Question

What is the percentage by mass of iron (Fe) in the mineral hematite (Fe $_{2}$ O $_{3}$ )?

Answer 1

Hint : As we know the formula for atomic composition of hematite ore i.e., (Fe $_{2}$ O $_{3}$ )and then it would be easy for us to find the mass composition with the help of atomic masses of the element.

Complete Step By Step Answer:
As we know that the hematite ore has formula (Fe $_{2}$ O $_{3}$ )which tells us that every sample of substance contains $2$ atoms of iron (Fe) and 3 atoms of oxygen (O).
We know the atomic masses of the Fe and O from the periodic table i.e., $55\cdot 85$ grams and $16$ grams respectively. From their atomic weight we came to know that Fe atoms are heavier i.e., having more mass composition than O atoms.
When we take just two atoms of Fe and three atoms O it combines to form (Fe $_{2}$ O $_{3}$ )molecule having Molecular weight of (Fe $_{2}$ O $_{3}$ ) is $2 \times 55\cdot 85+3 \times 16\cdot 0=160$ gram which represents the minimum amount of hematite (as we take the least amount i.e., one atom of hematite ore).
Percentage composition by mass of any compound =molar mass of atom \times $100$ /molar mass of the complete molecule.
Percentage composition by mass of iron (Fe)= $2 \times$ (molar mass of iron) \times $100$ /molecular mass of the molecule (Fe $_{2}$ O $_{3}$ ). $\Rightarrow$ $2 \times 55\cdot 85/160=69\cdot 81$ in mass of iron (Fe).
Therefore, $69\cdot 81$ by mass iron is present in hematite (Fe $_{2}$ O $_{3}$ ).

Note :
To find the percentage composition you may know a few things like: molecular mass of the atoms/elements and how to calculate the molecular mass of an unknown compound, you have to memorize the formula of percentage composition and how to apply in the question. By this information you can calculate percentage composition by mass easily.