1. Structure of the Atom Size:0.1 to 0.5 nanometers (1 nm = 1 x 10 -9 m) Nucleus Electron clouds ~ 5 x 10 -15 m ~ 2 x 10 -10 m Nucleus- consists of protons and neutrons

2. Iron atoms on a copper crystal

3. Iron atoms on a copper crystal Carbon monoxide molecules on platinum crystal

4. Iron atoms on a copper crystal Xenon atoms on nickel Carbon monoxide molecules on platinum crystal

5. Atomic number:number of protons in an atom. Symbol: Z Atomic mass: mass of an atom in a.m.u. (  number of protons + neutrons in an atom). Symbol: A

6. Atomic number:number of protons in an atom. Symbol: Z Atomic mass: mass of an atom in a.m.u. (  number of protons + neutrons in an atom). Symbol: A Atome-p+nZA (a.m.u.) H11011.008 Li33437.016 C666612.000 U929214692238.05

7. Conventional Notation for Chemical Symbols Cl 35 17 - 2 Mass numberNumber and sign of charge

8. Conventional Notation for Chemical Symbols Cl 35 17 - 2 Mass numberNumber and sign of charge Atomic number Number of atoms in entity

9. Conventional Notation for Chemical Symbols Cl 35 17 - 2 Mass numberNumber and sign of charge Atomic number Number of atoms in entity Examples: He 4 2 OH - CO 3 2-

10. Ions:atoms or molecules which have gained or lost one or more electrons Isotopes: atoms of the same element which have different masses, i.e. different numbers of neutrons

11. Ions:atoms or molecules which have gained or lost one or more electrons Isotopes: atoms of the same element which have different masses, i.e. different numbers of neutrons Examples of ions:H + H - NO 3 -

12. Ions:atoms or molecules which have gained or lost one or more electrons Isotopes: atoms of the same element which have different masses, i.e. different numbers of neutrons Examples of ions:H + H - NO 3 - Examples of isotopes:H 1 1 H 1 2 H 3 1 Hydrogen Deuterium Tritium

13. ElementsCompounds AtomsMolecules Identical atoms

14. ElementsCompounds AtomsMolecules IonsIsotopes Identical atoms ± e - ± n ± e - ± n

15. (Chemical) Atomic Mass: The average of the isotope masses of an element, weighted to reflect their relative natural abundances.

16. Example: Chlorine has two naturally occurring isotopes, 35 Cl (A = 34.97 a.m.u.) and 37 Cl (A = 36.96 a.m.u.). The respective natural abundances of these isotopes are 75.5% and 24.5%. Atomic mass of Cl = (34.97 x 75.5) + (36.96 x 24.5) 100 =35.45 a.m.u.

17. (Chemical) Atomic Mass: The average of the isotope masses of an element, weighted to reflect their relative natural abundances. Example: Chlorine has two naturally occurring isotopes, 35 Cl (A = 34.97 a.m.u.) and 37 Cl (A = 36.96 a.m.u.). The respective natural abundances of these isotopes are 75.5% and 24.5%. Atomic mass of Cl = (34.97 x 75.5) + (36.96 x 24.5) 100 =35.45 a.m.u. Molecular Mass:The sum of the atomic masses of all of the constituent atoms in a molecule.

18. (Chemical) Atomic Mass: The average of the isotope masses of an element, weighted to reflect their relative natural abundances. Example: Chlorine has two naturally occurring isotopes, 35 Cl (A = 34.97 a.m.u.) and 37 Cl (A = 36.96 a.m.u.). The respective natural abundances of these isotopes are 75.5% and 24.5%. Atomic mass of Cl = (34.97 x 75.5) + (36.96 x 24.5) 100 =35.45 a.m.u. Molecular Mass:The sum of the atomic masses of all of the constituent atoms in a molecule. Example: Molecular mass of H 2 ? (A.M. of H = 1.008 a.m.u.) M.M. of H 2 = 2 x 1.008 = 2.016 a.m.u.

19. Example 2:M.M. of carbon monoxide, CO? (A.M. of C = 12.01 a.m.u., A.M. of O = 15.99 a.m.u) M.M. of CO = 12.01 + 15.99 = 29.01 a.m.u.

20. Example 2:M.M. of carbon monoxide, CO? (A.M. of C = 12.01 a.m.u., A.M. of O = 15.99 a.m.u) M.M. of CO = 12.01 + 15.99 = 29.01 a.m.u. Example 3:M.M. of lead nitrate, Pb(NO 3 ) 2 ? (Pb =207.2, N = 14.01, O = 15.99) M.M. of Pb(NO 3 ) 2 = 207.2 + (14.01 x 2) + (15.99 x 6) = 331.16 a.m.u.

21. Chemical Accounting 1: The Mole Relating Number of Chemical Entities to their Mass

22. Consider the atomic mass table: ElementAtomic mass (of one atom) H1.008 a.m.u. He4.003 a.m.u. Li6.941 a.m.u. Be9.012 a.m.u. C 12.000 a.m.u....etc.

