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 man’s 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?