Molybdate reagent precipitates ammonium phosphomolybdate

Reference no: EM13176022

The man in the vat. Once upon a time, a workman at a dye factory fell into a vat containing a hot concentrated mixture of sulfuric and nitric acids, and he dissolved! Because nobody witnessed the accident, it was necessary to prove that he fell in so that the manâ€^TMs wife could collect his insurance money. The man weighed 70kg, and a human body contains about 6.3 parts per thousand phosphorus. The acid in the vat was analyzed for phosphorus to see if it contained a dissolved human.

a) The vat had 8.00 x 10^3 L of liquid, and 100.0mL were analyzed. If the man did fall into the vat, what is the expected quantity of phosphorus in 100.0mL?

b) The 100.0mL sample was treated with a molybdate reagent that precipitates ammonium phosphomolybdate, (NH4)3 [P(Mo12O40)] x 12H2O. This substance was dried at 110 degree C to remove waters of hydration and heated to 400 degree C until it reached a constant composition corresponding to the formula P2O5 x 24MoO3, which weighed 0.3718g. When a fresh mixture of the same acids (not from the vat) was treated in the same manner, 0.0331g of P2O5 x 24MoO3 (FM 3596.46) was produced. This blank determination gives the amount of phosphorus in the starting reagents. The P2O5 x 24MoO3 that could have come from the dissolved man is therefore 0.3718 â€" 0.0331 = 0.3387g. How much phosphorus was present in the 100.0mL sample? Is this quantity consistent with a dissolved man?

Reference no: EM13176022

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