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Chapter 3

Chapter 3. Chemical Equations and Stoichiometry. 3.1 F ormulae of Compounds 3.2 Derivation of Empirical Formulae 3.3 Derivation of Molecular Formulae 3.4 Chemical Equations 3.5 Calculations Based on Equations 3.6 Simple Titrations. 3.1 Formulae of Compounds (SB p.54).

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Chapter 3

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  1. Chapter 3 Chemical Equations and Stoichiometry 3.1 Formulae of Compounds 3.2Derivation of Empirical Formulae 3.3Derivation of Molecular Formulae 3.4 Chemical Equations 3.5 Calculations Based on Equations 3.6 Simple Titrations

  2. 3.1 Formulae of Compounds (SB p.54) ratio of no. of atoms Formulae of Compounds How can you describe the composition of compound X? 1st way = by chemical formula C?H?

  3. 3.1 Formulae of Compounds (SB p.54) carbonatoms hydrogen atoms How can you describe the composition of compound X? Compound X 2nd way = by percentage by mass Mass of carbon atoms inside = …. g Mass of hydrogen atoms inside = …. g

  4. 3.1 Formulae of Compounds (SB p.54) Compound Empirical formula Molecular formula Structural formula (a) Propene CH2 C3H6 (b) Nitric acid HNO3 HNO3 (c) Ethanol C2H6O C2H5OH (d) Glucose C6H12O6 C6H12O6 Check Point 3-1 Give the empirical molecular and structural formula for the following compounds Answer

  5. 3.1 Formulae of Compounds (SB p.55) The different types of formulae of some compounds

  6. 3.2 Derivation of Empirical Formulae (SB p.56) Solution: The relative molecular mass of CO2 = 12.0 + 2 x 16.0 = 44.0 Mass of carbon in 2.93 g of CO2 = 2.93 g x 12.0/44.0 = 0.80 g The relative molecular mass of H2O = 2 x 1.0 + 16.0 = 18.0 Mass of hydrogen in 1.80 g of H2O = 1.80 g x 2.0/18.0 = 0.20 g Example 3-1 A hydrogen was burnt completely in excess oxygen. It was found that 1.00 g of the hydrocarbon gives 2.93 g of carbon dioxide and 1.8 g of water. Find the empirical formula of the hydrocarbon.? (R.a.m.* : H = 1.0, C = 12.0, O = 16.0) Answer

  7. 3.2 Derivation of Empirical Formulae (SB p.57) Solution: (cont’d) Let the empirical formula of the hydrocarbon be CxHy. Mass of carbon in CxHy = Mass of carbon in CO2 Mass of hydrogen in CxHy = Mass of hydrogen in H2O The simplest whole number ratio of x and y can be determined by the following the steps in the below table.

  8. 3.2 Derivation of Empirical Formulae (SB p.57)

  9. 3.2 Derivation of Empirical Formulae (SB p.57) Solution: Mass of compound X = 0.46g Mass of carbon in compound X = 0.88 g x 12.0/44.0 = 0.24 g Mass of hydrogen in compound X = 0.54 g x 2.0/18.0 = 0..06g Mass of oxygen in compound X = 0.46 g – 0.24 g – 0.06 g = 0.16 g Example 3-2 Compound X is known to contain carbon, hydrogen and oxygen only. When it is burnt completely in excess oxygen, carbon dioxide and water are given out as the only products. It is found that 0.46 g of compound X gives 0.88 g of carbon dioxide and 0.54 g of water. Find the empirical formula of compound X. (R.a.m.* : H = 1.0, C = 12.0, O = 16.0) Answer

  10. 3.2 Derivation of Empirical Formulae (SB p.57) Solution: (cont’d) Let the empirical formula of compound X be CxHyOz. Therefore, the empirical formula of compound X is C2H6O.

  11. 3.1 Formulae of Compounds (SB p.58) Check Point 3-2 (a) 5 g of sulphur forms 10 g of an oxide on burning.What is the empirical formula of the oxide? (R.a.m. : O = 16.0, S = 32.1) (b) 19.85 f of element M combines with 25.61 g of oxygen to form an oxide. If the relative atomic mass of M is 331.0, find the empirical formula of the oxide. (R.a.m. : O =16.0) (c) Determine the empirical formula of copper(II) oxide using the following results. Experimental results: Mass of test tube = 21.430 g Mass of test tube + Mass of copper(II) oxide = 23.321g Mass of test tube + Mass of copper = 22.940g (R.a.m. : Cu = 63.5, O = 16.0) (a)Mass of sulphur = 5 g Mass of oxygen = (10 – 5) g The empirical formula of the sulphur oxide is SO2. Answer

  12. 3.1 Formulae of Compounds (SB p.58) (b) The empirical formula of the oxide is M2O5.

