Question

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 pipet 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)PO4*6H2O. 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?

Answer 1

You get 0.757 g of MgNH4PO4.6H₂O after drying.

Molar mass of MgNH4PO4.6H₂O = 245.4 g/mol

No of moles of MgNH4PO4.6H₂O = 0.757/245.4 = 3.08*10^-3 mol

1 mole of MgNH4PO4.6H₂O contains 1 mol of P.

So 0.757 g of MgNH4PO4.6H₂O contains 3.08*10^-3 mol of P.

Mass of P present in MgNH4PO4.6H₂O = 30.97 * 3.08*10^-3 = 0.0955 g of P ....[molar mass of P = 30.97 G]

0.757 g of MgNH4PO4.6H₂O contains 3.08*10^-3 mol of P

2 moles of P are there in 1 mol of P₂O₅.

So, 0.757 g of MgNH4PO4.6H₂O contains (3.08*10^-3 /2) or 1.54*10^-3 mol of P₂O₅.

Molar mass of P₂O₅ = 141.9 g

0.757 g of MgNH4PO4.6H₂O contains (1.54*10^-3 *141.9) or 0.2185 g P₂O₅.

The % of P₂O₅ in sample = (0.2185/0.422 ) *100

= 51.8 %