23. Consider the atomic mass table: ElementAtomic mass (of one atom) H1.008 a.m.u. He4.003 a.m.u. Li6.941 a.m.u. Be9.012 a.m.u. C 12.000 a.m.u....etc. Gives: Relative mass of one atom of a given element w.r.t. to mass of one atom of one element (carbon)

24. Practical work requires a table which gives: (a)relative mass of a given large number of atoms of each element w.r.t. mass of the same number of C atoms; (b)a value of 12 g for this given large number of C atoms.

25. Practical work requires a table which gives: (a)relative mass of a given large number of atoms of each element w.r.t. mass of the same number of C atoms; (b)a value of 12 g for this given large number of C atoms. Thus this “practical” table should look like............... ElementMass (of n atoms) H1.008 g He4.003 g Li6.941 g Be9.012 g C 12.000 g...etc. where n is the given large number of atoms.

26. What value of n will allow us to “scale up” from the a.m.u. scale (single atoms) to the gram scale (large numbers of atoms)? n = 6.022 x 10 23 entities = Avogadros Number (N) = 1 mole (mol) Now have a system for relating large numbers of atoms to their total mass, and for comparing the relative masses of any given large number of atoms of different elements:

27. What value of n will allow us to “scale up” from the a.m.u. scale (single atoms) to the gram scale (large numbers of atoms)? n = 6.022 x 10 23 entities = Avogadros Number (N) = 1 mole (mol) Now have a system for relating large numbers of atoms to their total mass, and for comparing the relative masses of any given large number of atoms of different elements: ElementAtomic Mass Scale“Mole Mass Scale” No. ofMassNo. ofMass Atoms(a.m.u.)Atoms (g) H 11.0081 mole1.008 He 14.0031 mole4.003 Li 16.9411 mole6.941 Be 19.0121 mole9.012...etc.....etc.....etc..

28. What value of n will allow us to “scale up” from the a.m.u. scale (single atoms) to the gram scale (large numbers of atoms)? n = 6.022 x 10 23 entities = Avogadros Number (N) = 1 mole (mol) Now have a system for relating large numbers of atoms to their total mass, and for comparing the relative masses of any given large number of atoms of different elements: ElementAtomic Mass Scale“Mole Mass Scale” No. ofMassNo. ofMass Atoms(a.m.u.)Atoms (g) H 11.0081 mole1.008 He 14.0031 mole4.003 Li 16.9411 mole6.941 Be 19.0121 mole9.012...etc.....etc.....etc.. 6.022 x 10 23 atoms

29. This system can also be used to determine the masses of large numbers of other chemical entities (e.g. molecules, ions, etc.). Example 1:Mass of 0.5 mole of plutonium (Pu) atoms?

30. This system can also be used to determine the masses of large numbers of other chemical entities (e.g. molecules, ions, etc.). Example 1:Mass of 0.5 mole of plutonium (Pu) atoms? From table of atomic masses: Pu = 244. This means that 1 atom of Pu has a mass of 244 a.m.u. and that 1 mole of Pu has a mass of 244 g. Since 1 mole Pu = 244 g, then 0.5 mole Pu has a mass of = 244 x 0.5 g. Answer: 122 g

31. Example 2:Mass of 1 mole of silicon carbide (SiC)?

32. A molecule of SiC consists of one atom of silicon and one atom of carbon bonded together. Thus 1 mole SiC consists of 1 mole Si + 1 mole C. From atomic mass table, Si = 28.086 and C = 12.011. So the mass of 1 mole SiC = 28.086 + 12.011 g

33. Example 2:Mass of 1 mole of silicon carbide (SiC)? A molecule of SiC consists of one atom of silicon and one atom of carbon bonded together. Thus 1 mole SiC consists of 1 mole Si + 1 mole C. From atomic mass table, Si = 28.086 and C = 12.011. So the mass of 1 mole SiC = 28.086 + 12.011 g Answer: 40.097 g

34. Example 3: Mass of 2 moles of SO 4 2- ions? (O = 15.999 and S = 32.06) 1 SO 4 2- ion contains 1 S and 4 O’s, thus: 2 mol SO 4 2- contains 2 mol sulphur and 4 x 2 mol oxygen Mass of 2 mol S = 2 x 32.06 g = 64.12 g Mass of 4 x 2 mol O = 4 x 2 x 15.99 g = 127.92 g Mass of 2 moles of sulphate ions = 192.04 g

35. Molarity Molarity = moles of solute litres of solution

36. Molarity Molarity = moles of solute litres of solution Example: What is the molarity of a solution made by dissolving 2.355 g of sulfuric acid (H 2 SO 4 ) in water and diluting to a final volume of 50.0 ml?

37. Molarity Molarity = moles of solute litres of solution Example: What is the molarity of a solution made by dissolving 2.355 g of sulfuric acid (H 2 SO 4 ) in water and diluting to a final volume of 50.0 ml? Molecular weight of H 2 SO 4 = 2.0 x 1 + 32.1 + 16.0 x 4 = 98.1 a.m.u. Molar mass of H 2 SO 4 = 98.1 g/mol

38. Molarity Molarity = moles of solute litres of solution Example: What is the molarity of a solution made by dissolving 2.355 g of sulfuric acid (H 2 SO 4 ) in water and diluting to a final volume of 50.0 ml? Molecular weight of H 2 SO 4 = 2.0 x 1 + 32.1 + 16.0 x 4 = 98.1 a.m.u. Molar mass of H 2 SO 4 = 98.1 g/mol 2.355 x 1/98.1 = 0.024 Units g x mol g -1 = mol => 0.024 mol H 2 SO 4 Molarity = moles of solute/ Litres of solution = 0.024 mol/ 0.05 L = 0.48 M

