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Chemical Reactions
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  • 1. CHEMICAL REACTIONS 1. Describe chemical change in terms of formulae, equations and valency. 2. Recognise oxidation and reduction in terms of electron transfer. 3. Recognise oxidation and reduction in terms of transfer of oxygen and hydrogen. 4. Explain the mechanism of dissolving and explain the formation of a precipitate. 5. Predict the identity of a precipitate in a precipitation reaction. 6. Describe thermal decomposition of metal hydroxides, carbonates and hydrogen carbonates. 7. Show an understanding of atomic an molecular masses. 8. Calculate percentage composition of compounds. 9. Calculate empirical formulae from data relating to percentage composition. 10.Calculate mass ratios of chemicals involved in chemical reactions. Thursday, 16 September 2010
  • 2. Thursday, 16 September 2010
  • 3. Thursday, 16 September 2010
  • 4. Thursday, 16 September 2010
  • 5. LANGUAGE Thursday, 16 September 2010
  • 6. Term Ans Definition GLOSSARY 1 Match the term with its definition > ans’s only in BOB A. Valency B. Ion C. Reactants D. Products E. Oxidation F. Reduction G. Half - equation Thursday, 16 September 2010
  • 7. Term Definition GLOSSARY 1- HANDOUT A. Ion B. Precipitate C. Species D. Redox 4. a term that describes any oxidation-reduction reaction Thursday, 16 September 2010
  • 8. Term Ans Definition GLOSSARY 2 Match the term with its definition > ans’s only in BOB Thursday, 16 September 2010
  • 9. Term Definition GLOSSARY 2- HANDOUT Thursday, 16 September 2010
  • 10. FORMULAE AND EQUATIONS Thursday, 16 September 2010
  • 11. FORMULAE Thursday, 16 September 2010
  • 12. VALENCY Copy Valency is a term that allows us to predict how an atom will bond with other atoms. It is given as a number. Two ways of thinking about valency: 1. For ionic compounds it is the number of electrons that an atom will gain or lose in order for it to gain an octet/duet of electrons. 2. For covalent compounds it is the number of bonds that it will form that will allow it to gain an octet/duet. Examples Oxygen, O has a valency of 2. (it will gain two electrons to form an octet when it forms an ionic compound or share two electrons to form an octet when it forms a covalent compound) • MgO (the Oxygen atom has a 2- charge in this compound) • CO2 (Each O atom has formed 2 bonds with carbon in this compound) Repeat the above illustration for Phosphorus. Use Phosphorus trichloride and Sodium phosphide as the compounds ESA: p61 - 63 Ex 8A Thursday, 16 September 2010
  • 13. WRITING FORMULAE Copy Examples Use your table of ions to write formluae for the following ionic compounds: Sodium Fluoride _____________ Magnesium Oxide __________ Sodium Oxide _______________ Zinc Chloride _____________ Lead Nitrate ______________ Potassium Sulphate ____________ Calcium Hydrogen Carbonate __________ Aluminium Oxide _______ THE PROCESS FOR WRITING FORMULAE Thursday, 16 September 2010
  • 14. WRITING FORMULAE Copy Examples Use your table of ions to write formluae for the following ionic compounds: Sodium Fluoride _____________ Magnesium Oxide __________ Sodium Oxide _______________ Zinc Chloride _____________ Lead Nitrate ______________ Potassium Sulphate ____________ Calcium Hydrogen Carbonate __________ Aluminium Oxide _______ Method 1: Working with ratios THE PROCESS FOR WRITING FORMULAE Example Write the formula for Calcium bicarbonate Thursday, 16 September 2010
  • 15. WRITING FORMULAE Copy Examples Use your table of ions to write formluae for the following ionic compounds: Sodium Fluoride _____________ Magnesium Oxide __________ Sodium Oxide _______________ Zinc Chloride _____________ Lead Nitrate ______________ Potassium Sulphate ____________ Calcium Hydrogen Carbonate __________ Aluminium Oxide _______ Method 1: Working with ratios THE PROCESS FOR WRITING FORMULAE Example 1. Work out the ratio of ions Write the formula that will cause a balance for Calcium of positive & negative bicarbonate charges Thursday, 16 September 2010
  • 16. WRITING FORMULAE Copy Examples Use your table of ions to write formluae for the following ionic compounds: Sodium Fluoride _____________ Magnesium Oxide __________ Sodium Oxide _______________ Zinc Chloride _____________ Lead Nitrate ______________ Potassium Sulphate ____________ Calcium Hydrogen Carbonate __________ Aluminium Oxide _______ Method 1: Working with ratios THE PROCESS FOR WRITING FORMULAE Example 1. Work out the ratio of ions Write the formula for Calcium that will cause a balance 1 Ca2+ HCO3 - of positive & negative bicarbonate 1 : 2 charges Thursday, 16 September 2010
  • 17. WRITING FORMULAE Copy Examples Use your table of ions to write formluae for the following ionic compounds: Sodium Fluoride _____________ Magnesium Oxide __________ Sodium Oxide _______________ Zinc Chloride _____________ Lead Nitrate ______________ Potassium Sulphate ____________ Calcium Hydrogen Carbonate __________ Aluminium Oxide _______ Method 1: Working with ratios THE PROCESS FOR WRITING FORMULAE Example 1. Work out the ratio of ions Write the formula for Calcium that will cause a balance 1 Ca2+ HCO3 - of positive & negative bicarbonate 1 : 2 charges 2. Writing it out without the charges Thursday, 16 September 2010
  • 18. WRITING FORMULAE Copy Examples Use your table of ions to write formluae for the following ionic compounds: Sodium Fluoride _____________ Magnesium Oxide __________ Sodium Oxide _______________ Zinc Chloride _____________ Lead Nitrate ______________ Potassium Sulphate ____________ Calcium Hydrogen Carbonate __________ Aluminium Oxide _______ Method 1: Working with ratios THE PROCESS FOR WRITING FORMULAE Example 1. Work out the ratio of ions Write the formula for Calcium that will cause a balance 1 Ca2+ HCO3 - of positive & negative bicarbonate 1 : 2 charges 2. Writing it out without the charges 2 Ca HCO3 Thursday, 16 September 2010
  • 19. WRITING FORMULAE Copy Examples Use your table of ions to write formluae for the following ionic compounds: Sodium Fluoride _____________ Magnesium Oxide __________ Sodium Oxide _______________ Zinc Chloride _____________ Lead Nitrate ______________ Potassium Sulphate ____________ Calcium Hydrogen Carbonate __________ Aluminium Oxide _______ Method 1: Working with ratios THE PROCESS FOR WRITING FORMULAE Example 1. Work out the ratio of ions Write the formula for Calcium that will cause a balance 1 Ca2+ HCO3 - of positive & negative bicarbonate 1 : 2 charges 2. Writing it out without the charges 2 Ca HCO3 3. Put the subscripts in place to reflect the ratio Thursday, 16 September 2010
  • 20. WRITING FORMULAE Copy Examples Use your table of ions to write formluae for the following ionic compounds: Sodium Fluoride _____________ Magnesium Oxide __________ Sodium Oxide _______________ Zinc Chloride _____________ Lead Nitrate ______________ Potassium Sulphate ____________ Calcium Hydrogen Carbonate __________ Aluminium Oxide _______ Method 1: Working with ratios THE PROCESS FOR WRITING FORMULAE Example 1. Work out the ratio of ions Write the formula for Calcium that will cause a balance 1 Ca2+ HCO3 - of positive & negative bicarbonate 1 : 2 charges 2. Writing it out without the charges 2 Ca HCO3 3. Put the subscripts in place to reflect the 3 Ca(HCO3)2 ratio Thursday, 16 September 2010
  • 21. WRITING FORMULAE Copy Examples Use your table of ions to write formluae for the following ionic compounds: Sodium Fluoride _____________ Magnesium Oxide __________ Sodium Oxide _______________ Zinc Chloride _____________ Lead Nitrate ______________ Potassium Sulphate ____________ Calcium Hydrogen Carbonate __________ Aluminium Oxide _______ Method 1: Working with ratios THE PROCESS FOR WRITING FORMULAE Example 1. Work out the ratio of ions Write the formula for Calcium that will cause a balance 1 Ca2+ HCO3 - of positive & negative bicarbonate 1 : 2 charges 2. Writing it out without the charges 2 Ca HCO3 3. Put the subscripts in place to reflect the 3 Ca(HCO3)2 ratio Thursday, 16 September 2010
  • 22. WRITING FORMULAE Copy Examples Use your table of ions to write formluae for the following ionic compounds: Sodium Fluoride _____________ Magnesium Oxide __________ Sodium Oxide _______________ Zinc Chloride _____________ Lead Nitrate ______________ Potassium Sulphate ____________ Calcium Hydrogen Carbonate __________ Aluminium Oxide _______ Method 1: Working with ratios THE PROCESS FOR WRITING FORMULAE Example 1. Work out the ratio of ions Write the formula for Calcium that will cause a balance 1 Ca2+ HCO3 - of positive & negative bicarbonate 1 : 2 charges 2. Writing it out without the charges 2 Ca HCO3 3. Put the subscripts in place to reflect the 3 Ca(HCO3)2 ratio Note The “1” is never shown as a subscript Ions that are made up of groups must be bracketed if the subscript is “2” or more. Thursday, 16 September 2010
