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IGCSE CHEMISTRY LESSON 3. Section 1 Principles of Chemistry a)States of matter b)Atoms c)Atomic structure d)Relative formula mass e)Chemical formulae.

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Presentation on theme: "IGCSE CHEMISTRY LESSON 3. Section 1 Principles of Chemistry a)States of matter b)Atoms c)Atomic structure d)Relative formula mass e)Chemical formulae."— Presentation transcript:

1 IGCSE CHEMISTRY LESSON 3

2 Section 1 Principles of Chemistry a)States of matter b)Atoms c)Atomic structure d)Relative formula mass e)Chemical formulae and chemical equations f)Ionic compounds g)Covalent substances h)Metallic crystals i)Electrolysis

3 Lesson 3 a)Relative formula mass b)Chemical formulae and Chemical equations 1.18 write word equations and balanced chemical equations to represent the reactions studied in this specification 1.19 use the state symbols (s), (l), (g) and (aq) in chemical equations to represent solids, liquids, gases and aqueous solutions respectively 1.20 understand how the formulae of simple compounds can be obtained experimentally, including metal oxides, water and salts containing water of crystallisation 1.21 calculate empirical and molecular formulae from experimental data 1.22 calculate reacting masses using experimental data and chemical equations 1.23 carry out mole calculations using volumes and molar concentrations.

4 Finding chemical formulae – experimentally! We can conduct experiments with compounds to determine the mass (or percentage) there is of each element.

5 Finding chemical formulae – experimentally! For example, we could heat magnesium in air. Magnesium oxide will be formed

6 Finding chemical formulae – experimentally! For example, we could heat magnesium in air. Magnesium oxide will be formed The mass will increase because oxygen is added

7 Finding chemical formulae – experimentally! Mass of Magnesium (g) Mass of magnesium oxide (g) Mass of oxygen (g) 0.0250.042 0.0060.010 0.1600.269

8 Finding chemical formulae – experimentally! Mass of Magnesium (g) Mass of magnesium oxide (g) Mass of oxygen (g) 0.0250.0420.017 0.0060.0100.004 0.1600.2690.109

9 Finding chemical formulae – experimentally! Let’s just stop there a second! What do we mean by the term ‘empirical formula’?

10 Finding chemical formulae – experimentally! The empirical formula is the simplest whole number ratio of atoms in a compound.

11 Finding chemical formulae – experimentally! The empirical formula is the simplest whole number ratio of atoms in a compound. Let’s calculate the empirical formula for magnesium oxide.

12 Finding chemical formulae – experimentally! Mass of Magnesium (g) Mass of magnesium oxide (g) Mass of oxygen (g) 0.0250.0420.017 0.0060.0100.004 0.1600.2690.109

13 Empirical formula

14 ElementsMagnesiumOxygen Mass (g)0.1600.109 Relative atomic mass2416 Number of moles

15 Empirical formula ElementsMagnesiumOxygen Mass (g)0.1600.109 Relative atomic mass2416 Number of moles

16 Empirical formula ElementsMagnesiumOxygen Mass (g)0.1600.109 Relative atomic mass2416 Number of moles0.00670.0068

17 Empirical formula ElementsMagnesiumOxygen Mass (g)0.1600.109 Relative atomic mass2416 Number of moles0.00670.0068 Most simple ratio 0.0067 0.0068 0.0067

18 Empirical formula ElementsMagnesiumOxygen Mass (g)0.1600.109 Relative atomic mass2416 Number of moles0.00670.0068 Most simple ratio 0.0067 0.0068 0.0067 Reacting atoms11

19 Empirical formula ElementsMagnesiumOxygen Mass (g)0.1600.109 Relative atomic mass2416 Number of moles0.00670.0068 Most simple ratio 0.0067 0.0068 0.0067 Empirical formulaMgO

20 Empirical formula ElementsIronoxygen Mass (g) Relative atomic mass Number of moles Most simple ratio Reacting atoms

21 Empirical formula ElementsIronoxygen Mass (g)2.240.96 Relative atomic mass Number of moles Most simple ratio Reacting atoms

22 Empirical formula ElementsIronoxygen Mass (g)2.240.96 Relative atomic mass5616 Number of moles Most simple ratio Reacting atoms

23 Empirical formula ElementsIronoxygen Mass (g)2.240.96 Relative atomic mass5616 Number of moles0.040.06 Most simple ratio0.04/0.040.06/0.04 Reacting atoms

