How many atoms of oxygen are contained in 47.6 g of Al2(CO3)3? (Molar mass = 234 g/mol)

How many atoms of oxygen are contained in 47.6 g of Al2(CO3)3? (Molar mass = 234 g/mol)

A. 1.23 × 10^23 O atoms

B. 2.96 × 10^24 O atoms

C. 2.87 × 10^25 O atoms

D. 1.10 × 10^24 O atoms

E. 3.68 × 10^23 O atoms

Solution: We are given mass of Al2(CO3)3 and need to find the number of atoms.

Plan of action: Convert mass to moles using the molar mass given. Convert moles to molecules using Avogadro's number. Convert molecules of Al2(CO3)3 to atoms of oxygen.

47.6 g of Al2(CO3)3 * 1mol Al2(CO3)3 * 6.02*10^23 molecules Al2(CO3)3 * 9 atoms O

234g Al2(CO3)3 1 mol Al2(CO3)3 1 molecule Al2(CO3)3

D. 1.10 × 10^24 O atoms is the correct answer