Chem Q's Help!!!! plzzz

[pillowcase]

Quick Question!!
Need someone to explain whats happening and how to get it in simple terms (I am stuck on Part C and D)
Can someone explain it to me? thanks
its from 2006 chem Exam 1

Question 3

A soluble fertiliser contains phosphorus in the form of phosphate ions (PO₄^3-). to determine the PO₄^3- content by gravimetric analysis, 5.97g of the firtiliser powder was completely dissolved in water to make a volume of 250.0ml. A 20.00mL volume of this solution was pipetted into a concial flask and the PO₄^3- ions in the solution were precipitated as MgNH₄PO₄. The precipitate was filtered, washed with water and then converted by heating into Mg₂P₂O₇. The mass of Mg₂P₂O₇ was 0.0352g.

a/ calculate the amount, in mole, of Mg₂P₂O₇.
Did this correct
n(Mg₂P₂O₇)=0.00352/222.6=1.58*10^-4mol

b/ Calculate the amount, in mole, of phosphorus in the 20.00mL volume of solution.
Did this correct
given in the equation 2PO₄^2----->Mg₂P₂O₇, the ratio is 2:1, therefore n(p)=2*n(Mg₂P₂O₇)
=3.16*10^-4mol

c/ calculate the amount, in mole, of phosphorus in 5.97g of fertilizer.
Need help here
My teacher tried to explain it to me, but i"m not entire sure i get it, I need help  , can someone explain it to me? in a simple way if possible?

d/ Calculate the percentage of phosphate ions (PO₄^3-) by mass in the fertiliser.
Need help here too
Can someone help me here too?


Much thanks for those who help, , greatly appreciated

[DJA]

Here we are:

c) Think about it this way: First establish what you have calculated so far which is n(p)=3.16*10^-4mol. But note that this mol is from the 20.00mL volume of the solution which was pipetted for the precipitation reaction. It is a smaller amount of phosphorus than from the original fertiliser powder because we have only taken 20.00mL of the dissolved solution from the original stock solution of 250.0mL which contains ALL the fertiliser.
Therefire, to find the amount in mol of phosphorus in the original 5.97g of fertiliser, we need to multiply your calculated mol of fertiliser by a factor to get from 20mL solution back to 250mL solution.

Do a bit of maths:
n(P)_original= 250.0/25.00 * 3.16*10^-4mol = 3.95*10^-3 mol

This value is the mol in the original.

d) So note that the formula of PO₄^3- has one P atom in it. Hence we can say n(P) = n(PO₄^3-)
Therefore  n(PO₄^3-) = 3.95*10^-3 mol

Then mass = mol*molar mass of PO₄^3-
m(PO₄^3-) = 3.95*10^-3 * 95.0 = 0.376 g

To find percentage of phosphate ions by MASS, you use the idea that there are 0.376 g of phosphate ions in a total 5.97g of fertiliser powder. Hence the percentage mass of phosphate ions given this original fertiliser mass would just be:

0.376/5.97 * 100 = 6.29%


Hope this helped  Ask if you need clarification

[pillowcase]

thankyou , that clarified things heaps, thankyou so much