24 Empirical formula ElementsIronoxygen Mass (g)2.240.96 Relative atomic mass5616 Number of moles0.040.06 Most simple ratio0.04/0.040.06/0.04 Reacting atoms11.5

25 Empirical formula ElementsIronoxygen Mass (g)2.240.96 Relative atomic mass5616 Number of moles0.040.06 Most simple ratio0.04/0.040.06/0.04 Reacting atoms23

26 Empirical formula ElementsIronoxygen Mass (g)2.240.96 Relative atomic mass5616 Number of moles0.040.06 Most simple ratio0.04/0.040.06/0.04 Empirical formulaFe 2 O 3

27 Empirical formula ElementsMagnesiumOxygen Mass (%) Relative atomic mass Number of moles Most simple ratio Reacting atoms

28 Empirical formula ElementsMagnesiumOxygen Mass (%)6040 Relative atomic mass Number of moles Most simple ratio Reacting atoms

29 Empirical formula ElementsMagnesiumOxygen Mass (%)6040 Relative atomic mass2416 Number of moles Most simple ratio Reacting atoms

30 Empirical formula ElementsMagnesiumOxygen Mass (%)6040 Relative atomic mass2416 Number of moles2.5 Most simple ratio Reacting atoms

31 Empirical formula ElementsMagnesiumOxygen Mass (%)6040 Relative atomic mass2416 Number of moles2.5 Most simple ratio2.5/2.5 Reacting atoms

32 Empirical formula ElementsMagnesiumOxygen Mass (%)6040 Relative atomic mass2416 Number of moles2.5 Most simple ratio2.5/2.5 Reacting atoms11

33 Empirical formula ElementsMagnesiumOxygen Mass (%)6040 Relative atomic mass2416 Number of moles2.5 Most simple ratio2.5/2.5 Empirical formulaMgO

34 Empirical formula ElementsCalciumCarbonOxygen Mass (g) Relative atomic mass Number of moles Most simple ratio Reacting atoms

35 Empirical formula ElementsCalciumCarbonOxygen Mass (g)10312 Relative atomic mass Number of moles Most simple ratio Reacting atoms

36 Empirical formula ElementsCalciumCarbonOxygen Mass (g)10312 Relative atomic mass 401216 Number of moles Most simple ratio Reacting atoms

37 Empirical formula ElementsCalciumCarbonOxygen Mass (g)10312 Relative atomic mass 401216 Number of moles 0.25 0.75 Most simple ratio Reacting atoms

38 Empirical formula ElementsCalciumCarbonOxygen Mass (g)10312 Relative atomic mass 401216 Number of moles 0.25 0.75 Most simple ratio 0.25/0.25 0.75/0.25 Reacting atoms

39 Empirical formula ElementsCalciumCarbonOxygen Mass (g)10312 Relative atomic mass 401216 Number of moles 0.25 0.75 Most simple ratio 0.25/0.25 0.75/0.25 Reacting atoms113

40 Empirical formula ElementsCalciumCarbonOxygen Mass (g)10312 Relative atomic mass 401216 Number of moles 0.25 0.75 Most simple ratio 0.25/0.25 0.75/0.25 Empirical formula CaCO 3

41 Lesson 3 a)Relative formula mass b)Chemical formulae and Chemical equations 1.18 write word equations and balanced chemical equations to represent the reactions studied in this specification 1.19 use the state symbols (s), (l), (g) and (aq) in chemical equations to represent solids, liquids, gases and aqueous solutions respectively 1.20 understand how the formulae of simple compounds can be obtained experimentally, including metal oxides, water and salts containing water of crystallisation 1.21 calculate empirical and molecular formulae from experimental data 1.22 calculate reacting masses using experimental data and chemical equations 1.23 carry out mole calculations using volumes and molar concentrations.

42 ? What is ‘water’ of crystallisation?

43 ?

44 ? = water found in the crystal structure of a compound (eg. copper sulphate) but not directly bonded to the compound, eg hydrated calcium chloride crystals CaCl 2.2H 2 O

45 ? What is ‘water’ of crystallisation? When a hydrated salt is heated, the water is evaporated off, and we can use the change in mass to calculate how many molecules of water there would be in the empirical formula.