39. Example: What is the concentration of water in 1 litre of water?

40. Molecular weight of H 2 O = 2.0 x 1 + 16.0 = 18.0 a.m.u.

41. Example: What is the concentration of water in 1 litre of water? Molecular weight of H 2 O = 2.0 x 1 + 16.0 = 18.0 a.m.u. Density of water = 1.0 g/cm 3 or 1.0 kg/L Mass of H 2 O in 1 L = 1 kg molarity = moles of solute/ litres of solution = 1000/18.0 g L -1 g -1 mol = 55.5 mol L -1 = 55.5 M

42. Example: The molecular mass of sugar (C 12 H 22 O 11 ) How many molecules of sugar are there in 1 spoon (0.025 kg) and what is the concentration of sugar in 1 cup of coffee (100 ml)?

43. Example: The molecular mass of sugar (C 12 H 22 O 11 ) How many molecules of sugar are there in 1 spoon (0.025 kg) and what is the concentration of sugar in 1 cup of coffee (100 ml)? Molecular mass of sugar = 12 x 12.00 + 22 x 1.01 + 11 x 16.00 = 342.22 a.m.u.

44. Example: The molecular mass of sugar (C 12 H 22 O 11 ) How many molecules of sugar are there in 1 spoon (0.025 kg) and what is the concentration of sugar in 1 cup of coffee (100 ml)? Molecular mass of sugar = 12 x 12.00 + 22 x 1.01 + 11 x 16.00 = 342.22 a.m.u. Mass of 1 mole of sugar = 342.22 g mol -1

45. Example: The molecular mass of sugar (C 12 H 22 O 11 ) How many molecules of sugar are there in 1 spoon (0.025 kg) and what is the concentration of sugar in 1 cup of coffee (100 ml)? Molecular mass of sugar = 12 x 12.00 + 22 x 1.01 + 11 x 16.00 = 342.22 a.m.u. Mass of 1 mole of sugar = 342.22 g mol -1 # of moles of sugar = 0.025 x 1000/342.22 = 0.073 mole

46. Example: The molecular mass of sugar (C 12 H 22 O 11 ) How many molecules of sugar are there in 1 spoon (0.025 kg) and what is the concentration of sugar in 1 cup of coffee (100 ml)? Molecular mass of sugar = 12 x 12.00 + 22 x 1.01 + 11 x 16.00 = 342.22 a.m.u. Mass of 1 mole of sugar = 342.22 g # of moles of sugar = 0.025 x 1000/342.22 = 0.073 mole # of molecules of sugar = = 0.073 x 6.022 x 10 23 mole molecule mole -1

47. Example: The molecular mass of sugar (C 12 H 22 O 11 ) How many molecules of sugar are there in 1 spoon (0.025 kg) and what is the concentration of sugar in 1 cup of coffee (100 ml)? Molecular mass of sugar = 12 x 12.00 + 22 x 1.01 + 11 x 16.00 = 342.22 a.m.u. Mass of 1 mole of sugar = 342.22 g # of moles of sugar = 0.025 x 1000/342.22 = 0.073 mole # of molecules of sugar = = 0.073 x 6.022 x 10 23 mole molecule mole -1 = 4.4 x 10 22 molecules

48. 2000 Exam How many moles of copper atoms are there in a copper coin of mass 2.5 g given that the atomic mass of copper is 63.55 a.m.u. and that the coin contains 55% copper by mass?

49. 2000 Exam How many moles of copper atoms are there in a copper coin of mass 2.5 g given that the atomic mass of copper is 63.55 a.m.u. and that the coin contains 55% copper by mass? (a) 0.04(b) 0.02(c) 0.01(d) 0.06(e) 0.03

50. 2000 Exam How many moles of copper atoms are there in a copper coin of mass 2.5 g given that the atomic mass of copper is 63.55 a.m.u. and that the coin contains 55% copper by mass? (a) 0.04(b) 0.02(c) 0.01(d) 0.06(e) 0.03 Mass of copper = 0.55 x 2.5 g = 1.38 g

51. 2000 Exam How many moles of copper atoms are there in a copper coin of mass 2.5 g given that the atomic mass of copper is 63.55 a.m.u. and that the coin contains 55% copper by mass? (a) 0.04(b) 0.02(c) 0.01(d) 0.06(e) 0.03 Mass of copper = 0.55 x 2.5 g = 1.38 g # of moles of copper = 1.38/ 63.55 (g/ g mol -1 )

52. 2000 Exam How many moles of copper atoms are there in a copper coin of mass 2.5 g given that the atomic mass of copper is 63.55 a.m.u. and that the coin contains 55% copper by mass? (a) 0.04(b) 0.02(c) 0.01(d) 0.06(e) 0.03 Mass of copper = 0.55 x 2.5 g = 1.38 g # of moles of copper = 1.38/ 63.55 (g/ g mol -1 ) = 0.02 moles

