What is the percentage composition of the iron in Fe(C2H3O2)2?
The molar mass of Fe(C2H3O2)2 is:
Fe: 1 x 55.85 = 55.85 g/mol
C: 2 x 12.01 = 24.02 g/mol
H: 6 x 1.01 = 6.06 g/mol
O: 4 x 16.00 = 64.00 g/mol
Total: 149.93 g/mol
The mass of iron in one mole of Fe(C2H3O2)2 is 55.85 g/mol.
The percent composition of iron in Fe(C2H3O2)2 is:
(55.85 g/mol / 149.93 g/mol) x 100% = 37.25%
To determine the percentage composition of iron (Fe) in Fe(C2H3O2)2, we need to calculate the mass of iron in one mole of the compound and then divide it by the molar mass of the compound.
The molar mass of Fe(C2H3O2)2 is calculated as follows:
Iron (Fe) has a molar mass of 55.845 g/mol.
Acetic acid (C2H3O2) has a molar mass of 60.052 g/mol, which consists of 2 carbon atoms (2 * 12.01 g/mol), 4 hydrogen atoms (4 * 1.008 g/mol), and 2 oxygen atoms (2 * 16.00 g/mol).
Thus, the molar mass of Fe(C2H3O2)2 is:
55.845 g/mol (Fe) + 2 * (2 * 12.01 g/mol (C) + 4 * 1.008 g/mol (H) + 2 * 16.00 g/mol (O)) = 183.1 g/mol.
To determine the mass of iron in one mole of Fe(C2H3O2)2, we look at the molecular formula, which shows that there is one iron atom per formula unit of Fe(C2H3O2)2. Therefore, the mass of iron in one mole of Fe(C2H3O2)2 is 55.845 g/mol.
Finally, to find the percentage composition of iron, we divide the mass of iron (55.845 g/mol) by the molar mass of Fe(C2H3O2)2 (183.1 g/mol) and multiply by 100:
(55.845 g/mol / 183.1 g/mol) * 100 = 30.5%.
Therefore, the percentage composition of iron in Fe(C2H3O2)2 is 30.5%.