      CGB IGCSE Chemistry Chapter 6 Review Answers           Home Assignments Tips and Hints Practice Problems Answers to Practice Problems   Chapter 6 Review Answers Quantitative Chemistry   Relative Atomic Mass (Ar)   The relative atomic mass of an element is the average mass of the naturally occurring atoms of the element, using a scale where an atom of carbon-12 has a mass of exactly 12.   Isotopes are naturally occurring atoms of the same element that have different masses.  Isotopes have the same number of protons but a different number of neutrons.               Example:  Find the relative atomic mass of Lithium.   Look on the periodic table and find Lithium.  In the periodic table the top number is the relative atomic mass.  In this case the relative atomic mass of Li is 7 Ar.  If you are using a different periodic table, the relative atomic mass is always bigger than the atomic number.   1.      Find the relative atomic mass of the following atoms. a.       Iron Fe = 56 Ar   b.      Cobalt Co = 59 Ar   c.       Radium Ra = 226 Ar   d.      Sulphur S = 32 Ar   Relative Formula Mass             The relative formula mass (Mr) of a substance is the sum of the relative atomic masses of the elements present in the formula unit.   If the substance is only made of simple molecules, this mass may also be called the relative molecular mass (Mr)               Example :  Find the relative formula mass (Mr) of Sulphuric Acid.                           Sulphuric acid has the formula H2SO4   Step 1:  Determine the relative atomic mass for all the atoms in the formula H = 1 Ar S = 32 Ar O = 16 Ar   Step 2:  Multiply the relative atomic masses of the atoms by the number of atoms in the formula              H = 1 Ar x 2 = 2 S = 32 Ar x 1 = 32 O = 16 Ar x 4 = 64   Step 3:  Add the totals for each individual element to determine the relative formula mass 2 + 32 + 64 = 98 Mr   The relative formula mass of Sulphuric acid is 98 Mr   2.      Find the relative formula mass (Mr) of the following compounds: a.       CaCO3 Ca = 40 Ar x 1 = 40 C = 12 Ar x 1 = 12 O = 16 Ar x 3 = 48   40 + 12 + 48 = 100 Mr The relative formula mass (Mr) of calcium carbonate is 100 Mr   b.      NaCl Na = 23 Cl = 35.5 23 + 35.5 = 58.5 Mr The relative formula mass (Mr) of sodium chloride is 58.5 Mr   c.       Al2O3 Al = 27 Ar x 2 = 54 O = 16 Ar x 3 = 48 54 + 48 = 102 Mr The relative formula mass (Mr) of aluminum oxide is 102 Mr   d.      Fe2O3 Fe = 56 Ar x 2 = 112 O = 16 Ar x 3 = 48 112 + 48 = 160 Mr The relative formula mass (Mr) of Iron(III)oxide is 160 Mr   Percentage by mass   Percentage by mass is determined by first finding the relative formula mass (Mr) then dividing the total mass of one element by the relative formula mass.   Example:  Determine the percentage by mass of Iron in Iron(II) oxide.                           Step one:  Determine the relative formula mass of iron(II) oxide.                                     The formula for iron(II)oxide is FeO                                     Fe = 56                                     O = 16                                     56 + 16 = 72 Mr                           Step two : Determine the mass of iron in the compound                                     Fe = 56 Ar                                     There is one atom of Fe in FeO                                     Fe = 56 x 1 = 56                                     Step three : Divide the mass of iron by the relative formula mass of iron(II)oxide                                     Fe/FeO = 56/72 = 0.78                           Step four: Multiply result by 100                                     0.78 x 100 = 78 %                           The percentage by mass of iron in iron oxide is 78%   3.      