  • 23. WRITING FORMULAE Copy Examples Use your table of ions to write formluae for the following ionic compounds: NaF Sodium Fluoride _____________ Magnesium Oxide __________ Sodium Oxide _______________ Zinc Chloride _____________ Lead Nitrate ______________ Potassium Sulphate ____________ Calcium Hydrogen Carbonate __________ Aluminium Oxide _______ Method 1: Working with ratios THE PROCESS FOR WRITING FORMULAE Example 1. Work out the ratio of ions Write the formula for Calcium that will cause a balance 1 Ca2+ HCO3 - of positive & negative bicarbonate 1 : 2 charges 2. Writing it out without the charges 2 Ca HCO3 3. Put the subscripts in place to reflect the 3 Ca(HCO3)2 ratio Note The “1” is never shown as a subscript Ions that are made up of groups must be bracketed if the subscript is “2” or more. Thursday, 16 September 2010
  • 24. WRITING FORMULAE Copy Examples Use your table of ions to write formluae for the following ionic compounds: NaF Sodium Fluoride _____________ Magnesium Oxide __________ MgO Sodium Oxide _______________ Zinc Chloride _____________ Lead Nitrate ______________ Potassium Sulphate ____________ Calcium Hydrogen Carbonate __________ Aluminium Oxide _______ Method 1: Working with ratios THE PROCESS FOR WRITING FORMULAE Example 1. Work out the ratio of ions Write the formula for Calcium that will cause a balance 1 Ca2+ HCO3 - of positive & negative bicarbonate 1 : 2 charges 2. Writing it out without the charges 2 Ca HCO3 3. Put the subscripts in place to reflect the 3 Ca(HCO3)2 ratio Note The “1” is never shown as a subscript Ions that are made up of groups must be bracketed if the subscript is “2” or more. Thursday, 16 September 2010
  • 25. WRITING FORMULAE Copy Examples Use your table of ions to write formluae for the following ionic compounds: NaF Sodium Fluoride _____________ Magnesium Oxide __________ MgO Sodium Oxide Na2O _______________ Zinc Chloride _____________ Lead Nitrate ______________ Potassium Sulphate ____________ Calcium Hydrogen Carbonate __________ Aluminium Oxide _______ Method 1: Working with ratios THE PROCESS FOR WRITING FORMULAE Example 1. Work out the ratio of ions Write the formula for Calcium that will cause a balance 1 Ca2+ HCO3 - of positive & negative bicarbonate 1 : 2 charges 2. Writing it out without the charges 2 Ca HCO3 3. Put the subscripts in place to reflect the 3 Ca(HCO3)2 ratio Note The “1” is never shown as a subscript Ions that are made up of groups must be bracketed if the subscript is “2” or more. Thursday, 16 September 2010
  • 26. WRITING FORMULAE Copy Examples Use your table of ions to write formluae for the following ionic compounds: NaF Sodium Fluoride _____________ Magnesium Oxide __________ MgO Sodium Oxide Na2O _______________ ZnCl2 Zinc Chloride _____________ Lead Nitrate ______________ Potassium Sulphate ____________ Calcium Hydrogen Carbonate __________ Aluminium Oxide _______ Method 1: Working with ratios THE PROCESS FOR WRITING FORMULAE Example 1. Work out the ratio of ions Write the formula for Calcium that will cause a balance 1 Ca2+ HCO3 - of positive & negative bicarbonate 1 : 2 charges 2. Writing it out without the charges 2 Ca HCO3 3. Put the subscripts in place to reflect the 3 Ca(HCO3)2 ratio Note The “1” is never shown as a subscript Ions that are made up of groups must be bracketed if the subscript is “2” or more. Thursday, 16 September 2010
  • 27. WRITING FORMULAE Copy Examples Use your table of ions to write formluae for the following ionic compounds: NaF Sodium Fluoride _____________ Magnesium Oxide __________ MgO Sodium Oxide Na2O ZnCl2 _______________ Zinc Chloride _____________ Pb(NO3)2 Lead Nitrate ______________ Potassium Sulphate ____________ Calcium Hydrogen Carbonate __________ Aluminium Oxide _______ Method 1: Working with ratios THE PROCESS FOR WRITING FORMULAE Example 1. Work out the ratio of ions Write the formula for Calcium that will cause a balance 1 Ca2+ HCO3 - of positive & negative bicarbonate 1 : 2 charges 2. Writing it out without the charges 2 Ca HCO3 3. Put the subscripts in place to reflect the 3 Ca(HCO3)2 ratio Note The “1” is never shown as a subscript Ions that are made up of groups must be bracketed if the subscript is “2” or more. Thursday, 16 September 2010
  • 28. WRITING FORMULAE Copy Examples Use your table of ions to write formluae for the following ionic compounds: NaF Sodium Fluoride _____________ Magnesium Oxide __________ MgO Sodium Oxide Na2O ZnCl2 _______________ Zinc Chloride _____________ Pb(NO3)2 Lead Nitrate ______________ K2SO4 Potassium Sulphate ____________ Calcium Hydrogen Carbonate __________ Aluminium Oxide _______ Method 1: Working with ratios THE PROCESS FOR WRITING FORMULAE Example 1. Work out the ratio of ions Write the formula for Calcium that will cause a balance 1 Ca2+ HCO3 - of positive & negative bicarbonate 1 : 2 charges 2. Writing it out without the charges 2 Ca HCO3 3. Put the subscripts in place to reflect the 3 Ca(HCO3)2 ratio Note The “1” is never shown as a subscript Ions that are made up of groups must be bracketed if the subscript is “2” or more. Thursday, 16 September 2010
  • 29. WRITING FORMULAE Copy Examples Use your table of ions to write formluae for the following ionic compounds: NaF Sodium Fluoride _____________ Magnesium Oxide __________ MgO Sodium Oxide Na2O ZnCl2 _______________ Zinc Chloride _____________ Pb(NO3)2 Lead Nitrate ______________ K2SO4 Potassium Sulphate ____________ Ca(HCO3)2 Calcium Hydrogen Carbonate __________ Aluminium Oxide _______ Method 1: Working with ratios THE PROCESS FOR WRITING FORMULAE Example 1. Work out the ratio of ions Write the formula for Calcium that will cause a balance 1 Ca2+ HCO3 - of positive & negative bicarbonate 1 : 2 charges 2. Writing it out without the charges 2 Ca HCO3 3. Put the subscripts in place to reflect the 3 Ca(HCO3)2 ratio Note The “1” is never shown as a subscript Ions that are made up of groups must be bracketed if the subscript is “2” or more. Thursday, 16 September 2010
  • 30. WRITING FORMULAE Copy Examples Use your table of ions to write formluae for the following ionic compounds: NaF Sodium Fluoride _____________ Magnesium Oxide __________ MgO Sodium Oxide Na2O ZnCl2 _______________ Zinc Chloride _____________ Pb(NO3)2 Lead Nitrate ______________ K2SO4 Potassium Sulphate ____________ Ca(HCO3)2 Al2O3 Calcium Hydrogen Carbonate __________ Aluminium Oxide _______ Method 1: Working with ratios THE PROCESS FOR WRITING FORMULAE Example 1. Work out the ratio of ions Write the formula for Calcium that will cause a balance 1 Ca2+ HCO3 - of positive & negative bicarbonate 1 : 2 charges 2. Writing it out without the charges 2 Ca HCO3 3. Put the subscripts in place to reflect the 3 Ca(HCO3)2 ratio Note The “1” is never shown as a subscript Ions that are made up of groups must be bracketed if the subscript is “2” or more. Thursday, 16 September 2010
  • 31. Copy Method 2: The swop and drop method 1. Write the ions down next to each other. 2. Drop each number down (without the charge) and swop it to make the subscript of the other element Example Write the formula for Aluminium Oxide The valencies If you are given the formula then think backwards to determine the valencies. Thursday, 16 September 2010
  • 32. Copy Method 2: The swop and drop method 1. Write the ions down next to each other. 2. Drop each number down (without the charge) and swop it to make the subscript of the other element Example Write the formula for Aluminium Oxide The valencies 1 Al O 3+ 2- If you are given the formula then think backwards to determine the valencies. Thursday, 16 September 2010
  • 33. Copy Method 2: The swop and drop method 1. Write the ions down next to each other. 2. Drop each number down (without the charge) and swop it to make the subscript of the other element Example Write the formula for Aluminium Oxide The valencies 1 Al O 3+ 2- If you are given the formula then think backwards to determine the valencies. Thursday, 16 September 2010
  • 34. Copy Method 2: The swop and drop method 1. Write the ions down next to each other. 2. Drop each number down (without the charge) and swop it to make the subscript of the other element Example Write the formula for Aluminium Oxide The valencies 1 Al O 3+ 2- If you are given the formula then think backwards to determine the valencies. Thursday, 16 September 2010
  • 35. Copy Method 2: The swop and drop method 1. Write the ions down next to each other. 2. Drop each number down (without the charge) and swop it to make the subscript of the other element Example Write the formula for Aluminium Oxide The valencies 1 Al O 3+ 2- 2 Al2O3 If you are given the formula then think backwards to determine the valencies. Thursday, 16 September 2010
  • 36. FORMULAE FOR SIMPLE IONS +1 +2 +3 _,, -1 H* Mg2* Al3+ c1- gz- hydrogen magnesium aluminium chloride oxide Li* Ca2* Fe3* oH- COr'- lithium calcium iron(III) hydroxide carbonate Na+ Fe2* No,* Soo'- sodium iron(II) nitrate sulfate K+ Cu2* HCO3- potasslum copper(II) hydrogen carbonate PO43- Zn2* Phosphate zinc Pb2+ lead Thursday, 16 September 2010
  • 37. UTILIZING BONDING POTENTIAL Valency can be shown using bonding symbols Examples Phosphorus has a valency of 3 and so has the potential to form 3 covalent bonds. Sometimes written: P Thursday, 16 September 2010
