91Ó°ÊÓ

The Molar mass of potassium nitrate, ${\mathrm{KNO}}_3$, is calculated below.

$\begin{array}{c}{\mathrm{M}}_{\mathrm{Ó\neg }}\text{ }\mathrm{of}\text{ â¶Ä‰}{\mathrm{KNO}}_3={\mathrm{M}}_{\mathrm{Ó\neg }}\text{ }\mathrm{of}\text{ â¶Ä‰}\mathrm{K}+{\mathrm{M}}_{\mathrm{Ó\neg }}\text{ }\mathrm{of}\text{ â¶Ä‰}\mathrm{N}+3×\left({\mathrm{M}}_{\mathrm{Ó\neg }}\text{ }\mathrm{of}\text{ â¶Ä‰}\mathrm{O}\right)\\ =39.10\text{ }\frac{\mathrm{g}}{\mathrm{mol}}+14.01\text{ }\frac{\mathrm{g}}{\mathrm{mol}}+3×\left(16.00\text{ }\frac{\mathrm{g}}{\mathrm{mol}}\right)\\ =101.11\text{ }\frac{\mathrm{g}}{\mathrm{mol}}\end{array}$

In 1 mol of ${\mathrm{KNO}}_3$, the mass is 101.11g and the number of moles of N atoms is 1. Since there is 1 mole of N atoms, the mass percentage of N is as follows.

Mass % of N $=\frac{1\text{ }\mathrm{mol}\text{ â¶Ä‰}\mathrm{N}\left(\frac{14.01\text{ }\mathrm{g}\text{ }\mathrm{N}}{\mathrm{mol}\text{ â¶Ä‰}\mathrm{N}}\right)}{101.11\text{ }\mathrm{g}\text{ â¶Ä‰}{\mathrm{KNO}}_3}×100=13.86\mathrm{\%}$.

The molar mass of ammonium nitrate, ${\mathrm{NH}}_4{\mathrm{NO}}_3$, is calculated below.

$\begin{array}{c}{\mathrm{M}}_{\mathrm{Ó\neg }}\text{ }\mathrm{of}\text{ â¶Ä‰}{\mathrm{NH}}_4{\mathrm{NO}}_3=2×\left({\mathrm{M}}_{\mathrm{Ó\neg }}\text{ }\mathrm{of}\text{ â¶Ä‰}\mathrm{N}\right)+4×\left({\mathrm{M}}_{\mathrm{Ó\neg }}\text{ }\mathrm{of}\text{ â¶Ä‰}\mathrm{H}\right)+3×\left({\mathrm{M}}_{\mathrm{Ó\neg }}\text{ }\mathrm{of}\text{ â¶Ä‰}\mathrm{O}\right)\\ =2×\left(14.01\text{ }\frac{\mathrm{g}}{\mathrm{mol}}\right)+4×\left(1.01\text{ }\frac{\mathrm{g}}{\mathrm{mol}}\right)+3×\left(16.00\text{ }\frac{\mathrm{g}}{\mathrm{mol}}\right)\\ =80.06\text{ }\frac{\mathrm{g}}{\mathrm{mol}}\end{array}$

In 1 mol of ${\mathrm{NH}}_4{\mathrm{NO}}_3$, the mass is 80.06g and the number of moles of N atoms is 2. Since there is 2 mole of N atoms, the mass percentage of N is as follows.

Mass % of N $=\frac{2\text{ }\mathrm{mol}\text{ â¶Ä‰}\mathrm{N}\left(\frac{14.01\text{ }\mathrm{g}\text{ }\mathrm{N}}{\mathrm{mol}\text{ â¶Ä‰}\mathrm{N}}\right)}{80.06\mathrm{g}\text{ â¶Ä‰}{\mathrm{NH}}_4{\mathrm{NO}}_3}×100=35.00\mathrm{\%}$.

The molar mass of ammonium sulfate, ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$, is calculated below.

