Question

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

1) Calculate the formula weight of ferrous ammonium sulfate. (Units = g/mol)

2) Calculate the mass of Fe in 172 mg of ferrous ammonium sulfate.

3) 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 75 ppm in Fe.

4) Stock solution A is prepared by dissolving 173 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.

5) 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

formula mass = (56 + (14+4*1)*2) + (32+16*4)*2 + 6*(1*2+16) = 392 g/mol

1)

formula weight of ferrous ammonium sulfate 392.14

1)392.14 gm/mol of ferrous ammonium sulfate can have 56 gm

2)172gm of ferrous ammonium sulfate ........?

   172*56/392.14 = 24.56 gm of Fe can present

3)

75 ppm of Fe = 75 grams per million(10^6) grams of H2O

In 5 ml of Solution we have 2 * 75 / 10^6

= 150 * 10^-6 grams of Fe

= 150 micro grams((mu)g) Fe

4)

10^6* (173/500) = 346000 mg = 346grams of Fe

5)500 ml of solution has 173 mg

100 mL of Solution has 10 * 173 * 10^-3 / 500 grams

= 3.46 mg

diluted to 100 mL

3.46 * 10^-3 grams in 100 mL

x grams in 10^6 mL

x = 3460/100

= 34.60 ppm