  • 38. EQUATIONS INTERACTIVE Thursday, 16 September 2010
  • 39. Wri$ng  Chemical  Equa$ons Chemical  reac$ons  are  the  chemists  shorthand  way   of  describing  chemical  reac$ons. Thursday, 16 September 2010
  • 40. Wri$ng  Chemical  Equa$ons Chemical  reac$ons  are  the  chemists  shorthand  way   of  describing  chemical  reac$ons. The  equa$ons  can  be  either: Thursday, 16 September 2010
  • 41. Wri$ng  Chemical  Equa$ons Chemical  reac$ons  are  the  chemists  shorthand  way   of  describing  chemical  reac$ons. The  equa$ons  can  be  either:   1.  Word  equa$ons     Thursday, 16 September 2010
  • 42. Wri$ng  Chemical  Equa$ons Chemical  reac$ons  are  the  chemists  shorthand  way   of  describing  chemical  reac$ons. The  equa$ons  can  be  either:   1.  Word  equa$ons     or     2.  Balanced  symbol  equa$ons Thursday, 16 September 2010
  • 43. Wri$ng  Chemical  Equa$ons Chemical  reac$ons  are  the  chemists  shorthand  way   of  describing  chemical  reac$ons. The  equa$ons  can  be  either:   1.  Word  equa$ons     or     2.  Balanced  symbol  equa$ons Thursday, 16 September 2010
  • 44. Wri$ng  Chemical  Equa$ons Chemical  reac$ons  are  the  chemists  shorthand  way   of  describing  chemical  reac$ons. The  equa$ons  can  be  either:   1.  Word  equa$ons     or     2.  Balanced  symbol  equa$ons For  both  types  all  reactants  (star$ng  substances)  and   products  of  the  reac$on  (end  substances)  must  be   correctly    iden$fied.   Thursday, 16 September 2010
  • 45. Symbol  Equa$ons Each  reactant  and  product  is  represented  by  a   formula  that  represents  the  smallest  par$cle  that  can   take  part  in  the  reac$on. Thursday, 16 September 2010
  • 46. Symbol  Equa$ons Each  reactant  and  product  is  represented  by  a   formula  that  represents  the  smallest  par$cle  that  can   take  part  in  the  reac$on. The  formula  of  any  solid  element  is  simply  the   symbol  (i.e.  for  one  atom) Thursday, 16 September 2010
  • 47. Symbol  Equa$ons Each  reactant  and  product  is  represented  by  a   formula  that  represents  the  smallest  par$cle  that  can   take  part  in  the  reac$on. The  formula  of  any  solid  element  is  simply  the   symbol  (i.e.  for  one  atom) Thursday, 16 September 2010
  • 48. Symbol  Equa$ons Each  reactant  and  product  is  represented  by  a   formula  that  represents  the  smallest  par$cle  that  can   take  part  in  the  reac$on. The  formula  of  any  solid  element  is  simply  the   symbol  (i.e.  for  one  atom)   e.g.  Magnesium  Metal      use  Mg Thursday, 16 September 2010
  • 49. Thursday, 16 September 2010
  • 50. Balancing  Equa$ons Equa$ons  must  be  “balanced”.  This  is  based  on  the   fact  that  during  a  reac$on  no  atoms  are  destroyed,   nor  are  any  new  atoms  created. Therefore  the  numbers  of  each  type  of  atom  in  the   reactants  must  equal  the  number  of  each  type  of   atom  in  the  products.   Thursday, 16 September 2010
  • 51. Rules: Thursday, 16 September 2010
  • 52. Rules: 1. Name  all  of  the  reactants  and  products Thursday, 16 September 2010
  • 53. Rules: 1. Name  all  of  the  reactants  and  products 2. Write  the  correct  formula  for  each  substance   involved. Thursday, 16 September 2010
  • 54. Rules: 1. Name  all  of  the  reactants  and  products 2. Write  the  correct  formula  for  each  substance   involved. 3.  Balance  so  the  number  of  each  type  of  atoms   remains  the  same  during  the  reac$on.  Formula  can   not  be  changed,  so  balancing  must  be  done  by   changing  the  number  of  par$cles  involved. Thursday, 16 September 2010
  • 55. EQUATION 1 Potassium        +          Chlorine  (g)                  Potassium  chloride                    K                          +              Cl2                                                                KCl            2  K                          +              Cl2                                                          2  KCl i.e.  2  atoms  of  potassium  react  with  1  molecule  of   chlorine  to  produce  2  units  of  potassium  chloride.   Why  “units”? Thursday, 16 September 2010
  • 56. EQUATION 1 Potassium        +          Chlorine  (g)                  Potassium  chloride                    K                          +              Cl2                                                                KCl            2  K                          +              Cl2                                                          2  KCl i.e.  2  atoms  of  potassium  react  with  1  molecule  of   chlorine  to  produce  2  units  of  potassium  chloride.   Why  “units”? Thursday, 16 September 2010
  • 57. EQUATION 1 Potassium        +          Chlorine  (g)                  Potassium  chloride                    K                          +              Cl2                                                                KCl Thursday, 16 September 2010
  • 58. EQUATION 1 Potassium        +          Chlorine  (g)                  Potassium  chloride                    K                          +              Cl2                                                                KCl Thursday, 16 September 2010
  • 59. EQUATION 1 Potassium        +          Chlorine  (g)                  Potassium  chloride                    K                          +              Cl2                                                                KCl            2  K                          +              Cl2                                                          2  KCl Thursday, 16 September 2010
  • 60. EQUATION 1 Potassium        +          Chlorine  (g)                  Potassium  chloride                    K                          +              Cl2                                                                KCl            2  K                          +              Cl2                                                          2  KCl Thursday, 16 September 2010
  • 61. EQUATION 1 Potassium        +          Chlorine  (g)                  Potassium  chloride                    K                          +              Cl2                                                                KCl            2  K                          +              Cl2                                                          2  KCl i.e.  2  atoms  of  potassium  react  with  1  molecule  of   chlorine  to  produce  2  units  of  potassium  chloride.   Thursday, 16 September 2010
  • 62. EQUATION 1 Potassium        +          Chlorine  (g)                  Potassium  chloride                    K                          +              Cl2                                                                KCl            2  K                          +              Cl2                                                          2  KCl i.e.  2  atoms  of  potassium  react  with  1  molecule  of   chlorine  to  produce  2  units  of  potassium  chloride.   Why  “units”? Thursday, 16 September 2010
  • 63. EQUATION 2            Sodium              +      Oxygen  (g)                                          Sodium  oxide              Na                          +              O2                                                                Na2O        4  Na                          +            O2                                                              2  Na2O               Thursday, 16 September 2010
  • 64. EQUATION 2            Sodium              +      Oxygen  (g)                                          Sodium  oxide              Na                          +              O2                                                                Na2O        4  Na                          +            O2                                                              2  Na2O               Thursday, 16 September 2010
  • 65. EQUATION 2            Sodium              +      Oxygen  (g)                                          Sodium  oxide              Na                          +              O2                                                                Na2O Thursday, 16 September 2010
  • 66. EQUATION 2            Sodium              +      Oxygen  (g)                                          Sodium  oxide              Na                          +              O2                                                                Na2O Thursday, 16 September 2010
  • 67. EQUATION 2            Sodium              +      Oxygen  (g)                                          Sodium  oxide              Na                          +              O2                                                                Na2O        4  Na                          +            O2                                                              2  Na2O               Thursday, 16 September 2010