Determine the percentage by mass of sulphur in the following compounds: a.       H2SO4 H = 1 Ar x 2 = 2 S = 32 Ar x 1 = 32 O = 16 Ar x 4 = 64 2 + 32 + 64 = 98 Mr The relative formula mass of sulphuric acid is 98 Mr   Sulphur has a mass of 32 in this compound.   Percentage by mass of sulphur = 32/98 x 100 = 33%   b.      MgSO4 Mg = 24 Ar x 1 = 24 S = 32 Ar x 1 = 32 O = 16 Ar x 4 = 64 24 + 32 + 64 = 118 Mr The relative formula mass of magnesium sulphate is 118 Mr   Sulphur has a mass of 32 in this compound   Percentage by mass of sulphur = 32 / 118 x 100 = 27 %   c.       SO2 S = 32 Ar x 1 = 32 O = 16 Ar x 2 = 32 32 + 32 = 64 Mr The relative formula mass of sulphur dioxide is 64 Mr   Sulphur has a mass of 32 in this compound   Percentage by mass of sulphur = 32 / 64 x 100 = 50%   d.      ZnSO4 Zn = 65 Ar x 1 = 65 S = 32 Ar x 1 = 32 O = 16 Ar x 4 = 64 65 + 32 + 64 = 161 Mr The relative formula mass of zinc sulphate is 161 Mr   Sulphur has a mass of 32 in this compound   Percentage by mass of sulphur = 32 / 161 x 100 = 20%   Compound formation and chemical formulas               A particular compound always contains the same elements.             These elements are always present in the same proportions by mass.             It does not matter where the compound is found or how it is made.             These proportions cannot be changed.   We can determine how much of a compound will be made or how  much reactant we need to start with by using ratios.   Example:  Calcium carbonate when heated forms calcium oxide and carbon dioxide.  If 10 grams of calcium carbonate are heated how much calcium oxide is formed?                           Step one :  Write the balanced formula                                     CaCO3 à CaO + CO2                           Step two : Determine the relative formula mass (Mr) of calcium carbonate.                                     Ca = 40 Ar x 1 = 40                                     C = 12 Ar x 1 = 12                                     O = 16 Ar x 3 = 48                                       40 + 12 + 48 = 100 Mr                                     CaCO3 = 100 g                           Step three : Determine the relative formula mass (Mr ) of calcium oxide                                     Ca = 40 Ar x 1 = 40                                     O = 16 Ar x 1 = 16                                     40 + 16 = 56 Mr                                     CaO = 56 g                           Step four : Set up your ratio                                     100 g of CaCO3 produces 56 g CaO                                     10 g of CaCO3 produces X g CaO                                       cross multiply                                       X = 10 g x 56 g                                                 100g                                       X = 5.6 g                           So 10 g of CaCO3 when heated will produce 5.6 g of CaO   4.      Sodium reacts with chlorine gas to form sodium chloride.  If  you want to produce 20 g of sodium chloride, how much sodium do you need to start with? 2 Na + Cl2 à 2 NaCl   Na = 23 Ar 2 Na = 23 x 2 = 46 g   Na = 23 Ar x 1 = 23 Cl = 35.5 Ar x 1 = 35.5 23 + 35.5 = 58.5 Mr NaCl = 58.5 Mr 2 NaCl = 58.5 x 2 = 117 g   46 g of Na = 117 g of NaCl X g of Na = 20 g of NaCl   X g of Na = 20 g x 46 g                           117 g   X g of Na = 7.9 g   To make 20 g of NaCl you need to react 7.9 g of Na.   5.      