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The Mole Concept

Chapter 2. The Mole Concept. 2.1 The Mole 2.2 Ideal Gas Equation 2.3 Determination of Relative Molecular Mass 2.4 The Faraday and the Mole. for counting common objects. for counting particles like atoms, ions, molecules. What is “Mole”?. 2.1 The Mole (SB p.28).

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The Mole Concept

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  1. Chapter 2 The Mole Concept 2.1The Mole 2.2 Ideal Gas Equation 2.3 Determination of Relative Molecular Mass 2.4 The Faraday and the Mole

  2. for counting common objects for counting particles like atoms, ions, molecules What is “Mole”? 2.1 The Mole (SB p.28)

  3. Avogadro constant(the amount in 1 mole) How large is the amount in 1 mole? 2.1 The Mole (SB p.28) 6.02 x 1023 = 602000000000000000000000

  4. All the people in the world $ 6.02 x 1023 so that each get: $ 1000 note count at a rate of 2 notes/sec 2.1 The Mole (SB p.28) ? 2000 years

  5. 1 Number of moles = How to find the number of moles? 2.1 The Mole (SB p.29) or number of particles = number of moles x (6.02 x 1023)

  6. 2.1 The Mole (SB p.29) Why defining 6.02 x 1023 as the amount for one mole? 12 g carbon contains 6.02 x 102312C atoms The mole is the amount of substance containing as many particles as the number of atoms in 12 g of carbon-12.

  7. 2.1 The Mole (SB p.29) Molar mass ……. C atom 6.02 x 1023 6.02 x 1023 ……. Relative atomic masses H atom Relative mass 12 1 1 g Molar mass is the mass, in grams, of 1 mole of a substance, e.g. the molar mass of H atom is 1 g.

  8. 2.1 The Mole (SB p.29) Molar mass is the same as the relative atomic mass in grams. Molar mass is the same as the relative molecular mass in grams. Molar mass is the same as the formula mass in grams.

  9. 2.1 The Mole (SB p.30) ……. ……. 1 g (1 mole) 1 g (1 mole) mass molar mass Calculations using molar mass Question We have a sample of 2 g of hydrogen atoms.What is the number of moles?(Relative atomic mass of H = 1) Number of moles = = 2

  10. 2.1 The Mole (SB p.30) Number of moles = 2 or Mass = number of moles x molar mass

  11. 2.1 The Mole (SB p.30) Example 2-1 What is the mass of 0.2 mole of calcium carbonate? (R.a.m.* : C = 12.0, O = 16.0, Ca = 40.1) Solution: The chemical formula of calcium carbonate is CaCO3. Molar mass of calcium carbonate = (40.1 + 12.0 + 3 x 16.0) g mol-1 = 100.1 g mol-1 Answer

  12. 2.1 The Mole (SB p.30) Solution: (cont’d) Mass of calcium carbonate = Number of moles x Molar mass = 0.2 mol x 100.1 g mol-1 = 20.02g

  13. 2.1 The Mole (SB p.31) Solution: Number of gold atoms in 20g of gold coin = x 6.02 x 1023 mol-1 = 6.11 x 1023 Example 2-2 Calculate the number of gold atoms in 20g of gold atom. (R.a.m. : Au = 197.0) Answer

  14. 2.1 The Mole (SB p.31) Example 2-3 It is given that the molar mass of water is 18.0g mol-1. (a)What is the mass of 4 moles of water molecule? (b) How many molecules are there? (c) How many atoms are there? • Solution: • Mass of water • = Number of moles x Molar mass • = 4 mol x 18.0 g mol-1 • = 72.0 g Answer

  15. 2.1 The Mole (SB p.31) Solution: (cont’d) (b) There are 4 moles of water molecules. Number of water molecules = Number of moles x Avogadro constant = 4 mol x 6.02 x 1023 mol-1 = 2.408 x 1024

  16. 2.1 The Mole (SB p.31) Solution: (cont’d) (c) 1 water molecule has 3 atoms (including 2 hydrogen atoms and 1 oxygen atoms). 1 mole of water molecule has 3 moles of atoms. Thus, 4 moles of water molecules have 12 moles of atoms. Number of atoms = 12 mol x 6.02 x 1023 mol-1 = 7.224 x 1024

