Question

A solid mixture weighing 0.05466 g contained only ferrous ammonium sulfate (FeSO4 · (NH4)2SO4 · 6 H2O, FM 392.15) and ferrous chloride (FeCl2 · 6 H2O, FM 234.84). The sample was dissolved in 1 M H2SO4, and the Fe2+ required 13.28 mL of 0.01274 M Ce4+ for complete oxidation to Fe3+ (Ce4+ + Fe2+ → Ce3+ + Fe3+). Calculate the weight percent of Cl in the original sample

Answer 1

Let x g of ferrous chloride (FeCl2 · 6 H2O, FM 234.84) are present which corresponds to moles of ferrous ions.

The solid mixture weighs 0.05466 g. The mass of ferrous ammonium sulfate (FeSO4 · (NH4)2SO4 · 6 H2O, FM 392.15) will be 0.05466 - x g which corresponds to moles of ferrous ions.

Total number of moles of ferrous ions ......(2)

The number of moles of ferrous ions are equal to the number of moles of Ce(IV) ions which are equal to

......(1)

From (1) and (2)

The number of moles of ferrous chloride (FeCl2 · 6 H2O, FM 234.84)

The number of moles of chloride ions

The atomic mass of chlorine is 35.5 g/mol. THe mass of chloride ions = g.

the weight percent of Cl in the original sample %