February 24, 2017

Homework Help: chemistry

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.

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