  17. 2.1 The Mole (SB p.31) • Solution: • The chemical formula of magnesium chloride is MgCl2. • Molar mass of MgCl2 • = (24.3 + 35.5 x2) g mol-1 • = 95.3 g mol-1 • Number of moles of MgCl2 • = • = 0.105 mol Example 2-4 A magnesium chloride solution contains 10 g of magnesium chloride solid (a) Calculate the number of moles of magnesium chloride in the solution. (b) Calculate the number of magnesium ions in the solution. (c) Calculate the number of chloride ions in the solution. (d) Calculate the total number of ions in the solution. (R.a.m.: Mg = 24.3, Cl = 35.5) Answer

  18. 2.1 The Mole (SB p.31) Solution: (cont’d) (b) 1 mole of MgCl2 contains 1 mole of Mg2+ and 2 moles of Cl- . Therefore, 0.105 mole of MgCl2 contains 0.105 mol x 6.02 x 1023 mol-1. Number of Mg2+ ions = Number of moles of Mg2+ x Avogadro constant = 0.105 mol x 6.02 x 1023 mol-1 = 6.321 x 1022

  19. 2.1 The Mole (SB p.31) Solution: (cont’d) (c) 0.105 mole of MgCl2 contains 0.21 mole of Cl- . Number of Cl- ions = Number of moles of Cl- x Avogadro constant = 0.21 mol x 6.02 x 1023 mol-1 = 1.264 x 1023 (d) Total number of ions = 6.321 x 1022 + 1.264 x 1023 = 1.896 x 1023

  20. 2.1 The Mole (SB p.32) Solution: The chemical formula of carbon dioxide is CO2. Molar mass of CO2 = (12.0 + 16.0 x 2) g mol-1 = 44.0 g mol-1 ∵ Number of mole = = ∴ = Mass of a CO2 molecule= = 7.31 x 10-23 g Example 2-5 What is the mass of carbon dioxide molecule? (R.a.m. : C = 12.0, O = 16.0) Answer

  21. 2.1 The Mole (SB p.32) (a) Mass = No. of moles x Molar mass Mass of ZnS = 0.01 mol x (65.0 + 32.1) g mol-1 = 0.01 mol x 95.7 g mol-1 = 0.975g • Check Point 2-1 • Find the mass in grams of 0.01 mole of zinc sulphide.(R.a.m. : S = 32.1, Zn = 65.4) • Find the number of ions in 5.61 g of calcium oxide. (R.a.m. : O = 16.0, Ca = 40.1) • Find the number of atoms in 32.05g of sulphur dioxide. (R.a.m. : O 16.0, S = 32.1) • There is 4.80 g of ammonium carbonate. Find the • (i) number of moles of the compound, • (ii) number of moles of ammonium ions, • (iii) number of moles of hydrogen atoms, and • (v) number of hydrogen atoms. Answer

  22. 2.1 The Mole (SB p.32) (b) No. of moles of CaO = = 0.1 mol 1 CaO formula unit contains 1 Ca2+ ion and 1 O2- ion. No. of moles of ions = 0.1 mole x 2 = 0.2 mol No of ions = 0.2 mol x 6.02 x 1023 mol-1 = 1.204 x 1023

  23. 2.1 The Mole (SB p.32) (c) No. of moles of SO2 = = 0.5 mol 1 SO2 molecule contains 1 S atom and 2O atoms. No. of mole of atoms = 0.5 mole x 3 = 1.5 mol No of atoms = 1.5 mol x 6.02 x 1023 mol-1 = 9.03 x 1023

  24. 2.1 The Mole (SB p.32) • (d) Molar mass of (NH4)2CO3 = 96.0 g mol-1 • No. of mole of (NH4)2CO3 = • ∵ 1 mole (NH4)2CO3 gives 2 moles NH4+. • No. of moles of (NH4)2CO3 = 0.05 mol x 2 • = 0.1 mol

  25. 2.1 The Mole (SB p.32) (iii) ∵ 1 mole (NH4)2CO3 gives 1 mole CO32- . No. of moles of CO32- = 0.05 mol (iv) 1 (NH4)2CO3 formula unit contains 8H atoms. No. of moles of H atoms = 0.05 mol x 8 = 0.4 mol (v) No. of H atoms = 0.4mol x 6.02 x 1023 mol-1 = 2.408 x 1023

