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The Mole Concept

2. The Mole Concept. 2.1 The Mole 2.2 Molar Volume and Avogadro’s Law 2.3 Ideal Gas Equation 2.4 Determination of Molar Mass 2.5 Dalton’s Law of Partial Pressures. 2.1. The Mole. for counting common objects. for counting particles like atoms, ions, molecules. What is “mole”?.

thor-schroeder
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The Mole Concept

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  1. 2 The Mole Concept 2.1 The Mole 2.2 Molar Volume and Avogadro’s Law 2.3 Ideal Gas Equation 2.4 Determination of Molar Mass 2.5 Dalton’s Law of Partial Pressures

  2. 2.1 The Mole

  3. for counting common objects for counting particles like atoms, ions, molecules What is “mole”? 2.1 The mole (SB p.18)

  4. Avogadro constant(the amount in 1 mole) How large is the amount in 1 mole? 2.1 The mole (SB p.19) 6.02  1023 = 602 000 000 000 000 000 000 000

  5. All the people in the world $ 6.02  1023 so that each get: $ 1000 note count at a rate of 2 notes/sec 2.1 The mole (SB p.19) ? 2000 years

  6. 1 Number of moles = How to find the number of moles? 2.1 The mole (SB p.19) or number of particles = number of moles  (6.02  1023)

  7. 2.1 The mole (SB p.19) Why defining 6.02 x 1023 as the amount for one mole? 12 g carbon contains 6.02  102312C atoms The mole is the amount of substance containing as many particles as the number of atoms in 12 g of carbon-12.

  8. 2.1 The mole (SB p.19) ……. C atom 6.02  1023 6.02  1023 ……. Relative atomic masses H atom Molar mass 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.

  9. 2.1 The mole (SB p.20) Example 2-1A Example 2-1B Example 2-1C Example 2-1E Example 2-1D Check Point 2-1 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.

  10. 2.2 Molar Volume and Avogadro’s Law

  11. 2.2 Molar volume and Avogadro’s law (SB p.24) What is molar volume of gases? at 25oC & 1 atm (Room temp & pressure / R.T.P.)

  12. 2.2 Molar volume and Avogadro’s law (SB p.24) 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.

  13. 2.2 Molar volume and Avogadro’s law (SB p.24) Avogadro’s Law 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 Let's Think 1

  14. 2.2 Molar volume and Avogadro’s law (SB p.24) Example 2-2C Example 2-2B Example 2-2A Check Point 2-2 Example 2-2D Interconversions involving number of moles

  15. 2.3 Ideal Gas Equation

  16. 2.3 Ideal gas equation (SB p.27) Boyle’s law At constant temperature, the volume of a given mass of a gas is inversely proportional to the pressure exerted on it PV = constant

  17. 2.3 Ideal gas equation (SB p.28) Schematic diagrams explaining Boyle’s law

  18. 2.3 Ideal gas equation (SB p.28) A graph of volume against the reciprocal of pressure for a gas at constant temperature

  19. 2.3 Ideal gas equation (SB p.28) Charles’ law At a constant pressure, the volume of a given mass of a gas is directly proportional to the absolute temperature.

  20. 2.3 Ideal gas equation (SB p.28) Schematic diagrams explaining Charles’ law

  21. 2.3 Ideal gas equation (SB p.28) A graph of volume against absolute temperature for a gas at constant pressure

  22. Combining: V  n (Avogadro’s Law) V  (Boyle’s Law) V  T (Charles Law) V  V = where R is a constant (called the universal gas constant) 2.3 Ideal gas equation (SB p.27) Ideal gas equation PV = nRT

  23. 2.3 Ideal gas equation (SB p.29) 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  10-3 m3 mol-1 T = 0 oC or 273K

  24. 2.3 Ideal gas equation (SB p.29) 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 = = = 8.314 J K-1mol-1

  25. 2.3 Ideal gas equation (SB p.29) Relationship between the ideal gas equation and the individual gas laws

  26. Example 2-3B Example 2-3A Check Point 2-3 Example 2-3C 2.3 Ideal gas equation (SB p.30)

  27. 2.4 Determination of Molar Mass

  28. Mass of volatile liquid injected = 26.590 - 26.330 = 0.260 g 2.4 Determination of molar mass (SB p.32)

  29. Volume of trichloromethane vapour = 74.4 - 8.2 = 66.2 cm3 2.4 Determination of molar mass (SB p.32)

  30. Temperature = 273 + 99 = 372 K 2.4 Determination of molar mass (SB p.32)

  31. Pressure = 101325 Nm-2 2.4 Determination of molar mass (SB p.32)

  32. 2.4 Determination of molar mass (SB p.32) PV = nRT 101325 Nm-2  66.2  10-6 m3 = n  8.314 J K-1 mol-1  372 K n = 2.169  10-3 mol

  33. 2.4 Determination of molar mass (SB p.32) Molar mass = = = 119.87 g mol-1

  34. Example 2-4A Example 2-4B Check Point 2-4 2.4 Determination of molar mass (SB p.32) PV = nRT………..(1) n = ………..(2) Where m is the mass of the volatile substance M is the molar mass of the volatile substance Combing (1) and (2), we obtain PV = RT M =

  35. 2.5 Dalton’s Law of Partial Pressures

  36. PT = PA + PB + PC 2.5 Dalton’s law of partial pressures (SB p.35) Dalton’s Law of Partial Pressures In a mixture of gases which do not react chemically, the total pressure of the mixture is the sum of thepartial pressures of the component gases (the sum that each gas would exert as if it is present alone under the same conditions).

