Heating 0.75 g of an organic compound containing phosphorus, with conc. HNO3 in the Carius tube followed by reaction with magnesia mixture gives MgNH4PO4. This on ignition left a residue of Mg2P2O7 weighing 1.2545 g. Calculate the percentage of phosphorus in organic compound.

Question

Heating 0.75 g of an organic compound containing phosphorus, with conc. HNO3 in the Carius tube followed by reaction with magnesia mixture gives MgNH4PO4. This on ignition left a residue of Mg2P2O7 weighing 1.2545 g. Calculate the percentage of phosphorus in organic compound. (A) 58% (B) 47% (C) 75% (D) 25%

Tardigrade Admin · Accepted Answer

Mass of the organic compound = 0.75 g Mass of Mg2P2O7 = 1.2545 g Thus, percentage of P =((62/222))×(mass of Mg2P2O7/mass of org. compound) × 100 = (62/222) × (1.2545/0.75)×100 = 46.7 % ≈ 47 %

Q. Heating 0.75 g of an organic compound containing phosphorus, with conc. HNO3​ in the Carius tube followed by reaction with magnesia mixture gives MgNH4​PO4​. This on ignition left a residue of Mg2​P2​O7​ weighing 1.2545 g. Calculate the percentage of phosphorus in organic compound.

58%

47%

75%

25%

Mass of the organic compound =0.75g Mass of Mg2​P2​O7​=1.2545g Thus, percentage of P =(22262​)×massoforg.compoundmassofMg2​P2​O7​​×100 =22262​×0.751.2545​×100=46.7%≈47%

Mass of the organic compound $= 0.75 g$
Mass of $M g_{2} P_{2} O_{7} = 1.2545 g$
Thus, percentage of P
$= (\frac{62}{222}) \times \frac{ma ss o f M g _{2} P _{2} O _{7}}{ma ss o f or g . co m p o u n d} \times 100$
$= \frac{62}{222} \times \frac{1.2545}{0.75} \times 100 = 46.7% \approx 47%$