  • 68. EQUATION 3 Sodium              +      sulfuric                                  sodium      +      water   hydroxide                acid                                              sulfate    NaOH                +    H2SO4                                      Na2SO4      +      H2O  2  NaOH            +    H2SO4                                      Na2SO4      +    2  H2O Thursday, 16 September 2010
  • 69. EQUATION 3 Sodium              +      sulfuric                                  sodium      +      water   hydroxide                acid                                              sulfate    NaOH                +    H2SO4                                      Na2SO4      +      H2O  2  NaOH            +    H2SO4                                      Na2SO4      +    2  H2O Thursday, 16 September 2010
  • 70. EQUATION 3 Sodium              +      sulfuric                                  sodium      +      water   hydroxide                acid                                              sulfate    NaOH                +    H2SO4                                      Na2SO4      +      H2O Thursday, 16 September 2010
  • 71. EQUATION 3 Sodium              +      sulfuric                                  sodium      +      water   hydroxide                acid                                              sulfate    NaOH                +    H2SO4                                      Na2SO4      +      H2O Thursday, 16 September 2010
  • 72. EQUATION 3 Sodium              +      sulfuric                                  sodium      +      water   hydroxide                acid                                              sulfate    NaOH                +    H2SO4                                      Na2SO4      +      H2O  2  NaOH            +    H2SO4                                      Na2SO4      +    2  H2O Thursday, 16 September 2010
  • 73. ACID/CARBONATE REACTION EQUATION 4 Calcium        +    hydrochloric                          calcium      +      carbon      +      water carbonate          acid                            chloride              dioxide     CaCO3        +      HCl                                                CaCl2          +    CO2          +        H2O      CaCO3      +    2  HCl                                          CaCl2          +        CO2          +      H2O   Thursday, 16 September 2010
  • 74. ACID/CARBONATE REACTION EQUATION 4 Calcium        +    hydrochloric                          calcium      +      carbon      +      water carbonate          acid                            chloride              dioxide     CaCO3        +      HCl                                                CaCl2          +    CO2          +        H2O      CaCO3      +    2  HCl                                          CaCl2          +        CO2          +      H2O   Thursday, 16 September 2010
  • 75. ACID/CARBONATE REACTION EQUATION 4 Calcium        +    hydrochloric                          calcium      +      carbon      +      water carbonate          acid                            chloride              dioxide     CaCO3        +      HCl                                                CaCl2          +    CO2          +        H2O   Thursday, 16 September 2010
  • 76. ACID/CARBONATE REACTION EQUATION 4 Calcium        +    hydrochloric                          calcium      +      carbon      +      water carbonate          acid                            chloride              dioxide     CaCO3        +      HCl                                                CaCl2          +    CO2          +        H2O     Thursday, 16 September 2010
  • 77. ACID/CARBONATE REACTION EQUATION 4 Calcium        +    hydrochloric                          calcium      +      carbon      +      water carbonate          acid                            chloride              dioxide     CaCO3        +      HCl                                                CaCl2          +    CO2          +        H2O      CaCO3      +    2  HCl                                          CaCl2          +        CO2          +      H2O   Thursday, 16 September 2010
  • 78. EQUATION 5 sodium        +            sulfuric                          sodium      +      carbon      +      water bicarbonate          acid                            sulfate                    dioxide      NaHCO3    +    H2SO4                                  Na2SO4    +    CO2      +      H2O     2  NaHCO3    +    H2SO4                                Na2SO4  +  2  CO2  +    2  H2O     Thursday, 16 September 2010
  • 79. EQUATION 5 sodium        +            sulfuric                          sodium      +      carbon      +      water bicarbonate          acid                            sulfate                    dioxide      NaHCO3    +    H2SO4                                  Na2SO4    +    CO2      +      H2O     2  NaHCO3    +    H2SO4                                Na2SO4  +  2  CO2  +    2  H2O     Thursday, 16 September 2010
  • 80. EQUATION 5 sodium        +            sulfuric                          sodium      +      carbon      +      water bicarbonate          acid                            sulfate                    dioxide      NaHCO3    +    H2SO4                                  Na2SO4    +    CO2      +      H2O   Thursday, 16 September 2010
  • 81. EQUATION 5 sodium        +            sulfuric                          sodium      +      carbon      +      water bicarbonate          acid                            sulfate                    dioxide      NaHCO3    +    H2SO4                                  Na2SO4    +    CO2      +      H2O     Thursday, 16 September 2010
  • 82. EQUATION 5 sodium        +            sulfuric                          sodium      +      carbon      +      water bicarbonate          acid                            sulfate                    dioxide      NaHCO3    +    H2SO4                                  Na2SO4    +    CO2      +      H2O     2  NaHCO3    +    H2SO4                                Na2SO4  +  2  CO2  +    2  H2O   Thursday, 16 September 2010
  • 83. EQUATION 6 Ferric  oxide      +      carbon                                      iron  +  carbon  monoxide        Fe2O3                    +                  C       Fe            +                          CO          Fe2O3                  +            3  C                  2  Fe            +                  3  CO     Thursday, 16 September 2010
  • 84. EQUATION 6 Ferric  oxide      +      carbon                                      iron  +  carbon  monoxide        Fe2O3                    +                  C       Fe            +                          CO          Fe2O3                  +            3  C                  2  Fe            +                  3  CO     Thursday, 16 September 2010
  • 85. EQUATION 6 Ferric  oxide      +      carbon                                      iron  +  carbon  monoxide        Fe2O3                    +                  C       Fe            +                          CO Thursday, 16 September 2010
  • 86. EQUATION 6 Ferric  oxide      +      carbon                                      iron  +  carbon  monoxide        Fe2O3                    +                  C       Fe            +                          CO          Fe2O3                  +            3  C                  2  Fe            +                  3  CO Thursday, 16 September 2010
  • 87. EQUATION 7 Magnesium      +      steam                              magnesium  +    hydrogen                                                                                                                  hydroxide                            gas                Mg                    +            H2O        Mg(OH)2              +            H2                Mg                    +      2  H2O        Mg(OH)2              +            H2   Thursday, 16 September 2010
  • 88. EQUATION 7 Magnesium      +      steam                              magnesium  +    hydrogen                                                                                                                  hydroxide                            gas                Mg                    +            H2O        Mg(OH)2              +            H2                Mg                    +      2  H2O        Mg(OH)2              +            H2   Thursday, 16 September 2010
  • 89. EQUATION 7 Magnesium      +      steam                              magnesium  +    hydrogen                                                                                                                  hydroxide                            gas                Mg                    +            H2O        Mg(OH)2              +            H2 Thursday, 16 September 2010
  • 90. EQUATION 7 Magnesium      +      steam                              magnesium  +    hydrogen                                                                                                                  hydroxide                            gas                Mg                    +            H2O        Mg(OH)2              +            H2 Thursday, 16 September 2010
