Problem: How many atoms of oxygen are contained in 47.6 g of Al2(CO3)3? The molar mass of Al2(CO3)3 is 233.99 g/mol. A. 1.23 × 1023 O atoms B. 2.96 × 1024 O atoms C. 2.87 x 1025 O atoms D. 1.10 x 1024 O atoms E. 3.19 x 1024 O atoms

For this problem, we’re being asked to determine the number of oxygen (O) atoms contained in 47.6 g of Al­₂(CO₃)₃. The flow for this problem will be like this:

Mass of Al­₂(CO₃)₃ (molar mass of Al­₂(CO₃)₃) → Moles of Al­₂(CO₃)₃ (mole-to-mole comparison) → Moles of O (Avogadro’s number) → Atoms of O

• Mass Al­₂(CO₃)₃ = 47.6 g

• Molar mass Al­₂(CO₃)₃ = 233.99 g/mol

• Mole-to-mole comparison: From the molecular formula of butane, Al­₂(CO₃)₃, we can do a mole-to-mole comparison and see that there are 9 moles of O in 1 mole of Al­₂(CO₃)₃.

• Avogadro’s Number: Avogadro’s number gives us the number of entities present in 1 mole of anything: 6.022 × 10²³ O atoms in 1 mole O.

Now that we know the values we need, we can now determine the number of O atoms: