Answer:
44.91% of Oxygen in Iron (III) hydroxide
Step-by-step explanation:
To solve this question we must find the molar mass of Fe(OH)3 and the molar mass of the oxygen in this molecule. Percent composition will be:
Molar mass Oxygen / molar mass Fe(OH)3 * 100
Molar mass Fe(OH)3 and oxygen:
1Fe = 55.845g/mol*1 = 55.845
3O = 16.00g/mol*3 = 48.00 - Molar mass of Oxygen
3H = 1.008g/mol*3 = 3.024
55.845 + 48.00 + 3.024 =
106.869g/mol is molar mass of Fe(OH)3
% Composition of oxygen is:
48.00g/mol / 106.869g/mol * 100 =
44.91% of Oxygen in Iron (III) hydroxide