Question

Gravimetric Analysis by Precipitation of Complex The orthophosphate (PO43−) content of a 0.5185-g sample is isolated and weighed as ammonium phosphomolybdate precipitate, (NH4)3PO4·12MoO3. The weight of the precipitate is 1.0190 g.

What was the percentage, by mass, of phosphorus in the original sample? (Assume that all phophorus was present in the sample as orthophosphate.)

Calculate the equivalent percentage P2O5 in the original sample.

Answer 1

mass of (NH4)3PO4·12MoO3 ---> 1.0190 g

MW of (NH4)3PO4·12MoO3 = 149.0867 + 12*143.9382 = 1876.3451 g/mol

then

mol of precipitate = mass/MW = 1.0190/1876.3451 = 0.00054307 mol of precipitate

1 mol of precipitate = 1 mol of PO4-3

0.00054307 mol of precipitate = 0.00054307 mol of PO4-3

0.00054307 mol of PO4-3 = 0.00054307 mol of P

mass of P = mol*MW = 0.00054307*30.973 = 0.01682 g of P

mass of sample = 0.5185 g

% P in sampl e= mass of P / mass of sample ¨100% = 0.01682 /0.5185 *100 = 3.2439 %

euiqvalent % of P2O5 in sample:

0.00054307 mol of P = 1/2*0.00054307 = 0.000271535 mol of P2O5

mass of P2O5 = 141.944522*0.000271535 = 0.03854 g of P2O5

% mass = 0.03854/0.5185*100 = 7.432 % as P2O5