Answer :
To determine how many grams of iron are in 225 grams of Fe3O4 (iron(II,III) oxide or magnetite), we must first determine the molar mass of Fe3O4 and the fraction of this molar mass that is due to iron. Then we can calculate the mass of iron in 225 grams of Fe3O4.
Step 1: Calculate the molar mass of Fe3O4.
Fe3O4 is composed of 3 iron (Fe) atoms and 4 oxygen (O) atoms. The molar mass of an element is the mass of one mole of the element's atoms and is given on the periodic table. The molar mass of iron is approximately 55.845 grams per mole (g/mol), and the molar mass of oxygen is approximately 15.999 grams per mole.
Molar mass of Fe3O4 = (3 × molar mass of Fe) + (4 × molar mass of O)
= (3 × 55.845 g/mol) + (4 × 15.999 g/mol)
= 167.535 g/mol + 63.996 g/mol
= 231.531 g/mol
Step 2: Determine the mass fraction of iron in Fe3O4.
Now, we find out the mass fraction of iron in Fe3O4 by dividing the combined molar mass of the iron atoms by the molar mass of Fe3O4.
Mass fraction of iron in Fe3O4 = (combined molar mass of iron atoms) / (molar mass of Fe3O4)
= (3 × molar mass of Fe) / (molar mass of Fe3O4)
= (3 × 55.845 g/mol) / (231.531 g/mol)
= 167.535 g/mol / 231.531 g/mol
= 0.7234
This means that iron makes up approximately 72.34% of the mass of Fe3O4 by mass.
Step 3: Calculate the mass of iron in 225 grams of Fe3O4.
Now, we can simply multiply the mass of Fe3O4 by the mass fraction of iron to get the mass of iron in 225 grams of Fe3O4.
Mass of iron = mass fraction of iron × mass of Fe3O4
= 0.7234 × 225 g
= 162.765 g
Therefore, there are approximately 162.765 grams of iron in 225 grams of Fe3O4.
