Final answer:
To find the mass of oxygen in 15.5 g of Al(NO3)3, we calculate the molar mass of the compound and the fraction of this mass that is oxygen, then apply this fraction to the given mass. The calculation yields approximately 10.48 g of oxygen.
Step-by-step explanation:
To calculate the mass of oxygen contained in 15.5 g of Al(NO3)3, we must know the molar mass of both the compound and the oxygen element in it. Aluminum nitrate, Al(NO3)3, consists of one aluminum atom, three nitrogen atoms, and nine oxygen atoms. The molar mass of Al(NO3)3 is the sum of the molar masses of its constituent atoms.
The molar mass of aluminum (Al) is about 26.98 g/mol, that of nitrogen (N) is about 14.01 g/mol, and that of oxygen (O) is 16.00 g/mol. Thus, the molar mass of Al(NO3)3 can be calculated as:
(1 × 26.98) + (3 × 14.01) + (9 × 16.00) = 212.98 g/mol.
Next, we consider the proportion of the molar mass that is due to oxygen, which is (9 × 16.00) = 144 g/mol. The fraction of the compound's mass that is oxygen is then:
144 g/mol ÷ 212.98 g/mol = 0.676 (approximately).
To find the mass of oxygen in 15.5 g of Al(NO3)3, we multiply the total mass of the compound by the fraction that is oxygen:
15.5 g × 0.676 = 10.48 g.
Therefore, the mass of oxygen in 15.5 g of Al(NO3)3 is about 10.48 g.