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Question: Nitrogen percentage is the highest in which of the fertilizers: (A) Ammonium sulphate (B) \(CAN\...

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

Explanation

Solution

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 100100 .

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

Complete step-by-step answer: 1. Ammonium sulphate:
We know the molecular formula of ammonium sulphate is (NH4)2SO4{(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 =14×2=28 = 14 \times 2 = 28, 1414 is multiplied by two as two molecules of nitrogen are present.
Molecular weight of the compound=132 = 132
Putting the values in the formula %of(N)\% of(N) =M.W.of(N)M.W.of(Compound)×100 = \dfrac{{M.W.of(N)}}{{M.W.of(Compound)}} \times 100 we get:
2832×100=21.2%\dfrac{{28}}{{32}} \times 100 = 21.2\% , hence percentage of nitrogen in ammonium sulphate is 21.2%21.2\%
2. CANCAN: Calcium ammonium nitrate:
The molecular formula of CANCAN will be CaH4N4O9Ca{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 =14×4=56 = 14 \times 4 = 56, 1414 is multiplied by 44 as 44 molecules of nitrogen are present.
Molecular weight of the compound =244 = 244
Putting the values in the formula %of(N)\% of(N) =M.W.of(N)M.W.of(Compound)×100 = \dfrac{{M.W.of(N)}}{{M.W.of(Compound)}} \times 100 we get:
56244×100=22.9%\dfrac{{56}}{{244}} \times 100 = 22.9\% , hence percentage of nitrogen in CANCAN is 22.9%22.9\%
3. Urea:
The molecular formula for urea will be NH2CONH2N{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 =14×2=28 = 14 \times 2 = 28 ,1414 is multiplied by two as two molecules of nitrogen are present.
Molecular weight of the compound =60 = 60
Putting the values in the formula %of(N)\% of(N) =M.W.of(N)M.W.of(Compound)×100 = \dfrac{{M.W.of(N)}}{{M.W.of(Compound)}} \times 100 we get:
2860×100=46.6%\dfrac{{28}}{{60}} \times 100 = 46.6\% , hence urea contains 46.6%46.6\% of nitrogen in it.
4. Calcium cyanamide:
The molecular formula for calcium cyanamide will be CaCN2CaC{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 =14×2=28 = 14 \times 2 = 28 ,1414 is multiplied by two as two molecules of nitrogen are present.
Molecular weight of the compound =80 = 80
Putting the values in the formula %of(N)\% of(N) =M.W.of(N)M.W.of(Compound)×100 = \dfrac{{M.W.of(N)}}{{M.W.of(Compound)}} \times 100 we get:
2880×100=35%\dfrac{{28}}{{80}} \times 100 = 35\% , hence the percentage of nitrogen in calcium cyanamide is 35%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. NH2CONH2N{H_2}CON{H_2} the molecular weight =14+2+12+16+14+2=60 = 14 + 2 + 12 + 16 + 14 + 2 = 60. Here N=14,H=1×2,C=12,O=16N = 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 44.