Question

Nitrogen percentage is the highest in which of the fertilizers:
(A) Ammonium sulphate
(B) $CAN$
(C) Urea
(D) Calcium cyanamide

Answer 1

Hint: To determine the highest percentage of nitrogen in the given fertilizers we will apply the concept of percentage composition. We calculate the percentage composition of an element by dividing the mass of the single element of the compound to the total mass of the compound and then multiplying it by $100$ .

Formula used: $\% of(N)$ $ = \dfrac{{M.W.of(N)}}{{M.W.of(Compound)}} \times 100$$ ,$where $M.W.$ means molecular weight.

Complete step-by-step answer:1. Ammonium sulphate:
We know the molecular formula of ammonium sulphate is ${(N{H_4})_2}S{O_4}$ so to determine the percentage of nitrogen we need to find the molecular weight of nitrogen and also of the compound.
So, molecular weight of nitrogen $ = 14 \times 2 = 28$, $14$ is multiplied by two as two molecules of nitrogen are present.
Molecular weight of the compound$ = 132$
Putting the values in the formula $\% of(N)$$ = \dfrac{{M.W.of(N)}}{{M.W.of(Compound)}} \times 100$ we get:
$\dfrac{{28}}{{32}} \times 100 = 21.2\% $ , hence percentage of nitrogen in ammonium sulphate is $21.2\% $
2. $CAN$: Calcium ammonium nitrate:
The molecular formula of $CAN$ will be $Ca{H_4}{N_4}{O_9}$ , so in order to find the percentage of nitrogen we will determine the molecular weight of nitrogen in this compound. Also, we will find the molecular weight of the whole compound.
So, the molecular weight of nitrogen $ = 14 \times 4 = 56$, $14$ is multiplied by $4$ as $4$ molecules of nitrogen are present.
Molecular weight of the compound $ = 244$
Putting the values in the formula $\% of(N)$$ = \dfrac{{M.W.of(N)}}{{M.W.of(Compound)}} \times 100$ we get:
$\dfrac{{56}}{{244}} \times 100 = 22.9\% $ , hence percentage of nitrogen in $CAN$ is $22.9\% $
3. Urea:
The molecular formula for urea will be $N{H_2}CON{H_2}$ , so in order to find the percentage of nitrogen we will determine the molecular weight of nitrogen in this compound. Also, we will find the molecular weight of the whole compound.
So, the molecular weight of nitrogen $ = 14 \times 2 = 28$ ,$14$ is multiplied by two as two molecules of nitrogen are present.
Molecular weight of the compound $ = 60$
Putting the values in the formula $\% of(N)$$ = \dfrac{{M.W.of(N)}}{{M.W.of(Compound)}} \times 100$ we get:
$\dfrac{{28}}{{60}} \times 100 = 46.6\% $ , hence urea contains $46.6\% $ of nitrogen in it.
4. Calcium cyanamide:
The molecular formula for calcium cyanamide will be $CaC{N_2}$ , so in order to find the percentage of nitrogen we will determine the molecular weight of nitrogen in this compound. Also, we will find the molecular weight of the whole compound.
So, the molecular weight of nitrogen $ = 14 \times 2 = 28$ ,$14$ is multiplied by two as two molecules of nitrogen are present.
Molecular weight of the compound $ = 80$
Putting the values in the formula $\% of(N)$$ = \dfrac{{M.W.of(N)}}{{M.W.of(Compound)}} \times 100$ we get:
$\dfrac{{28}}{{80}} \times 100 = 35\% $ , hence the percentage of nitrogen in calcium cyanamide is $35\% $
From the above calculations we clearly see that the highest percentage of nitrogen is present in urea.

So, the correct answer is Option(C).

Note:To find the molecular weight of the compound we will add all the elements present in the compound and multiply each element with the number of atoms present in it. For example, in urea i.e. $N{H_2}CON{H_2}$ the molecular weight $ = 14 + 2 + 12 + 16 + 14 + 2 = 60$. Here $N = 14,H = 1 \times 2,C = 12,O = 16$ , hydrogen is multiplied by two as the compound contains two atoms of hydrogen, so the total number of hydrogen present will be $4$.