  26. 2.1 The Mole (SB p.33) What is Molar Volume of Gases? at 250C & 1 atm (Room temp & pressure / rtp)

  27. 2.1 The Mole (SB p.33) 22.4 dm3 22.4 dm3 22.4 dm3 22.4 22.4 22.4 at 00C & 1 atm (Standard temp & pressure / stp)

  28. 2.1 The Mole (SB p.33) V V V V v v v at any other fixed temp & pressure

  29. 2.1 The Mole (SB p.33) Avogadro’s Law Equal volumes of all gases at the same temperature and pressure contain the same number of molecules. Equal volumes of all gases at the same temperature and pressure contain the same number of moles of gases. So 1 mole of gases should have the same volume at the same temperature and pressure. V  n where n is the no. of moles of gas

  30. 2.1 The Mole (SB p.34) Solution: Molar mass of chloride gas (Cl2) = 35.5 x 2 g mol-1 = 71.0 g mol-1 Number of moles of Cl2 = = 0.05 mol Volume of Cl2 = Number of moles of Cl2 x Molar volume = 0.05 mol x 24.0 dm3mol-1 =1.2 dm3 Example 2-6 Find the volume occupied by 3.55 g of chloride gas at room temperature and pressure (Molar volume of gas at R.T.P. = 24.0 dm3 mol-1;R.a.m. : Cl = 35.5) Answer

  31. 2.1 The Mole (SB p.34) Solution: Molar volume of carbon dioxide gas at S.T.P. = 22.4 dm3 mol-1 Number of moles of CO2 = = 2 x 10 –4 mol Number of CO2 molecules = 2 x 10-4 mol x 6.02 x 1023 mol-1 =1.204 x 1020 Example 2-7 Find the number of molecules in 4.48 cm3 ofcarbon dioxide gas at standard temperature and pressure. (Molar volume of gas at S.T.P. = 22.4 dm3 mol-1; Avogrado constant = 6.02 x 1023 mol-1) Answer

  32. 2.1 The Mole (SB p.35) Solution: Molar mass of nitrogen gas (N2) = (14.0 + 14.0) g mol-1 = 28.0 g mol-1 ∵ Density = = ∴ Density of N2 = = 1.167 g dm-3 Example 2-8 The molar volume of nitrogen gas is found to be 24.0 dm3 mol-1 at room temperature and pressure. Find the density of nitrogen gas. (R.a.m. : N = 14.0) Answer

  33. 2.1 The Mole (SB p.35) Solution: Number of moles of the gas = 0.05 mol Molar mass of the gas = = 32 g mol-1 Relative molecular mass of the gas = 32 (no unit) Example 2-9 1.6 g of a gas occupies 1.2 dm3 at room temperature and pressure. What is the relative molecular mass of the gas? (Molar Volume of gas at R.T.P. = 24.0 dm3 mol-1) Answer

  34. 2.1 The Mole (SB p.35) • No. of moles of H2 = • = 0.3 mol • Volume = No. of moles x Molar volume • = 0.3 mol x 24.0 dm3 mol-1 • = 7.2 dm3 • Check Point 2-2 • Find the volume of 0.6 g of hydrogen gas at room temperature and pressure. (R.a.m. : H = 1.0; molar volume at R.T.P. = 24.0 dm3 mol-1) • Calculate the number of molecules in 4.48 dm3 of hydrogen at standard temperature and pressure. ( Molar volume at S.T.P. = 22.4 dm3 mol-1) • The molar volume of oxygen is 22.4 dm3 mol-1 at standard temperature and pressure. Find the density of oxygen in g cm-3 at S.T.P.. (R.a.m. : O = 16.0) • What mass of oxygen has the same number of moles as that in 3.2 g of sulphur dioxide? (R.a.m. : O = 16.0, S = 32.1) Answer

  35. 2.1 The Mole (SB p.35) (b) No. of moles of H2 = = 0.2 mol No of H2 molecules = 0.2 mol x 6.02 x 1023 mol-1 = 1.204 x 1023

  36. 2.1 The Mole (SB p.35) ( c) Density = = Molar mass of O2 = 16.0 x 2 g mol-1 =32.0 g mol-1 Molar volume of O2 = 22..4 dm3 mol-1 = 22 400 cm3 mol-1 Density = = 1.43 x 10-3 g cm-3

