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CGB IGCSE Chemistry
Answers - Practice Problems Quantitative Chemistry
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Quantitative Chemistry
 
1.  Find the relative Atomic Mass (Ar) for the following elements.
  • O  - 16Ar
  • C - 12Ar
  • S - 32Ar
  • P -31Ar
  • Mg - 24Ar
  • Fe - 56Ar

2.  Find the relative formula mass (Mr) for the following compounds.

  • H2O

         H = 1Ar   O = 16Ar

        2H = 2 x 1 = 2

        1O = 1 x 16 = 16

        2 + 16 = 18

        H2O = 18 Mr

  • H2SO4

           H = 1Ar

           S = 32 Ar

           O = 16 Ar

           2H = 2 x 1 = 2

           1S = 1 x 32 = 32

          4O = 4 x 16 = 64

          2 + 32 + 64 = 98

         H2SO4 = 98Mr

  • CaCO3

         Ca = 40Ar

         C = 12 Ar

         O = 16 Ar

         1 Ca = 1 x 40 = 40

         1 C = 1 x 12

         3 O = 3 x 16 = 48

        40 + 12 + 48 = 100

        CaCO3 = 100 Mr

  • MgO

        Mg = 24 Ar

        O = 16 Ar

        1 Mg = 1 x 24 = 24

       1 O = 1 x 16

       24 + 16 = 40

       MgO = 40 Mr

  • Na2CO3

         Na = 23 Ar

        C = 12 Ar

        O = 16 Ar

         2Na = 2 x 23 = 46

         1 C = 1 x 12 = 12

         3 O = 3 x 16 = 48

        46 + 12 + 48 = 106

        Na2CO3 = 106 Mr

  • Mg(NO3)2

          Mg = 24 Ar

           N = 14 Ar

          O = 16 Ar

         1 Mg = 1 x 24 = 24

         2 N = 2 x 14 = 28

         6 O = 6 x 16 = 96

         24 + 28 + 96 = 148

         Mg(NO3)2 = 148 Mr

  • FeCl3

        Fe = 56 Ar

        Cl = 35.5 Ar

        1 Fe = 1 x 56 = 56

        3 Cl = 3 x 35.5 = 111.5

        56 + 111.5 = 167.5

        FeCl3 = 167.5 Mr

  • CH3COOH

      C = 12 Ar

      H = 1 Ar

       O = 16 Ar

       2 C = 2 x 12 = 24

       4 H = 4 x 1 = 4

        2 O = 2 x 16 = 32

       24 + 4 + 32 = 60

       CH3COOH = 60 Mr

  • H3PO4

        H = 1 Ar

        P = 31 Ar

        O = 16 Ar

         3 H = 3 x 1 = 3

         1 P = 1 x 31 = 31

         4 O = 4 x 16 = 64

         3 + 31 + 64 = 98

          H3PO4 = 98 Mr

  • C6H8O7

        C = 12 Ar

        H = 1 Ar

        O = 16 Ar

        6 C = 6 x 12 = 72

        8 H = 8 x 1 = 8

        7 O = 7 x 16 = 112

       72 + 8 + 112 = 192

         C6H8O7 = 192 Mr

  • KNO3

             K = 39 Ar

        N = 14 Ar

        O = 16 Ar

        1 K = 1 x 39 = 39

        1 N = 1 x 14 = 14

        3 O = 3 x 16 = 48

        39 + 14 + 48 = 101

         KNO3 = 101 Mr

2.  Find the percentage of oxygen in the following compounds.  Which compound has the greatest percentage of oxygen? 

  • CaO

         First we need to find the Relative Atomic Mass of each element

          Ca = 40 Ar

          O = 16 Ar

          Then we need to find the Relative Formula mass of the compound

          1 Ca = 1 x 40 = 40

         1 O = 1 x 16 = 16

          40 + 16 = 56

          CaO = 56 Mr

           Then we need to divide the  Relative Atomic Mass of Oxygen by the Formula Mass of CaO and multiply by 100.

            16 / 56 = .29 x 100 = 29 %

            The percentage of oxygen in CaO is 29%

  • HNO3

           First we need to find the relative atomic masses of the elements.  

