Eggshells contain a high percentage by mass of calcium carbonate, CaCO₃. A student wants to find out what percentage of an eggshell is calcium carbonate and uses the following method. This method uses a known excess of acid to dissolve the eggshell. The amount of unreacted acid is then determined by titration with an alkali. Assume the acid only reacts with the CaCO₃ in the eggshell. step 1 Wash an empty eggshell with distilled water. step 2 Warm the eggshell in an oven for a few minutes until dry. step 3 Grind the eggshell into a powder. step 4 Weigh approximately 2 g of the eggshell powder into a conical flask using a balance which measures to three decimal places. step 5 Add 100 cm³ of 2.00 mol dm⁻³ hydrochloric acid to the conical flask. step 6 Loosely cover the conical flask and leave for two days. step 7 Filter the contents of the conical flask, with any rinsings, into a 250.0 cm³ volumetric flask and top-up to the mark using distilled water. step 8 Transfer 25.00 cm³ of the solution prepared in step 7 into a conical flask, add a few drops of thymol blue indicator and titrate against 1.00 mol dm⁻³ sodium hydroxide using a 50 cm³ burette. The calcium carbonate in the eggshell reacts with the excess hydrochloric acid as follows. CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l) The excess acid reacts with the sodium hydroxide solution as follows. NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)
- (a)(i)Suggest how the student could confirm the eggshell is completely dry in step 2.[1]
- (a)(ii)State why the eggshell is made into a powder in step 3 before making up the solution. Explain your answer.[1]
- (a)(iii)Suggest why the solution is left for two days in step 6 before being used.[1]
- (b)(i)The student uses exactly 2.136g of powdered eggshell and obtains the results shown in Table 1.1. Table 1.1 titration number | rough | 1 | 2 | 3 final burette reading/cm³ | 16.55 | 32.85 | 16.10 | 32.30 initial burette reading/cm³ | 0.00 | 16.55 | 0.10 | 16.10 titre/cm³ Complete Table 1.1. Calculate the mean titre. mean titre = ...................................... cm³[2]
- (b)(ii)Calculate the amount, in mol, of unreacted HCl(aq) in the solution prepared in step 7. Show your working. hydrochloric acid = ...................................... mol[2]
- (b)(iii)Calculate the amount, in mol, of CaCO₃ that reacts with the excess of acid. Use your answer to calculate the percentage by mass of CaCO₃ in the eggshell. Show your working. percentage by mass of CaCO₃ = ...................................... %[3]
- (c)Name a suitable piece of apparatus which could be used to transfer 25.00 cm³ of solution in step 8.[1]
- (d)In step 4, a conical flask is weighed using a balance accurate to three decimal places and the mass recorded. The eggshell is placed in the conical flask and the mass increases by 2.136g. Calculate the percentage error in measuring the mass of this eggshell. Show your working. percentage error = ...................................... %[1]
- (e)State the effect on the percentage by mass if the eggshell is not completely dried in step 2. Explain your answer.[1]
- (f)The student repeats the method using the same apparatus, but decides to use 0.100 mol dm⁻³ NaOH(aq) to reduce the risk of corrosion or damage to eyes. Explain how this introduces a weakness to the experimental procedure.[1]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2022). The mark or mark type is shown in brackets.
- [1] (reheat and) reweigh until mass remains constant
- [1] increases the rate of reaction And due to increase the surface area
- [1] to ensure the reaction is complete
- [2] M1: 16.55 16.30 16.00 16.20 M2: 16.25 cm³
- [2] M1 nNaOH 1.00 x 16.25/1000 = 0.01625 M2 nHCI unreacted = 0.01625 × 250 / 25 = 0.1625
- [3] M1 nHCl reacted [2.00 × 100/1000] – 0.1625 = 0.0375 M2 nCaCO3 reacted = 0.0375/2 = 0.01875 and mass CaCO3 reacted = 0.01875 × 100.1 = 1.876875g M3 %CaCO3 = [1.876875/2.136] × 100 = 87.9%
- [1] (25cm³) volumetric pipette
- [1] ((2 x 0.0005) / 2.136) × 100 (= 0.0468%)
- [1] the % by mass (of CaCO3 in eggshell sample) would be less (than the true value) and the (measured) mass (of eggshell) would be greater (than the true value)
- [1] volume of NaOH would be 162.5 cm³ (which is too large for a 50 cm³ burette)
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