Question

What is the percentage by mass of ammonium nitrate?

Answer 1

As we know that percentage composition of a molecule or a compound is the ratio of an amount of each element (or atom) to the total amount of individual elements (or atoms) in a molecule or a compound, which is then multiplied with 100.
Formula used: We will use the following formula:-
Percentage composition of each atom = $\dfrac{\text{Molar mass of the atom present}}{\text{Molar mass of the molecule}}$

Complete answer:
Let us first discuss about the percentage composition as follows:-
Percentage composition: It is the ratio of an amount of each element (or atom) to the total amount of individual elements (or atoms) in a molecule or a compound, which is then multiplied with 100.
-The mass of 1 mole of ammonium nitrate ($N{{H}_{4}}N{{O}_{3}}$):-
Molar mass of N = 14g/mol
Molar mass of H = 1g/mol
Molar mass of O = 16g/mol
So total molar mass of$N{{H}_{4}}N{{O}_{3}}$= [2(14) + 4(1) + 3(16)]g/mol = 80g/mol
-Percentage composition of nitrogen (N) in ammonium nitrate ($N{{H}_{4}}N{{O}_{3}}$):-
Total mass of nitrogen present in the molecule of $N{{H}_{4}}N{{O}_{3}}$= 2(14)g/mol = 28g/mol
So percent composition of N = $\dfrac{\text{Molar mass of the atom present}}{\text{Molar mass of the molecule}}$
$\begin{aligned} & \Rightarrow \dfrac{28g/mol}{80g/mol}\times 100\% \\\ & \Rightarrow 35\% \\\ \end{aligned}$
-Percentage composition of hydrogen (H) in ammonium nitrate ($N{{H}_{4}}N{{O}_{3}}$):-
Total mass of hydrogen present in the molecule of $N{{H}_{4}}N{{O}_{3}}$= 4(1)g/mol = 4g/mol
So percent composition of H = $\dfrac{\text{Molar mass of the atom present}}{\text{Molar mass of the molecule}}$
$\begin{aligned} & \Rightarrow \dfrac{4g/mol}{80g/mol}\times 100\% \\\ & \Rightarrow 5\% \\\ \end{aligned}$
-Percentage composition of oxygen (O) in ammonium nitrate ($N{{H}_{4}}N{{O}_{3}}$):-
Total mass of oxygen present in the molecule of $N{{H}_{4}}N{{O}_{3}}$= 3(16)g/mol = 48g/mol
So percent composition of O = $\dfrac{\text{Molar mass of the atom present}}{\text{Molar mass of the molecule}}$
$\begin{aligned} & \Rightarrow \dfrac{48g/mol}{80g/mol}\times 100\% \\\ & \Rightarrow 60\% \\\ \end{aligned}$

Note:
-Always remember to solve these questions along with the units so as to obtain an accurate result with minimum errors.
-Also we must remember and learn atomic numbers and molar masses of certain elements for time saving and easy solving.