           H = 1 Ar

            N = 14 Ar

            O = 16 Ar

            Then we need to find the mass of oxygen in the formula

             3 O = 3 x 16 = 48

             Then we need to find the relative formula mass of the compound

             1 H = 1 x 1 = 1

             1 N = 1 x 14 = 14

             3 O = 3 x 16 = 48

             1 + 14 + 48 = 63

              HNO3 = 63 Mr

             Then we need to divide the mass of oxygen in the formula by the relative formula mass of HNO3 and multiply by 100.

              48/63 = 0.68 x 100 = 68% oxygen by mass in HNO3

  • C2H5OH

        First we need to find the relative atomic masses for the elements in the compound

         C = 12 Ar

         H = 1Ar

         O = 16 Ar

         Then we need to find the mass of oxygen in the compound

         1 O = 1 x 16 = 16

          Then we need to find the relative formula mass of the compound.

          2 C = 2 x 12 = 24

          6 H = 6 x 1 = 6

          1 O = 1 x 16 = 16

          24 + 16 + 6 = 46

          C2H5OH = 46 Mr

          Then we need to divide the mass of oxygen in the compound by the relative formula mass of the compound and multiply by 100.

           16/46 = 0.35 x 100 = 35% oxygen by mass in C2H5OH

  • C6H12O

          First we need to find the relative atomic masses for the elements in the compound.

           C = 12 Ar

           H = 1 Ar

           O = 16 Ar

           Then we need to find the mass of oxygen in the compound.

            1 O = 1 x 16 = 16

           Then we need to find the relative formula mass of the compound

            6 C = 6 x 12 = 72

           12 H = 12 x 1 = 12

           1 O = 1 x 16 = 16

            72 + 12 + 16 = 100

            C6H12O = 100 Mr

           Then we need to divide the mass of oxygen by the mass of the compound and multiply by 100.

            16/100 = 0.16 x 100 = 16% oxygen by mass in C6H12O

The compound that has the greatest percentage of oxygen by mass is HNO3.

3.  Find the percentage of carbon in the following compounds.  Which compound has the least percentage of carbon?

  • CO2

        First we need to find the relative atomic masses of the elements in the compound.

         C = 12 Ar

         O = 16 Ar

         Then we need to find the mass of carbon in the compound.

          1 C = 1 x 12 = 12

         Then we need to find the relative formula mass of the compound.

         1 C = 12

         2 O = 2 x 16 = 32

        12 + 32 = 44

         CO2 = 44 Mr

         Then we need to divide the mass of carbon by the mass of the compound and multiply by 100.

         12/44= 0.27 x 100 = 27 % carbon by mass in CO2

  • C2H2

         First we need to find the relative atomic mass of the elements in the compound

          C = 12 Ar

         H = 1 Ar

         Then we need to find the mass of carbon in the compound.

          2 C = 2 x 12 = 24

          Then we need to find the relative formula mass of the compound.

          2 C = 2 x 12 = 24

          2 H = 2 x 1 = 2

          24 + 2 = 26

          C2H2 = 26 Mr

          Then we need to divide the mass of carbon by the mass of the compound.

           24/26 = 0.92 x 100 = 92% carbon by mass in C2H2

  • MgCO3

          First we need to find the relative atomic mass of the elements in the compound

           Mg = 24 Ar

            C = 12 Ar

            O = 16 Ar

         Then we need to find the mass of carbon in the compound.

           1 C = 1 x 12 = 12

          Then we need to find the relative formula mass of the compound.

          1 Mg = 1 x 24 = 24

          1 C = 1 x 12

          3 O = 3 x 16 = 48

          24 + 12 + 48 = 84

          MgCO3 = 84 Mr

          Then we need to divide the mass of carbon by the mass of the compound and multiply by 100.

           12 / 84 = 0.14 x 100 = 14% carbon by mass in MgCO3

         MgCO3 has the least percentage of carbon by mass

4. Sebastian needs to make 20 g of ZnCl2.  He knows he can make ZnCl2 by reacting Zn with HCl acid.

  • Balance the equation

              Zn + 2HCl --> ZnCl2 + H2

  • Determine the relative atomic masses and the relative formula masses for the reactants and the products.

