%3Cp%3Ethe%20phosphate%26nbsp%3Bin%26nbsp%3B5.00g%26nbsp%3Bof%26nbsp%3Bfertilize

Question

%3Cp%3Ethe%20phosphate%26nbsp%3Bin%26nbsp%3B5.00g%26nbsp%3Bof%26nbsp%3Bfertilizer%26nbsp%3Bsample%26nbsp%3Bwas%26nbsp%3Bprecipated%26nbsp%3Bas%26nbsp%3Bmagnesium%26nbsp%3Bammonium%26nbsp%3Bphosphate%2C%26nbsp%3BMgNH4PO4.%26nbsp%3BIgnition%26nbsp%3Bat%26nbsp%3B900%26nbsp%3Bdeg%26nbsp%3BCel%26nbsp%3Bconverted%26nbsp%3Bthe%26nbsp%3Bprecipitate%26nbsp%3Bto%26nbsp%3Bmagnesium%26nbsp%3Bpyrophosphate%2C%26nbsp%3BMgP2O7%26nbsp%3Bwhich%26nbsp%3Bweighed%26nbsp%3B.7511g%3C%2Fp%3E%3Cp%3E%3Cbr%3E%3C%2Fp%3E%3Cp%3E%26nbsp%3Ba)%26nbsp%3Bcalculate%26nbsp%3Bthe%26nbsp%3Bpercentage%26nbsp%3Bphosphorus%26nbsp%3Bin%26nbsp%3Bthe%26nbsp%3Bfertilizer%3C%2Fp%3E%3Cp%3E%3Cbr%3E%3C%2Fp%3E%3Cp%3Eb)%26nbsp%3Bif%26nbsp%3Bthe%26nbsp%3Bphosphorus%26nbsp%3Bwas%26nbsp%3Bknown%26nbsp%3Bto%26nbsp%3Bbe%26nbsp%3Bin%26nbsp%3Bthe%26nbsp%3Bform%26nbsp%3Bof%26nbsp%3BCa3(PO4)2%26nbsp%3Bin%26nbsp%3Bthe%26nbsp%3Boriginal%26nbsp%3Bsample%2C%26nbsp%3Bwhat%26nbsp%3Bis%26nbsp%3Bthe%26nbsp%3Bpercentage%26nbsp%3Bcalcium%26nbsp%3Bphospate%26nbsp%3Bin%26nbsp%3Bthe%26nbsp%3Boriginal%26nbsp%3Bsample%22%3F%3C%2Fp%3E%3Cp%3E%3Cbr%3E%3C%2Fp%3E%3Cp%3ECalcium%26nbsp%3Bphosphate%3D310.18%2C%26nbsp%3Bmagnessium%26nbsp%3Bammonium%26nbsp%3Bphosphate%3D137.32%2C%26nbsp%3Bmagnesium%26nbsp%3Bpyrophosphate%3D222.57%3C%2Fp%3E

Explanation / Answer

total weight of 1 mole of magnesium Pyrophosphate = 222.57g

total weight of one mole Phosphorus=30.97g

there are two mols of phosphorous in each magnesium Pyrophosphate.

i.e % mass of phosphorus in magnesium Pyrophosphate = 2* 30.97/222.57 = 0.278 = 27.8%.

So, mas of phosphorus in 0.7511 g = 0.7511 * 27.8% = 0.209g.

So, in total we have 0.105g. total weight of eertilizer = 5g

so% Phosphorus in FERTILIZER = (0.209/5) * 100% = 4.18%

2.

We know, each Ca3(PO4)2 has two moles of phosphate. Total mass of 1 mole Ca3(PO4)2 = 30.97*2 + 16.00*4*2 + 40.8 * 2 = 271.54.

mass of 2 moles Phosphorus = 30.97*2 = 61.94

So, % mass of Phosphorus in Ca3(PO4)2 = 61.94/271.54 = 22.8%

So, if we had 0.209g of Phosphorus, that means 22.8% of Ca3(PO4)2 = 0.209

or, 22.8m/100 = 0.209

or, m =0.9167

So, in the fertilizer, % of Ca3(PO4)2 =( 0.9167 / 5)* 100 = 18.33%

Hope this helps.

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