Question

Calculate the percentage composition of iron present in ferric oxide $\left( {F{e_2}{O_3}} \right)$ (Atomic mass of $Fe = 56$ )
A. $70\% $
B. $30\% $
C. $40\% $
D. $56.5\% $

Answer 1

Hint: percentage composition is defined as the ratio of amount of each element to the total amount of the compound or molecule multiplied by $100$ .It is given by the formula: Percentage composition $\% = \dfrac{{{m_e}}}{{{m_t}}} \times 100$ . Using the given data we will find the percentage of iron present in ferric oxide.

Complete step by step answer:
Percentage composition is given by the formula $\% = \dfrac{{{m_e}}}{{{m_t}}} \times 100$
Where
$\% = $ percentage composition of an element
${m_e} = $ mass of each element
${m_t} = $ total mass of compound or molecule
$100 = $ it is used to multiply to get a percentage.
Mass of iron present in ferric oxide $ = 56 \times 2$
Mass of iron present in ferric oxide $ = 112$
Now, we will calculate the molecular mass of each element in ferric oxide$\left( {F{e_2}{O_3}} \right)$.
Given: atomic mass of iron $Fe = 56$
Atomic mass of oxygen$\left( O \right) = 16$
So, the molecular mass is given by the formula: Molecular mass of molecule ${A_2}{B_3} = 2 \times $ atomic mass of $A + 3 \times $atomic mass of $B$ .
We will use this formula to calculate the molecular mass of ferric oxide $\left( {F{e_2}{O_3}} \right)$ ,
Molecular mass of ferric oxide $\left( {F{e_2}{O_3}} \right) = 2 \times $ of atomic mass of $Fe + 3 \times $atomic mass of $O$ .
Substituting the values of atomic masses in the above equation we get,
Molecular mass of ferric oxide $\left( {F{e_2}{O_3}} \right) = 2 \times 56 + 3 \times 16$
Molecular mass of ferric oxide $\left( {F{e_2}{O_3}} \right) = 112 + 48$
Molecular mass of ferric oxide $\left( {F{e_2}{O_3}} \right) = 160$
Now we know the amount of iron in ferric oxide $ = 112$ and the total mass of ferric oxide $ = 160$ . From these values we will calculate the percentage composition of iron in ferric oxide by using the formula given below: $\% = \dfrac{{{m_e}}}{{{m_t}}} \times 100$
Substituting the values in the above equation we get,
$\% = \dfrac{{112}}{{160}} \times 100$
$\% = 0.7 \times 100$
$\% = 70$
The percentage composition of iron in ferric oxide is $70\% $

So, the correct answer is “Option A”.

Note: Percentage composition gives us the amount of each element present in the compound .
It is very helpful in the chemical analysis of any type of compound.
The sum of all the mass percentages should be $100$ percent.