  • 91. EQUATION 7 Magnesium      +      steam                              magnesium  +    hydrogen                                                                                                                  hydroxide                            gas                Mg                    +            H2O        Mg(OH)2              +            H2                Mg                    +      2  H2O        Mg(OH)2              +            H2 Thursday, 16 September 2010
  • 92. EQUATION 8 zinc      +      sulfuric         zinc              +          hydrogen                                  acid                                                  sulfate                              gas Thursday, 16 September 2010
  • 93. EQUATION 8 zinc      +      sulfuric         zinc              +          hydrogen                                  acid                                                  sulfate                              gas      Zn      +        H2SO4       ZnSO4        +                  H2 Thursday, 16 September 2010
  • 94. EQUATION 8 zinc      +      sulfuric         zinc              +          hydrogen                                  acid                                                  sulfate                              gas      Zn      +        H2SO4       ZnSO4        +                  H2                 (balanced,   no  modifica2on  needed) Thursday, 16 September 2010
  • 95. EQUATION 9 Aluminium      +      oxygen     aluminium  oxide                Al                      +      O2                                                              Al2O3              4  Al                    +        3  O2                                                      2  Al2O3   Thursday, 16 September 2010
  • 96. EQUATION 9 Aluminium      +      oxygen     aluminium  oxide                Al                      +      O2                                                              Al2O3              4  Al                    +        3  O2                                                      2  Al2O3   Thursday, 16 September 2010
  • 97. EQUATION 9 Aluminium      +      oxygen     aluminium  oxide                Al                      +      O2                                                              Al2O3 Thursday, 16 September 2010
  • 98. EQUATION 9 Aluminium      +      oxygen     aluminium  oxide                Al                      +      O2                                                              Al2O3     Thursday, 16 September 2010
  • 99. EQUATION 9 Aluminium      +      oxygen     aluminium  oxide                Al                      +      O2                                                              Al2O3              4  Al                    +        3  O2                                                      2  Al2O3   Thursday, 16 September 2010
  • 100. NOTES Thursday, 16 September 2010
  • 101. OXIDATION - REDUCTION Thursday, 16 September 2010
  • 102. METALS ARE REDUCING AGENTS Thursday, 16 September 2010
  • 103. Copy OXIDATION and REDUCTION Definitions “LEO the lion Loss of Electrons is Oxidation goes GER” Gain of Electrons is Reduction • Oxidation is also the addition of oxygen or the removal of hydrogen. • Reduction is the removal of oxygen or the addition of hydrogen. • Oxidation and reduction always occur together Oxidants & Reductants • An oxidant is a substance that causes oxidation and is itself reduced. Example: In the reaction between Zinc and Oxygen, Oxygen is the oxidant. It has caused Zinc to gain oxygen. Zinc Oxide is produced. • A reductant is a substance that causes reduction and is itself oxidised. Example: In the reaction between Hydrogen and Copper oxide, Hydrogen is the reductant. It has caused Copper oxide to lose oxygen. Copper is produced. Thursday, 16 September 2010
  • 104. Half equations Copy REDOX REACTION TYPES Simple electron transfer Reactions that involve atoms losing or gaining electrons. Ions become atoms, atoms become ions. Example 1 Burning magnesium in oxygen - final equation: 2Mg + O2 --> 2MgO Thursday, 16 September 2010
  • 105. Half equations Copy REDOX REACTION TYPES Simple electron transfer Reactions that involve atoms losing or gaining electrons. Ions become atoms, atoms become ions. Example 1 Burning magnesium in oxygen - final equation: 2Mg + O2 --> 2MgO Magnesium loses electrons to form ions Thursday, 16 September 2010
  • 106. Half equations Copy REDOX REACTION TYPES Simple electron transfer Reactions that involve atoms losing or gaining electrons. Ions become atoms, atoms become ions. Example 1 Burning magnesium in oxygen - final equation: 2Mg + O2 --> 2MgO Magnesium loses electrons to form ions (Mg --> Mg2+ + 2e-) x 2 Thursday, 16 September 2010
  • 107. Half equations Copy REDOX REACTION TYPES Simple electron transfer Reactions that involve atoms losing or gaining electrons. Ions become atoms, atoms become ions. Example 1 Burning magnesium in oxygen - final equation: 2Mg + O2 --> 2MgO Magnesium loses electrons to form ions (Mg --> Mg2+ + 2e-) x 2 Oxygen gains electrons to form ions Thursday, 16 September 2010
  • 108. Half equations Copy REDOX REACTION TYPES Simple electron transfer Reactions that involve atoms losing or gaining electrons. Ions become atoms, atoms become ions. Example 1 Burning magnesium in oxygen - final equation: 2Mg + O2 --> 2MgO Magnesium loses electrons to form ions (Mg --> Mg2+ + 2e-) x 2 Oxygen gains electrons to form ions O2 + 4e- --> 2O2- Thursday, 16 September 2010
  • 109. Half equations Copy REDOX REACTION TYPES Simple electron transfer Reactions that involve atoms losing or gaining electrons. Ions become atoms, atoms become ions. Example 1 Burning magnesium in oxygen - final equation: 2Mg + O2 --> 2MgO Magnesium loses electrons to form ions (Mg --> Mg2+ + 2e-) x 2 Oxygen gains electrons to form ions O2 + 4e- --> 2O2- Thursday, 16 September 2010
  • 110. Half equations Copy REDOX REACTION TYPES Simple electron transfer Reactions that involve atoms losing or gaining electrons. Ions become atoms, atoms become ions. Example 1 Burning magnesium in oxygen - final equation: 2Mg + O2 --> 2MgO Magnesium loses electrons to form ions (Mg --> Mg2+ + 2e-) x 2 Oxygen gains electrons to form ions O2 + 4e- --> 2O2- These equations can only be added once the first equation has been doubled. This ensures that the number of electrons gained in the reaction equals the number of electrons lost. Thursday, 16 September 2010
  • 111. Half equations Copy REDOX REACTION TYPES Simple electron transfer Reactions that involve atoms losing or gaining electrons. Ions become atoms, atoms become ions. Example 1 Burning magnesium in oxygen - final equation: 2Mg + O2 --> 2MgO Magnesium loses electrons to form ions (Mg --> Mg2+ + 2e-) x 2 Oxygen gains electrons to form ions O2 + 4e- --> 2O2- These equations can only be added once the first equation has been doubled. This ensures that the number of electrons gained in the reaction equals the number of electrons lost. Adding the two equations and cancelling the charges gives the final equation. The final equation describes both the reduction and oxidation together: Thursday, 16 September 2010
  • 112. Half equations Copy REDOX REACTION TYPES Simple electron transfer Reactions that involve atoms losing or gaining electrons. Ions become atoms, atoms become ions. Example 1 Burning magnesium in oxygen - final equation: 2Mg + O2 --> 2MgO Magnesium loses electrons to form ions (Mg --> Mg2+ + 2e-) x 2 Oxygen gains electrons to form ions O2 + 4e- --> 2O2- These equations can only be added once the first equation has been doubled. This ensures that the number of electrons gained in the reaction equals the number of electrons lost. Adding the two equations and cancelling the charges gives the final equation. The final equation describes both the reduction and oxidation together: 2Mg + O2 + 4e- --> 2 Mg2+ + 2O2- + 4e- Thursday, 16 September 2010
  • 113. Half equations Copy REDOX REACTION TYPES Simple electron transfer Reactions that involve atoms losing or gaining electrons. Ions become atoms, atoms become ions. Example 1 Burning magnesium in oxygen - final equation: 2Mg + O2 --> 2MgO Magnesium loses electrons to form ions (Mg --> Mg2+ + 2e-) x 2 Oxygen gains electrons to form ions O2 + 4e- --> 2O2- These equations can only be added once the first equation has been doubled. This ensures that the number of electrons gained in the reaction equals the number of electrons lost. Adding the two equations and cancelling the charges gives the final equation. The final equation describes both the reduction and oxidation together: 2Mg + O2 + 4e- --> 2 Mg2+ + 2O2- + 4e- 2Mg + O2 --> 2MgO Thursday, 16 September 2010
  • 114. Spectator ions Copy Example 2 Zinc powder added to Copper Sulphate solution. Sulphate ion is a spectator ion. - final equation: Zn + CuSO4 --> Cu + ZnSO4 Zn --> Zn2+ + 2e- Oxidation - half equations: Cu2+ + 2e- --> Cu Reduction Zn + Cu2+ + 2e- --> Zn2+ + Cu + 2e- - Adding the two half equations and cancelling out the charges gives the final equation (above). In Summary Half equations are used to show how electrons are lost and gained in redox reactions (reactions involving reduction and oxidation) Thursday, 16 September 2010
