To determine the number of atoms of oxygen present in 300 grams of calcium carbonate, we need to use the molar mass and the stoichiometry of the compound.
The chemical formula of calcium carbonate is CaCO₃. It consists of one calcium atom (Ca), one carbon atom (C), and three oxygen atoms (O).
Calculate the molar mass of calcium carbonate:
- Molar mass of Ca = 40.08 g/mol
- Molar mass of C = 12.01 g/mol
- Molar mass of O = 16.00 g/mol
- Molar mass of CaCO₃ = (1 × 40.08 g/mol) + (1 × 12.01 g/mol) + (3 × 16.00 g/mol) = 100.09 g/mol
Determine the number of moles of calcium carbonate in 300 grams:
- Moles of CaCO₃ = Mass of CaCO₃ / Molar mass of CaCO₃
- Moles of CaCO₃ = 300 g / 100.09 g/mol = 2.997 mol (approximately)
Use the stoichiometry of the compound to find the number of moles of oxygen:
- For every 1 mole of CaCO₃, there are 3 moles of oxygen.
- Moles of oxygen = Moles of CaCO₃ × 3
- Moles of oxygen = 2.997 mol × 3 = 8.991 mol (approximately)
Convert the number of moles of oxygen to atoms:
- Avogadro's number states that 1 mole of any substance contains 6.022 × 10^23 entities (atoms, molecules, etc.).
- Number of atoms of oxygen = Moles of oxygen × Avogadro's number
- Number of atoms of oxygen = 8.991 mol × (6.022 × 10^23 atoms/mol) = 5.419 × 10^24 atoms (approximately)
Therefore, there are approximately 5.419 × 10^24 atoms of oxygen in 300 grams of calcium carbonate.