Question

Question: An ionic compound used as a chemical fertilizer has the composition (by mass) 48.46% O, 23.45% P, 21.21% N, 6.87% H. Give the name and chemical formula of the compound and draw Lewis diagrams for the two types of ions that make it up.

Short answer

The chemical formula of the compound is ${\left( {N{H_4}} \right)_2}HP{O_4}$ and the lewis diagrams for two types of ions are as follows,

Step-by-step solution

Step 1:

According to the question the compound contains 48.46% oxygen, 23.45% phosphorous, 21.21% nitrogen and 6.87% hydrogen.

Mass of N = 21.21 g

Mass of H = 6.87 g

Mass of P = 23.45 g

Mass of O = 48.46 g

Mass of H = 6.87 g

Mole of Nitrogen $\left(\mathrm{N}\right)=\frac{21.21\mathrm{g}}{14{\mathrm{gmol}}^{-1}}=1.51\mathrm{mol}$

Mole of Hydrogen $\left(\mathrm{H}\right)=\frac{6.87\mathrm{g}}{1{\mathrm{gmol}}^{-1}}=6.87\mathrm{mol}$

Mole of$\left(\mathrm{P}\right)=\frac{23.45\mathrm{g}}{31{\mathrm{gmol}}^{-1}}=0.75\mathrm{mol}$

Mole of $\left(\mathrm{O}\right)=\frac{48.46\mathrm{g}}{16{\mathrm{gmol}}^{-1}}=3.03\mathrm{mol}$

All can be divisible by least number to find the empirical values so,

For nitrogen$\left(\mathrm{N}\right)=\frac{1.51}{0.756}=2$

For$\left(\mathrm{H}\right)=\frac{6.87}{0.756}=9$

For$\left(\mathrm{P}\right)=\frac{0.756}{0.756}=1$

For$\left(\mathrm{O}\right)=\frac{3.03}{0.756}=4$

Hence, the empirical formula of the compound is ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{HPO}}_4$.

Step 2:

The lewis diagram of ${\mathrm{HPO}}_4^{2-}$is

The lewis diagram of ${\mathrm{NH}}_4^{+}$is,