Question

Calculate the percentage composition of iron and oxygen in ${\mathrm{Fe}}_2{\mathrm{O}}_3$.

Given : Atomic weight of Fe = 55.8u, O = 16u

Answer 1

Step 1: Given data

Atomic weight of Fe = 55.8u, O = 16u

Step 2: Finding the mass for the given data

The molecular mass of ${\mathrm{Fe}}_2{\mathrm{O}}_3$ = (Number of iron atoms x atomic mass of iron) + (Number of oxygen atoms x atomic mass of oxygen)

The molecular mass of ${\mathrm{Fe}}_2{\mathrm{O}}_3$ = $\left(2\times 55.8\right)+\left(3\times 16\right)$

The molecular mass of ${\mathrm{Fe}}_2{\mathrm{O}}_3$ = $111.6+48=159.6$u

Step 3: Finding the mass for each element

Mass of $\mathrm{Fe}$ = (Number of iron atoms x atomic mass of iron) $=\left(2\times 55.8\right)=111.6u$

Mass of $\mathrm{O}$ = (Number of oxygen atoms x atomic mass of oxygen) $=\left(3\times 16\right)=48u$

Step 4: Finding the % composition

% composition is the proportion by mass of each element present in a compound.

% composition of $\mathrm{Fe}=\frac{111.6}{159.6}=69.92\%$

% composition of $\mathrm{O}=\frac{48}{159.6}=30.07\%$

Thus, the percentage composition of iron and oxygen in ${\mathrm{Fe}}_2{\mathrm{O}}_3$ is $69.92\%\mathrm{and}30.07\%$respectively.