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

To find the percentage of CaCO₃, first convert the mass of CaO back to the mass of CaCO₃. Given the mass of CaO is 3.87 g and the molar mass of CaO is 56.1 g/mol, we can find the number of moles of CaO:

$n(CaO) = \frac{m(CaO)}{M_{CaO}} = \frac{3.87 \, g}{56.1 \, g \, mol^{-1}} = 0.0690 \, mol$

By stoichiometry, since 1 mol of CaCO₃ produces 1 mol of CaO, $n(CaCO₃) = 0.0690 \, mol$

Now, we calculate the mass of CaCO₃: $m(CaCO₃) = n(CaCO₃) \times M_{CaCO₃} = 0.0690 \, mol \times 100.1 \, g \, mol^{-1} = 6.91 \, g$

Finally, the percentage of CaCO₃ in the rock sample is: $\% \text{CaCO}_3 = \left( \frac{m(CaCO₃)}{m_{sample}} \right) \times 100 = \left( \frac{6.91 \, g}{8.64 \, g} \right) \times 100 \approx 79.8\%$

One possible explanation for the difference in the percentages is that during the chemical analysis with ammonium oxalate, some of the calcium ions might not have precipitated completely, leading to a lower mass of CaO obtained and, consequently, a lower calculated percentage of CaCO₃ in the rock sample.