RI Planning Experiments 2 Tutorial - Ans to Self-check Questions

-15- Solutions to Self-Check Questions Question 1 (a) Cu(OH)2nCuCO3  (n+1) CuO + H2O + nCO2 [1] (b) 1. Using an analytical/weighing balance, we igh and record the mass of a clean, empty and dry crucible. 2. Weigh out accurately about 5.00 g of solid Cu(OH) 2nCuCO3 into the crucible. Record the total mass of the crucible and the solid Cu(OH)2nCuCO3. 3. Using a Bunsen burner, heat the crucible and its contents gently at first, and then heat strongly for 10 minutes. 4. Cool and weigh the crucible and its contents and record the total mass. 5. Repeat the heating-cooling-weighing pr ocess until constant mass is obtained. [1] weigh using an analytical balance and record mass of empty crucible, and total mass of crucible with solid [1] heat crucible and contents gently then strongly using a Bunsen burner [1] allow crucible and contents to cool down before weighing and recording the mass [1] repeat the heating-cooling-weighing process until constant mass is obtained. Tabulation of results Mass of empty crucible / g A Mass of crucible and Cu(OH)2nCuCO3 / g B Mass of crucible and its contents after first heating / g after second heating / g after third heating / g C D D Amount of CuO obtained = DA 63.5 + 16.0 = DA 79.5 mol [1] Amount of Cu in Cu(OH)2nCuCO3 = amount of CuO obtained % by mass of Cu = 63.5 × (DA) 79.5 × (BA) × 100% [1] Question 2 (a) ±0.05 x 2 23.50 x 100% = ±0.426% [1] (b) CO2(g) will dissolve / is soluble in water, resulting in a lower volume of CO2(g) measured. [1] (c) The percentage purity calculated will be lower than expected. [1] The smaller the volume of CO2(g), the smaller the amount of CO2 calculated, which in turn gives a smaller amount of Na2CO3 calculated in FA 1 and hence lower percentage purity. [1] [1] Table must be constructed with correct headers and units.

-16- (d)(i) [1] Collection of gas using a gas syringe [1] FA 1 in small test-tube in conical flask (d)(ii) With a gas syringe of 100 cm3 capacity, assume that 75.0 cm3 of CO2 gas is collected. Amount of CO2 gas = 75.0 / 24000 = 3.125 x 10–3 mol [1] Na2CO3(s) + 2HCl(aq)  CO2(g) + 2NaCl(aq) + H2O(l) Amount of Na2CO3 = 3.125 x 10–3 mol Molar mass of Na2CO3 = 2(23.0) + 12.0 + 3(16.0) = 106.0 g mol–1 Mass of Na2CO3 = (3.125 x 10–3)(106.0) = 0.3313 g Assuming that FA 1 contains 90% by mass of Na2CO3, mass of FA 1 to be used = 0.3313 / 90 x 100 = 0.368 g [1] Question 3 (a) Unlike HC l which is a strong acid, CH 3COOH is a weak acid and does not dissociate completely in aqueous solution. As energy is required to further ionise the unionised CH3COOH, this resulted in a less exothermic enthalpy change of neutralisation. (b) Pre-calculations Let volume of FA 1 or FA 2 used be 40 cm3. n(HCl) = 40/1