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)3233.99 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

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Mole Concept concept. You can view video lessons to learn Mole Concept. Or if you need more Mole Concept practice, you can also practice Mole Concept practice problems.