53. 2000 Exam How many moles of copper atoms are there in a copper coin of mass 2.5 g given that the atomic mass of copper is 63.55 a.m.u. and that the coin contains 55% copper by mass? (a) 0.04(b) 0.02(c) 0.01(d) 0.06(e) 0.03 Mass of copper = 0.55 x 2.5 g = 1.38 g # of moles of copper = 1.38/ 63.55 (g/ g mol -1 ) = 0.02 moles

55. http://www.ul.ie/~ces/Resource.html B3053 CH4701 2C Mon 16.00 CH4701 2B Fri 10.00 CH4701 2A Wed 12.00 CH4711 2A Wed 10.00 CH4721 2A Th 10.00

56. Example What mass of NaOH and HCl are require to make 29.22 g of NaCl? NaOH + HCl ——>NaCl + H 2 O(1) Atomic masses: Na 22.99 a.m.u., Cl 35.45 a.m.u., O 15.999 a.m.u. and H 1.008 a.m.u.

57. Example What mass of NaOH and HCl are require to make 29.22 g of NaCl? NaOH + HCl ——>NaCl + H 2 O(1) Atomic masses: Na 22.99 a.m.u., Cl 35.45 a.m.u., O 15.999 a.m.u. and H 1.008 a.m.u. 1 molecule NaOH + 1 molecule HCl ——> 1 molecule NaCl + 1 molecule H 2 O 1 mole NaOH + 1 mole HCl ——>1 mole NaCl + 1 mole H 2 O

58. Example What mass of NaOH and HCl are require to make 29.22 g of NaCl? NaOH + HCl ——>NaCl + H 2 O(1) Atomic masses: Na 22.99 a.m.u., Cl 35.45 a.m.u., O 15.999 a.m.u. and H 1.008 a.m.u. 1 molecule NaOH + 1 molecule HCl ——> 1 molecule NaCl + 1 molecule H 2 O 1 mole NaOH + 1 mole HCl ——>1 mole NaCl + 1 mole H 2 O x mole NaOH + x mole HCl ——>x mole NaCl + x mole H 2 O where x is the number of moles present in 29.22 g NaCl

59. Example What mass of NaOH and HCl are require to make 29.22 g of NaCl? NaOH + HCl ——>NaCl + H 2 O(1) Atomic masses: Na 22.99 a.m.u., Cl 35.45 a.m.u., O 15.999 a.m.u. and H 1.008 a.m.u. 1 molecule NaOH + 1 molecule HCl ——> 1 molecule NaCl + 1 molecule H 2 O 1 mole NaOH + 1 mole HCl ——>1 mole NaCl + 1 mole H 2 O x mole NaOH + x mole HCl ——>x mole NaCl + x mole H 2 O where x is the number of moles present in 29.22 g NaCl Molecular mass of NaCl = 22.99 + 35.45 = 58.44 a.m.u. 1 mole of NaCl has a mass of 58.44 g

60. X = 1 x 29.22/58.44 = 0.5 mole

61. Calculate the mass of 0.5 mole of NaOH and of HCl

62. X = 1 x 29.22/58.44 = 0.5 mole Calculate the mass of 0.5 mole of NaOH and of HCl Mass of 0.5 mole of NaOH = 0.5 x (22.99 + 15.999 + 1.008) = 20.00g Mass of 0.5 mole HCl = 0.5 x (1.008 + 35.45) = 18.23 g

63. X = 1 x 29.22/58.44 = 0.5 mole Calculate the mass of 0.5 mole of NaOH and of HCl Mass of 0.5 mole of NaOH = 0.5 x (22.99 + 15.999 + 1.008) = 20.00g Mass of 0.5 mole HCl = 0.5 x (1.008 + 35.45) = 18.23 g Answer: 20.00 g NaCl and 18.23 g HCl.

64. Q. 8 2001 Exam The alcohol content of a beer is 5% by weight. How many moles of ethanol (C 2 H 5 OH) are there in 500 g of beer? (a)5.4(b) 0.54(c) 0.44(d) 5 (b)none of the above

65. Q. 8 2001 Exam The alcohol content of a beer is 5% by weight. How many moles of ethanol (C 2 H 5 OH) are there in 500 g of beer? (a)5.4(b) 0.54(c) 0.44(d) 5 (b)none of the above Mass of alcohol = 0.05 x 500 g = 25 g C 2 H 5 OH mass of one molecule = (2 x 12) + (5 x 1) + 16 + 1 = 46 amu

66. Q. 8 2001 Exam The alcohol content of a beer is 5% by weight. How many moles of ethanol (C 2 H 5 OH) are there in 500 g of beer? (a)5.4(b) 0.54(c) 0.44(d) 5 (b)none of the above Mass of alcohol = 0.05 x 500 g = 25 g C 2 H 5 OH mass of one molecule = (2 x 12) + (5 x 1) + 16 + 1 = 46 amu Mass of one mole of ethanol = 46 g mol -1 # of moles = 25/46 g/ g mol -1 = 0.54 moles

67. Q. 8 2001 Exam The alcohol content of a beer is 5% by weight. How many moles of ethanol (C 2 H 5 OH) are there in 500 g of beer? (a)5.4(b) 0.54(c) 0.44(d) 5 (b)none of the above Mass of alcohol = 0.05 x 500 g = 25 g C 2 H 5 OH mass of one molecule = (2 x 12) + (5 x 1) + 16 + 1 = 46 amu Mass of one mole of ethanol = 46 g mol -1 # of moles = 25/46 g/ g mol -1 = 0.54 moles

68. Early Chemical Concepts and Their Present Day Uses

69. Basic idea of the atom: Democritus (400 B.C.) Alchemists (Middle Ages), Lavoisier (1743 - 1794).

70. Basic idea of the atom: Democritus (400 B.C.) Alchemists (Middle Ages), Lavoisier (1743 - 1794). Viewed the notion of atoms from a philosophical point of view, i.e. did not try to prove their existence.

