1. Find the molar mass of aluminum carbonate (Al₂(CO₃)₃):

* Aluminum (Al): 26.98 g/mol

* Carbon (C): 12.01 g/mol

* Oxygen (O): 16.00 g/mol

* Molar mass of Al₂(CO₃)₃ = (2 * 26.98) + (3 * 12.01) + (9 * 16.00) = 233.99 g/mol

2. Calculate the moles of aluminum carbonate:

* Moles = mass / molar mass

* Moles = 47.6 g / 233.99 g/mol = 0.203 moles Al₂(CO₃)₃

3. Determine the moles of oxygen atoms:

* Each mole of Al₂(CO₃)₃ contains 9 moles of oxygen atoms.

* Moles of oxygen = 0.203 moles Al₂(CO₃)₃ * 9 moles O / 1 mole Al₂(CO₃)₃ = 1.83 moles O

4. Convert moles of oxygen to atoms of oxygen:

* Avogadro's number: 6.022 x 10²³ atoms/mol

* Atoms of oxygen = 1.83 moles O * 6.022 x 10²³ atoms/mol = 1.10 x 10²⁴ atoms O

Therefore, there are approximately 1.10 x 10²⁴ atoms of oxygen in 47.6g of aluminum carbonate.