Question

A student weighed 0.2053 g of ferrous ammonium sulfate hexahydrate into a small beaker, dissolved the solid in deionized water, performed a quantitative transfer into a 50-mL volumetric flask, added 5 drops of 6 M sulfuric acid, and diluted to the mark with deionized water. What was the iron concentration in this solution in mg/L? Report your answer to 1 decimal place and do not report units.

Answer 1

ferrous ammonium sulfate hexahydrate = Fe(NH₄)₂(SO₄)₂.6H₂O

Molar mass of ferrous ammonium sulfate hexahydrate = 392.13 g/mole

Thus, moles of ferrous ammonium sulfate hexahydrate in 0.2053 g of it = mass/molar mass = 5.22*10^-4

Thus, moles of Fe²⁺ = moles of ferrous ammonium sulfate hexahydrate = 5.22*10^-4

Now, molar mass of Fe = 56 g/mole

THus, mass of Fe²⁺ = moles*molar mass = 0.029 g = 29 mg

Thus, concentration of Fe²⁺ = mass/volume of solution in litres = 29/0.05584.9 mg/L