ratio 3 4 3 C3H4O3

  19. 3.3 Derivation of Molecular Formulae

  20. 3.3 Derivation of molecular formulae (SB p.49) What is molecular formulae? Molecular formula = (Empirical formula)n

  21. 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

  22. 473 K, 1.00 atm CnH2n+2(l) CnH2n+2(g) 3.28 dm3 12.0 g Q.15 = 142 g mol1 Relative molecular mass = 142

  23. 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

  24. 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

  25. Structural formula : bond-line structure

  26. Na : Al : S : Calculate the % by mass of the constituent elements of soda alum. Na2SO4·Al2(SO4)3·24H2O

  27. H : O : Calculate the % by mass of the constituent elements of soda alum. Na2SO4·Al2(SO4)3·24H2O

  28. 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)

  29. 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+416.0) = M – 83.0 % by mass of C in [PdCxHyNz](ClO4)2 % by mass of C in [PdCxHyNz](SCN)2

  30. Solving by simultaneous equations, x = 14, M = 557 % by mass of H in [PdCxHyNz](ClO4)2 y = 28 557 = 106.0 + 12.014 + 1.028 + 14.0z + 2(35.5+416.0) z = 4

  31. 3.4 Chemical Equations

  32. 3.4 Chemical equations (SB p.53) Chemical equations a A + b B  c C + d D a, b, c, d are stoichiometric coefficients

  33. 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

  34. 2.43 g excess ? 3.4 Chemical equations (SB p.53) Example (a) Excess oxygen 2Mg(s) + O2(g)  2MgO(s)

  35. 3.4 Chemical equations (SB p.53) Example (a) Excess oxygen 2Mg(s) + O2(g)  2MgO(s) 2.43 g excess ? = 4.03 g

  36. 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 ?

  37. 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

  38. 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

  39. P4 (s) + 5O2(g) 2P2O5(s) Q.16 4.00 g 6.00 g

  40. excess limiting reactant P4(s) + 5O2(g) 2P2O5(s) 0.0323 mol 0.188 mol O2 is in excess

  41. P4(s) + 5O2(g) 2P2O5(s) Q.16 0.0323 mol 20.0323 mol = 9.17 g

  42. Q.17 = 2 : 3 2Al(s) + 3Cu2+(aq)  2Al3+(aq) + 3Cu(s)

  43. 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

  44. CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) 2.8 dm3 25C 1.65 atm 35.0 dm3 31C 1.25 atm ? dm3 125C 2.50 atm O2 is in excess and CH4 is the limiting reactant

  45. CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) 0.189 mol 0.189 mol ? dm3 125C 2.50 atm = 2.47 dm3

  46. 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

  47. 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)

  48. 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

  49. Gay Lussac’s law a A(g) + b B(g)  c C(g) + d D(g) At fixed T & P

  50. CxHy(g) + O2(g)  xCO2(g) + H2O(?) Determination of Molecular Formula from Reacting Volumes of Gases For CxHy

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