Question

Long ago, a workman at a dye factory fell into a vat containing hot, concentrated sulfuric and nitric acids. He dissolved completely! Since no one 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 weighted 70 kg and a human body contains ~6.3 parts per thousand (mg/g) phosphorous. The acid in the vat was analyzed of phosphorous to see whether it contained a dissolved human.

The vat contained 8.00x10³ L of liquid and a 100.0 mL sample was analyzed. If the man did fall into the vat, what is the expected quantity of phosphorous in 100.0 mL?

The 100.0 mL sample was treated with a molybdate reagent that precipitated ammonium phosphomolybdate (NH₃)[P(Mo₁₂O₄₀)]*12H₂O. This substance was dried at 110^oC to remove waters of hydration and heated to 400^oC until it reached the constant composition P₂O₅*24MoO₃, which weighed 0.3718 g. When a fresh mixture of the same acids (not from the vat) was treated in the same manner, 0.0331 g of P₂O₅*24MoO₃ (FM 3596.46) was produced. This blank determination gives the amount of phosphorous in the starting reagents. The P₂O₅*24MoO₃ that could have come from the dissolved man is therefore 0.3387 g. How much phosphorous was present in the 100.0 mL sample? Is this quantity consistent with a dissolved man? please show work.

Answer 1

Given

Mass of man = 70 Kg

Phosphorous content from man's body is 6.3 mg/g (= 6.3 g/Kg) of body weight

so total phosphorous content = 6.3 (g/Kg) * 70 Kg = 441 g of phosphorous

Given

Mass of P2O5.24MoO3 present in 100 ml sample taken = 0.3718 g

Mass of P2O5.24MoO3 that could have come from dissolved man is 0.3387 g

Molar mass of P2O5.24MoO3 = 3597 g/mol

Molar mass of Phosphorous = 31 g/mol

2 mole of P is present in P2O5.24MoO3 so molar mass of phosphorous * 2 will be used

Mass of Phosphorous in 100 ml sample = (0.3718 g * 2 * molar mass of Phosphorous) / MM of P2O5.24MoO3

= (0.3718 g * 2 * 31 g/mol) / 3597 g/mol = 6.4 * 10^-3 g

3.2 * 10^-3 g of phosphorous was present in 100 ml of sample

Mass of Phosphorous could have came from man dissolved = (0.3387 g * 2 * molar mass of Phosphorous) / MM of P2O5.24MoO3

= (0.3387 g * 2 * 31 g/mol) / 3597 g/mol = 5.8 * 10^-3 g

so 2.9 * 10^-3 g in 100 ml (0.1 L) of sample so in 8 * 10³ L of sample

5.8 * 10^-3 * 8 * 10³ * 10= 464 g of phosphorous

it is almost consistent with amount of phosphorous present in human body. which is 441 g