Answer
The percent of $Fe_{3}O_4$ in the ore equals to $4.79\%$.
Work Step by Step
We can calculate the mass of $Fe_{3}O_4$ required to obtain $1.56g$ of iron:
$m(Fe_3O_4) = m(Fe)\times \frac{Mr(Fe_3O_4)}{3\times Ar(Fe)}=1.56g\times \frac{232}{3\times 56}=2.15g$
Hence, the percent of $Fe_{3}O_4$ in the ore equals to:
$\alpha = \frac{2.15g}{45g}=4.79\%$
