Question

Mass of small beaker	112.64 g
Mass of Fertilizer and Beaker	123.03 g
Mass of fertilizer sample	10.39 g
Mass of filter paper	0.44 g
Mass of filter paper and dry MgNH4PO4*6H2O precipitate	0.96 g
Mass of dry MgNH4PO4*6H2O precipitate	0.52 g
Moles of MgNH4PO4*6H2O	0.0021 mol
Moles of Phosphorous	0.0021 mol
Mass of Phosphorous	0.0065 g
% P in fertilizer sample	.62 %
Moles of P2O5	mol
Mass of P2O5	g
% as P2O5 in fertilizer sample	%

Answer 1

It is assumed that P₂O₅ is the source of P in the fertilizer. Assume a hypothetical decomposition of P₂O₅ (without charges on the ions).

P₂O₅ -------> 2 P + 5 O

As per the stoichiometric equation above,

1 mole P₂O₅ = 2 moles P.

We have deduced earlier that the fertilizer contained 0.0021 mole P; therefore, mole(s) of P₂O₅ = (0.0021 mole P)*(1 mole P₂O₅/2 mole P) = 0.00105 mole P₂O₅ (ans)

Molar mass of P₂O₅ = (2*30.9738 + 5*15.9994) g/mol = 141.9446 g/mol.

Mass of P₂O₅ in the fertilizer = (0.00105 mole)*(141.9446 g/mol) = 0.14904 g ≈ 0.149 g (ans).

% P₂O₅ in the fertilizer = (0.149 g)/(10.39 g)*100 = 1.434% (ans).