Question

What is the percentage by mass of sodium nitrate?

Answer 1

We very well know that percentage composition of a molecule or a compound or a complex is the amount of each element or atom divided by the total amount of individual elements or atoms in that particular molecule or compound and this 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 begin with discussion of percentage composition as follows:-
Percentage composition: It refers to the ratio of an amount of each element to the total amount of individual elements in a particular molecule, compound or complex which is further multiplied with 100. Mathematically it is represented as follows:-
Percentage composition of each atom = $\dfrac{\text{Molar mass of the atom present}}{\text{Molar mass of the molecule}}$
-The mass of 1 mole of sodium nitrate ($NaN{{O}_{3}}$):-
Molar mass of Na = 23 g/mol
Molar mass of N = 14 g/mol
Molar mass of O = 16g/mol
So total molar mass of$NaN{{O}_{3}}$= [(23) + (14) + 3(16)]g/mol = 85 g/mol
-Percentage composition of sodium (Na) in sodium nitrate ($NaN{{O}_{3}}$):-
Total mass of sodium present in the molecule of $NaN{{O}_{3}}$= 23 g/mol
So percent composition of Na = $\dfrac{\text{Molar mass of the atom present}}{\text{Molar mass of the molecule}}$
$\begin{aligned} & \Rightarrow \dfrac{23g/mol}{85g/mol}\times 100\% \\\ & \Rightarrow 27.05\% \\\ \end{aligned}$
-Percentage composition of nitrogen (N) in sodium nitrate ($NaN{{O}_{3}}$):-
Total mass of nitrogen present in the molecule of $NaN{{O}_{3}}$= 14 g/mol
So percent composition of N = $\dfrac{\text{Molar mass of the atom present}}{\text{Molar mass of the molecule}}$
$\begin{aligned} & \Rightarrow \dfrac{14g/mol}{85g/mol}\times 100\% \\\ & \Rightarrow 16.47\% \\\ \end{aligned}$
-Percentage composition of oxygen (O) in sodium nitrate ($NaN{{O}_{3}}$):-
Total mass of oxygen present in the molecule of $NaN{{O}_{3}}$= 3(16) g/mol = 48 g/mol
So percent composition of O = $\dfrac{\text{Molar mass of the atom present}}{\text{Molar mass of the molecule}}$
$\begin{aligned} & \Rightarrow \dfrac{48g/mol}{85g/mol}\times 100\% \\\ & \Rightarrow 56.47\% \\\ \end{aligned}$

Note:
-Remember that the sum of mass percentage of all the elements present in a molecule or compound is always equal to hundred.
-Also try to remember and learn atomic numbers and molar masses of important elements from the periodic table for time saving.