Question

The phosphorus present in an organic compound of mass 75 g is oxidized to phosphoric acid by heating with fuming nitric acid. The phosphoric acid so obtained is precipitated as \[\text{MgN}{{\text{H}}_{\text{4}}}\text{P}{{\text{O}}_{\text{4}}}\], which on ignition is converted into$\text{M}{{\text{g}}_{\text{2}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{7}}}$.If the mass $\text{M}{{\text{g}}_{\text{2}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{7}}}$is 50 g. the percentage composition of phosphorus present in the sample of the organic compound is:
A) \[19.5\]
B) \[18.5\]
C) \[20.0\]
D) \[18.6\]

Answer 1

Hint: the known mass of compound heated in presence of nitric acid which oxidizes to phosphoric acid and then precipitated. The percentage of phosphorus in the compound can be determined as:
${\scriptstyle{}^{\text{0}}/{}_{\text{0}}}\text{ of P = }\dfrac{\text{Atomic mass of P}}{\text{molar mass of precipitated compound}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{Given mass of precipitate }}{\text{Known mass of compound }}\text{ }\!\!\times\!\!\text{ 100}$

Complete step by step answer:
Let’s first consider the data given in the problem.
$\begin{align}
  & \text{Mass of organic compound = W g = 75 g} \\
 & \text{Mass of M}{{\text{g}}_{\text{2}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{7}}}\text{ obtained = }{{\text{W}}_{\text{1}}}\text{ g} \\
\end{align}$
Here, the phosphorus in the unknown organic compound is oxidized to phosphoric acid $\text{ }{{\text{H}}_{\text{3}}}\text{P}{{\text{O}}_{\text{4}}}$ by heating the compound in presence of nitric acid$\text{ HN}{{\text{O}}_{\text{3}}}$. The reaction is as follows,
\[\begin{matrix}
   \text{Phosphorus-organic compound} & \xrightarrow{\text{HN}{{\text{O}}_{\text{3}}}} & {{\text{H}}_{\text{3}}}\text{P}{{\text{O}}_{\text{4}}} & \to \text{ MgN}{{\text{H}}_{\text{4}}}\text{P}{{\text{O}}_{\text{4}}}\text{(}\downarrow \text{) } \\
\end{matrix}\xrightarrow{\text{ignition}}\text{M}{{\text{g}}_{\text{2}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{7}}}\]
The obtained phosphoric acid is then precipitated as\[\text{MgN}{{\text{H}}_{\text{4}}}\text{P}{{\text{O}}_{\text{4}}}\], which on ignition is converted into the$\text{M}{{\text{g}}_{\text{2}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{7}}}$.
We are given that the percentage of composition of phosphorus in the sample is$14{\scriptstyle{}^{0}/{}_{0}}$.
From the above data given we know that the one molecule $\text{M}{{\text{g}}_{\text{2}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{7}}}$contains the two phosphorus atom. Thus from stoichiometry, we can write,
$\text{1 molecule of M}{{\text{g}}_{\text{2}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{7}}}\text{ =2 atoms of P}$
This equation can be written in terms of molecular weight. that is,
$\text{ Molecular mass of M}{{\text{g}}_{\text{2}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{7}}}\text{ = 2 (Molar mass of P)}$
Let’s first calculate the molecular mass of$\text{M}{{\text{g}}_{\text{2}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{7}}}$.
$\begin{align}
  & \text{molecular mass of M}{{\text{g}}_{\text{2}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{7}}}\text{ = 2(24) + 2(31) + 7(16)} \\
 & \text{ = 48 + 62 + 112} \\
 & \text{ = 222 g mo}{{\text{l}}^{\text{-1}}} \\
\end{align}$

From stoichiometry, we can write,
$\begin{align}
  & \text{ Molecular mass of M}{{\text{g}}_{\text{2}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{7}}}\text{ = 2 (Molar mass of P)} \\
 & \text{ 222 g mo}{{\text{l}}^{\text{-1}}}\text{ M}{{\text{g}}_{\text{2}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{7}}}\text{ = 31 g mo}{{\text{l}}^{\text{-1}}}\text{ of 2P} \\
\end{align}$
Since we know that, each $\text{M}{{\text{g}}_{\text{2}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{7}}}$is associated with the two phosphorus atoms.
Then the mass of phosphorus in ${{\text{W}}_{\text{1}}}\text{ g}$of the $\text{M}{{\text{g}}_{\text{2}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{7}}}$is
$\begin{align}
  & \text{222 g M}{{\text{g}}_{\text{2}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{7}}}\text{ = 62 g mo}{{\text{l}}^{\text{-1}}}\text{ of 2P} \\
 & \text{ }{{\text{W}}_{\text{1}}}\text{ g M}{{\text{g}}_{\text{2}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{7}}}\text{ = mass g of P} \\
\end{align}$
$\text{ mass g of P = }\dfrac{62}{222}\text{ }\times \text{ }{{\text{W}}_{\text{1}}}\text{ g}$
Now, we can calculate the percent composition of phosphorus in the organic sample by taking the ratio of the mass of phosphorus to the known mass of the organic compound. the percentage can be calculated as follows:
$\begin{align}
  & \text{ }{\scriptstyle{}^{\text{0}}/{}_{\text{0}}}\text{ of P in compound = }\dfrac{\text{mass of P}}{\text{mass of organic compound}}\text{ }\!\!\times\!\!\text{ 100} \\
 & \text{ = }\dfrac{\dfrac{\text{62}}{\text{222}}\text{ }\!\!\times\!\!\text{ }{{\text{W}}_{\text{1}}}\text{ }}{\text{W}}\text{ }\!\!\times\!\!\text{ 100} \\
 & \text{ = }\dfrac{\text{(0}\text{.279) }\!\!\times\!\!\text{ }{{\text{W}}_{\text{1}}}\text{ }}{\text{W}}\text{ }\!\!\times\!\!\text{ 100 } \\
 & \text{ = }\dfrac{\text{(0}\text{.279) }\!\!\times\!\!\text{ 50 }}{\text{75}}\text{ }\!\!\times\!\!\text{ 100 } \\
 & \text{ = }\dfrac{\text{ 13}\text{.95 }}{\text{75}}\text{ }\!\!\times\!\!\text{ 100 } \\
 & \text{ = (0}\text{.186) }\!\!\times\!\!\text{ 100 } \\
 & {\scriptstyle{}^{\text{0}}/{}_{\text{0}}}\text{ of P in compound = 18}\text{.6 }{\scriptstyle{}^{\text{0}}/{}_{\text{0}}} \\
 & \text{ } \\
\end{align}$
Therefore the percentage composition of phosphorus in the given organic compound is $\text{18}\text{.6 }{\scriptstyle{}^{0}/{}_{0}}$.

Hence, (D) is the correct option.

Note: In solving such a question remember to get the proper stoichiometric relation between the molecular formula of compound and atom of interest. This is also called the carius method for the quantitative estimation of phosphorus.