  13. 3.1 Formulae of Compounds (SB p.58) (c) Mass of Cu = (22.940 - 21.430) g = 1.51g Mass of O = (23.321 - 22.940) = 0.381 g The empirical formula of the oxide is CuO.

  14. 3.2 Derivation of Empirical Formulae (SB p.58) Determination of Empirical Formula From Combustion by Mass Composition by mass Empirical formula

  15. 3.2 Derivation of Empirical Formulae (SB p.58) Solution: Let the empirical formula of the hydrocarbon be CxHy, and the mass of the compound be 100 g. Mass of carbon in the compound = 75 g Mass of hydrogen in the compound=(100 –75) g = 25 g Therefore, the empirical formula of the hydrocarbon is CH4. Example3-3Compound A contains carbon and hydrogen only. It is found that the compound contains 75% carbon by mass. Determine its empirical formula. (Relative atomic masses: C=12, H=1 ) Answer

  16. 3.2 Derivation of Empirical Formulae (SB p.59) Solution: Let the mass of phosphorus chloride be 100g. Then, Mass of phosphorus in the compound = 22.55g Mass of chloride in the compound = 77.45g Therefore, the empirical formula of the phosphorus chloride is PCl3. Example 3-4 The percentage by mass of phosphorus and chlorine in a sample of a phosphorus chloride are 22.55% and 77.45% respectively. Find the empirical formula of the chloride. (R.a.m. : P = 31.0, Cl = 35.5) Answer

  17. 3.2 Derivation of Empirical Formulae (SB p.59) • Let the mass of vitamin C analyzed be 100g. • The empirical formula of vitamin C is C3H4O3. • Check Point 3-3 • Find the empirical formula of vitamin C if it consists of 40.9% caarbon, 54.5% oxygen and 4.6% hydrogen by mass. ( R.a.m.: C = 12.0, H = 1.0, O = 16.0) • Each 325 mg tablet of aspirin consists of 195.0 mg carbon 14.6 mg hydrogen and 115.4mg oxygen. Determine the empirical formula of aspirin. (R.a.m. : C= 12.0, H = 1.0, O = 16.0) Answer

  18. 3.2 Derivation of Empirical Formulae (SB p.59) (b) In order to facilitate calculation, the masses of the elements are multiplied by 1000 first. The empirical formula of aspirin is C9H8O4.

  19. 3.3 Derivation of Molecular Formulae (SB p.60) What is Molecular Formulae? Molecular formula ? = (Empirical formula)n

  20. 3.3 Derivation of Molecular Formulae (SB p.60) Empirical formula Molecular mass Molecular formula

  21. 3.3 Derivation of Molecular Formulae (SB p.60) Solution: Let the empirical formula of the hydrocarbon be CxHy. Mass of carbon in the hydrocarbon = 14.6g x 12.0/44.0 = 4.0g Mass of hydrogen in the hydrocarbon = 9.0g x 2.0/18.0 = 1.0g Example 3-5 A hydrogen was burnt completely in excess oxygen. It was found that 5.00 g of the hydrocarbon gives 14.6 g of carbon dioxide and 9.0 g of water. Given that the relative molecular mass of the hydrocarbon is 30.0, determine its molecular formula. hydrocarbon.? (R.a.m.* : H = 1.0, C = 12.0, O = 16.0) Answer

  22. 3.3 Derivation of Molecular Formulae (SB p.60) Solution: (cont’d) Therefore, the empirical formula of the hydrocarbon is CH3. The molecular formula of the hydrocarbon is (CH3)n. Relative molecular mass of (CH3)n = 30.0 n x (12.0 + 1.0 x 3) = 30.0 n= 2 Therefore, the molecular formula of the hydrocarbon is C2H6.

  23. 3.3 Derivation of Molecular Formulae (SB p.61) Solution: Let the empirical formula of the hydrocarbon be CxHyOz. Mass of carbon in the compound = 44.44g Mass of hydrogen in the compound = 6.18g Mass of oxygen in the compound = 49.38g Example 3-6 Compound X is known to contain 44.44% carbon, 6.18% hydrogen and 49.38% oxygen by mass. A typical analysis shows that it has a relative molecular mass of 162.0. Find its molecular formula(R.a.m.* : H = 1.0, C = 12.0, O = 16.0) Answer

  24. 3.3 Derivation of Molecular Formulae (SB p.61) Solution(cont’d) The empirical formula of compound X is C6H10O5. The molecular formula of compound X is (C6H10O5)n. Relative molecular mass of (C6H10O5)n = 162.0 n x (12.0 x 6 + 1.0 x 10 + 16.0 x 5) = 162.0 n = 1 Therefore, the molecular formula of compound is C6H10O5.

