Rust occurs on iron when the metal is exposed to oxygen in the air in a process called oxidation. The chemical name for rust is iron (III) oxide. The formula is Fe2O3

Question

. Calculate the percentage composition of iron(III) oxide. Type your answers in the appropriate spaces below. Please carry your answers out to the hundredths place, or two decimal places. The units are designated for you.

(4 points)
% Fe

% O

Answer 1

To calculate the percentage composition of iron(III) oxide (Fe\(_2\)O\(_3\)), we need to determine the molar mass of Fe\(_2\)O\(_3\) and then find the contribution of each element to this mass.

1. Calculate the molar mass of Fe\(_2\)O\(_3\):

   • Atomic mass of iron (Fe): approximately 55.85 g/mol
   • Atomic mass of oxygen (O): approximately 16.00 g/mol

   Molar mass calculation: \[ \text{Molar mass of Fe}_2\text{O}_3 = (2 \times 55.85 , \text{g/mol}) + (3 \times 16.00 , \text{g/mol}) \] \[ = (111.70 , \text{g/mol}) + (48.00 , \text{g/mol}) = 159.70 , \text{g/mol} \]

2. Calculate the percentage composition:

   • Percentage of Iron (Fe): \[ % \text{Fe} = \left( \frac{2 \times 55.85 , \text{g/mol}}{159.70 , \text{g/mol}} \right) \times 100 \] \[ = \left( \frac{111.70 , \text{g/mol}}{159.70 , \text{g/mol}} \right) \times 100 \approx 69.94% \]

   • Percentage of Oxygen (O): \[ % \text{O} = \left( \frac{3 \times 16.00 , \text{g/mol}}{159.70 , \text{g/mol}} \right) \times 100 \] \[ = \left( \frac{48.00 , \text{g/mol}}{159.70 , \text{g/mol}} \right) \times 100 \approx 30.06% \]

3. Final Answers:

   • % Fe = 69.94 %
   • % O = 30.06 %

You can enter these values as requested.