Some rocks were thought to consist of insoluble silica (SiO2) and calcium carbonate (CaCO3; molar mass 100.1 g mol-1) - VCE - SSCE Chemistry - Question 3 - 2007 - Paper 1

Using the mass of CaO obtained after heating:

1. Calculate the number of moles of CaO:

   [n(CaO) = \frac{3.87 \text{ g}}{56.1 \text{ g mol}^{-1}} \approx 0.0690 \text{ mol}]

2. Knowing that each mole of CaO corresponds to one mole of CaCO3, we can find the moles of CaCO3: [n(CaCO3) = 0.0690 \text{ mol}]

3. Calculate the mass of CaCO3:

   [m(CaCO3) = n(CaCO3) \times 100.1 \text{ g mol}^{-1} = 0.0690 \times 100.1 \approx 6.91 \text{ g}]

4. Finally, we can calculate the percentage of CaCO3 in the original rock sample:

   [% \text{ of CaCO3} = \left( \frac{6.91}{8.64} \right) \times 100 \approx 79.9%]

One possible explanation for the difference in percentages calculated in parts a. and b. is that the ammonium oxalate method may not have precipitated all the calcium ions present in the original sample. Some calcium may have remained in solution or been lost during the filtration process, leading to a lower measured amount of CaCO3 compared to the expected value derived from the direct calculation.