Final answer:
To find the percent composition of oxygen in Manganese (III) nitrate, Mn(NO3)3, we need to determine the mass of oxygen in the compound and divide it by the molar mass of the compound. Manganese (III) nitrate consists of one manganese ion (Mn3+) and three nitrate ions (NO3-). The percent composition of oxygen in Mn(NO3)3 is 16.45%.
Step-by-step explanation:
To find the percent composition of oxygen in Manganese (III) nitrate, Mn(NO3)3, we need to determine the mass of oxygen in the compound and divide it by the molar mass of the compound.
Manganese (III) nitrate consists of one manganese ion (Mn3+) and three nitrate ions (NO3-). Each nitrate ion contains one oxygen atom, so there are three oxygen atoms in the compound.
The molar mass of Mn(NO3)3 can be calculated by adding the atomic masses of manganese (Mn), nitrogen (N), and oxygen (O). The atomic masses are Mn = 54.94 g/mol, N = 14.01 g/mol, and O = 16.00 g/mol. So the molar mass of Mn(NO3)3 is 54.94 + (3 * 14.01) + (3 * 16.00) = 291.85 g/mol.
The mass of oxygen in the compound is (3 * 16.00) = 48.00 g/mol. To determine the percent composition of oxygen, we divide the mass of oxygen by the molar mass of the compound and multiply by 100:
Percent composition of oxygen = (48.00 g/mol / 291.85 g/mol) * 100 = 16.45%