Example Calculation - 1

This worked example brings together a mole calculation and a percentage yield.Question: 3.00 g of magnesium ribbon is heated in a crucible in excess oxygen. The mass of magnesium oxide collected is 4.50 g. Calculate the percentage yield. (A_r: Mg = 24.3, O = 16.0)2Mg(s) + O₂(g) → 2MgO(s)Step 1: moles of the limiting reactant. Oxygen is in excess, so magnesium is limiting.amount of Mg = 3.00 ÷ 24.3 = 0.1235 molStep 2: mole ratio. Mg : MgO = 2 : 2 = 1 : 1, so at most 0.1235 mol of MgO can form.Step 3: theoretical yield. M_r of MgO = 24.3 + 16.0 = 40.3theoretical yield = 0.1235 × 40.3 = 4.98 gStep 4: percentage yield.percentage yield = (4.50 ÷ 4.98) × 100 = 90.4%Explaining the result: the yield is below 100% because some magnesium may not have reacted completely (a layer of oxide can coat the ribbon), some magnesium can react with nitrogen in the air to form magnesium nitride, and some white oxide smoke escapes when the lid is lifted.Exam tips: keep the unrounded value of the moles in your calculator for the next step, use the balanced equation for the ratio even when it looks obvious, and give the final answer to 3 significant figures to match the data.