  • 115. Copy Example 3 (class prac) A bright steel nail is placed in 2 mL of copper sulphate solution (0.1 molL-1) in a test tube. Observation: - half equations: - final equation: Example 4 (class prac) A 2 cm strip of Magnesium ribbon is placed in 2 mL of Lead nitrate solution in a test tube. Observation: - half equations: - final equation: Thursday, 16 September 2010
  • 116. Copy Example 5 (class prac) A 2 cm strip of Magnesium ribbon is placed in 2 mL of Silver nitrate solution in a test tube. Observation: - half equations: - final equation: Example 6 (class prac) A small (1 - 2 cm strip) of Copper foil is placed in 2 mL of Lead nitrate solution in a test tube. Observation: - half equations: - final equation: More redox reaction pracs (ABA P34: “Reactions 4 & 5) Thursday, 16 September 2010
  • 117. WHEN DISCUSSING CHEMISTRY OF REDOX REACTION ALWAYS INCLUDE A DISCUSSION OF ELECTRON TRANSFER IN ADDITION TO APPLICATION OF OTHER PRINCIPLES OF REDUCTION/OXIDATION Thursday, 16 September 2010
  • 118. PRECIPITATION Thursday, 16 September 2010
  • 119. Calcium Chloride solution (clear & colourless) added to Silver Nitrate solution (clear & colourless) white precipitate. Thursday, 16 September 2010
  • 120. Silver Nitrate solution (clear & colourless) added to Sodium Sulphide solution (clear & colourless) black precipitate Thursday, 16 September 2010
  • 121. Copy WHAT IS A PRECIPITATE? A precipitate may form when one ionic solution is added to another. It is not possible to see through a solution when a precipitate forms in it. The solution becomes cloudy. The precipitate will eventually settle to the bottom of the container. Definition A precipitation reaction has occurred when a solid forms as a result of a reaction between two solutions. The solid forms because it is an insoluble product of the reaction. Thursday, 16 September 2010
  • 122. HOW PRECIPITATES FORM Example A solution of Copper Sulphate is added to a solution of Sodium Carbonate. Copper ions spread through the water The copper sulphate solution contains: Sulphate ions spread through the water Sodium ions spread through the water The sodium carbonate solution contains: Carbonate ions spread through the water Forces of attraction between the Copper ions and Sulphate ions are weak enough for the water molecules to get between the ions. The same is true for the sodium and carbonate ions. The same cannot be said for the Copper ions that collide with the Carbonate ions in this mixture. A bond forms that is stronger than the attraction that water has for the ions and so a precipitate is formed. The Sodium and sulphate ions are dispersed and free to move amongst the water molecules. They are spectator ions. Copper Sulphate + Sodium Carbonate --> Copper Carbonate + Sodium Sulphate Ion equation: Complete formula equation: Thursday, 16 September 2010
  • 123. SOLUBILITY RULES “You can predict whether or not a compound is soluble by using the following simple rules:” 1. All nitrates are soluble. 2. All group 1 metal compounds and ammonium compounds are soluble. 3. All chlorides are soluble except for silver chloride and lead chloride 4. All sulfates are soluble except for barium sulfate and lead sulfate. 5. All carbonates are insoluble except group 1 and ammonium compounds 6. All hydroxides and oxides are insoluble (exception Sodium & Potassium) Thursday, 16 September 2010
  • 124. Thursday, 16 September 2010
  • 125. Thursday, 16 September 2010
  • 126. Thursday, 16 September 2010
  • 127. THERMAL DECOMPOSITION Thursday, 16 September 2010
  • 128. EQUATIONS OF DECOMPOSITION REACTIONS A decomposition reaction is one in which a compound breaks down to form simpler compounds. A thermal decomposition requires heat. Word equations for simple decomposition reactions 1. The thermal decomposition of carbonates, Eg ... Calcium Carbonate --> Calcium Oxide + Carbon Dioxide Copper Carbonate --> Copper Oxide + Carbon Dioxide 2. The thermal decomposition of sodium bicarbonate, Eg... Sodium Bicarbonate --> Sodium Carbonate + Water + Carbon Dioxide 3. The decomposition of hydroxides Copper Hydroxide --> Copper Oxide + Water Note Not all metal carbonates, hydrogen carbonates and hydroxides will decompose on heating. Position on the reactivity series is an important factor Thursday, 16 September 2010
  • 129. REACTIVITY SERIES OF METALS Most reactive K Na Li In a simple electron exchange reaction (redox), a metal higher in the series will Ca donate electrons to one that is lower in Mg the series Al Zn Example Fe Sn Zn(s) + CuSO4(aq) ---> ZnSO4(aq) + Cu(s) Pb Grey Blue Colourless orange-brown Cu Hg Ag Least reactive Au Thursday, 16 September 2010
  • 130. CALCULATIONS Thursday, 16 September 2010
  • 131. ATOMIC & MOLECULAR MASSES The mole Consider the formation of Carbon Monoxide: C + O ---> CO This is often understood as meaning “1 atom of C” + “1 atom of O” ---> “1 molecule of CO” But working with single atoms when you are performing calculations involving mass means working with masses that are too small. The chemist needs to find a larger unit so that masses in calculations are realistic. This unit is called the mole. A mole is a unit that consists of 6.023 x 1023 objects. Atomic masses “This means that 1 mole of hydrogen atomic 1 (6.023 x 1023 atoms) has a mass of 1.0 g.” number H Relative atomic 1.0 mass, Ar “The atomic mass of Carbon is 12 gmol-1.” Thursday, 16 September 2010
  • 132. Mr = Relative molecular mass = the number of grams per mole for the compound. A formula gives us the mole ratio of each type of atom in a compound. Molecular masses can be calculated from relative atomic mass values and formulae. Example The formula for Methane is CH4. This means that in a sample of Methane there are 4 moles of hydrogen atoms for every mole of carbon. Mr(CH4) = the number of grams per mole of methane molecules. The periodic table tells us that: C = 12.0g per mole H = 1.0g per mole So ...... Mr(CH4) = 12.0 + 4 x 1.0 = 16 gmol-1 Thursday, 16 September 2010
  • 133. Example: A question from the 2003 NCEA examination paper Thursday, 16 September 2010
  • 134. Example: A question from the 2003 NCEA examination paper Mr(FeCO3) = 55.9 + 12 + (3 x 16) = 115.9 gmol-1 Thursday, 16 September 2010
  • 135. Example: A question from the 2003 NCEA examination paper Mr(FeCO3) = 55.9 + 12 + (3 x 16) = 115.9 gmol-1 Mr[Fe(NO3)3] = 55.9 + (3 x 14) + (9 x 16) = 241.9 gmol-1 Thursday, 16 September 2010
  • 136. Example: A question from the 2003 NCEA examination paper Mr(FeCO3) = 55.9 + 12 + (3 x 16) = 115.9 gmol-1 Mr[Fe(NO3)3] = 55.9 + (3 x 14) + (9 x 16) = 241.9 gmol-1 Mr(Fe2O3) = 2 x 55.9 + 3 x 16 = 159.8 gmol-1 Thursday, 16 September 2010
  • 137. A question from the 2005 NCEA examination paper “Try these questions out for yourself” ESA: Ex 12A Thursday, 16 September 2010
  • 138. A question from the 2005 NCEA examination paper “Try these questions out for yourself” 159.6 gmol-1 ESA: Ex 12A Thursday, 16 September 2010
  • 139. A question from the 2005 NCEA examination paper “Try these questions out for yourself” 159.6 gmol-1 106 gmol-1 ESA: Ex 12A Thursday, 16 September 2010
  • 140. A question from the 2005 NCEA examination paper “Try these questions out for yourself” 159.6 gmol-1 106 gmol-1 261 gmol-1 ESA: Ex 12A Thursday, 16 September 2010
  • 141. PERCENTAGE COMPOSITION Percentage composition of an element/s in a compound is the ratio of the mass per mole of the element/s to the mass per mole of the compound expressed as a percentage. Example Consider that you need to calculate the percentage of Sulfur in Copper Sulphate. The relative atomic masses are: Cu = 63.5 gmol-1 S = 32.1 gmol-1 O = 16.0 gmol-1 Mr (CuSO4) = 159.6 gmol-1 (from previous question) Ar (S) = 32.1 gmol-1 Percentage Sulfur = 32.1 = 20.1% 159.6 Thursday, 16 September 2010
  • 142. YIELD The following equation is useful: m Mr = m m = mass (g) N where Mr N N = number of moles (mol) Example Calculate the mass of water that can be made from 3 kg of Hydrogen. H = 1.0 gmol-1 O = 16.0 gmol-1 Mr (H2O) = 18 gmol-1 (from previous question) Process 1. Write the equation for the production of water from Hydrogen. 2. Calculate the number of moles of Hydrogen (H2) in 3 kg. 3. The equation tells us that for every mole of H2 there is one mole of H2O produced. 4. The number of moles of H2O and its relative molecular mass can be used to calculate the mass of H2O produced. Thursday, 16 September 2010
  • 143. Answer m Mr N Example: A question from the 2003 NCEA examination paper Thursday, 16 September 2010
  • 144. Answer H2 + O ---> H2O m Mr N Example: A question from the 2003 NCEA examination paper Thursday, 16 September 2010
  • 145. Answer H2 + O ---> H2O N(H2) = m = 3000 g = 1500 mol Mr 2 gmol-1 m Mr N Example: A question from the 2003 NCEA examination paper Thursday, 16 September 2010
  • 146. Answer H2 + O ---> H2O N(H2) = m = 3000 g = 1500 mol Mr 2 gmol-1 The equation tells us that for every mole of H2 there is one mole of H2O produced... so 1500 mol of H2O are produced. m Mr N Example: A question from the 2003 NCEA examination paper Thursday, 16 September 2010
