Edexcel IGCSEChemistryGeneral Applications11 Marks
This question is about the determination of the formula of a metal oxide. A student heats a known mass of magnesium in a crucible. The student records the mass of the magnesium and the mass of the magnesium oxide formed. [Figure 4 referenced - implicitly, this experiment often uses a crucible and lid, but no diagram is shown in the question paper for Q17 directly, rather the description is given.]
- (a)(i)State one safety precaution the student should take during this experiment.[1]
- (a)(ii)Explain why it is necessary to lift the lid of the crucible occasionally during heating.[2]
- (a)(iii)Explain why the student heats the crucible to constant mass.[2]
- (b)(i)A student performs the experiment using zinc. The student records these results. Mass of empty crucible = 15.36 g Mass of crucible + zinc = 16.01 g Mass of crucible + zinc oxide = 16.41 g Calculate the mass of zinc used.[1]
- (b)(ii)Calculate the mass of oxygen that reacted.[1]
- (b)(iii)Calculate the amount in moles of zinc atoms that reacted. [Aᵣ: Zn = 65, O = 16][1]
- (b)(iv)Calculate the amount in moles of oxygen atoms that reacted. [Aᵣ: Zn = 65, O = 16][1]
- (b)(v)Deduce the empirical formula of zinc oxide.[2]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme. The mark or mark type is shown in brackets.
- B1 for (a)(i) wear eye protection / safety goggles / use fume cupboard / avoid contact with product
- M1 for (a)(ii) (to allow) oxygen / air (to enter)
- A1 for (a)(ii) (to allow) magnesium to react / burn / oxidise (completely)
- M1 for (a)(iii) (to ensure) all magnesium has reacted / oxidised / no unreacted magnesium
- A1 for (a)(iii) (or to ensure) all water has been removed (if magnesium has been exposed to air) / all product formed / all reaction has finished
- B1 for (b)(i) 16.01 – 15.36 = 0.65 (g)
- B1 for (b)(ii) 16.41 – 16.01 = 0.40 (g)
- B1 for (b)(iii) 0.65 / 65 = 0.010 (mol)
- B1 for (b)(iv) 0.40 / 16 = 0.025 (mol)
- M1 for (b)(v) Zn : O = 0.010 : 0.025 = 1 : 2.5 (or correct ratio based on ECF)
- A1 for (b)(v) Zn₂O₅ (or correct empirical formula based on ECF)
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