Question 111 marks
Question
Potassium alum is a hydrated salt containing aluminium ions, potassium ions and sulfate ions. 1 mol of hydrated potassium alum contains 12 mol of water of crystallisation. Hydrated potassium alum decomposes when heated, losing its water of crystallisation and becoming anhydrous. You will determine the formula of potassium alum by heating the hydrated salt until it becomes anhydrous. FA 1 is hydrated potassium alum. (a) Method • Weigh a crucible with its lid. Record the mass in the space for results. • Add all of the FA 1 to the crucible. • Weigh the crucible and lid with FA 1. Record the mass. • Calculate and record the mass of FA 1 added. • Place the crucible on the pipe-clay triangle. Heat the crucible and contents gently for approximately 2 minutes with the lid on. • Remove the lid. Heat the crucible and contents strongly for approximately 5 minutes. • Replace the lid and leave the crucible and residue to cool for at least 5 minutes. While the crucible is cooling, you may begin work on Question 2 or Question 3. • Reweigh the crucible and contents with the lid on. Record the mass. • Remove the lid. Heat the crucible and contents strongly for a further 2 minutes. • Replace the lid and leave the crucible and residue to cool for at least 5 minutes. • Reweigh the crucible and residue with the lid on. Record the mass. • Calculate and record the mass of residue obtained. Results I II III IV V [5] (b) Calculations (i) Calculate the amount, in mol, of water of crystallisation lost during the thermal decomposition of FA 1. amount of H2O lost = …… mol [1] (ii) Use the information given and your answer to (b)(i) to determine the amount, in mol, of potassium alum used. amount of potassium alum = …… mol [1] (iii) Calculate the relative formula mass, Mr, of anhydrous potassium alum. Mr = …… [1] (iv) Anhydrous potassium alum contains aluminium ions, potassium ions and sulfate ions. 1 mol of potassium alum also contains 1 mol of aluminium ions. Use the Mr you have calculated in (b)(iii) to suggest the formula of anhydrous potassium alum. Show your working. The formula = …… . [1] (c) (i) The uncertainty in a single balance reading for a two decimal place balance is 0.01 g. Calculate the maximum percentage error in your measurement of the mass of the residue of anhydrous potassium alum. Show your working. maximum percentage error = ……% [1] (ii) A student obtains a higher value for the relative formula mass, Mr, of anhydrous potassium alum than expected. The student incorrectly suggests that this is because some of the anhydrous potassium alum residue decomposes to aluminium oxide and potassium oxide during strong heating. Explain why the student’s suggestion is not correct. …… [1] [Total: 11]
Mark scheme
1(a) I Six unambiguous headings for balance readings AND calculated values AND correct units 5 • (mass of) crucible (and lid) • (mass of) crucible (and lid) and FA 1 • (mass of) crucible (and lid) and residue / contents after first heating • (mass of) crucible (and lid) and residue / contents after second heating • (mass of) FA 1 • (mass of) residue • Units: / g, (g), in g (or g with every entry) II Recording of all weighings • All four weighings recorded to same number of decimal places (two or three). • Fourth reading is within +0.02 and –0.05 g of third reading. III Correct calculation of masses Correct masses of FA 1 and residue AND both answers to min 2 sf. Accuracy marks Calculate candidate’s mass ratio (to 2 d.p.) = mass FA 1 / mass of residue. IV award if ratio is within the range 1.70–2.00 (inclusive) V award if ratio is within the range 1.80–1.90 (inclusive) 1(b)(i) Correct calculation of amount of water 1 amount of H2O = (mass of FA 1 – mass of residue) / 18 AND answer to 2–4 s.f. 1(b)(ii) Correct use of amount of water 1 amount of potassium alum, FA 1 = (b)(i) / 12 AND answer to 2–4 s.f. 1(b)(iii) Correct use of amount of anhydrous potassium alum 1 Mr of anhydrous alum = mass of residue / (b)(ii) AND answer to 2–4 s.f. 1(b)(iv) Formula is Al K(SO4)2 1 AND (correct) working which shows use of (b)(iii) AND use of 27(.0) or 258.3. 1(c)(i) Correct expression 1 (Uncertainty U (for 2 d.p. balance) = 0.01) % error in weighing FA 1 = 2 × U / mass of residue × 100 AND answer given. 1(c)(ii) If anhydrous potassium alum decomposes: 1 mass loss will be higher / too high OR no longer is only water lost OR (calculated) moles of water / (b)(i) will be higher AND (calculated) amount / moles of residue / (anhydrous) potassium alum / (b)(ii) will be higher / too high. (so Mr of residue / (b)(iii) will be too low).