  37. 2.1 The Mole (SB p.35) (d) No. of moles of SO2 = = 0.05 mol No of moles of O2 = 0.05 mol Mass = No. of moles x Molar mass Mass of O2 = 0.05 mol x 16.0g mol-1 = 1.6g

  38. 2.2 Ideal Gas Equation (SB p.36) Gas Laws Boyle’s law states that: At constant temperature, the volume of a given mass of a gas is inversely proportional to the pressure exerted on it PV = constant

  39. 2.2 Ideal Gas Equation (SB p.36)

  40. 2.2 Ideal Gas Equation (SB p.37) Charles’ Law states that: At a constant pressure, the volume of a given mass of a gas is directly proportional to the absolute temperature.

  41. 2.2 Ideal Gas Equation (SB p.37)

  42. 2.2 Ideal Gas Equation (SB p.37) The Boyle’s law and Charles’ law gives the ideal gas equation Combining: V  n (Avogadro’s Law) V  1/P (Boyle’s Law) V  T (Charles Law) V  nT/P V = RnT/P where R is a constant (called the Universal Gas Constant) PV = nRT

  43. 2.2 Ideal Gas Equation (SB p.37) For one mole of an ideal gas at standard temperature and pressure, P = 760 mmHg, 1 atm or 101 325 Nm-2 (Pa) V = 22.4 dm3 mol-1 or 22.4 x 10-3 m3 mol-1 T = 0 oC or 273K By substituting the values of P, V and T in S.I. Units into the equation, the value of ideal gas constant can be found. R = PV/T = = 8.314 JK-1mol-1

  44. 2.2 Ideal gas equation (SB p.38) Solution: As the number of moles of the gas is fixed, PV/T should be a constant = P2 = 876.17 mmHg. Note: All temperature values used in gas laws are on the Kelvin scale. Example 2-10 A 500 cm3 sample of a gas in a sealed container at 700 mm Hg and 25oC is heated to 100oC. What is the final pressure of the gas? Answer

  45. 2.2 Ideal gas equation (SB p.38) Example 2-11 A reaction vessel of 500 cm3 is filled with oxygen at 25oC and the final pressure exerted on it is 101 325 Nm-2. How many moles of oxygen are there? (Ideal gas constant = 8.314 J K-1 mol-1) Solution: PV = nRT 101 325 Nm-2 x 500 x 10-6 m3 = n x 8.314 J K-1 mol-1 x (273 + 25) K n = 0.02 mol There is 0.02 mole of oxygen in the reaction vessel. . Answer

  46. 2.2 Ideal gas equation (SB p.39) Solution: Applying the equation, T = = = 1523.4 K The highest temperature it can be safely heated to is 1250.4oC. Example 2-12 A 5 dm3 vessel can withstand a maximum internal pressure of 50 atm. If 2 moles of nitrogen gas is pumped into the vessel, what is the highest temperature it can be safely heated to? (1 atm = 101 325 Nm-2, ideal gas constant = 8.314 J K-1 mol-1) Answer

  47. 2.2 Ideal gas equation (SB p.39) • T2 = 586 K • Check Point 2-3 • A reaction vessel is filled with a gas at 20oC and 5atm. If the vessel can withstand a maximum internal pressure of 10 atm, what is the highest temperature it can be safely heated to? • A balloon is filled with helium at 25oC. The pressure exerted and the volume of balloon are found to be 1.5 atm and 450 cm3 respectively. How many moles of helium have been introduced into the balloon? ( 1 atm = 101 325 Nm-2; ideal gas constant = 8.314 J K-1 mol-1) • (c) 25.8 cm3 sample of a gas has a pressure of 690 mm Hg and a temperature of 17oC. What is the volume if the pressure is changed to 1.85 atm and the temperature to 345 K? ( 1 atm = 760 mmHg) Answer

  48. 2.2 Ideal gas equation (SB p.39) (b) PV= nRT 1.5 x 101 325 Nm-2 x 450 x 10-6 m3 = n x 8.314 J K-1 mol-1 x (273+25) K n = 0.0276 mol

  49. 2.2 Ideal gas equation (SB p.39) (c ) V=15.06cm3

  50. 2.3 Determination of Relative Molecular Mass (SB p.39) Mass of volatile liquid injected = 26.590 - 26.330 = 0.260 g

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