  37. 2.5 Dalton’s law of partial pressures (SB p.35) Consider a mixture of gases A, B and C occupying a volume V. It consists of nA, nB and nC moles of each gas. The total number of moles of gases in the mixture ntotal = nA + nB + nC If the equation is multiplied by RT/V, then ntotal (RT/V) = nA (RT/V) + nB (RT/V) + nC (RT/V) i.e. Ptotal = PA + PB + PC (so Dalton’s Law is a direct consequence of the Ideal Gas Equation)

  38. Example 2-5A Example 2-5B Example 2-5C Check Point 2-5 Example 2-5D 2.5 Dalton’s law of partial pressures (SB p.35) Besides, the partial pressure of each component gas can be calculated from the Ideal gas law. PA = nA(RT/V) and Ptotal =ntotal(RT/V) i.e. PA= (nA/ntotal) Ptotal PA = xA Ptotal

  39. The END

  40. 2.1 The mole (SB p.20) Back Example 2-1A What is the mass of 0.2 mol of calcium carbonate? Answer The chemical formula of calcium carbonate is CaCO3. Molar mass of calcium carbonate = (40.1 + 12.0 + 16.0  3) g mol-1 = 100.1 g mol-1 Mass of calcium carbonate = Number of moles  Molar mass = 0.2 mol  100.1 g mol-1 = 20.02 g

  41. Molar mass of gold = 197.0 g mol-1 Number of moles = = 0.1015 mol Number of gold atoms = 0.1015 mol  6.02  1023 mol-1 = 6.11  1022 2.1 The mole (SB p.21) Back Example 2-1B Calculate the number of gold atoms in a 20 g gold pendant. Answer

  42. 2.1 The mole (SB p.21) Example 2-1C • It is given that the molar mass of water is 18.0 g mol-1. • What is the mass of 4 moles of water molecules? • How many molecules are there? • How many atoms are there? Answer

  43. 2.1 The mole (SB p.21) Example 2-1C • Mass of water = Number of moles  Molar mass • = 4 mol  18.0 g mol-1 • = 72.0 g • There are 4 moles of water molecules. • Number of water molecules • = Number of moles  Avogadro constant • = 4 mol  6.02  1023 mol-1 • = 2.408  1024

  44. 2.1 The mole (SB p.21) Back Example 2-1C • 1 water molecule has 3 atoms (i.e. 2 hydrogen atoms and 1 oxygen atom). • 1 mole of water molecules has 3 moles of atoms. • Thus, 4 moles of water molecules have 12 moles of atoms. • Number of atoms = 12 mol  6.02  1023 mol-1 • = 7.224  1024

  45. The chemical formula of magnesium chloride is MgCl2. • Molar mass of MgCl2 = (24.3 + 35.5  2) g mol-1 = 95.3 g mol-1 • Number of moles of MgCl2 = • = 0.105 mol 2.1 The mole (SB p.22) Example 2-1D • A magnesium chloride solution contains 10 g of magnesium chloride solid. • Calculate the number of moles of magnesium chloride in the solution. Answer

  46. 2.1 The mole (SB p.22) Example 2-1D • (b) Calculate the number of magnesium ions in the solution. Answer • 1 mole of MgCl2 contains 1 mole of Mg2+ ions and 2 moles of Cl- ions. • Therefore, 0.105 mol of MgCl2 contains 0.105 mol of Mg2+ ions. • Number of Mg2+ ions • = Number of moles of Mg2+ ions  Avogadro constant • = 0.105 mol  6.02  1023 mol-1 • = 6.321  1022

  47. 2.1 The mole (SB p.22) Example 2-1D • (c) Calculate the number of chloride ions in the solution. Answer • 0.105 mol of MgCl2 contains 0.21 mol of Cl- ions. • Number of Cl- ions • = Number of moles of Cl- ions  Avogadro constant • = 0.21 mol  6.02  1023 mol-1 • = 1.264  1023

  48. 2.1 The mole (SB p.22) Back Example 2-1D • (d) Calculate the total number of ions in the solution. Answer • Total number of ions • = 6.321  1022 + 1.264  1023 • = 1.896  1023

  49. 2.1 The mole (SB p.23) The chemical formula of carbon dioxide is CO2. Molar mass of CO2 = (12.0 + 16.0  2) g mol-1 = 44.0 g mol-1 Number of moles = = = Mass of a CO2 molecule = = 7.31  10-23 g Example 2-1E Back • What is the mass of a carbon dioxide molecule? Answer

  50. 2.1 The mole (SB p.23) Check Point 2-1 • Find the mass in grams of 0.01 mol of zinc sulphide. Answer • Mass = No. of moles  Molar mass • Mass of ZnS = 0.01 mol  (65.4 + 32.1) g mol-1 • = 0.01 mol  97.5 g mol-1 • = 0.975 g

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