Chemistry Analytical

Question

A standard solution for Fe²⁺ determination can be prepared from ferrous ammonium sulfate (Fe(NH₄)₂(SO₄)2^.6H₂O).

Calculate the mass of Fe in 136 mg of ferrous ammonium sulfate.

Calculate the mass of ferrous ammonium sulfate you would need to add to a 5.00 mL volumetric flask to make a solution that is 55 ppm in Fe.

Stock solution A is prepared by dissolving 194 mg of ferrous ammonium sulfate in water and diluting to volume in a 500-mL volumetric flask. Calculate the concentration of iron, in ppm, in stock solution A.

Stock solution B is prepared by diluting 10.00 mL of stock solution A to 100.00 mL. Calculate the concentration of iron, in ppm, in stock solution B.

Answer 1

molar mass Fe(NH₄)₂(SO₄)₂^.6H₂O = 392.1 g / mole

136 mg x 1 mmol / 392.1 mg = 0.3469 mmoles Fe(NH₄)₂(SO₄)₂^.6H₂O

mass of Fe in 0.3469 mmoles Fe(NH₄)₂(SO₄)₂^.6H₂O = 0.3469 mmol Fe x 55.85 mg/mol = 19.4 mg Fe

% Fe in 0.3469 mmoles Fe(NH₄)₂(SO₄)₂^.6H₂O = 14.2% (55.8 g / 392 g = 0.142)

55 ppm Fe = 55 mg Fe / L x 0.005 L = 0.275 mg Fe

(0.142) (x mg) = 0.275 mg

x = 1.94 mg 0.3469 mmoles Fe(NH₄)₂(SO₄)₂^.6H₂O needed in 5 ml to make 55 ppm

In stock A, you have 10 times the 1.94 mg (194 mg) but it's also in 10 times the volume (500 ml vs 5.00 ml). So, the [Fe] in ppm = 55 ppm.

In stock B, you have diluted stock A by a factor of 10 (10 ml up to 100 ml = 10x dilution). The [Fe] will thus be 1/10 that in stock A = 5.5 ppm. Done a more conventional way, we have...

(10 ml)(55 ppm) = (100 ml)(x ppm) and x = 5.5 ppm