71. Basic idea of the atom: Democritus (400 B.C.) Alchemists (Middle Ages), Lavoisier (1743 - 1794). Viewed the notion of atoms from a philosophical point of view, i.e. did not try to prove their existence. Robert Boyle (1627 - 91) Disproved Aristotelian approach (based on a priore theories).

72. Basic idea of the atom: Democritus (400 B.C.) Alchemists (Middle Ages) Lavoisier (1743 - 1794). Viewed the notion of atoms from a philosophical point of view, i.e. did not try to prove their existence. Robert Boyle (1627 - 91) Disproved Aristotelian approach (based on a priore theories). Introduced the “Scientific Method” as the underlying philosophy of science.

74. Observation The Scientific Method

75. Observation Hypothesis/Theory The Scientific Method

76. Observation Hypothesis/Theory Prediction The Scientific Method

77. Observation Hypothesis/Theory Prediction Experiment The Scientific Method

78. Observation Hypothesis/Theory Prediction Experiment Do results agree with theory? yes Scientific Law Modify theory no The Scientific Method

79. John Dalton (1767 - 1844) Used the Scientific Method to develop a basic atomic theory:

80. John Dalton (1767 - 1844) Used the Scientific Method to develop a basic atomic theory: 1. Matter is made up of atoms which are indivisible and indestructible.

81. John Dalton (1767 - 1844) Used the Scientific Method to develop a basic atomic theory: 1. Matter is made up of atoms which are indivisible and indestructible. 2. Atoms of a given element are identical both in mass and in chemical proportions.

82. John Dalton (1767 - 1844) Used the Scientific Method to develop a basic atomic theory: 1. Matter is made up of atoms which are indivisible and indestructible. 2. Atoms of a given element are identical both in mass and in chemical proportions. 3. Atoms of different elements have different masses and different chemical properties.

83. John Dalton (1767 - 1844) Used the Scientific Method to develop a basic atomic theory: 1. Matter is made up of atoms which are indivisible and indestructible. 2. Atoms of a given element are identical both in mass and in chemical proportions. 3. Atoms of different elements have different masses and different chemical properties. 4. Atoms of different elements combine in simple whole numbers to form compounds.

84. John Dalton (1767 - 1844) Used the Scientific Method to develop a basic atomic theory: 1. Matter is made up of atoms which are indivisible and indestructible. 2. Atoms of a given element are identical both in mass and in chemical proportions. 3. Atoms of different elements have different masses and different chemical properties. 4. Atoms of different elements combine in simple whole numbers to form compounds. 5. When a compound is decomposed, the recovered atoms are unchanged and can form the same or new compounds.

85. Gay-Lussac (1778 - 1850) “Gases react in simple whole number units of volume, and where the products are gases they also have simple whole number units of volume.”

86. Gay-Lussac (1778 - 1850) “Gases react in simple whole number units of volume, and where the products are gases they also have simple whole number units of volume.” Thus: 2 volumes H 2 + 1 volume O 2 2 volumes H 2 O

87. Gay-Lussac (1778 - 1850) “Gases react in simple whole number units of volume, and where the products are gases they also have simple whole number units of volume.” Thus: 2 volumes H 2 + 1 volume O 2 2 volumes H 2 O 3 volumes H 2 + 1 volume N 2 2 volumes NH 3

88. Gay-Lussac (1778 - 1850) “Gases react in simple whole number units of volume, and where the products are gases they also have simple whole number units of volume.” Thus: 2 volumes H 2 + 1 volume O 2 2 volumes H 2 O 3 volumes H 2 + 1 volume N 2 2 volumes NH 3 Note: this relationship is only true if pressure and temperature are constant throughout.

89. Avogadro (1776 - 1856) Avogadro’s Hypothesis (1811): "Equal volumes of gases under the same conditions of temperature and pressure contain the same number of particles (i.e. atoms or molecules)."

90. Avogadro (1776 - 1856) Avogadro’s Hypothesis (1811): "Equal volumes of gases under the same conditions of temperature and pressure contain the same number of particles (i.e. atoms or molecules)." Volume of 1 mole of any gaseous pure substance at STP (273.15K, 101.3 kPa) = 22.41 dm 3 = Gas Molar Volume Thus for any gas at STP: 1 mole (= atomic or molecular mass in grams) occupies 22.41 dm 3.

