Question
How the phosphorus is analysed in organic compound? Explain.
✓
Answer
For the estimation of phosphorus. When the quantity of organic compound is heated with fumed nitric acid, the phosphorus present in it gets oxidized into phosphoric acid $\left( H _3 PO _4\right)$. On adding ammonia and ammonium molybdate solution, a precipitate of ammonium phosphomolybdate $\left[\left( NH _4\right)_3 PO _4 \cdot 12 MoO _3\right]$ is obtained or by adding magnesia mixture to phosphoric acid it is precipitated in form of $MgNH _4 PO _4$, on the combustion of which $Mg _2 P _2 O _7$ is obtained.
Let the mass of organic compound $= m _1 g$ and the mass of ammonium phosphomolybdate is $m _1 g$. Molar mass of $\left( NH _4\right)_3 PO _4 \cdot 12 MoO _3=1877 gm$
Hence $\%$ of phosphorus $=\frac{31 \times m _1 \times 100}{1877 \times m } \%$
When phosphorus is analysed in form of $Mg _2 P _2 O _7$,
then $\%$ of phosphorus $=\frac{62 \times m_1 \times 100}{222 \times m}$
Here mass of $Mg _2 P _2 O _7=222 u$
Mass of organic compound $= m _1 g$
Mass of formed $Mg _2 P _2 O _7= m$, and mass of two phosphorus atoms in $Mg _2 P _2 O _7$ is 62 .
Let the mass of organic compound $= m _1 g$ and the mass of ammonium phosphomolybdate is $m _1 g$. Molar mass of $\left( NH _4\right)_3 PO _4 \cdot 12 MoO _3=1877 gm$
Hence $\%$ of phosphorus $=\frac{31 \times m _1 \times 100}{1877 \times m } \%$
When phosphorus is analysed in form of $Mg _2 P _2 O _7$,
then $\%$ of phosphorus $=\frac{62 \times m_1 \times 100}{222 \times m}$
Here mass of $Mg _2 P _2 O _7=222 u$
Mass of organic compound $= m _1 g$
Mass of formed $Mg _2 P _2 O _7= m$, and mass of two phosphorus atoms in $Mg _2 P _2 O _7$ is 62 .
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