Gravimetric analysis of phosphorous

[Julia78]

I just finished an experiment where I precipitated 2.30 g of MgNH4PO4 6H20.

The gravimetric analysis of phosphorous in this experiment is based on the
precipitation of phosphorous as MgNH4PO4 . 6H2O from a solution that contains the
monohydrogen phosphate ion (HPO42- ), ammonium ions, and magnesium ions. The
balanced reaction is:
5 H2O + HPO42- + NH41+ + Mg2+ + OH1- =====> MgNH4PO4 . 6H2O I need to calculate percent yield of MgNH4PO4 . 6H2O obtained in my experiment using 10-52-10 plant food label as a guide to determine the theoretical yield of MgNH4PO4 . 6H2O.

So, the plant food contains:
10% N
52% P2O5
10% MgO

In my experiment I used 1.2 grams of 10-52-10 plant food ( + 4 grams MgSO4 7H2O) and obtained 2.30 g of MgNH4PO4 6H20
I know that % yield = actual yield/ theoretical yield x 100%
But, how do I calculate actual yield from the data I have?

So far I calculated:
air dry weight of MgNH4PO4 . 6H2O = 2.30 g
Grams of phosphorusin MgNH4PO4 6H20 = 0.290g
%P in plant food = 24.16
% P2O5 in equivalent to %P above = 55.45
% p2O5 in 10-52-10 plant food = 52

When do I do next?? Any help is greatly appreciated?

 

[Borek]

Something is strange here - gravimetric analysis is not to calculate percent yield - it is based on the assumption that percent yield is exactly 100%, and it is done to check what is the amount of analyte in the sample.

But if you really need it... How many moles of P in the sample? How many moles of phosphate should be precipitated if all phosphorus was converted to phosphoric acid? What should be the mass of the precipitate?

--

 

[Julia78]

So, the amount of phosphorus:
0.290 g P = 0.00936 moles P
Amount of phosphate --> from the reaction above...
(2.30 g MgNH4PO4 . 6H2O)(1 mol/ 245.45 g) (1 mol PO4 2-/ 1 mol MgNH4PO4 . 6H2O) = 0.00957 moles

I still do not understand how to obtain theoretical weight of precipitate?

No matter how I calculate the mass of precipitate I obtain 2.30 g, which is the actual yield I obtained in my experiment...

 

[Borek]

You should calculate amount of phosphorus in the sample from its mass and the fact it contains 52% by mass P₂O₅, if you calculate it from the amount of the precipitated salt and use it to calculate yield, you will just prove that 1=1.

If you have one atom of phosphorus, how many molecules of MgNH₄PO₄ can you make? From dozen atoms? From a mole?

 

[Julia78]

THANK YOU! I finnally got it!