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In a gravimetric determination of phosphorus, 0.248g an organic compound was strongly heated in a Carius tube with concentrated nitric acid. Phosphoric acid so produced was precipitated as MgNH_(4)PO_(4) which on ignition yielded 0.444g of Mg_(2)P_(2)O_(7). Find the percentage of phosphorus in the compound. (Mg= 24, P= 31, O= 16) |
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Answer» Solution :MOLES of P in `Mg_(2)P_(2)O_(7)=2 xx` moles of `Mg_(2)P_(2)O_(7)` `=2 xx (0.444)/(222)= 0.004g` Weight of `P= 0.004 xx 31g` = 0.124g percentage of `P= (0.124)/(0.248)xx 100= 50%` |
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