Question

In diammonium hydrogen phosphate, ${{(N{{H}_{4}})}_{2}}HP{{O}_{4}}$, the percentage of:
(This question has multiple correct options.)
(A) ${{P}_{2}}{{O}_{5}}$ is 53.78%
(B) $N{{H}_{3}}$ is 25.76%
(C) P is maximum
(D) N is minimum

Answer 1

To solve this question we first need to determine the molecular mass of diammonium hydrogen phosphate, ${{(N{{H}_{4}})}_{2}}HP{{O}_{4}}$. The molecular mass of one mole of a compound can be calculated by taking the sum of atomic masses of all the elements present in the molecule.

Complete answer:
Now, we know that the atomic mass of
N = 14.01 u
H = 1.010 u
P = 30.97 u
O = 16.00 u
Diammonium hydrogen phosphate, ${{(N{{H}_{4}})}_{2}}HP{{O}_{4}}$, can also be written as ${{H}_{9}}{{N}_{2}}{{O}_{4}}P$.
So, the molar mass of diammonium hydrogen phosphate (${{(N{{H}_{4}})}_{2}}HP{{O}_{4}}$) will be

& {{M}_{{{(N{{H}_{4}})}_{2}}HP{{O}_{4}}}}=2\times {{M}_{N}}+9\times {{M}_{H}}+4\times {{M}_{O}}+{{M}_{P}} \\\ & {{M}_{{{(N{{H}_{4}})}_{2}}HP{{O}_{4}}}}=2\times 14.01+9\times 1.01+4\times 16+30.97 \\\ & {{M}_{{{(N{{H}_{4}})}_{2}}HP{{O}_{4}}}}=132.06\text{ g/mol} \\\ \end{aligned}$$ Now, the formula to calculate the percentage composition of a constituent in a compound is given by: $$%\text{ composition = }\dfrac{\text{mass of molecule in 1 mole of compound}}{\text{molar mass of compound}}\times 100$$ So, the percentage compositions of elements N, H, O, and P are as follows $$\begin{aligned} & (N)=\dfrac{2\times 14.01}{132.06}\times 100=21.21 \\\ & (H)=\dfrac{9\times 1.01}{132.06}\times 100=6.87 \\\ & (O)=\dfrac{4\times 16}{132.06}\times 100=48.46 \\\ & (P)=\dfrac{30.97}{132.06}\times 100=23.45 \\\ \end{aligned}$$ From the above calculations, we can see that the percentage compositions of elements in ${{(N{{H}_{4}})}_{2}}HP{{O}_{4}}$ are N = 21.21% H = 6.870% O = 48.46% P = 23.45 % Now, diammonium hydrogen phosphate is produced as follows $${{P}_{2}}{{O}_{5}}+4N{{H}_{3}}+3{{H}_{2}}O\to 2{{(N{{H}_{4}})}_{2}}HP{{O}_{4}}$$ Hence, we can say that one mole of diammonium hydrogen phosphate (${{(N{{H}_{4}})}_{2}}HP{{O}_{4}}$) consists of a half mole of phosphorus pentoxide (${{P}_{2}}{{O}_{5}}$). Also, it can be said that one mole of diammonium hydrogen phosphate (${{(N{{H}_{4}})}_{2}}HP{{O}_{4}}$) consists of two moles of ammonia ($N{{H}_{3}}$). So, the percentage compositions of ${{P}_{2}}{{O}_{5}}$ and $N{{H}_{3}}$ will be $\begin{aligned} & ({{P}_{2}}{{O}_{5}})=\dfrac{0.5\times [2\times 30.97+5\times 16]}{132.06}\times 100=53.78 \\\ & (N{{H}_{3}})=\dfrac{2\times [14.01+3\times 1.01]}{132.06}\times 100=25.76 \\\ \end{aligned}$ From the above calculations, we can see that the percentage compositions of ${{P}_{2}}{{O}_{5}}$ and $N{{H}_{3}}$ in ${{(N{{H}_{4}})}_{2}}HP{{O}_{4}}$ are ${{P}_{2}}{{O}_{5}}$ = 53.78% $N{{H}_{3}}$ = 25.76% **So, the correct answers are option (A) and option (B).** **Note:** It should be noted that upon dissociation at a temperature of $100{}^\circ C$ and dissociation pressure of 5mmHg, diammonium hydrogen phosphate dissociates as follows $${{(N{{H}_{4}})}_{2}}HP{{O}_{4}}(s)\rightleftarrows N{{H}_{3}}(g)+(N{{H}_{4}}){{H}_{2}}P{{O}_{4}}(s)$$ The structure of diammonium hydrogen phosphate salt is as follows.