Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are 69.9 and 30.1 respectively

General
1

determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are 69.9 and 30.1 respectively

2

Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are 69.9 and 30.1 respectively

Answer: To determine the molecular formula of an oxide of iron given the mass percent of iron (Fe) and oxygen (O), we need to follow these steps:

  1. Calculate the mass of each element in 100 grams of the compound.

    • Mass of iron (Fe) = 69.9 grams
    • Mass of oxygen (O) = 30.1 grams

  2. Convert these masses to moles by dividing by their respective atomic masses.

    • Molar mass of Fe = 55.845 g/mol
    • Molar mass of O = 16.00 g/mol
    \text{Moles of Fe} = \frac{69.9 \text{ g}}{55.845 \text{ g/mol}} \approx 1.252 \text{ moles}
    \text{Moles of O} = \frac{30.1 \text{ g}}{16.00 \text{ g/mol}} \approx 1.881 \text{ moles}

  3. Find the simplest whole number ratio of the moles of Fe to O.

    • Simplify the mole ratio by dividing by the smallest number of moles:
    \text{Ratio of Fe to O} = \frac{1.252}{1.252} : \frac{1.881}{1.252} \approx 1 : 1.503
    • To convert this to the nearest whole number ratio, multiply both parts of the ratio by 2 (since 1.503 is approximately \dfrac{3}{2}):
    1 \times 2 : 1.503 \times 2 \approx 2 : 3

  4. Determine the empirical formula based on this ratio.

    • The simplest whole number ratio of Fe to O is 2:3, so the empirical formula is Fe_2O_3.

  5. Verify the mass percentage from the empirical formula to ensure accuracy.

    • Molar mass of Fe_2O_3:

      2 \times 55.845 \text{ g/mol} + 3 \times 16.00 \text{ g/mol} = 111.69 \text{ g/mol} + 48.00 \text{ g/mol} = 159.69 \text{ g/mol}

    • Mass percentage of Fe in Fe_2O_3:

      \left(\frac{111.69 \text{ g/mol}}{159.69 \text{ g/mol}}\right) \times 100 \approx 69.9\%

    • Mass percentage of O in Fe_2O_3:

      \left(\frac{48.00 \text{ g/mol}}{159.69 \text{ g/mol}}\right) \times 100 \approx 30.1\%

Thus, the calculated mass percentages confirm the empirical formula is accurate.

Conclusion: The molecular formula of the iron oxide in which the mass percent of iron is 69.9% and the mass percent of oxygen is 30.1% is \boxed{Fe_2O_3}.