Answer
Mass percent of Fe = $80.51$%
Mass percent of Fe$_{2}$O$_{3}$ = $19.49$%
Work Step by Step
Using the chemical equation: MnO$_{4}$$^{-} + 5$Fe$^{2+} +8$H$_{3}$O$^{+}$ $\rightarrow$ Mn$^{2+} + 5$Fe$^{3+} + 12$H$_{2}$O
Given that we used $37.50mL$ of $0.04240M$ of KMnO$_{4}$
$\frac{0.03750L}{1}\times\frac{0.04240 mol\space KMnO_{4}}{1 L}\times\frac{5 mol\space Fe^{2+} }{1mol\space KMnO_{4}}\times\frac{1mol\space Fe}{1 mol\space Fe^{2+}}\times\frac{55.8g\space Fe}{1 mol \space Fe} = 0.44361g\space Fe$
Using this we can calculate the mass percent of Fe in the sample:
$\frac{ 0.44361g\space Fe}{0.5510g\space sample}\times100 = 80.51$% of Fe
Now we can subtract the mass percent of Fe from 100% to get the mass percent of Fe$^{3+}$:
$100$% - $80.51$% = $19.49$% of Fe$^{3+}$ or Fe$_{2}$O$_{3}$
