Question

You wish to prepare 250.0250.0 mL of a 400.0400.0 ppm w/v Fe2+Fe2+ ( MW=55.845 g/molMW=55.845 g/mol ) solution. How many grams of ferrous ammonium sulfate hexahydrate (Fe(NH4)2(SO4)2⋅6H2O(Fe(NH4)2(SO4)2⋅6H2O , MW=392.14 g/molMW=392.14 g/mol ) are needed to prepare this solution? Assume the final solution has a density of 1.00 g/mL.

Answer 1

You need 0.702g of ferrous ammonium sulfate hexahydrate to prepare the desired solution.

Step-by-step explanation:

First, you want to prepare 250.0mL of a 400.0ppm of Fe²⁺ (That is mg/L, w/v):

0.250L * (400mg / L) = 100mg Fe²⁺ = 0.1g Fe²⁺

Molar mass of Fe is 55.845g. moles are:

0.1g Fe²⁺ * (1mol / 55.845g) =

1.79x10⁻³ moles Fe²⁺

The Fe²⁺ comes from ferrous ammonium sulfate hexahydrate, as 1 mole of this salt contains 1 mole of Fe²⁺, moles of the salt you need are:

1.79x10⁻³ moles Fe(NH₄)₂(SO₄)₂.6H₂O.

To convert these moles to grams you must use molar weight, MW, thus:

1.79x10⁻³ moles Fe(NH₄)₂(SO₄)₂.6H₂O * (392.14g / 1mol) =

You need 0.702g of ferrous ammonium sulfate hexahydrate to prepare the desired solution.