Q3.27P - Chemistry: Molecular Nature Of Matter And Change

Question

The effectiveness of a nitrogen fertilizer is determined mainly by its mass % N. Rank the following fertilizer, most effective first: potassium nitrate; ammonium nitrate; ammonium sulfate; urea,$\mathbf{CO}\left({\mathrm{NH}}_2\right)\mathbf{.}$

Step-by-Step Solution

Answer

The Effectiveness of the given fertilizers is

Urea > ammonium nitrate > ammonium sulfate > potassium nitrate.

Step 1: To Calculate the molar mass of potassium nitrate, KNO 3

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

$\begin{array}{c}{\mathrm{M}}_{\mathrm{\omega }}\mathrm{of}{\mathrm{KNO}}_3={\mathrm{M}}_{\mathrm{\omega }}\mathrm{of}\mathrm{K}+{\mathrm{M}}_{\mathrm{\omega }}\mathrm{of}\mathrm{N}+3\times \left({\mathrm{M}}_{\mathrm{\omega }}\mathrm{of}\mathrm{O}\right)\\ =39.10\frac{\mathrm{g}}{\mathrm{mol}}+14.01\frac{\mathrm{g}}{\mathrm{mol}}+3\times \left(16.00\frac{\mathrm{g}}{\mathrm{mol}}\right)\\ =101.11\frac{\mathrm{g}}{\mathrm{mol}}\end{array}$

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 $=\frac{1\mathrm{mol}\mathrm{N}\left(\frac{14.01\mathrm{g}\mathrm{N}}{\mathrm{mol}\mathrm{N}}\right)}{101.11\mathrm{g}{\mathrm{KNO}}_3}\times 100=13.86\mathrm{\%}$.

Step 2: To Calculate the molar mass of ammonium nitrate, NH 4 NO 3

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

$\begin{array}{c}{\mathrm{M}}_{\mathrm{\omega }}\mathrm{of}{\mathrm{NH}}_4{\mathrm{NO}}_3=2\times \left({\mathrm{M}}_{\mathrm{\omega }}\mathrm{of}\mathrm{N}\right)+4\times \left({\mathrm{M}}_{\mathrm{\omega }}\mathrm{of}\mathrm{H}\right)+3\times \left({\mathrm{M}}_{\mathrm{\omega }}\mathrm{of}\mathrm{O}\right)\\ =2\times \left(14.01\frac{\mathrm{g}}{\mathrm{mol}}\right)+4\times \left(1.01\frac{\mathrm{g}}{\mathrm{mol}}\right)+3\times \left(16.00\frac{\mathrm{g}}{\mathrm{mol}}\right)\\ =80.06\frac{\mathrm{g}}{\mathrm{mol}}\end{array}$

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 $=\frac{2\mathrm{mol}\mathrm{N}\left(\frac{14.01\mathrm{g}\mathrm{N}}{\mathrm{mol}\mathrm{N}}\right)}{80.06\mathrm{g}{\mathrm{NH}}_4{\mathrm{NO}}_3}\times 100=35.00\mathrm{\%}$.

Step 3: To Calculate the molar mass of ammonium sulfate, ( NH 4 ) 2 SO 4

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

$\begin{array}{c}{\mathrm{M}}_{\mathrm{\omega }}\mathrm{of}{\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4=2\times \left({\mathrm{M}}_{\mathrm{\omega }}\mathrm{of}\mathrm{N}\right)+8\times \left({\mathrm{M}}_{\mathrm{\omega }}\mathrm{of}\mathrm{H}\right)+{\mathrm{M}}_{\mathrm{\omega }}\mathrm{of}\mathrm{S}+4\times \left({\mathrm{M}}_{\mathrm{\omega }}\mathrm{of}\mathrm{O}\right)\\ =2\times \left(14.01\frac{\mathrm{g}}{\mathrm{mol}}\right)+8\times \left(1.01\frac{\mathrm{g}}{\mathrm{mol}}\right)+32.07\frac{\mathrm{g}}{\mathrm{mol}}+4\times \left(16.00\frac{\mathrm{g}}{\mathrm{mol}}\right)\\ =132.17\frac{\mathrm{g}}{\mathrm{mol}}\end{array}$

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 $=\frac{2\mathrm{mol}\mathrm{N}\left(\frac{14.01\mathrm{g}\mathrm{N}}{\mathrm{mol}\mathrm{N}}\right)}{132.17\mathrm{g}{\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4}\times 100=21.20\mathrm{\%}$.

Step 4: To Calculate the molar mass of urea, CO ( NH 2 ) 2

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

$\begin{array}{c}{\mathrm{M}}_{\mathrm{\omega }}\mathrm{of}\mathrm{CO}{\left({\mathrm{NH}}_2\right)}_2={\mathrm{M}}_{\mathrm{\omega }}\mathrm{of}\mathrm{C}+{\mathrm{M}}_{\mathrm{\omega }}\mathrm{of}\mathrm{O}+2\times \left({\mathrm{M}}_{\mathrm{\omega }}\mathrm{of}\mathrm{N}\right)+4\times \left({\mathrm{M}}_{\mathrm{\omega }}\mathrm{of}\mathrm{H}\right)\\ =12.01\frac{\mathrm{g}}{\mathrm{mol}}+16.00\frac{\mathrm{g}}{\mathrm{mol}}+2\times \left(14.01\frac{\mathrm{g}}{\mathrm{mol}}\right)+4\times \left(1.01\frac{\mathrm{g}}{\mathrm{mol}}\right)\\ =60.07\frac{\mathrm{g}}{\mathrm{mol}}\end{array}$

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 $=\frac{2\mathrm{mol}\mathrm{N}\left(\frac{14.01\mathrm{g}\mathrm{N}}{\mathrm{mol}\mathrm{N}}\right)}{60.07\mathrm{g}\mathrm{CO}{\left({\mathrm{NH}}_2\right)}_2}\times 100=46.65\mathrm{\%}$.

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.