Posted by smiley on Friday, August 31, 2007 at 2:57pm.
The phosphorus in a completely dried fertilizer sample weighing 0.4498 grams was precipitated as MgNH4PO4*6H2O. The precipitate weight was determined to be 236.8 mg and ignited to Mg2P2O7. The weight of the ignited precipitate was 268.9 mg. Calculate the percent phosphorus.
My work:
((268.9 mg Mg2P2O7 x ((2*molecular weight of phosphorus)/molecular weight of Mg2P2O7))/449.8 mg)* 100
= 33.28%
Am I right? Or should the weight of MgNH4PO4*6H2O be included. If the weight of MgNH4PO4*6H2O should be included, please show a sample calculation. Thanks in advance.

chemistry  ~christina~, Friday, August 31, 2007 at 5:30pm
If my thinking is that this is a simple problem and doesn't have anything more complicated to it.
I like working in grams so I'll use grams. I think your missing some info though in the equation you used.
.2689g Mg2P2O7 (1mol Mg2P2O7/222.54g)(2mol P/1mol Mg2P2O7)(30.97g/1molP)=
(0.07484gP/ 0.4498g)*100 = _________
check my work and if you have any questions just ask 
chemistry  DrBob222, Friday, August 31, 2007 at 6:26pm
((268.9 mg Mg2P2O7 x ((2*molecular weight of phosphorus)/molecular weight of Mg2P2O7))/449.8 mg)* 100
= 33.28%
If you run the numbers correctly you should have arrived at the correct answer. P is 30.97 and Mg2P2O7 is 222.55, but for some reason your answer is exactly twice what it should be. The mass of the MgNH4PO4*6H2O in the problem is extraneous information and is not needed to solve the problem. Christina has worked the problem correctly. Look at what you substituted for the molar masses or post your work and we can locate the error for you.