Calculate the percentage composition of oxygen in calcium trioxocarbonate(IV) [Ca=40, C=12, O=16]

  • A 16
  • B 48
  • C 40
  • D 12
jamb 2018chemistry

The correct answer is B. 48

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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