Question

What is the percentage by mass of each element in $FeC{{O}_{3}}$?

Answer 1

Approach this question by the concept of percentage composition of a compound. As we know, the percentage composition of a compound or a complex is the mass of each element or atom divided by the total mass of the compound and it is then multiplied with 100.
Formula used:
We will use the following formula:-
Percentage composition of each atom = $\dfrac{\text{Molar mass of the atom present}}{\text{Molar mass of the molecule}}\times 100\%$

Complete answer:
Let us first discuss about percentage composition as follows:-
Percentage composition: It is generally referred to the ratio of mass of each element to the total mass of individual elements in a particular compound or complex and it is further multiplied with 100. Mathematically it can be represented as follows:-
Percentage composition of each atom = $\dfrac{\text{Molar mass of the atom present}}{\text{Molar mass of the molecule}}\times 100\%$

-The molar mass of 1 mole of ferrous carbonate ($FeC{{O}_{3}}$):-
Molar mass of Fe = 56 g/mol
Molar mass of C = 12 g/mol
Molar mass of O = 16g/mol
So total molar mass of$FeC{{O}_{3}}$= [(56) + (12) + 3(16)]g/mol = 116 g/mol

-Percentage composition of iron (Fe) in ferrous carbonate ($FeC{{O}_{3}}$):-
Total mass of iron present in the molecule of $FeC{{O}_{3}}$= 56 g/mol
So percent composition of Fe = $\dfrac{\text{Molar mass of the atom present}}{\text{Molar mass of the molecule}}\times 100\%$
$\begin{aligned} & \Rightarrow \dfrac{56g/mol}{116g/mol}\times 100\% \\\ & \Rightarrow 48.27\% \\\ \end{aligned}$

-Percentage composition of carbon (C) in ferrous carbonate ($FeC{{O}_{3}}$):-
Total mass of carbon present in the molecule of $FeC{{O}_{3}}$= 12 g/mol
So percent composition of C = $\dfrac{\text{Molar mass of the atom present}}{\text{Molar mass of the molecule}}\times 100\%$
$\begin{aligned} & \Rightarrow \dfrac{12g/mol}{116g/mol}\times 100\% \\\ & \Rightarrow 10.34\% \\\ \end{aligned}$

-Percentage composition of oxygen (O) in ferrous carbonate ($FeC{{O}_{3}}$):-
Total mass of oxygen present in the molecule of $FeC{{O}_{3}}$= 3(16) g/mol = 48 g/mol
So percent composition of O = $\dfrac{\text{Molar mass of the atom present}}{\text{Molar mass of the molecule}}\times 100\%$
$\begin{aligned} & \Rightarrow \dfrac{48g/mol}{116g/mol}\times 100\% \\\ & \Rightarrow 41.37\% \\\ \end{aligned}$

Note:
-Always remember that the sum of mass percentage of all the elements present in a compound or a complex will always be equal to 100.
-Also try to learn the atomic numbers and molar masses of various important elements from the periodic table in order to save time while solving these questions.