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In a gravimetric determination of phosph...

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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To find the percentage of phosphorus in the organic compound, we will follow these steps: ### Step 1: Calculate the molecular weight of Mg₂P₂O₇ The molecular formula of the compound formed is Mg₂P₂O₇. We need to calculate its molecular weight using the atomic weights provided. - Atomic weight of Magnesium (Mg) = 24 g/mol - Atomic weight of Phosphorus (P) = 31 g/mol - Atomic weight of Oxygen (O) = 16 g/mol ...
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In the Carius method of estimation (a gravimetric determination) of phosphorus, 0.248 g of an organic compound gave a precipitate of Mg_(2)NH_(4)PO_(4) which on ignition yielded 0.444 g of Mg_(2)P_(2)O_(7) . What is the percentage of phosphorus in the compound? Strategy: Find the molar mass of Mg_(2)P_(2)O_(7) through atomic masses. Calculate the moles of Mg_(2)P_(2)O_(7) to get the moles of P and finally find the mass of P to get the percentage in the o .c .

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