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

To find the empirical formula of the iron oxide, we follow these steps:

1. Determine the mass of each element in the compound. Let's assume we have a $100\ g$ sample of the iron oxide. Since it contains $69.9\%$ iron, the mass of iron ($Fe$) is $69.9\ g$.
The remaining mass must be oxygen ($O$).
Mass of $O = 100\ g - 69.9\ g = 30.1\ g$.

2. Convert the masses of the elements into moles using their molar masses.
Moles of $Fe = \frac{\text{Mass of } Fe}{\text{Molar mass of } Fe} = \frac{69.9\ g}{56\ g\ mol^{-1}} \approx 1.248\ mol$.
Moles of $O = \frac{\text{Mass of } O}{\text{Molar mass of } O} = \frac{30.1\ g}{16\ g\ mol^{-1}} \approx 1.881\ mol$.

3. Determine the simplest whole-number ratio of the atoms. To do this, divide the number of moles of each element by the smallest number of moles calculated in step 2 (which is $1.248$).
Ratio of $Fe = \frac{1.248}{1.248} = 1$.
Ratio of $O = \frac{1.881}{1.248} \approx 1.5$.

4. Convert these decimal ratios into whole numbers by multiplying both values by the same smallest possible integer (in this case, $2$).
Ratio of $Fe = 1 \times 2 = 2$.
Ratio of $O = 1.5 \times 2 = 3$.

Thus, the simplest whole-number ratio of $Fe$ atoms to $O$ atoms is $2:3$, making the empirical formula $Fe_{2}O_{3}$.