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If the fertilizers listed below are priced according to their nitrogen content, which will be the least expensive per 50 kg bag?
(A) Urea, ${(N{H_2})_2}CO$
(B) Ammonia, $N{H_3}$
(C) Ammonium Nitrate, $N{H_4}N{O_3}$
(D) Guanidine, $HNC{(N{H_3})_2}$


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Last updated date: 20th Jun 2024
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Answer
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Hint: To find the nitrogen content in the compound, we can use the following formula
\[\% N = \dfrac{{{\text{Weight of N in the compound}}}}{{{\text{Molecular weight of the compound}}}} \times 100\]
Atomic weight of nitrogen is $14gmmo{l^{ - 1}}$.

Complete step by step solution:
We will find the % nitrogen in all the compounds. In order to find this, we will first find the molecular weight of all the given compounds. Finally from the %N composition, we will be able to predict the cheapest fertilizer.
Molecular weight of ${(N{H_2})_2}CO$ = 2(Atomic weight of N) + 4 (Atomic weight of H) + Atomic weight of C + Atomic weight of O
Molecular weight of ${(N{H_2})_2}CO$ = 2(14) + 4(1) + 12 + 16 = 60$gmmo{l^{ - 1}}$
Molecular weight of $N{H_3}$ = Atomic weight of N + 3(Atomic weight of H)
Molecular weight of $N{H_3}$ = 14 + 3(1) = 17$gmmo{l^{ - 1}}$
Molecular weight of $N{H_4}N{O_3}$ = 2(Atomic weight of N) + 4(Atomic weight of H) + 3(Atomic weight of O)
Molecular weight of $N{H_4}N{O_3}$ = 2(14) + 4(1) + 3(16) = 80 $gmmo{l^{ - 1}}$
Molecular weight of $HNC{(N{H_3})_2}$ = 3(Atomic weight of N) + 7(Atomic weight of H) + Atomic weight of C
Molecular weight of $HNC{(N{H_3})_2}$ = 3(14) + 7(1) + 12 = 61$gmmo{l^{ - 1}}$
Now, we will find the %Nitrogen in all the given compounds using the below given formula.
\[\% N = \dfrac{{{\text{Weight of N in the compound}}}}{{{\text{Molecular weight of the compound}}}} \times 100{\text{ }}....{\text{(1)}}\]
 For Urea:
We know that the weight of N in the compound is $28gmmo{l^{ - 1}}$. So, we can write the equation (1) as
\[\% N = \dfrac{{28}}{{60}} \times 100 = 46.66\% \]
For Ammonia:
Here, the weight of N in the compound is $14gmmo{l^{ - 1}}$. So, we can write the equation (1) as
\[\% N = \dfrac{{14}}{{17}} \times 100 = 82.35\% \]
For Ammonium nitrate:
In this compound, the total weight of N in the compound is 28$gmmo{l^{ - 1}}$. So, we can write the equation (1) as
\[\% N = \dfrac{{28}}{{80}} \times 100 = 35\% \]
For Guanidine:
Here, the weight of N in the compound is 42$gmmo{l^{ - 1}}$. So, we can write the equation (1) as
\[\% N = \dfrac{{42}}{{61}} \times 100 = 68.85\% \]
So, from the above results we can say that ammonium nitrate has the least % of nitrogen in its structure. So, as the prices are set according to the nitrogen content, it will be the least expensive amongst all.

Therefore, the correct answer is (C).

Note: Note that the % nitrogen content is not dependent on the weight of the sample. So, no matter the weight of the sample, % nitrogen will remain the same in all samples with different weights of the same compound.