In one analysis the mass of dolomite used was 3.72 g - VCE - SSCE Chemistry - Question 8 - 2005 - Paper 1

Since each mole of calcium carbonate (CaCO₃) produces one mole of calcium oxide (CaO), the moles of CaCO₃ is also 0.0221 moles.

The molar mass of calcium carbonate (CaCO₃) is approximately 100.09 g/mol. Thus, the mass of CaCO₃ is calculated as:

$\text{mass of CaCO₃} = \text{moles of CaCO₃} \times \text{molar mass of CaCO₃} = 0.0221\, moles \times 100.09\, g/mol \approx 2.21\, g$

To find the percentage of calcium carbonate in the dolomite sample, use the formula:

$\text{percentage of CaCO₃} = \left( \frac{\text{mass of CaCO₃}}{\text{mass of dolomite}} \right) \times 100 = \left( \frac{2.21\, g}{3.72\, g} \right) \times 100 \approx 59.5\%$

Thus, the percentage of calcium carbonate in the dolomite sample is closest to option D: 59.5.