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`.

Explanation

Molar mass of `MgP_(2)O_(7) = (2 xx 24 + 2 xx 31 + 7 xx 16) g`
`= 222 g`
`n_(Mg_(2)P_(2)O_(7)) = ("Mass"_(Mg_(2)P_(2)O_(7)))/("Molar mass"_(Mg_(2)P_(2)O_(7))) = (0.444 g)/(222 g mol^(-1))`
`n_(P) = 2n_(Mg_(2)P_(2)O_(7)) = 2 xx (0.44)/(222) mol`
`Mass_(P) = n_(P) xx "Molar mass"_(P) = 2 xx (0.44)/(222) xx 31 g`
`%` of `P` in o .c. `= ("Mass"_(P))/(Mass_(o .c.)) xx 100%`
`= (2 xx 0.444 xx 31)/(222 xx 0.248) xx 100% = 50%`