The percentage of phosphorus in a detergent product may bedetermined by sample t

Question

The percentage of phosphorus in a detergent product may bedetermined by sample treatment including the precipitation ofmagnesium ammonium phosphate(MgNH4PO4*6H2O; 245.42 g/mol),using an excess of both magnesium and ammonium ions. The separatedand washed precipitate is then strongly heated to decompose it tomagnesium pyrophosphate (Mg2P2O7;222.57 g/mol) which is weighed. If a 3.00 gram sample of detergentyields 0.2550 grams of Mg2P2O7,calculate the percentage of phosphorus (30.97 g/mol) in thedetergent.

The answer should be 2.37%, but I really need to learn how to solvethis. Any help would be appreciated, thanks.

Explanation / Answer

First you need to find the percent of phosphorus in themagnesium pyrophosphate. By divideing the molecular weight ofthe two P atoms by the weight of the whole molecule itis 2*30.97/222.57*100= 27.8%. So that means that of the .2550 g only 27.8% of that isphosphorus. Multiply to find the grams of P in the sample .2550*.278=.071 g. Which is the number of grams of phosphorus there was in thedetergent ( assuming that all reactions ran to completion. iereactants were in excess.) Divide that by the total weight of the detergent to find thepercent weight. .071/3*100=2.37 %

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