Phosphate in a 0.2711g sample was precipitates giving 1.1682g of (NH4)2PO4•12MoO3(molar mass=1876.6g/mol).Find the percent P ?

Question

(atomic mass=30.97g/mol) and percentage P2O5(molar mass=141.95g/mol)in the sample.

Update:

Please Note that if in case the formal was given was wrong then probably our teacher made a mistake. Thanks to "Roger the Mole" for his valuable effort.

Answer 1

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There's difficulty at the beginning.  The molar mass of "(NH4)2PO4•12MoO3" is actually 1858.5467 g/mol.  The difference between that and the value given in the question is 18.05 g/mol, which is very close to the molar mass of water.  Is the given formula missing a molecule of water of hydration?

But continuing the calculation by assuming the given molar mass is correct and the formula is wrong:

(1.1682 g) / (1876.6 g/mol) x (1 mol P / 1 mol precipitate) x (30.97376 g P/mol) /

(0.2711 g total) = 0.071123 = 7.1122% P by mass

(1.1682 g) / (1876.6 g/mol) x (1 mol P / 1 mol precipitate) x (1 mol P2O5 / 2 mol P) x

(141.95 g P2O5/mol) / (0.2711 g total) = 0.16298 = 16.298% P2O5 by mass