Iron(II)sulphate reacts with sodium hydroxide to form a light green precipitate of iron(II)hydroxide and sodium sulphate.  If  100 g of iron(II)sulphate is reacted with excess sodium hydroxide, how much precipitate is produced? FeSO4 + 2NaOH à Fe(OH)2 + Na2SO4   Fe = 56 Ar x 1 = 56 S = 32 Ar x 1 = 32 O = 16 Ar x 4 = 64 56 + 32 + 64 = 152 Mr FeSO4 = 152 Mr   Fe = 56 Ar x 1 = 56 O = 16 Ar x 2 = 32 H = 1 Ar x 2 = 2 56 + 32 + 2 = 90 Mr Fe(OH)2 = 90 Mr   152 g of FeSO4 = 90 g of Fe(OH)2 100 g of FeSO4 = X g of Fe(OH)2   X g of Fe(OH)2 = 100 x 90                                  152   X g of Fe(OH)2 = 59 g   100 g of FeSO4 reacts to form 59 g of Fe(OH)2   6.      Calcium carbonate reacts with hydrochloric acid to form calcium chloride, carbon dioxide and water.  How much calcium carbonate must you start with to form 40 g of carbon dioxide? CaCO3 + 2HCl à CaCl2 + CO2 + H2O   Ca = 40 Ar x 1 = 40 C = 12 Ar x 1 = 12 O = 16 Ar x 3 = 48 40 + 12 + 48 = 100 Mr CaCO3 = 100 Mr   C = 12 Ar x 1 = 12 O = 16 Ar x 2 = 32 12 + 32 = 44 Mr CO2 = 44 Mr   100 g CaCO3 = 44 g CO2 X g CaCO3 = 40 g CO2   X g CaCO3 = 40 x 100                               44   X g CaCO3 = 91 g   91 g of CaCO3 reacts with HCl to form 40 g of CO2   7.      Magnesium burns in air to form Magnesium oxide.  How much magnesium oxide will be produced if 56 g of magnesium is burned? 2Mg + O2 à 2MgO   Mg = 24 Ar 2 Mg = 2 x 24 = 48 g   Mg = 24 Ar x 1 = 24 O = 16 Ar x 1 = 16 24 + 16 = 40 Mr MgO = 40 Mr 2 MgO = 40 x 2 = 80 g   48 g Mg = 80 g MgO 56 g Mg = X g MgO   X g MgO = 56 x 80                        48   X g MgO = 93 g   56 g Mg burns in oxygen to form 93 g of MgO   The mole               One mole of a substance has a mass equal to its relative formula mass in grams.   One mole of a substance contains 6.02 x 1023 per mole (Avogadro constant) of atoms, molecules, or formula units, depending on the substance considered.               Example : Determine the mass of one mole of potassium atoms.                         Step one : Determine the relative atomic mass of potassium                                     K = 39 Ar                                     1 mole K = 39 g   8.      Determine the mass of one mole of the following atoms: a.       Lithium Li = 7 Ar 1 mole Li = 7 g   b.      Carbon C = 12 Ar 1 mole C = 12 g   c.       Oxygen O = 16 Ar 1 mole O = 16 g   d.      Aluminum Al = 27 Ar 1 mole Al = 27 g   Example : Determine the mass of one mole of hydrogen molecules.             Step one : Determine the formula of hydrogen molecules                         H2              Step two : Determine the relative formula mass                         H = 1 Ar x 2 = 2 Mr                         1 mole H2 = 2 g   9.      Determine the mass of one mole of the following molecules a.       Oxygen O2 O = 16 Ar x 2 = 32 Mr 1 mole O2 = 32 g   b.      Chlorine Cl2 Cl = 35.5 Ar x 2 = 71 Mr 1 mole Cl2 = 71 g   c.       Bromine Br2 Br = 80 Ar x 2 = 160 Mr 1 mole Br2 = 160 g   d.      Fluorine Fl2 Fl = 19 Ar x 2 = 38 Mr 1 mole Fl2 = 38 g     Example : Determine the mass of one mole of water             Step one : Determine the formula                         H2O             Step two : Determine the relative formula mass                         H = 1 Ar x 2 = 2                         O = 16 Ar x 1 = 16                         2 + 16 = 18 Mr                         1 mole H2O = 18 g   10.  Determine the mass of one mole of the following compounds: a.       Magnesium oxide MgO   Mg = 24 Ar x 1 = 24 O = 16 Ar x 1 = 16 24 + 16 = 40 Mr 1 mole MgO = 40 g   b.      Sodium hydroxide NaOH   Na = 23 Ar x 1 = 23 O = 16 Ar x 1 = 16 H = 1 Ar x 1 = 1 23 + 16 + 1 = 40 Mr 1 mole NaOH = 40 g   c.       