46 ? What is ‘water’ of crystallisation? When a hydrated salt is heated, the water is evaporated off, and we can use the change in mass to calculate how many molecules of water there would be in the empirical formula. eg. MgSO 4.nH 2 O What will the value of ‘n’ be?

47 Finding chemical formulae of hydrated salts– experimentally! Mass of hydrated magnesium sulphate (g) Mass of anhydrous magnesium sulphate (g) Mass of water (g) 9.444.584.86

48 Finding chemical formulae of hydrated salts– experimentally! Mass of hydrated magnesium sulphate (g) Mass of anhydrous magnesium sulphate (g) Mass of water (g) 9.444.584.86

49 Empirical formula CompoundsMgSO 4 H2OH2O Mass (g)4.584.86 Relative molecular mass Number of moles Most simple ratio Reacting molecules

50 Empirical formula CompoundsMgSO 4 H2OH2O Mass (g)4.584.86 Relative molecular mass 12018 Number of moles Most simple ratio Reacting molecules

51 Empirical formula CompoundsMgSO 4 H2OH2O Mass (g)4.584.86 Relative molecular mass 12018 Number of moles0.0390.27 Most simple ratio Reacting molecules

52 Empirical formula CompoundsMgSO 4 H2OH2O Mass (g)4.584.86 Relative molecular mass 12018 Number of moles0.0390.27 Most simple ratio0.039/0.0390.27/0.039 Reacting molecules

53 Empirical formula CompoundsMgSO 4 H2OH2O Mass (g)4.584.86 Relative molecular mass 12018 Number of moles0.0390.27 Most simple ratio0.039/0.0390.27/0.039 Reacting molecules16.93 (7)

54 Empirical formula CompoundsMgSO 4 H2OH2O Mass (g)4.584.86 Relative molecular mass 12018 Number of moles0.0390.27 Most simple ratio0.039/0.0390.27/0.039 Empirical formulaMgSO 4.7H 2 O

55 Phew! Take a break!

56 Molecular formula

57 Molecular formula shows the actual number of atoms in a compound and is a simple multiple of the empirical formula.

58 Molecular formula Eg. Empirical formula = CH 2 Molecular formula = C 3 H 6 (empirical formula x 3)

59 Molecular formula Eg. Find the molecular formula of a compound which has the empirical formula CH 2 and a relative molecular mass of 56g. Mass of empirical formula CH 2 = 12 + 2 = 14g

60 Molecular formula Eg. Find the molecular formula of a compound which has the empirical formula CH 2 and a relative molecular mass of 56g. Mass of empirical formula CH 2 = 12 + 2 = 14g Divide relative molecular mass by empirical mass = 56 / 14 = 4

61 Molecular formula Eg. Find the molecular formula of a compound which has the empirical formula CH 2 and a relative molecular mass of 56g. Mass of empirical formula CH 2 = 12 + 2 = 14g Divide relative molecular mass by empirical mass = 56 / 14 = 4 So the molecular formula = 4 x (CH 2 ) = C 4 H 8

62 Lesson 3 a)Relative formula mass b)Chemical formulae and Chemical equations 1.18 write word equations and balanced chemical equations to represent the reactions studied in this specification 1.19 use the state symbols (s), (l), (g) and (aq) in chemical equations to represent solids, liquids, gases and aqueous solutions respectively 1.20 understand how the formulae of simple compounds can be obtained experimentally, including metal oxides, water and salts containing water of crystallisation 1.21 calculate empirical and molecular formulae from experimental data 1.22 calculate reacting masses using experimental data and chemical equations 1.23 carry out mole calculations using volumes and molar concentrations.

63 Lesson 3 a)Relative formula mass b)Chemical formulae and Chemical equations 1.18 write word equations and balanced chemical equations to represent the reactions studied in this specification 1.19 use the state symbols (s), (l), (g) and (aq) in chemical equations to represent solids, liquids, gases and aqueous solutions respectively 1.20 understand how the formulae of simple compounds can be obtained experimentally, including metal oxides, water and salts containing water of crystallisation 1.21 calculate empirical and molecular formulae from experimental data 1.22 calculate reacting masses using experimental data and chemical equations 1.23 carry out mole calculations using volumes and molar concentrations.

64 End of Lesson 3 (part 1) In this lesson we have covered: Finding chemical formulae – experimentally! Empirical formula Water of crystallisation and hydrated salts Molecular formula

65


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