Correct Answer: Option B

Explanation

Calcium trioxocarbonate (IV) = CaCO\(_3\)

  Percentage of Oxygen = \(\frac{\text{Molar mass of 3O}}{\text{Molar mass of} CaCO_3}\) × 100%

  Percentage of Oxygen = \(\frac{(3 \times 16)}{(40 + 12 + 48)}\)  x 100%

  Percentage of Oxygen = \(\frac{48}{100}\) x 100%

  Percentage of Oxygen = 48%

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