Question

An oxide of iron contains $69.9\%$ iron, its empirical formula, is:
(Given : Molar mass of $Fe$ and $O$ are $56$ and $16\ g\ mol^{-1}$ respectively.)

• A. $FeO$
• B. $Fe_{2}O_{3}$
• C. $Fe_{3}O_{4}$
• D. $FeO_{3}$

Hint:

Find the number of moles for iron and oxygen by dividing their mass percentages by their respective atomic weights, then find the simplest whole-number ratio.

Answer 1

The Correct Option is B

We can solve this by calculating the molar ratio of the constituents directly from their mass percentages.

Concept: The empirical formula represents the simplest whole-number ratio of moles of atoms present in one mole of a compound.

Step 1: Write down the mass percentage of each element.
Percentage of $Fe = 69.9\%$
Percentage of $O = 100\% - 69.9\% = 30.1\%$

Step 2: Divide the percentages by the respective atomic masses to find the relative number of moles.
Relative moles of $Fe = \frac{69.9}{56} = 1.248$
Relative moles of $O = \frac{30.1}{16} = 1.881$

Step 3: Find the simplest molar ratio by dividing each by the lowest value ($1.248$).
$$Fe : O = \frac{1.248}{1.248} : \frac{1.881}{1.248}$$
$$Fe : O = 1 : 1.507$$

Step 4: As the ratio contains a non-integer ($1.5$), we multiply both sides of the ratio by $2$ to obtain the nearest whole numbers.
$$Fe : O = (1 \times 2) : (1.5 \times 2) = 2 : 3$$

The resulting empirical formula is $Fe_{2}O_{3}$.