O-Level Chemical Calculations: worked solution

Question

Given that the measured (actual) mass of calcium carbonate obtained is 1.32 g and its theoretical mass is 1.25 g, calculate the percentage yield of calcium carbonate. Express your answer to three significant figures. [3]

Step 1 - recall what percentage yield compares. It compares the mass you actually got with the mass the equation predicts you should get: percentage yield = (actual yield / theoretical yield) x 100% Step 2 - pick out each quantity from the question. Actual (measured) mass of calcium carbonate = 1.32 g Theoretical mass of calcium carbonate = 1.25 g Step 3 - substitute. Both masses are in grams, so the units cancel and no conversion is needed. percentage yield = (1.32 g / 1.25 g) x 100% Step 4 - divide first, then multiply by 100. 1.32 / 1.25 = 1.056 1.056 x 100 = 105.6 Step 5 - round to three significant figures, as the question demands. 105.6 has four significant figures: 1, 0, 5 and 6. The fourth figure is 6, which is 5 or more, so the third figure (5) rounds up to 6. 105.6% = 106% (3 s.f.) A yield above 100% is not an arithmetic slip: it means the solid weighed was still damp or contained an impurity, so it weighed more than pure calcium carbonate would.