  25. 3.3 Derivation of Molecular Formulae (SB p.61) Water of Crystallization Derived from Composition by Mass

  26. 3.3 Derivation of Molecular Formulae (SB p.61) Example 3-7 The chemical formula of hydrated copper(II) sulphate is known to be CuSO4.xH2O. It is found that the percentage of water by mass in the compound is 36%. Find x.(R.a.m. : H=1.0, O=16.0, S=32.1, Cu=63.5) Solution: Let Relative molecular mass of CuSO4xH2O = 63.5 + 32.1 + 16.0 x 4 + (1.0x2 = 16.0)x = 159.6 + 18x Relative molecular mass of water of crystallization =18x 18x/(159.6 + 18x) = 36/100 1800x=5745.6 + 648 x 1152x= 5745.6 x = 4.99  5 Therefore, the chemical formula of hydrated copper(II) sulphate is CuSO4 5H2O Answer

  27. 3.3 Derivation of Molecular Formulae (SB p.63) • (i) Let the mass of compound Z be 100g. • The empirical formula of compound Z is CH2O. • Check Point 3-4 • Find Compound Z is the major component of a healthy drink. It contains 40.00% carbon, 6.67% hydrogen and 53.33% oxygen. • (i) Find the empirical formula of compound Z. • (ii) If the relative molecular mass of compound Z is 180, finds its molecular formula.(R.a.m. : C= 12.0, H = 1.0, O = 16.0) Answer

  28. 3.3 Derivation of Molecular Formulae (SB p.63) (a) (ii) Let the molecular formula of compound Z be (CH2O)n. n x (12.0 = 1.0 x 2 = 16.0) = 180 30n = 180 n = 6 The molecular formula of Z is C6H12O6.

  29. 3.3 Derivation of Molecular Formulae (SB p.63) • (b) • Since the chemical formula of (NH4)2Sx is (NH4)2S3, the value of x is 3. Check Point 3-4 (b) (NH4)2Sx contains 72.72% sulphur by mass is water. Find the value of x. (R.a.m.: H = 1.0, N = 14.0, O = 16.0) (c) In the compound MgSO4nH2O, 51.22% by mass is water. Find the value of n. (R.a.m.: H = 1.0, O = 16.0, Mg = 24.3, S = 32.1) Answer

  30. 3.3 Derivation of Molecular Formulae (SB p.63) • (c) • Since the chemical formula of MgSO4nH2O • is MgSO47H2O , the value of x is 7.

  31. 3.3 Derivation of Molecular Formulae (SB p.63) Example 3-8 The chemical formula of ethanoic acid is CH3COOH. Calculate the percentages by mass of carbon, hydrogen and oxygen by mass respectively. (R.a.m. : C=12.0, H=1.0, O=16.0 ) Solution: Relative molecular mass of CH3COOH = 12.0 x 2 + 1.0 x 4 + 16.0 x 2 = 60.0 % by mass of C = 12.0 x 2/ 60.0 x 100%= 40.00% % by mass of H = 1.0 x 4 /60.0 x 100% = 6.67% % by mass of O = 16.0 x 2/60.0 x 100% = 53.33% The percentage by mass of carbon, hydrogen and oxygen are 40.00%, 6.67% and 53.33% respectively. Answer

  32. 3.3 Derivation of Molecular Formulae (SB p.63) Example 3-9 Calculate the mass of iron metal in a sample of 20g of hydrated iron (II) sulphate, FeSO47H2O. (R.a.m. : Fe = 55.8 , H=1.0, O=16.0 ) Solution: Relative molecular mass of FeSO4·7H2O = 55.8 + 32.1 + 16.0 x 4 + (1.0x2+16.0) x 7=277.9 % by mass of Fe = 55.8/277.9 x 100% = 20.08% Mass of Fe = 20g x 20.08% = 4.02g Answer

