Question

Which of the following compounds contains the highest percentage of nitrogen?

(a) Ca (NO₃)₂

(b) CaCN₂

(c) (NH₄)₂SO₄

(Ca=40,N=14,O=16.C=12,H=1,S=32)

Answer 1

(a) $\mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2$

Step 1: Write all the given values

$\mathrm{Ca}$=40, $\mathrm{N}$=14, $\mathrm{O}$=16

Step 2: Write the formula to find the molecular mass

The molecular mass of $\mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2$

= 1 atom of $\mathrm{Ca}$ + 2 atoms of $\mathrm{N}$ + 6 atoms of $\mathrm{O}$

= 40 + 2 x 14 + 6 x 16

= 164 u

Step 3: Write the formula to find the percentage

Percentage of N in $\mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2$ = $\frac{MassofN}{MassofNinCa{\left(N{O}_3\right)}_2}$ x 100

= $\frac{2x14}{164}\times 100$

= $\frac{2800}{164}$

= $17.08\%$

Step 4: Write the final answer

Hence, the Percentage of $\mathrm{N}$ in $\mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2$ is 17.08 %

(b) ${\mathrm{CaCN}}_2$

Step 1: Write all the given values

$\mathrm{Ca}$=40, $\mathrm{C}$= 12 $\mathrm{N}$=14,

Step 2: Write the formula to find the molecular mass

The molecular mass of ${\mathrm{CaCN}}_2$

= 1 atom of $\mathrm{Ca}$ + 1 atom of $\mathrm{C}$ + 2 atoms of $\mathrm{N}$

= 40 + 12 + 2 x 14

= 40 + 12 + 28

= 80 u

Step 3: Write the formula to find the percentage

Percentage of $\mathrm{N}$ in ${\mathrm{CaCN}}_2$ = $\frac{MassofN}{MassofNinCaC{N}_2}$ x 100

= $\frac{14x100}{80}$

= 17.5 %

Percentage of $\mathrm{N}$ in ${\mathrm{CaCN}}_2$ = 2 x 17.5 = 35 %

Step 4: Write the final answer

Hence, the Percentage of $\mathrm{N}$ in ${\mathrm{CaCN}}_2$ is 35 %

(c) ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$

Step 1: Write all the given values

$\mathrm{N}$=14, $\mathrm{H}$=1

$\mathrm{O}$=16 $\mathrm{S}$=32

Step 2: Write the formula to find the molecular mass

The molecular mass of ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$

= 2 atoms of $\mathrm{N}$ + 8 atoms of $\mathrm{H}$ + 1 atoms of $\mathrm{S}$ + 4 atoms of $\mathrm{O}$

= 2 x 14 + 8 x 1 + 1 x 32 + 4 x 16

28 + 8 + 32 + 64

= 132 u

Step 3: Write the formula to find the percentage

Percentage of $\mathrm{N}$ in ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$ = $\frac{MassofN}{Massof{\left(N{H}_4\right)}_{2}S{O}_4}$ x 100

= $\frac{14x100}{132}$

= 10.60 %

The percentage of $\mathrm{N}$ in ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$ is 2 x 10.60 = 21.21 %

Step 4: Write the final answer

Hence, the Percentage of $\mathrm{N}$ in ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$ is 21.21%

Thus, ${\mathrm{CaCN}}_2$ (35% of Nitrogen) contains the highest percentage of nitrogen.