$\begin{array}{c}{\mathrm{M}}_{\mathrm{Ó\neg }}\text{ }\mathrm{of}\text{ â¶Ä‰}{\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4=2×\left({\mathrm{M}}_{\mathrm{Ó\neg }}\text{ }\mathrm{of}\text{ â¶Ä‰}\mathrm{N}\right)+8×\left({\mathrm{M}}_{\mathrm{Ó\neg }}\text{ }\mathrm{of}\text{ â¶Ä‰}\mathrm{H}\right)+{\mathrm{M}}_{\mathrm{Ó\neg }}\text{ }\mathrm{of}\text{ â¶Ä‰}\mathrm{S}+4×\left({\mathrm{M}}_{\mathrm{Ó\neg }}\text{ }\mathrm{of}\text{ â¶Ä‰}\mathrm{O}\right)\\ =2×\left(14.01\text{ }\frac{\mathrm{g}}{\mathrm{mol}}\right)+8×\left(1.01\text{ }\frac{\mathrm{g}}{\mathrm{mol}}\right)+32.07\text{ }\frac{\mathrm{g}}{\mathrm{mol}}+4×\left(16.00\text{ }\frac{\mathrm{g}}{\mathrm{mol}}\right)\\ =132.17\text{ }\frac{\mathrm{g}}{\mathrm{mol}}\end{array}$

In 1 mol of ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$, the mass is 132.17g and the number of moles of N atoms is 2. Since there is 2 mole of N atoms, the mass percentage of N is as follows.

Mass % of N $=\frac{2\text{ }\mathrm{mol}\text{ â¶Ä‰}\mathrm{N}\left(\frac{14.01\text{ }\mathrm{g}\text{ }\mathrm{N}}{\mathrm{mol}\text{ â¶Ä‰}\mathrm{N}}\right)}{132.17\mathrm{g}\text{ â¶Ä‰}{\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4}×100=21.20\mathrm{\%}$.

The molar mass of urea, $\mathrm{CO}{\left({\mathrm{NH}}_2\right)}_2$, is calculated below.

$\begin{array}{c}{\mathrm{M}}_{\mathrm{Ó\neg }}\text{ }\mathrm{of}\text{ â¶Ä‰}\mathrm{CO}{\left({\mathrm{NH}}_2\right)}_2={\mathrm{M}}_{\mathrm{Ó\neg }}\text{ }\mathrm{of}\text{ â¶Ä‰}\mathrm{C}+{\mathrm{M}}_{\mathrm{Ó\neg }}\text{ }\mathrm{of}\text{ â¶Ä‰}\mathrm{O}+2×\left({\mathrm{M}}_{\mathrm{Ó\neg }}\text{ }\mathrm{of}\text{ â¶Ä‰}\mathrm{N}\right)+4×\left({\mathrm{M}}_{\mathrm{Ó\neg }}\text{ }\mathrm{of}\text{ â¶Ä‰}\mathrm{H}\right)\\ =12.01\text{ }\frac{\mathrm{g}}{\mathrm{mol}}+16.00\text{ }\frac{\mathrm{g}}{\mathrm{mol}}+2×\left(14.01\text{ }\frac{\mathrm{g}}{\mathrm{mol}}\right)+4×\left(1.01\text{ }\frac{\mathrm{g}}{\mathrm{mol}}\right)\\ =60.07\text{ }\frac{\mathrm{g}}{\mathrm{mol}}\end{array}$

In 1 mol of $\mathrm{CO}{\left({\mathrm{NH}}_2\right)}_2$, the mass is 60.07g and the number of moles of N atoms is 2. Since there is 2 mole of N atoms, the mass percentage of N is as follows.

Mass % of N $=\frac{2\text{ }\mathrm{mol}\text{ â¶Ä‰}\mathrm{N}\left(\frac{14.01\text{ }\mathrm{g}\text{ }\mathrm{N}}{\mathrm{mol}\text{ â¶Ä‰}\mathrm{N}}\right)}{60.07\mathrm{g}\text{ â¶Ä‰}\mathrm{CO}{\left({\mathrm{NH}}_2\right)}_2}×100=46.65\mathrm{\%}$.

Therefore, Below is the trend of the effectiveness of the given fertilizer, from the most to the least.

Urea > ammonium nitrate > ammonium sulfate > potassium nitrate.