WJEC Chemistry for AS Level Student Book: 2nd Edition (Draft)

 1.3 Chemical calculations

Worked examples What is the amount of sodium present in 0.23g of sodium? A r of sodium = 23.0 Molar mass of sodium = 23.0gmol −1 Amount of sodium =  m M  =  0.23g 23.0gmol −1  = 0.01mol If you need 0.05mol of sodium hydroxide, what mass of the substance do you have to weigh out? Molar mass of NaOH = 40gmol −1 Mass of sample =  n × M  = 0.05 × 40  = 2g Calculating reacting masses In a chemical reaction, reactants change into products. If we are given the mass of reactants, we can find out the mass of products formed as long as we have a balanced chemical equation for the reaction. An equation tells us not only what substances react together but also what amounts of substances, in moles, react together. The ratio between amounts in moles of reactants and products are called the stoichiometric ratios (mole ratios). E.g. the equation for the burning of magnesium in oxygen to produce magnesium oxide 2Mg + O 2 →  2MgO tells us that 2 moles of magnesium react with 1 mole of oxygen to produce 2 moles of magnesium oxide. What mass of magnesium oxide forms if we burn 1.215g of magnesium in an excess of oxygen? To calculate this, the following route is used: Step 1 Change the mass of Mg into amount of moles (divide by molar mass). Step 2 Use the balanced equation to state the mole ratio of Mg :MgO, hence deduce the moles of MgO. Step 3 Change the amount in moles of MgO to mass (multiply by molar mass).

Knowledge check (a) Calculate the mass of 0.020mol of sodium carbonate. (b) Calculate the amount (in moles) of 1.36g calcium carbonate. 6

Key term

Stoichiometry  is the molar relationship between the amounts of reactants and products in a chemical reaction.

Study point Excess oxygen implies that there is more than enough oxygen present to react with all the magnesium burnt in it. When the reaction is complete, all the magnesium has reacted and some oxygen remains.

Step 1 Amount in moles of Mg = 1.215 24.3  = 0.050mol Step 2 The mole ratio from the equation is 2Mg : 2MgO i.e. 1 : 1 therefore 0.050mol Mg gives 0.050mol MgO Step 3 Molar mass of MgO = 24.3 + 16 = 40.3 Mass MgO = 0.05 × 40.3 = 2.015g 37 DRAFT Stretch & challenge When iron ore (iron(III) oxide) is reduced by carbon monoxide in a blast furnace, iron is produced. Calculate how much iron is produced from 1kg of iron ore. Knowledge check In a laboratory, iron can be produced by reducing iron(III) oxide with aluminium. Fe 2 O 3 + 2Al → 2Fe + Al 2 O 3 What mass of iron(III) oxide is needed to produce 1.20 g of iron? 7