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

FREE Expert Solution

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:


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Problem Details

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

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