  33. 3.3 Derivation of Molecular Formulae (SB p.63) • Molar mass of K2Cr2O7 • = (39.1x2+52.0+16.0x7) g mol-1 = 294.2 g mol-1 • % by mass of K • = 39.1 x 2 g mol-1/294.2 g mol-1 x 100% = 26.58% • % by mass of Cr • = 52.0 x 2 g mol-1/294.2g mol-1 x 100% =35.25% • % by mass of O • = 16.0 x 7 g mol-1/294.2g mol-1 x 100% = 38.07% • Check Point 3-5 • (a) Calculate percentages by mass of potassium, chromium and oxygen in potassium chromate (VI), K2Cr2O7.(R.a.m. : K = 39.1 . Cr = 52.0, O = 16.0) • (b) Find the mass of metal and water of crystallization in • 100 g of Na2SO4·10H2O; • 70g of Fe2O3·8H2O. • (R.a.m.: H = 1.0, O = 16.0, Na = 23, S = 32.1, Fe = 55.8) Answer

  34. 3.3 Derivation of Molecular Formulae (SB p.63) (b)( i) Molar mass of Na2SO4·10H2O= 322.1 g mol-1 Mass of Na = 23.0 x 2 g mol-1/ 322.1 g mol-1 x 100g = 14.28 g Mass of H2O = 18.0 x 10 g mol-1/ 322.1 g mol-1 x 100g = 14.28 g (ii) Molar mass of Fe2O3·8H2O= 303.6 g mol-1 Mass of Fe = 55.8 x 2 g mol-1/303.6g mol-1 x 70g = 25.73 g Mass of H2O = 18.0 x 8 g mol-1/303.6g mol-1 x 70g = 33.20 g

  35. 3.4 Chemical Equations (SB p.64) mole ratios Chemical Equations a A + b B  c C + d D (can also be volume ratios for gases) Stoichiometry = relative no. of moles of substances involved in a chemical reaction.

  36. 3.4 Chemical Equations (SB p.64) Check Point 3-6 Give the chemical equations for the following reactions: (a) Zinc + steam zinc oxide + hydrogen (b) Magnesium + silver nitrate silver + magnesium nitrate (c) Butane + oxygen carbon dioxide + water • Zn(s) + H2O(g) ZnO(s) + H2(g) • (b) Mg(s) + 2 AgNO3(aq) 2Ag(s) + Mg(NO3)2(aq) • (c) 2C4H10(g) + 13O2(g) 8CO2(g) + 10H2O(l) Answer

  37. 3.5 Calculations Based on Equations (SB p.65) Calculations Based on Equations Calculations involving Reacting Masses

  38. 3.5 Calculations Based on Equations (SB p.65) Solution: CuO(s) + H2(g) Cu(s) + H2O(l) As the mole ratio of Cu : CuO is 1 : 1, the number of moles of Cu formed is the same as the number of moles of CuO reduced. Number of moles of CuO reduced = 12.45/ (63.5 + 13.0) g mol-1 = 0.157 mol Number of mole of Cu formed = 0.157 mol Mass of Cu / 63.5 g mol-1 = 0.157 Mass of Cu = 0.157 mol x 63.5 g mol-1= 9.97g Therefore, the mass of copper formed in the reaction is 9.97g. Example 3-10 Calculate the mass of copper formed when 12.45g of copper(II) oxide is completely reduced by hydrogen. (R.a.m. : H=1.0, O=16.0, Cu = 63.5 ) Answer

  39. 3.5 Calculations Based on Equations (SB p.65) Example 3-11 Sodium hydrogencarbonate decomposes according to the following equation. 2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(l) In order to obtain 240 cm3 of CO2 at room temperature and pressure, what is the minimum amount of sodium hydrogencarbonate required? (R.a.m. : H = 1.0, C =12.0, O = 16.0, Na = 23.0; molar volume of gas at R.T.P. = 24.0 dm3mol-1) Solution: Number of moles of CO2 formed = 240cm3/ 24000cm3 mol-1 = 0.01 mol From the equation, 2 moles of NaHCO3(s) will form 1 mole of CO2(g). Number of moles of NaHCO3 required = 0.01 x 2 = 0.02 mol Mass of NaHCO3 required = 0.02 mol x(23.0 + 1.0 + 16.0 x 3) g mol-1 = 0.02 mol x 84.0g mol-1 = 1.68 g Therefore, the minimum amount of sodium hydrogencarbonate required is 1.68g. Answer

  40. 3.5 Calculations Based on Equations (SB p.66) Calculations Based on Equations Calculations involving Volumes of Gases

  41. 3.5 Calculations Based on Equations (SB p.66) Solution: Number of moles of CO2 formed C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(l) 2 mol : 7 mol : 4 mol : 6 mol (from equation) 2 volumes: 7 volumes : 4 volumes : - (by Avogadro’s law) It can be judged from the equation that the mole ratio of CO2 : C2H6 is 4 :2, and the volume ratio of CO2 : C2H6 should also be 4:2. Let x be the volume of CO2(g) formed x /20cm3 = 4/2 x = 40 cm3 Therefore, the volume of CO2 formed is 40 cm3. Example 3-12 Calculate the volume of carbon dioxide formed when 20 cm3 of ethane and 70 cm3 of oxygen are exploded, assuming all volumes are measured at room temperature and pressure. Answer

  42. 3.5 Calculations Based on Equations (SB p.67) Solution: Let the molecular formula of the hydrocarbon be CxHy. Volume of hydrogen reacted = 10 cm3 Volume of O2(g) unreacted = 50 cm3 Volume of O2(g) reacted = 30 cm3 Volume of CO2(g) formed = 20 cm3 CxHy + (x + y/4) O2 CO2 + y/2 H2O 1 volume : (x + y/4) volumes : x volumes Volume of CO2 (g)/ volume of CxHy(g) = 20 cm3/ 10cm3 = 2 X =2 Volume of O2(g) / volume of CxHy(g) =(x + y/4) / 1= 30/ 10 (x + y/4)= 3 Y = 2 Molecular formula is C2H4 Example 3-13 10 cm3 of a gaseous hydrocarbon was mixed with 80cm3 of oxygen which was in excess. The mixture was exploded and then cooled. The volume left was 70cm3. Upon passing the resulting gaseous mixture through concentrated sodium hydroxide solution ( to absorb carbon dioxide), the volume of the residual gas became 50 cm3. Find the molecular formula of the hydrocarbon. Answer

  43. 3.5 Calculations Based on Equations (SB p.68) • No. of moles of H2 = No. of moles of Mg • Volume of H2/ 24.0 dm3 mol-1 = 2.43 g / 24.3 g mol-1 • Volume of H2 =2.4 dm3 • (b) 1/3 x no. of moles of Cl2 =1/2x no. of moles of PCl3 • 1/3 x mass of Cl2 / (35.5 x 2) g mol-1 • = 1/2 x 100g / (31.0 + 35.5 + 3 ) g mol-1 • Mass of Cl2 =77.45g • Check Point 3-7 • Find the volume of hydrogen produced at R.T.P. when 2.43g of magnesium reacts with excess hydrochloric acid. (R.a.m. : Mg = 24.3; molar volume of gas at R.T.P. = 24.0 dm3mol-1. • Find the minimum mass of chlorine required to produced 100 g of phosphorus trichloride ( PCl3). • 20 cm3 of a gaseous hydrocarbon and 150 cm3 of oxygen were exploded in a closed vessel. After cooling, 110 cm3 of gases remained. After passing through a solution of concentrated sodium hydroxide, the volume left was 50 cm3 .Determine the molecular of the hydrocarbon. • Calculate the volume of carbon dioxide formed when 5cm3 of methane burns in excess oxygen, assuming all volumes are measured at room temperature and pressure. Answer

  44. 3.5 Calculations Based on Equations (SB p.68) (c)Volume of CxHy used = 20 cm3 Volume of CO2 formed = 60 cm3 Volume of O2 used = 100 cm3 Volume of CxHy : volume of CO2 = 1 : x = 20 : 60 x = 3 Volume of CxHy : volume of O2 = 1 : x + y/4 = 20 : 100 x + y/4 = 5 3 + y/4 = 5 y = 8

  45. 3.5 Calculations Based on Equations (SB p.68) (d)Volume of CxHy used = 20 cm3 It can be judged from the equation that the mole ratio of CO2 : CH4 is 1:1, the volume ratio of CO2 : CH4 should also be 1:1. x / 5 = 1/1 x = 5 The volume of carbon dioxide gas is 5 cm3.

  46. 3.6 Simple Titrations (SB p.68) Simple Titrations Acid-Base Titrations Acid-Base Titrationswith Indicators Acid-Base Titrationswithout Indicators (to be discussed in later chapters)

  47. 3.6 Simple Titrations (SB p.69) Copper(II) sulphate solution Finding the concentration of a solution Copper(II) sulphate + solute Water solvent solution

  48. 3.6 Simple Titrations (SB p.69) Finding the concentration of a solution ~50 cm3

  49. 3.6 Simple Titrations (SB p.69) Finding the concentration of a solution ~50 cm3

  50. 3.6 Simple Titrations (SB p.69) Finding the concentration of a solution ~50 cm3

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