Question

To determine the amount of iron in a dietary supplement, a random sample of 15 tablets weighing a total of 20.622 g was ground into a fine powder. A 3.0202-g sample of the fine powder was dissolved in 150 mL of water along with 1.2 g of urea, NH₂CONH₂. The solution was heated, which made the solution slowly turn basic as hydroxide was formed by decomposition of urea. As the solution turned basic, solid Fe(OH)₃ formed.

NH₂CONH₂ + H₂O → 2 OH^- + 2 NH₄⁺ + CO₂
Fe³⁺ + 3 OH^- → Fe(OH)₃(s)

Fe(OH)₃ forms gelatinous precipitates that are hard to filter unless the crystals form slowly--that is why the OH^- was formed in situ with a slow reaction instead of being dumped in as NaOH. The precipitate was filtered, rinsed, and heated in a furnace to convert the Fe(OH)₃ into Fe₂O₃, yielding 0.1695 g. Report the iron content of the dietary supplement as g FeSO₄7H₂O (FM 392.14) per tablet. (Hint: Think about what happens to the iron atoms.)

=? g ferrous ammonium sulfate per tablet

Answer 1

Given

Mass of 15 tablets = 20.622 gm

Mass of 1 tablet = 20.622 / 15 = 1.3748 gm

Mass of sample taken for experiment = 3.0202 gm

Mass of Fe₂O₃= 0.1695 gm

1 mole Fe₂O₃ 2 mole Fe

Molar mass of Fe₂O₃= ( 2x 55.85) + ( 3 x 16)= 159.7 gm / mole

i e 159.7 gm Fe₂O₃   111.7 gm Fe

0.1695 gm Fe₂O₃  111.7 x 0.1695 / 159.7

0.1185 gm Fe

3.0202 gm sample    0.1185 gm Fe

1.3748 gm sample 0.1185 x 1.3748 / 3.0202

   0.05394 gm Fe

1 mole Fe   1 mole ferrous ammoniumsulphate

55.85 gm Fe   392.14 gm FAS

0.05394 gm Fe 392.14 x 0.05394 / 55.85

   0.3787 gm FAS

Ans : Amount of FAS in 1 tablet = 0.3787 gm