Question

The compounds ${\left({\mathrm{NH}}_4\right)}_{\mathbf{2}}{\mathbf{HPO}}_{\mathbf{4}}$ and ${\mathbf{NH}}_{\mathbf{4}}\left(H_2{\mathrm{PO}}_4\right)$ are water-soluble fertilizers.

(a) Which has the higher mass $\mathbf{\%}$ of $\mathbf{P}$?

(b) Why might the one with the lower mass $\mathbf{\%}$ of$\mathbf{P}$ be preferred?

Short answer

(a) ${\mathrm{NH}}_4\left({\mathrm{H}}_2{\mathrm{PO}}_4\right)$Has greater mass percentage than $\mathrm{P}$.

(b) The one with the lower mass $\%$of$\mathrm{P}$ be preferred, because of the phosphorus cycle's imbalance.

Step-by-step solution

Definition of water-soluble fertilizers

Water soluble fertilisers are nutrients that are completely soluble in water and are administered by Foliar Spray & Fertilizers procedures, allowing nutrients to be rapidly absorbed and utilised in the plant system without evaporation, leaching, or runoff.

Compute the molar masses of each compound

(a)

Compute the molar masses of each combination, then divided the atomic mass of$\mathrm{P}$by the molar mass of the compound to get the mass percent of $\mathrm{P}$.

Let’s calculate the molar masses of ${\left({\text{NH}}_4\right)}_2{\text{HPO}}_4$

$$\begin{gathered} M\left({\left(N{H}_4\right)}_{2}HP{O}_4\right)=2A\left(N\right)+9A\left(H\right)+A\left(P\right)+4A\left(O\right) \\ M\left({\left(NH\right)}_{2}HP{O}_4\right)=132.06gmo{l}^{-1} \\ w\left(P,{\left(N{H}_4\right)}_{2}HP{O}_4\right)=\frac{A\left(P\right)}{M\left({\left(NH\right)}_{2}HHO{O}_4\right)} \\ w\left(P{\left(N{H}_4\right)}_{2}HPO\right)=\frac{30.974}{132.06}\times 100\% \\ w\left(P,{\left(N{H}_4\right)}_{2}HP{O}_4\right)=23.45\% \end{gathered}$$

Then calculate the molar masses of ${\text{NH}}_4\left({\text{H}}_2{\text{PO}}_4\right)$

$$\begin{gathered} M\left(N{H}_4\left(H_{2}P{O}_4\right)\right)=A\left(N\right)+6A\left(H\right)+A\left(P\right)+4A\left(O\right) \\ M\left(N{H}_4\left(H_{2}P{O}_4\right)\right)=115.03gmo{l}^{-1} \\ w\left(P,N{H}_4\left(H_{2}P{O}_4\right)\right)=\frac{A\left(P\right)}{M\left(N{H}_4\left(H_{2}P{O}_4\right)\right)} \\ w\left(P,N_4\left(H_{2}P{O}_4\right)\right)=\frac{30.974}{115.03}·100\text{\%} \\ w\left(P,N_4\left(H_2{P}_4\right)\right)=26.93\text{\%} \end{gathered}$$

From the calculation, we observed that ${\text{NH}}_4\left({\text{H}}_2{\text{PO}}_4\right)$has higher mass percentage than $\mathrm{P}$.

Reason to prefer lower mass percentage of  P

b)

The overuse of soluble phosphate fertilisers has become a major source of phosphorus cycle imbalance.

The fertiliser with the lowest mass percentage of $\mathrm{P}$ is preferred to reduce that undesirable effect.