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0.2 of an organic compound containing phosphorus gave 1.877 g of ammonium phosphomolybdate by usual analysis. Calculate the percentage of phosphorus in the organic compound.

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`underset(1877g)((NH_(3))_(3)PO_(4). 12MoO_(3)) -= underset(31g)(P)`
`:. % P = (31)/(1877) xx (1.877)/(0.2) = 15.5`

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