Question

Show the calculation for determing mass % of P₂5O₅ in your fertilizer sample (starting from the MgNH₄PO_6​​ ​​* 6H2O product mass you will measure = 3 g).

Answer 1

1 mole of P₂O₅ whose molar mass is 141.95 g/mol, with its 2 P's produces 2 molar masses of MgNH4PO4*6H2O , whose molar mass is 245.40 g/mol

therefore 2 x 245.40 g of MgNH4PO4*6H2O is produced from 141.95 g of P2O5

hence 3 g of MgNH4PO4*6H2O is produced from 141.95 x 3/ (2 x 245.40) g of P2O5 = 0.868 g of P2O5

Therefore % of P2O5 in fertilizer sample = (0.868/ wt. of fertilizer taken) x 100

[wt. of fertilizer taken is not given, so the final result could not be calculated]