Q22.64CP - Chemistry: Molecular Nature Of Matter And Change

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?

Step-by-Step Solution

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 1: 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.

Step 2: 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 $\mathrm{P}$ .

Step 3: 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.