How would I find the percent of P2O5 present in a sample of KH2PO4 I know that phosphorus is 22.761% by mass of KH2PO4. I weighed out 0.422g of KH2PO4, then filtered it with water. Then I added 10 mL of 0.4 M MgSO4 to the filtrate. Then I added 1 mL of 6 M NH3 slowly with a disposable pipette while stirring. Then I touched the stirring rod to red litmus paper. Then I continued adding 1 mL NH3 at a time until the litmus paper turned blue, and then I added 0.5 mL more of NH3. Then I let the solution stand for 15 minutes to precipitate the insoluble Mg(NH4)PO46H2O. Then I filtered the precipitate with suction filtration and by gravity and washed the precipitate with two 5 mL portions of isopropyl alcohol. I sucked air through the filter for about 20 minutes and then scraped the precipitate onto a pre-weighed watch glass. Then I let it dry for about half an hour. I weighed the dry precipitate in a watch glass. The mass of the product should be used to determine the percentage of P2O5 in the sample. The mass of the product without the watch glass was 0.757. So how do I determine the mass of phosphorus atoms present in the Mg(NH4)PO4 6H2O and the percent of P2O5 in the sample
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