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

Hint: 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
$$\begin{array}{rl}& M_{{(N{H}_4)}_{2}HP{O}_4}=2\times M_N+9\times M_H+4\times M_O+M_P\\ & M_{{(N{H}_4)}_{2}HP{O}_4}=2\times 14.01+9\times 1.01+4\times 16+30.97\\ & M_{{(N{H}_4)}_{2}HP{O}_4}=132.06\text{ g/mol}\end{array}$$
Now, the formula to calculate the percentage composition of a constituent in a compound is given by:
$$$$
So, the percentage compositions of elements N, H, O, and P are as follows
$$\begin{array}{rl}& (N)={\displaystyle \frac{2\times 14.01}{132.06}}\times 100=21.21\\ & (H)={\displaystyle \frac{9\times 1.01}{132.06}}\times 100=6.87\\ & (O)={\displaystyle \frac{4\times 16}{132.06}}\times 100=48.46\\ & (P)={\displaystyle \frac{30.97}{132.06}}\times 100=23.45\end{array}$$
From the above calculations, we can see that the percentage compositions of elements in ${(N{H}_4)}_{2}HP{O}_4$ are
N = 21.21%
H = 6.870%
O = 48.46%
P = 23.45 %
Now, diammonium hydrogen phosphate is produced as follows
$$P_2{O}_5+4N{H}_3+3H_{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
$\begin{array}{rl}& (P_2{O}_5)={\displaystyle \frac{0.5\times [2\times 30.97+5\times 16]}{132.06}}\times 100=53.78\\ & (N{H}_3)={\displaystyle \frac{2\times [14.01+3\times 1.01]}{132.06}}\times 100=25.76\end{array}$
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.