      CGB IGCSE Chemistry Chapter 6 Review           Home Assignments Tips and Hints Practice Problems Answers to Practice Problems   Chapter 6 Review 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 b.      Cobalt c.       Radium d.      Sulphur   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 b.      NaCl c.       Al2O3 d.      Fe2O3   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 b.      MgSO4 c.       SO2 d.      ZnSO4   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? 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? 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? 7.      Magnesium burns in air to form Magnesium oxide.  How much magnesium oxide will be produced if 56 g of magnesium is burned?   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 b.      Carbon c.       Oxygen d.      Aluminum   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 b.      Chlorine c.       Bromine d.      Fluorine   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 b.      Sodium hydroxide c.       Potassium bromide d.      Sodium chloride   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 b.      4 moles of potassium nitrate c.       0.5 moles of calcium oxide d.      3.4 moles of iron(II)carbonate   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 b.      4 moles of hydrogen gas c.       5.6 moles of nitric acid d.      0.02 moles of ethane   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 b.      Lithium carbonate c.       Calcium sulphate d.      Iron(III)nitrate         Enter supporting content here