A-LevelChemistryAnalytical techniques9701/52 · Feb/Mar 2022 · Q19 Marks

A student has a sample of copper(II) sulfate crystals, CuSO₄•xH₂O. The student wants to show that the value of x is 5. The student uses the following method. step 1 Weigh a clean crucible on a balance reading to two decimal places. Record the mass. step 2 Place the sample of CuSO₄•xH₂O into the crucible. Record the mass. step 3 Heat the crucible gently for about 1 minute then strongly for about 4 minutes. step 4 Weigh the crucible and contents. Record the mass.

(a)Identify the instruction that is missing between step 3 and step 4.[1]
(b)Explain why gentle heating takes place in step 3.[1]
(c)Name the apparatus that should be used to hold the crucible during heating.[1]
(d)The method is incomplete. State the step(s) that should be carried out to complete the method.[1]
(e)(i)The student records their results in Table 1.1. Table 1.1 | | mass/g | |---|---| | mass of crucible | 13.60 | | mass of crucible + contents before heating | 21.09 | | mass of crucible + contents at the end of experiment | 17.94 | Calculate the experimental value of x from these results. experimental value of x =[3]
(e)(ii)Suggest why the experimental value of x varies from the expected value of 5. If you were unable to obtain an answer to (e)(i), use the experimental value x = 6.9. This is not the correct answer.[1]
(f)The empty crucible weighs 13.60g. Calculate the percentage error in this measurement. Show your working. percentage error =[1]

Mark scheme

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

[1] allow (crucible / contents) to cool
[1] to prevent spitting of crystals
[1] pipe-clay triangle
[1] heat to constant mass
[3] M1: (mass H2O) = 3.15 (g) AND (mass CuSO4) = 4.34 (g)
M2: n(H2O) = 3.15 / 18 = 0.175 AND n(CuSO4) = 4.34 / 159.6 = 0.02719...
M3: 0.02719.../0.02719... AND 0.175/0.02719... = 6.44
[1] If value of x in 1(e)(i) is above 5: CuSO4 must have decomposed / crystals were wet before heating / some crystals were lost when heated (e.g. during spitting) /impurity that decomposes or evaporates
If value of x in 1(e)(i) is below 5: did not heat it enough / not all water driven off/ unreactive impurity
[1] [(2 × 0.005) / 13.60] × 100 = 0.0735 (%)

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