91. Avogadro (1776 - 1856) Avogadro’s Hypothesis (1811): "Equal volumes of gases under the same conditions of temperature and pressure contain the same number of particles (i.e. atoms or molecules)." Volume of 1 mole of any gaseous pure substance at STP (273.15K, 101.3 kPa) = 22.41 dm 3 = Gas Molar Volume Thus for any gas at STP: 1 mole (= atomic or molecular mass in grams) occupies 22.41 dm 3. Examples:1 mol of H 2 (M.M.= 2.02) has a mass of 2.02 g and at STP occupies 22.41 dm 3.

92. Avogadro (1776 - 1856) Avogadro’s Hypothesis (1811): "Equal volumes of gases under the same conditions of temperature and pressure contain the same number of particles (i.e. atoms or molecules)." Volume of 1 mole of any gaseous pure substance at STP (273.15K, 101.3 kPa) = 22.41 dm 3 = Gas Molar Volume Thus for any gas at STP: 1 mole (= atomic or molecular mass in grams) occupies 22.41 dm 3. Examples:1 mol of H 2 (M.M.= 2.02) has a mass of 2.02 g and at STP occupies 22.41dm 3. 0.5 mol of CO (M.M.= 28.01) has a mass of 14.005 g and at STP occupies 11.205 dm 3.

93. Avogadro's Hypothesis implies gas density is proportional to molecular mass. Allowed the early chemists to deduce molecular mass from gas density measurements.

94. Avogadro's Hypothesis implies gas density is proportional to molecular mass. Allowed the early chemists to deduce molecular mass from gas density measurements. Cannizzaro (1858) Put all of the foregoing hypotheses together into a method for determining accurate atomic and molecular masses, but only for gaseous compounds. Postulated that over a large number of compounds of a particular element, at least one will have only one atom of the element per molecule. Mass of the element in 22.4 L of this compound is the atomic mass.

95. CompoundMass of Mass of contained gas/ g hydrogen/ g Hydrogen (H 2 )22 Methane (CH 4 ) 164 Ethane (C 2 H 6 )306 Water (H 2 O)182 Hydrogen sulfide (H 2 S)342 Hydrogen cyanide (HCN)271 Hydrogen chloride (HCl)361 Ammonia (NH 3 )173 Pyridine (C 5 H 5 N)795 Minimum mass of hydrogen = 1 g Atomic mass of hydrogen = 1 a.m.u.

96. Dulong and Petit (1819) Found a correlation between atomic mass and specific heat of solid elements: Atomic mass x specific heat  25 JK -1 mo1 -1 where atomic mass is in grams per mole and specific heat in JK -1 g -1.

97. Dulong and Petit (1819) Found a correlation between atomic mass and specific heat of solid elements: Atomic mass x specific heat  25 JK -1 mo1 -1 where atomic mass is in grams per mole and specific heat in JK -1 g -1. ElementExperimental values of heat capacity Al24.3 J K -1 mol -1 Fe25.2 Ni26.0 Ag25.5 Au25.2 Pb26.8 For Bi, heat capacity = 0.123 J => atomic mass of 211 a.m.u. measured atomic mass = 209 a.m.u.

98. Determination of Empirical and Molecular Formulae

99. Empirical formula: the relative number of atoms of each element in a molecule or formula unit

100. Determination of Empirical and Molecular Formulae Empirical formula: the relative number of atoms of each element in a molecule or formula unit Molecular formula: the actual number of atoms of each element in a molecule.

101. Determination of Empirical and Molecular Formulae Empirical formula: the relative number of atoms of each element in a molecule or formula unit Molecular formula: the actual number of atoms of each element in a molecule. For ethane: molecular formula = C 2 H 6, empirical formula = CH 3.

102. Determination of Empirical and Molecular Formulae Empirical formula: the relative number of atoms of each element in a molecule or formula unit Molecular formula: the actual number of atoms of each element in a molecule. For ethane: molecular formula = C 2 H 6, empirical formula = CH 3. Elemental composition data Empirical formula Molecular formula Molecular mass

103. Procedure: 1. If % composition data are given, assume 100 g of compound present. If direct weights are given, use these data.

104. Procedure: 1. If % composition data are given, assume 100 g of compound present. If direct weights are given, use these data. 2. Determine the number of moles of each element.

105. Procedure: 1. If % composition data are given, assume 100 g of compound present. If direct weights are given, use these data. 2. Determine the number of moles of each element. 3. Divide each value in point 2 by the smallest of them.

106. Procedure: 1. If % composition data are given, assume 100 g of compound present. If direct weights are given, use these data. 2. Determine the number of moles of each element. 3. Divide each value in point 2 by the smallest of them. 4. If necessary, multiply all the numbers by the smallest integer possible to obtain approximate whole number values for each element.

107. Procedure: 1. If % composition data are given, assume 100 g of compound present. If direct weights are given, use these data. 2. Determine the number of moles of each element. 3. Divide each value in point 2 by the smallest of them. 4. If necessary, multiply all the numbers by the smallest integer possible to obtain approximate whole number values for each element. 5. Write these whole numbers as subscripts of the element symbols to give an empirical formula.

108. Procedure: 1. If % composition data are given, assume 100 g of compound present. If direct weights are given, use these data. 2. Determine the number of moles of each element. 3. Divide each value in point 2 by the smallest of them. 4. If necessary, multiply all the numbers by the smallest integer possible to obtain approximate whole number values for each element. 5. Write these whole numbers as subscripts of the element symbols to give an empirical formula. 6. If the molecular weight is known, determine how many empirical formula weights are required to obtain the molecular weight. Use this factor to multiply the number of atoms of each element in the empirical formula to give the molecular formula.

