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0.12 gm of an organic compound containin...

`0.12` gm of an organic compound containing phosphorus gave `0.22` gm of `Mg_2P_2O_7` by the usual analysis. Calculate the percentage of phosphorus in the compound.

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precentage of phosphorus
`=(62)/(222)xx("mass of "Mg_2P_2O_7)/("mass of compound")xx100`
`=(62)/(222)xx(0.22)/(0.12)xx100=51.20%`
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Estimation of phosphorous: Second method: A known mass of compound is heated with fuming HNO_(3) or sodium peroxide (Na_(2)O_(2) in Carius tube which converts phosphorous to H_(3)PO_(4) . Magnesia mixture (MgCl_(2)+NH_(4)Cl) is then added, which gives the percipate of magnesium ammonium phosphate (MgNH_(4).PO_(4)) which on heating gives magnesium pyrophosphate (Mg_(2)P_(2)O_(7)) , which is weighted. Percentage of P =("Atomic mass of P")/("Molecular mass of " Mg_(2)P_(2)O_(7)) xx ("Molecular mass of "Mg_(2)P_(2)O_(7) xx 100)/("Mass of compound")=(62)/(222) xx("Mass of " Mg_(2)P_(2)O_(7)xx 100)/("Mass of compound") 0.12 gm of and organic compound containing phosphorus gave 0.22 gm of Mg_(2)P_(2)O_(7) by the usual analysis. Calculate the percentage of phosphorus in the compound.

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