         Relative Atomic Masses

            Zn = 65 Ar

            H = 1 Ar

            Cl = 35.5 Ar

         Relative Formula Masses

             HCl

              1 H =  1 x 1 = 1

              1 Cl = 1 x 35.5 = 35.5

               1 + 35.5 = 36.5

            HCl = 36.5 Mr

             ZnCl2

             1 Zn = 1 x 56 = 56

             2 Cl = 2 x 35.5 = 71

             56 + 71 = 127

             ZnCl2 = 127 Mr

             H2

             2 H = 2 x 1 = 2

             H2 = 2 Mr

             check that you calculations are correct by substituting the atomic and formula masses in your equation.

               Zn + 2HCl --> ZnCl2 + H2

               56 Ar + 2 x 36.5 Mr = 127 Mr + 2 Ar

                56 + 73 = 127 + 2

                 129 = 129

                 The relative formula and atomic masses are correct because mass cannot be created or destroyed - therefore the mass of the reactants must equal the mass of the products.

  • Determine the ratio of Zn to ZnCl2

         According to our relative formula and atomic masses we need 56 g of Zn to form 127 g of ZnCl2.  Therefore our ratio is

                                                           56 g Zn = 127 g ZnCl2

  • Determine the ratio of HCl to ZnCl2

        According to our relative formula masses ( and our equation) that 73 g of HCl are needed to make 127 g of ZnCl2.  Therefore our ratio is

                                                         73 g HCl = 127 g ZnCl2

  • Determine how much Zn and HCl you need to form 20 g of ZnCl2.

        To determine how much Zn we need to start with we need to cross multiply.

        First we need to set up our ratio

                                                          56 g Zn = 127 g ZnCl2

                                                            X g Zn = 20 g ZnCl2

         Then we need to cross multiply

                        (56 g Zn x 20 g ZnCl2) / 127 g ZnCl2 = 8.8 g of Zn

                        If you look at the equation, the ZnCl2 cross each other out to leave you with Zn.

           Therefore we need 8.8 g of Zn to form 20 g of ZnCl2.

To determine how much HCl we need to form 20 g of ZnCl2 we need to cross multiply.

First we have to set up our ratio.

                                                 73 g HCl = 127 g ZnCl2

                                                  X g HCl = 20 g ZnCl2

            Then we need to cross multiply

                                 (73 g HCl x 20 g ZnCl2) / 127 g ZnCl2 = 11.5 g HCl

            11.5 g of HCl are needed to make 20 g of ZnCl2  

      

5.  Jane wants to make 100 g of Hg.  To do this she is going to heat HgO to form Hg and O2.

  • What type of reaction is this?

         This is a decomposition reaction.

  • Balance the equation

              2 HgO --> 2 Hg + O2

  • Determine the relative formula and relative atomic masses of the reactants and of the products.

         Relative Atomic Masses

               Hg = 201 Ar

                O = 16 Ar

          Relative Formula Masses

              HgO

              1 Hg = 1 x 201 = 201

              1 O = 1 x 16 = 16

              201 + 16 = 217

              HgO = 127 Mr

              O2

              2 O = 2 x 16 = 32

              to make sure that your relative atomic and formula masses are correct substitute them for the formulas in your equation.

               2HgO --> 2Hg + O2

               2 x 217 --> 2 x 201 + 32

               434 --> 402 + 32

                434 --> 434

      The relative formula and atomic masses are correct because mass cannot be created or destroyed and the mass of the reactants must equal the mass of the products.

  • Determine the ratio of HgO to Hg

        According to our relative atomic and formula masses and our equation, 434 g of HgO are needed to make 402 g of Hg.  Therefore our ratio is

                                          434 g HgO = 402 g  Hg

  • Determine how much HgO Jane needs to heat to form 100g of Hg.

         To determine how much HgO Jane needs to start with to form 100g of Hg we need to cross multiply.  First we have to set up a ratio.

                                          434 g HgO = 402 g Hg

                                             X g HgO = 100 g Hg

            Then we need to cross multiply

                                   (434 g HgO x 100 g Hg) / 402 g  Hg = 108 g HgO

            Jane needs to heat 108 g HgO to form 100 g Hg.

If a handout is available online (e.g., a newspaper article) I might include the appropriate link to the information students need on this page.