Calcium trioxocarbonate (IV) = CaCO3_{3}3
Percentage of Oxygen = Molar mass of 3OMolar mass of CaCO3\frac{\text{Molar mass of 3O}}{\text{Molar mass of }\mathrm{CaCO}_3}Molar mass of CaCO3Molar mass of 3O × 100%
Percentage of Oxygen = (3×16)(40+12+48)\frac{(3 \times 16)}{(40 + 12 + 48)}(40+12+48)(3×16) x 100%
Percentage of Oxygen = 48100\frac{48}{100}10048 x 100%
Percentage of Oxygen = 48%
