Calculate the percentage composition of oxygen in calcium trioxocarbonate(IV) [Ca=40, C=12, O=16]
- 16
-
48
- 40
- 12
The correct answer is: 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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