A-LevelChemistryAnalytical techniques9701/33 · Oct/Nov 2022 · Q210 Marks

2 In this experiment you will determine the percentage by mass of magnesium in a hydrated salt, MgX•7H2O, where X represents the anion. You will measure the loss of mass when a sample of the hydrated salt is heated to form the anhydrous salt. FA 5 is a pure sample of MgX•7H2O

(a)Method • Weigh the crucible with its lid. Record the mass. • Add all of FA 5 to the crucible. • Reweigh the crucible with its lid and FA 5. Record the mass. • Support the crucible in the pipe-clay triangle on top of the tripod. • Heat the crucible with the lid on gently for about 1 minute. • Remove the lid and then heat strongly for a further 4 minutes. • Replace the lid and allow the crucible to cool for at least 5 minutes. While the crucible is cooling you may wish to begin work on Question 3. • When the crucible is cool enough to handle, reweigh the crucible with its lid and its contents. Record the mass. : Calculate and record the mass of FA 5 used. : Calculate and record the mass of water lost.[5]
(b)(i)Calculate the amount, in mol, of water lost. amount of H2O lost = ...................................... mol [1][1]
(b)(ii)Calculate the percentage by mass of magnesium in FA 5. Show your working. percentage by mass of Mg = .......................................... % [2][2]
(c)Suggest two assumptions that must be made for this experiment to give an accurate value of the percentage of Mg. [2][2]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2022). The mark or mark type is shown in brackets.

[5] I table of data must show all of the following: • (mass of) crucible + lid • (mass of) crucible + lid + FA 5 • (mass of) crucible + lid + residue / contents / MgX • (mass of) FA 5 • (mass of) water lost / mass lost Il recording of data • unit / g, (g) or in grams is covering each entry • all three balance readings are recorded to the same number of decimal points (2 or 3 dp). III correctly calculates • mass of FA 5 • mass of water lost IV, V award IV if δ < 0.20 award V if δ ≤ 0.10 where δ is the difference between the supervisor's and candidate's mass ratio = mass of FA 5 / mass of water lost and calculated correct to 2 dp the theoretical value of 1.96 is used if there is no supervisor's value available
[1] correctly calculates amount of H2O lost = mass of water lost / 18 (mol) AND answer to 2 – 4 sf
[2] correctly uses M1: amount of Mg = (b)(i) / 7 (mol) M2: percentage of Mg = [{(b)(i) x 24.3} / (7 × mass of FA 5)] × 100 (%) AND answer to 2 – 4 sf
[2] M1: all the water is lost. OR heating was for long enough to reach constant mass. (owtte) M2: MgX / residue is thermally stable / does not decompose. OR the mass loss is due to the water only.

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