  • 147. Answer H2 + O ---> H2O N(H2) = m = 3000 g = 1500 mol Mr 2 gmol-1 The equation tells us that for every mole of H2 there is one mole of H2O produced... so 1500 mol of H2O are produced. m(H2O) = Mr x N = 18 gmol-1 x 1500 mol = 27000 g = 27 kg m Mr N Example: A question from the 2003 NCEA examination paper Thursday, 16 September 2010
  • 148. Example: A question from the 2005 NCEA examination paper “Try this question out for yourself” ABA: P54 Q.2, 5 & 6 (PTO) Thursday, 16 September 2010
  • 149. Example: A question from the 2005 NCEA examination paper “Try this question out for yourself” Answer 1. N(NaHCO3) = m = 0.5 g = 5.95 x 10-3 mol Mr 84 gmol-1 2. The equation tells us that for every mole of NaHCO3 there is half a mole of Na2CO3 produced... so 2.975 x 10-3 mol of Na2CO3 are produced. 3. m(Na2CO3) = Mr x N = 106 gmol-1 x 2.975 x 10-3 mol = 0.315 g ABA: P54 Q.2, 5 & 6 (PTO) Thursday, 16 September 2010
  • 150. ABA: P54 Q.2, 5 & 6 Thursday, 16 September 2010
  • 151. ANSWERS Thursday, 16 September 2010
  • 152. ANSWERS N(C3H8) = m = 660 g = 15 mol Mr 44 gmol-1 Thursday, 16 September 2010
  • 153. ANSWERS N(C3H8) = m = 660 g = 15 mol Mr 44 gmol-1 The equation tells us that for every mole of C3H8 that is burnt, there are 3 moles of CO2 produced. Thursday, 16 September 2010
  • 154. ANSWERS N(C3H8) = m = 660 g = 15 mol Mr 44 gmol-1 The equation tells us that for every mole of C3H8 that is burnt, there are 3 moles of CO2 produced. Therefore N(CO2) = 3 x 15 = 45 mol Thursday, 16 September 2010
  • 155. ANSWERS N(C3H8) = m = 660 g = 15 mol Mr 44 gmol-1 The equation tells us that for every mole of C3H8 that is burnt, there are 3 moles of CO2 produced. Therefore N(CO2) = 3 x 15 = 45 mol m(CO2) = Mr x N = 44 gmol-1 x 45 mol = 1980 g = 1.98 kg Thursday, 16 September 2010
  • 156. ANSWERS Thursday, 16 September 2010
  • 157. ANSWERS (a) 2NaHCO3 ---> Na2CO3 (s) + H2O (l) + CO2 (g) Thursday, 16 September 2010
  • 158. ANSWERS (a) 2NaHCO3 ---> Na2CO3 (s) + H2O (l) + CO2 (g) (b) N(NaHCO3) = m = 6.5 g = 7.74 x 10-2 mol Mr 84 gmol -1 Thursday, 16 September 2010
  • 159. ANSWERS (a) 2NaHCO3 ---> Na2CO3 (s) + H2O (l) + CO2 (g) (b) N(NaHCO3) = m = 6.5 g = 7.74 x 10-2 mol Mr 84 gmol -1 The equation tells us that for every mole of NaHCO3 that decomposes, half a mole of Na2CO3 is produced. Thursday, 16 September 2010
  • 160. ANSWERS (a) 2NaHCO3 ---> Na2CO3 (s) + H2O (l) + CO2 (g) (b) N(NaHCO3) = m = 6.5 g = 7.74 x 10-2 mol Mr 84 gmol -1 The equation tells us that for every mole of NaHCO3 that decomposes, half a mole of Na2CO3 is produced. Therefore N(Na2CO3) = 0.5 x 7.74 x 10-2 = 3.87 x 10-2 mol Thursday, 16 September 2010
  • 161. ANSWERS (a) 2NaHCO3 ---> Na2CO3 (s) + H2O (l) + CO2 (g) (b) N(NaHCO3) = m = 6.5 g = 7.74 x 10-2 mol Mr 84 gmol -1 The equation tells us that for every mole of NaHCO3 that decomposes, half a mole of Na2CO3 is produced. Therefore N(Na2CO3) = 0.5 x 7.74 x 10-2 = 3.87 x 10-2 mol m(Na2CO3) = Mr x N = 106 gmol-1 x 3.87 x 10-2 mol = 4.102 g Thursday, 16 September 2010
  • 162. EMPIRICAL FORMULAE The empirical formula gives the simplest ratio of one atom to another in a compound. It is calculated from the percentage composition (by mass) of a compound. Example Calculate the empirical formula of a compound of iron and sulfur that contains 63.6% iron. Process 1. Consider that there is 100 g of the compound and calculate the number of grams of the element from the percentage. 2. Calculate the number of moles of that element in 100g of the compound using the elements relative atomic mass. 3. Repeat steps 2 & 3 for the other element/s and write the folmula from the mole ratio that is evident in your answers Thursday, 16 September 2010
  • 163. EMPIRICAL FORMULAE The empirical formula gives the simplest ratio of one atom to another in a compound. It is calculated from the percentage composition (by mass) of a compound. Example Calculate the empirical formula of a compound of iron and sulfur that contains 63.6% iron. In 100 g of the compound, there will be 63.6 g of iron. Process 1. Consider that there is 100 g of the compound and calculate the number of grams of the element from the percentage. 2. Calculate the number of moles of that element in 100g of the compound using the elements relative atomic mass. 3. Repeat steps 2 & 3 for the other element/s and write the folmula from the mole ratio that is evident in your answers Thursday, 16 September 2010
  • 164. EMPIRICAL FORMULAE The empirical formula gives the simplest ratio of one atom to another in a compound. It is calculated from the percentage composition (by mass) of a compound. Example Calculate the empirical formula of a compound of iron and sulfur that contains 63.6% iron. In 100 g of the compound, there will be 63.6 g of iron. The number of moles of iron in 100 g of the compound, N = m = 63.6 = 1.138 mol Ar 55.9 Process 1. Consider that there is 100 g of the compound and calculate the number of grams of the element from the percentage. 2. Calculate the number of moles of that element in 100g of the compound using the elements relative atomic mass. 3. Repeat steps 2 & 3 for the other element/s and write the folmula from the mole ratio that is evident in your answers Thursday, 16 September 2010
  • 165. EMPIRICAL FORMULAE The empirical formula gives the simplest ratio of one atom to another in a compound. It is calculated from the percentage composition (by mass) of a compound. Example Calculate the empirical formula of a compound of iron and sulfur that contains 63.6% iron. In 100 g of the compound, there will be 63.6 g of iron. The number of moles of iron in 100 g of the compound, N = m = 63.6 = 1.138 mol Ar 55.9 The number of moles of sulfur in 100 g of the compound, N = m = 36.4 = 1.134 mol Ar 32.1 Process 1. Consider that there is 100 g of the compound and calculate the number of grams of the element from the percentage. 2. Calculate the number of moles of that element in 100g of the compound using the elements relative atomic mass. 3. Repeat steps 2 & 3 for the other element/s and write the folmula from the mole ratio that is evident in your answers Thursday, 16 September 2010
  • 166. EMPIRICAL FORMULAE The empirical formula gives the simplest ratio of one atom to another in a compound. It is calculated from the percentage composition (by mass) of a compound. Example Calculate the empirical formula of a compound of iron and sulfur that contains 63.6% iron. In 100 g of the compound, there will be 63.6 g of iron. The number of moles of iron in 100 g of the compound, N = m = 63.6 = 1.138 mol Ar 55.9 The number of moles of sulfur in 100 g of the compound, N = m = 36.4 = 1.134 mol Ar 32.1 For every mole of Iron there is almost exactly one mole of Sulfur therefore the formula must be FeS Process 1. Consider that there is 100 g of the compound and calculate the number of grams of the element from the percentage. 2. Calculate the number of moles of that element in 100g of the compound using the elements relative atomic mass. 3. Repeat steps 2 & 3 for the other element/s and write the folmula from the mole ratio that is evident in your answers Thursday, 16 September 2010
  • 167. Example: A question from the 2005 NCEA examination paper “Try this question out for yourself” Note that in this question you do not have to get the masses from percentage composition. Thursday, 16 September 2010
  • 168. Example: A question from the 2005 NCEA examination paper “Try this question out for yourself” Note that in this question you do not have to get the masses from percentage composition. The number of moles of iron in 8.65 g of iron: N = m = 8.65 = 0.155 mol Ar 55.9 Thursday, 16 September 2010
  • 169. Example: A question from the 2005 NCEA examination paper “Try this question out for yourself” Note that in this question you do not have to get the masses from percentage composition. The number of moles of iron in 8.65 g of iron: N = m = 8.65 = 0.155 mol Ar 55.9 The number of moles of oxygen in 3.72 g of oxygen: N = m = 3.72 = 0.233 mol Ar 16 Thursday, 16 September 2010
  • 170. Example: A question from the 2005 NCEA examination paper “Try this question out for yourself” Note that in this question you do not have to get the masses from percentage composition. The number of moles of iron in 8.65 g of iron: N = m = 8.65 = 0.155 mol Ar 55.9 The number of moles of oxygen in 3.72 g of oxygen: N = m = 3.72 = 0.233 mol Ar 16 If you divide 0.233 by 0.155 you get a ratio of 1.5:1. The simplest whole number ratio is 3:2 which means that for every 3 moles of oxygen there are two moles of iron. The formula must be Fe2O3. Thursday, 16 September 2010
  • 171. Alternatively when given the g per mole of a compound and the % composition then first calculate the mass of each element in that many g (not 100g sample). This will allow you to divide each mass to find how many moles of each element are in 1 mole of the compound. THIS NEGATES THE NEED TO CONVERT THE RATIO TO A SIMPLEST FORM AND THEN TO TEST IT FOR CONSISTENCY WITH THE ORIGINAL Mr GIVEN IN THE QUESTION !! BETTER METHOD: GIVES THE ANSWER DIRECTLY. Thursday, 16 September 2010