Potassium bromide KBr   K = 39 Ar x 1 = 39 Br = 80 Ar x 1 = 80 39 + 80 = 119 Mr 1 mole KBr = 119 g   d.      Sodium chloride NaCl   Na = 23 Ar x 1 = 23 Cl = 35.5 Ar x 1 = 35.5 23 + 35.5 = 58.5 Mr 1 mole NaCl = 58.5 g   Example : Determine the mass of two moles of zinc chloride.             Step one : Determine the formula                         ZnCl2                         Step two : Determine the relative formula mass                         Zn = 65 Ar x 1 = 65                         Cl = 53.5 Ar x 2 = 71                         65 + 71 = 136 Mr                         ZnCl2 = 136 Mr                         1 mole ZnCl2 = 136 g               Step three : Multiply the mass of one mole by the number of moles                         136 g x 2 moles = 272 g                         mole                           2 moles ZnCl2 = 272 g   11.  Determine the mass of the following compounds: a.       2 moles of magnesium sulphate MgSO4   Mg = 24 Ar x 1 = 24 S = 32 Ar x 1 = 32 O = 16 Ar x 4 = 64 24 + 32 + 64 = 120 Mr 1 mole MgSO4 = 120 g   120 g x 2 moles = 240 g mole   2 moles MgSO4 = 240 g   b.      4 moles of potassium nitrate KNO3   K = 39 Ar x 1 = 39 N = 14 Ar x 1 = 14 O = 16 Ar x 3 = 48 39 + 14 + 48 = 101 Mr 1 mole KNO3 = 101 g   101 g x 4 moles = 404 g mole   4 moles KNO3 = 404 g   c.       0.5 moles of calcium oxide CaO   Ca = 40 Ar x 1 = 40 O = 16 Ar x 1 = 16 40 + 16 = 56 Mr 1 mole CaO = 56 g   56 g x 0.5 moles = 28 g mole   0.5 moles CaO = 28 g   d.      3.4 moles of iron(II)carbonate FeCO3   Fe = 56 Ar x 1 = 56 C = 12 Ar x 1 = 12 O = 16 Ar x 3 = 48 56 + 12 + 48 = 116 Mr 1 mole FeCO3 = 116 g   116 g x 3.4 moles = 394 g mole   3.4 moles FeCO3 = 394 g     Example :  How many molecules are present in 0.5 moles of water?             Step one:  Multiply the number of moles by the avogadro constant.                         0.5 moles x 6.02 x 1023 molecules = 3.01 x 1023 molecules                                                             mole   12.  Determine the number of molecules or atoms in: a.       1 mole of calcium atoms 1 mole x 6.02 x 1023 atoms = 6.02 x 1023 atoms                          mole   b.      4 moles of hydrogen gas 4 moles x 6.02 x 1023 molecules = 24 x 1023 molecules                               mole   c.       5.6 moles of nitric acid 5.6 moles x 6.02 x 1023 molecules = 34 x 1023 molecules                               mole d.      0.02 moles of ethane 0.02 moles x 6.02 x 1023 molecules = .12 x 1023 molecules                               mole Working out chemical formulas using moles               Compounds always forms in fixed ratios.   Example : Determine the number of moles of copper and the number of moles of oxygen in copper(II)oxide.                         Step one : Determine the formula of copper(II)oxide The formula of Copper(II)oxide is CuO, as determined by the charges on the ions. Step two : Using the small numbers after the symbol for each atom to determine the number of moles.  Remember, if there is no number then there is one mole.                                     CuO                                     Cu = 1 mole                                     O = 1 mole   13.  Determine the number of moles of each element in the following compounds: a.       Potassium oxide K2O K = 2 moles O = 1 mole   b.      Lithium carbonate Li2CO3 Li = 2 moles C = 1 mole O = 3 moles   c.       Calcium sulphate CaSO4 Ca = 1 mole S = 1 mole O = 4 moles   d.      Iron(III)nitrate Fe(NO3)3 Fe = 1 mole N = 3 moles O = 9 moles         Enter supporting content here