Question

A 0.6505 g sample of ferrite ore is dissolved in nitric acid. All of the iron present is oxidized to iron(III). The solution is filtered and made basic with addition of ammonium hydroxide. The iron precipitates as the iron(III) hydroxide hydrated solid. The precipitate is collected in a crucible and ignited to produce Fe2O3. What is the mass % of iron in the sample if the analysis produced 0.3010 g Fe2O3?

Answer 1

Molar mass of Fe2O3 = 159.69 g/mol
mass of Fe2O3 = 0.3010 g
number of moles of Fe2O3 = mass/ molar mass
                                                        = 0.3010/159.69
                                                         =1.885*10^-3 mol
Since 1 mol of Fe2O3 has 2 moles of Fe3+,
Number of moles of Fe3+ = 2*number of moles of Fe2O3
            = 2*1.885*10^-3 mol
            =3.77*10^-3 mol
So number of moles of iron present in original sample = 3.77*10^-3 mol
Molar mass of iron = 56 g
Mass of iron present in original sample = number of moles * molar mass
                                                                                 = 3.77*10^-3 mol * 56
                                                                                  =0.2111 g
mass % of iron = mass of iron * 100 / total mass
                               = 0.2111 * 100 / 0.6505
                                = 32.5 %
Answer: 32.5 %