  • 172. MASS IN CHEMICAL REACTIONS The same principles used in calculating yields applies to calculating other masses in chemical reactions. Example What is the percentage purity of a 50 g sample of limestone which produces 21 g of Carbon dioxide on heating? Equation: CaCO3(s) ---> CaO(s) + CO2(g) Answer Thursday, 16 September 2010
  • 173. MASS IN CHEMICAL REACTIONS The same principles used in calculating yields applies to calculating other masses in chemical reactions. Example What is the percentage purity of a 50 g sample of limestone which produces 21 g of Carbon dioxide on heating? Equation: CaCO3(s) ---> CaO(s) + CO2(g) Answer N(CO2) = m = 21 g = 0.477 mol Mr 44 gmol-1 Thursday, 16 September 2010
  • 174. MASS IN CHEMICAL REACTIONS The same principles used in calculating yields applies to calculating other masses in chemical reactions. Example What is the percentage purity of a 50 g sample of limestone which produces 21 g of Carbon dioxide on heating? Equation: CaCO3(s) ---> CaO(s) + CO2(g) Answer N(CO2) = m = 21 g = 0.477 mol Mr 44 gmol-1 The equation tells us that for every mole of CO2 there is one mole of CaCO3 required... so 0.477 mol of CaCO3 are required. Thursday, 16 September 2010
  • 175. MASS IN CHEMICAL REACTIONS The same principles used in calculating yields applies to calculating other masses in chemical reactions. Example What is the percentage purity of a 50 g sample of limestone which produces 21 g of Carbon dioxide on heating? Equation: CaCO3(s) ---> CaO(s) + CO2(g) Answer N(CO2) = m = 21 g = 0.477 mol Mr 44 gmol-1 The equation tells us that for every mole of CO2 there is one mole of CaCO3 required... so 0.477 mol of CaCO3 are required. m(CaCO3) = Mr x N = 100.1 gmol-1 x 0.477 mol = 47.7 g Thursday, 16 September 2010
  • 176. MASS IN CHEMICAL REACTIONS The same principles used in calculating yields applies to calculating other masses in chemical reactions. Example What is the percentage purity of a 50 g sample of limestone which produces 21 g of Carbon dioxide on heating? Equation: CaCO3(s) ---> CaO(s) + CO2(g) Answer N(CO2) = m = 21 g = 0.477 mol Mr 44 gmol-1 The equation tells us that for every mole of CO2 there is one mole of CaCO3 required... so 0.477 mol of CaCO3 are required. m(CaCO3) = Mr x N = 100.1 gmol-1 x 0.477 mol = 47.7 g % Purity = 47.7 = 95.4% 50 Thursday, 16 September 2010
  • 177. PRACTICAL Thursday, 16 September 2010
  • 178. OXIDATION Thursday, 16 September 2010
  • 179. The oxidation of Zinc ADDITION OF OXYGEN Blue bunsen flame Look for a change on the Zinc powder surface of the powder xxxxxxxxxx Observation Gauze Word & Symbol equations: + Oxygen + O2 Oxidant Reductant (Substance which does the oxidising) (Substance which does the reducing) Thursday, 16 September 2010
  • 180. The oxidation of Hydrochloric acid REMOVAL OF HYDROGEN Blue litmus paper (moistened with water) test-tube Look for a change Concentrated in the colour of Hydrochloric acid the litmus paper Observation Manganese Dioxide Word & Symbol equations: Manganese dioxide + Hydrochloric acid Manganese chloride + Water + Chlorine + + + Oxidant Reductant (Substance which does the oxidising) (Substance which does the reducing) Thursday, 16 September 2010
  • 181. The oxidation of Zinc LOSS OF ELECTRONS shake the test-tube Look for a change in the colour the solution from side to side Copper Sulphate solution and the appearance of a coloured solid Zinc powder Observation Word & Symbol equations: Zinc + Zinc Sulphate + + + Reductant Oxidant (Substance which does the oxidising) (Substance which does the reducing) Carry out the following oxidation reactions: 1.Copper Nitrate solution added to powdered Zinc 2.Silver Nitrate solution added to Copper metal For each reaction write word and symbol equations in addition to identifying the oxidant and reductant Thursday, 16 September 2010
  • 182. The oxidation of Sulphur ADDITION OF OXYGEN 2 1 UI Gas jar 1. Light the sulphur in the spoon Deflagrating spoon 2. Place the spoon into the jar Pure oxygen 3. Swirl the water (allowing the gas to dissolve in the water. Sulphur 4. Add 3/4 drops of Universal indicator Wate solution (UI) Word & Symbol equations: + + Oxidant Reductant (Substance which does the oxidising) (Substance which does the reducing) Thursday, 16 September 2010
  • 183. REDUCTION Thursday, 16 September 2010
  • 184. The reduction of copper oxide REMOVAL OF OXYGEN Copper Oxide Carbon Powder Test - tube holder Look for a change where the two powders meet Bunsen burner Word & Symbol equations: + Carbon + Carbon Dioxide + C + Oxidant Reductant (Substance which does the oxidising) (Substance which does the reducing) Thursday, 16 September 2010
  • 185. The reduction of copper oxide ADDITION OF HYDROGEN Copper Oxide Excess powder Hydrogen burning Hydrogen gas Hot bunsen Look for a change flame on the surface of the powder Word & Symbol equations: + Hydrogen + Water + + Oxidant Reductant (Substance which does the oxidising) (Substance which does the reducing) Thursday, 16 September 2010
  • 186. The reduction of Iodine GAIN OF ELECTRONS 2 or 3 drops of water Fine mixture of aluminium powder and iodine powder • •• •• ••••• ••• ••• • • ••••• ••• • •••• • ••••• • •• • Mortar Word & Symbol equations: + + Oxidant Reductant (Substance which does the oxidising) (Substance which does the reducing) Thursday, 16 September 2010
  • 187. More reduction GAIN OF ELECTRONS For each of the following reactions. 1. Complete the word equation 2. Write the symbol equation 3. State the substance that has been reduced and explain why Magnesium + Lead Nitrate solution + + + has been reduced because Potassium Iodide solution + Chlorine solution + + + has been reduced because Thursday, 16 September 2010
  • 188. More reduction GAIN OF ELECTRONS For each of the following reactions. 1. Complete the word equation 2. Write the symbol equation 3. State the substance that has been reduced and explain why Magnesium + Lead Nitrate solution Magnesium Nitrate solution + Lead Mg + Pb(NO3)2 Mg(NO3)2 + Pb Lead has been reduced because it has gained electrons Potassium Iodide solution + Chlorine solution + + + has been reduced because Thursday, 16 September 2010
  • 189. THERMAL DECOMPOSITION Thursday, 16 September 2010
  • 190. THE DECOMPOSITION OF HYDROGEN PEROXIDE 1 Add 20 to 30 mL of 2 Hydrogen Peroxide to Pe Hy roxi Manganese dioxide (enough dr de Lower a glowing splint og en to line the bottom of a into the neck of the conical flask) conical flask Conical flask Manganese dioxide Observation Word equation: Water + Balanced formula equation: Thursday, 16 September 2010
  • 191. Theory A decomposition reaction is one in which a compound breaks down to form simpler compounds. A thermal decomposition requires heat. H H H O O O O H H O O O O H H H Thursday, 16 September 2010
  • 192. THERMAL DECOMPOSITION OF COPPER CARBONATE Take care with suck back: the delivery Test tube holder tube needs to be withdrawn before the Boiling tube boiling tube is Copper Carbonate withdrawn from the (powder) flame Bunsen burner Limewater Mat Repeat with Sodium Bicarbonate Observation Word equation: Copper carbonate ---> Copper Oxide + Carbon dioxide Balanced formula equation: Link observations to the chemical species present Thursday, 16 September 2010
  • 193. DECOMPOSITION OF Steps ZINC HYDROXIDE Observation Equations Thursday, 16 September 2010
  • 194. MORE DECOMPOSITION Steps 1 2 Add 1 mL of Copper Sulphate solution to a 1 test tube CU SO OH 4 Na Add dilute sodium hydroxide drop by drop 2 until no further reaction occurs. 3 Heat the mixture gently over a bunsen flame 3 Observation Equations Thursday, 16 September 2010
  • 195. OBSERVATIONS & CLASSIFICATION Thursday, 16 September 2010
  • 196. MAKING OBSERVATIONS Thursday, 16 September 2010
  • 197. Oxides Properties/appearance carbon dioxide colourless, sweet smelling gas carbon monoxide colourless, odourless gas nitrogen dioxide a reddish-brownish gas that has a sharp smell sulphur dioxide colourless gas that has a sharp smell copper oxide black solid lithium oxide Grey solid magnesium oxide white solid iron (II) oxide black solid (not rust) iron (III) oxide reddish-brown solid zinc oxide yellow powder Thursday, 16 September 2010
  • 198. Metal compounds Properties/appearance group 1 metal carbonates white solids group 2 metal carbonates white solids copper carbonate green solid copper sulphate blue crystals copper nitrate blue crystals copper chloride green solid Iron (II) Hydroxide Practically white (Green tinge if traces of oxygen are present) Iron (III) Hydroxide Brown Calcium Hydroxide White Zinc Hydroxide White Iron Nitrate Pale green Barium Sulphate White solids Thursday, 16 September 2010
  • 199. CLASSIFYING REACTIONS Thursday, 16 September 2010
  • 200. Classify each reaction by writing R, P or D in the boxes (right) Thursday, 16 September 2010
  • 201. Classify each reaction by writing R, P or D in the boxes (right) Thursday, 16 September 2010