109. A sample of vitamin C of mass 8.00 g was analysed and Found to contain 3.27 g of carbon, 0.366 g of hydrogen and 4.36 g of oxygen. What is the molecular formula of vitamin C? The molar mass of vitamin C is 176.14 g mol -1.

110. A sample of vitamin C of mass 8.00 g was analysed and Found to contain 3.27 g of carbon, 0.366 g of hydrogen and 4.36 g of oxygen. What is the molecular formula of vitamin C? The molar mass of vitamin C is 176.14 g mol -1. 1. Mass % of carbon = 3.27g/ 8.00 g x 100 = 40.9% Mass % of hydrogen = 0.366 g/ 8.00 g x 100 = 4.58% Mass % of oxygen = 4.36 g/ 8.00 g x 100 = 54.5%

111. A sample of vitamin C of mass 8.00 g was analysed and Found to contain 3.27 g of carbon, 0.366 g of hydrogen and 4.36 g of oxygen. What is the molecular formula of vitamin C? The molar mass of vitamin C is 176.14 g mol -1. 1. Mass % of carbon = 3.27g/ 8.00 g x 100 = 40.9% Mass % of hydrogen = 0.366 g/ 8.00 g x 100 = 4.58% Mass % of oxygen = 4.36 g/ 8.00 g x 100 = 54.5% In 100 g, have 40.9 g C, 4.58 g H and 54.5 g O.

112. A sample of vitamin C of mass 8.00 g was analysed and Found to contain 3.27 g of carbon, 0.366 g of hydrogen and 4.36 g of oxygen. What is the molecular formula of vitamin C? The molar mass of vitamin C is 176.14 g mol -1. 1. Mass % of carbon = 3.27g/ 8.00 g x 100 = 40.9% Mass % of hydrogen = 0.366 g/ 8.00 g x 100 = 4.58% Mass % of oxygen = 4.36 g/ 8.00 g x 100 = 54.5% In 100 g, have 40.9 g C, 4.58 g H and 54.5 g O. 2. # of moles C = 40.9 g /12.01 g mol -1 = 3.41 mol # of moles H = 4.58 g /1.008 g mol -1 = 4.54 mol # of moles O = 54.5 g /16.00 g mol -1 = 3.41 mol

113. A sample of vitamin C of mass 8.00 g was analysed and Found to contain 3.27 g of carbon, 0.366 g of hydrogen and 4.36 g of oxygen. What is the molecular formula of vitamin C? The molar mass of vitamin C is 176.14 g mol -1. 1. Mass % of carbon = 3.27g/ 8.00 g x 100 = 40.9% Mass % of hydrogen = 0.366 g/ 8.00 g x 100 = 4.58% Mass % of oxygen = 4.36 g/ 8.00 g x 100 = 54.5% In 100 g, have 40.9 g C, 4.58 g H and 54.5 g O. 2. # of moles C = 40.9 g /12.01 g mol -1 = 3.41 mol # of moles H = 4.58 g /1.008 g mol -1 = 4.54 mol # of moles O = 54.5 g /16.00 g mol -1 = 3.41 mol 3. 3.41 C: 4.54 H : 3.41 O divide by 3.41 =>1 C: 1.33 H : 1.33 O

114. A sample of vitamin C of mass 8.00 g was analysed and Found to contain 3.27 g of carbon, 0.366 g of hydrogen and 4.36 g of oxygen. What is the molecular formula of vitamin C? The molar mass of vitamin C is 176.14 g mol -1. 1. Mass % of carbon = 3.27g/ 8.00 g x 100 = 40.9% Mass % of hydrogen = 0.366 g/ 8.00 g x 100 = 4.58% Mass % of oxygen = 4.36 g/ 8.00 g x 100 = 54.5% In 100 g, have 40.9 g C, 4.58 g H and 54.5 g O. 2. # of moles C = 40.9 g /12.01 g mol -1 = 3.41 mol # of moles H = 4.58 g /1.008 g mol -1 = 4.54 mol # of moles O = 54.5 g /16.00 g mol -1 = 3.41 mol 3. 3.41 C: 4.54 H : 3.41 O divide by 3.41 =>1 C: 1.33 H : 1.33 O 4. Multiply by 3 3 C: 4 H : 3 O

115. 5. Empirical formula is C 3 H 4 O 3. Molar mass of empirical formula: 3 x 12.00 + 4 x 1.008 + 3 x 16.00 = 88.032 g mol –1

116. 5. Empirical formula is C 3 H 4 O 3. Molar mass of empirical formula: 3 x 12.00 + 4 x 1.008 + 3 x 16.00 = 88.032 g mol -1 6. # of empirical formula units per molecule = 174.14/ 88.06 g g -1 mol mol -1 = 2 Molecular formula is C 6 H 8 O 6

117. 5. Empirical formula is C 3 H 4 O 3. Molar mass of empirical formula: 3 x 12.00 + 4 x 1.008 + 3 x 16.00 = 88.032 g mol -1 6. # of empirical formula units per molecule = 174.14/ 88.06 g g -1 mol mol -1 = 2 Molecular formula is C 6 H 8 O 6 The molar mass of oxalic acid is 90 g mol -1 and its empirical formula is CHO 2. What is its molecular formula?

