Answer is: empirical formula is Fe₂O₃.
m(Fe) = 7,50 g.
m(iron oxide) = 10,71 g.
n(Fe) = m(Fe) ÷ M(Fe).
n(Fe) = 7,50 g ÷ 55,85 g/mol = 0,134 mol.
m(O) = m(iron oxide) - m(Fe).
m(O) = 10,71 g - 7,50 g = 3,21 g.
n(O) = 3,21 g ÷ 16 g/mol = 0,20 mol.
n(Fe) : n(O) = 0,134 mol : 0,2 mol = 2 : 3.
