Explain the Carius method and its principle for the estimation of phosphorus in an organic compound.

(N/A) Principle: $A$ known mass of an organic compound is heated with fuming nitric acid,which oxidizes the phosphorus present in the compound to phosphoric acid $(H_{3}PO_{4})$.
Method $1$: The phosphoric acid is precipitated as ammonium phosphomolybdate,$(NH_{4})_{3}PO_{4} \cdot 12MoO_{3}$,by adding ammonia and ammonium molybdate.
Reaction: $P$ $\xrightarrow{\text{fuming } HNO_{3}, \Delta} H_{3}PO_{4}$ $\xrightarrow{NH_{3} / (NH_{4})_{2}MoO_{4}} (NH_{4})_{3}PO_{4} \cdot 12MoO_{3(s)}$
Calculation: If $m$ is the mass of the organic compound and $m_{1}$ is the mass of ammonium phosphomolybdate,then:
$\% P = \frac{31}{1877} \times \frac{m_{1}}{m} \times 100$
Method $2$: Alternatively,phosphoric acid is precipitated as $MgNH_{4}PO_{4}$ by adding magnesia mixture $(Mg^{2+} + NH_{4}OH)$,which on ignition yields magnesium pyrophosphate $(Mg_{2}P_{2}O_{7})$.
Reaction: $MgNH_{4}PO_{4} \xrightarrow{\text{ignition}} Mg_{2}P_{2}O_{7(s)} + H_{2}O + N_{2}$
Calculation: If $m_{1}$ is the mass of $Mg_{2}P_{2}O_{7}$ obtained:
$\% P = \frac{62}{222} \times \frac{m_{1}}{m} \times 100$