118. 5. Empirical formula is C 3 H 4 O 3. Molar mass of empirical formula: 3 x 12.00 + 4 x 1.008 + 3 x 16.00 = 88.032 g mol -1 6. # of empirical formula units per molecule = 174.14/ 88.06 g g -1 mol mol -1 = 2 Molecular formula is C 6 H 8 O 6 The molar mass of oxalic acid is 90 g mol -1 and its empirical formula is CHO 2. What is its molecular formula? 1. 90 g mol -1

119. 5. Empirical formula is C 3 H 4 O 3. Molar mass of empirical formula: 3 x 12.00 + 4 x 1.008 + 3 x 16.00 = 88.032 g mol -1 6. # of empirical formula units per molecule = 174.14/ 88.06 g g -1 mol mol -1 = 2 Molecular formula is C 6 H 8 O 6 The molar mass of oxalic acid is 90 g mol -1 and its empirical formula is CHO 2. What is its molecular formula? 1. 90 g mol -1 5. Empirical formula is CHO 2. Molar mass of empirical formula: 12.01 + 1.008 + 2 x 16.00 = 45.018 g mol -1

120. 5. Empirical formula is C 3 H 4 O 3. Molar mass of empirical formula: 3 x 12.00 + 4 x 1.008 + 3 x 16.00 = 88.032 g mol -1 6. # of empirical formula units per molecule = 174.14/ 88.06 g g -1 mol mol -1 = 2 Molecular formula is C 6 H 8 O 6 The molar mass of oxalic acid is 90 g mol -1 and its empirical formula is CHO 2. What is its molecular formula? 1. 90 g mol -1 5. Empirical formula is CHO 2. Molar mass of empirical formula: 12.01 + 1.008 + 2 x 16.00 = 45.018 g mol -1 6. # of empirical formula units per molecule = 90/ 45.018 g g -1 mol mol -1 = 2 Molecular formula is C 2 H 2 O 4

121. Example: A compound contains 5.6 g N, 1.6 g H, 20.6 g Cr and 22.2 g O. Find its empirical formula. (At. masses: N = 14.0, H = 1.01, Cr = 52.0, O = 16.0)

122. Example: A compound contains 5.6 g N, 1.6 g H, 20.6 g Cr and 22.2 g O. Find its empirical formula. (At. masses: N = 14.0, H = 1.01, Cr = 52.0, O = 16.0) Solution: 1. Use direct mass.

123. Example: A compound contains 5.6 g N, 1.6 g H, 20.6 g Cr and 22.2 g O. Find its empirical formula. (At. masses: N = 14.0, H = 1.01, Cr = 52.0, O = 16.0) Solution: 1. Use direct mass. 2 # moles of each element: N = 5.6/14.0 = 0.4 mol N atoms H = 1.6/1.01 = 1.58 mol H atoms Cr = 20.6/52.0 = 0.4 mol Cr atoms O = 22.2/16.0 = 1.39 mol O atoms

124. Example: A compound contains 5.6 g N, 1.6 g H, 20.6 g Cr and 22.2 g O. Find its empirical formula. (At. masses: N = 14.0, H = 1.01, Cr = 52.0, O = 16.0) Solution: 1. Use direct mass. 2 # moles of each element: N = 5.6/14.0 = 0.4 mol N atoms H = 1.6/1.01 = 1.58 mol H atoms Cr = 20.6/52.0 = 0.4 mol Cr atoms O = 22.2/16.0 = 1.39 mol O atoms 3. Divide by smallest (0.4): N = 1, H = 3.95, Cr = 1, O = 3.48

125. Example: A compound contains 5.6 g N, 1.6 g H, 20.6 g Cr and 22.2 g O. Find its empirical formula. (At. masses: N = 14.0, H = 1.01, Cr = 52.0, O = 16.0) Solution: 1. Use direct mass. 2 # moles of each element: N = 5.6/14.0 = 0.4 mol N atoms H = 1.6/1.01 = 1.58 mol H atoms Cr = 20.6/52.0 = 0.4 mol Cr atoms O = 22.2/16.0 = 1.39 mol O atoms 3. Divide by smallest (0.4): N = 1, H = 3.95, Cr = 1, O = 3.48 4. Multiply by 2 to obtain approx. whole nos.: N2, H8, Cr2, O7

126. Example: A compound contains 5.6 g N, 1.6 g H, 20.6 g Cr and 22.2 g O. Find its empirical formula. (At. masses: N = 14.0, H = 1.01, Cr = 52.0, O = 16.0) Solution: 1. Use direct mass. 2 # moles of each element: N = 5.6/14.0 = 0.4 mol N atoms H = 1.6/1.01 = 1.58 mol H atoms Cr = 20.6/52.0 = 0.4 mol Cr atoms O = 22.2/16.0 = 1.39 mol O atoms 3. Divide by smallest (0.4): N = 1, H = 3.95, Cr = 1, O = 3.48 4. Multiply by 2 to obtain approx. whole nos.: N2, H8, Cr2, O7 5. Write as subscripts: N 2 H 8 Cr 2 O 7