Question

The iron (55.847 g/mol) in a 600.0 mL sample of a natural water was determined by precipitation of the cation as hydrated Fe2O3. The precipitate was filtered, washed, and ignited in a crucible to form Fe2O3 (MW = 159.69 g/mol). The mass of the product was 0.1245 g. Calculate the ppm iron in the water.

Remember that there are 2 moles of Fe in each mole of Fe2O3 and that ppm is the same as mg/L for dilute aqueous solutions.

Answer 1

Fe₂O₃ molar mass = 159.69 g/mol

Fe₂O₃ mass obtained = 0.1245 g.

So, number of moles of Fe₂O₃ = (0.1245/159.69) mol = 7.8 x 10^-4 mol.

Now, In 1 mole of Fe₂O₃ of 2 moles of Fe is present.

In 7.8 x 10^-4 mole of Fe₂O₃ of 1.56 x 10^-3 moles of Fe is present.

Again,

Mass of 1 mole of Fe = 55.847 g

So, mass of 1.56 x 10^-3 moles of Fe = (1.56 x 10^-3 x 55.847) g = 8.71 x 10^-2 g = 87.1 mg

Now, in 600 mL (=0.6 L) of sample contains 87.1 mg of iron .

So, in 1L of sample iron content will be (87.1/0.6) mg = 145.17 mg of iron.

Hence 145.17 ppm iron in the water.