Question

Find the mass of the nitrogen contained in $1kg$ of
(i) $KN{O_3}$
(ii) $N{H_4}N{O_3}$
(iii) ${\left( {N{H_4}} \right)_2}HP{O_4}$

Answer 1

The mass of nitrogen depends on the moles of the nitrogen present in the compound. The moles are directly related to the number of nitrogen in the chemical formula.

Complete answer:
The amount of each of the given salts is $1kg$. The molar mass of each of the salts is determined first. Then the actual moles of the compound will be evaluated. The count of the number of nitrogen on each compound is calculated from the moles of each compound present in $1kg$.
(i) $KN{O_3}$. This compound is known as potassium nitrate. It consists of one potassium atom, one nitrogen atom and three oxygen atoms. The molar mass of $KN{O_3}$ = atomic mass of $K$ + atomic mass of $N$ + $3{\text{ }} \times$ atomic mass of $O$ .
= $39 + 14 + 3 \times 16 = 101$.
Moles of $KN{O_3}$ = $\dfrac{{amount{\text{ }}of{\text{ }}KN{O_3}}}{{molar{\text{ }}mass{\text{ }}of{\text{ }}KNO3}} = \dfrac{{1000}}{{101}} = 9.9moles$.
Thus the mass of nitrogen in $9.9moles$ of $KN{O_3}$ = $9.9 \times 14{\text{ }} = \;\;138.6g$ .
(ii) $N{H_4}N{O_3}$. This compound is known as ammonium nitrate. It consists of two nitrogen atoms, four hydrogen atoms and three oxygen atoms. The molar mass of $N{H_4}N{O_3}$ = $2{\text{ }} \times$ atomic mass of $N$ + $4{\text{ }} \times$ atomic mass of $H$ + $3{\text{ }} \times$ atomic mass of $O$.
= $2 \times 14 + 4 \times 1 + 3 \times 16 = 80g$ .
Moles of $N{H_4}N{O_3}$ = $\dfrac{{amount{\text{ }}of{\text{ }}N{H_4}N{O_3}}}{{molar{\text{ }}mass{\text{ }}of{\text{ }}N{H_4}N{O_3}}} = \dfrac{{1000}}{{80}} = 12.5moles$.
Thus the mass of nitrogen present in $12.5moles$ of $N{H_4}N{O_3}$ = $12.5 \times (2 \times 14) = 350g$ .
(iii) ${\left( {N{H_4}} \right)_2}HP{O_4}$. This compound is known as ammonium phosphate. It consists of two nitrogen atoms, nine hydrogen atoms, one phosphorous atom and four oxygen atoms. The molar mass of ${\left( {N{H_4}} \right)_2}HP{O_4}$ = $2{\text{ }} \times$ atomic mass of $N$ + $9{\text{ }} \times$ atomic mass of $H$ + atomic mass of $P$ + $4{\text{ }} \times$ atomic mass of $O$ .
=$2 \times 14 + 9 \times 1 + 31 + 4 \times 16 = 132g$.
Moles of ${\left( {N{H_4}} \right)_2}HP{O_4}$ = $\dfrac{{amount{\text{ }}of{\text{ }}{{(N{H_4})}_2}P{O_4}}}{{molar{\text{ }}mass{\text{ }}of{\text{ }}{{(N{H_4})}_2}P{O_4}}} = \dfrac{{1000}}{{132}} = 7.6moles$.
Thus the mass of nitrogen present in $7.6moles$ of ${\left( {N{H_4}} \right)_2}HP{O_4}$ = $7.6 \times (2 \times 14) = 212.8g$.

Note: The mole of a substance is the ratio of the weight of the substance and the molar mass or atomic mass of the substance. Thus one mole of a substance means the weight taken of the substance is equal to the molar mass of the substance.