Q.Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.

Concept: Empirical Formula from Percentage Composition

The empirical formula represents the simplest whole-number ratio of atoms in a compound. We convert mass percentages to moles, then find the smallest integer ratio.

Step 1: Assume 100 g of compound, so we have 69.9 g Fe and 30.1 g O₂.

Step 2: Convert to moles using atomic masses (Fe = 56 u, O = 16 u):

Step 3: Divide by the smallest (1.248) to get the ratio:

Multiplying by 2 gives whole numbers: Fe₂O₃.

Why this approach works

An empirical formula tells us the simplest whole-number ratio of atoms in a compound. Mass percentages alone don't reveal this ratio because different elements have different atomic masses—69.9 g of iron contains far fewer atoms than 69.9 g of oxygen. We convert mass to moles (which count particles) using molar masses, then scale to the smallest integers.

Step-by-step solution

1. Assume a 100 g sample

This makes the arithmetic transparent: 69.9% iron means 69.9 g Fe, and 30.1% oxygen means 30.1 g O in our sample.

2. Convert each mass to moles

The molar mass of iron is $M_{\mathrm{Fe}} = 56 \, \mathrm{g/mol}$, and for oxygen $M_{\mathrm{O}} = 16 \, \mathrm{g/mol}$.

3. Find the mole ratio

Divide both by the smaller number of moles to get the ratio:

4. Convert to whole numbers

The ratio $1 : 1.507$ is close to $1 : 1.5 = 1 : \frac{3}{2}$. Multiply both sides by 2 to clear the fraction:

This tells us there are 2 iron atoms for every 3 oxygen atoms.

5. Write the empirical formula

The simplest whole-number ratio Fe : O = 2 : 3 gives us $\mathrm{Fe_2O_3}$.

Method: Percentage Composition to Empirical Formula

This method uses the mass percentages of each element to find the simplest whole-number mole ratio — the empirical formula.

Steps

Step 1: Assume 100 g of the compound

This converts percentages directly into grams.

• Mass of iron (Fe) = $69.9 \, \text{g}$
• Mass of oxygen (O) = $30.1 \, \text{g}$

Step 2: Convert mass to moles

Use atomic masses:

• Fe = $55.85 \, \text{g/mol}$
• O = $16.00 \, \text{g/mol}$

Step 3: Find the simplest whole-number ratio

Divide each mole value by the smallest number of moles (here, 1.25):

Step 4: Convert to whole numbers

Multiply both by 2 to clear the decimal:

Step 5: Write the empirical formula

The simplest ratio is $\text{Fe}_2\text{O}_3$.

Final Answer: The empirical formula is $\mathbf{Fe_2O_3}$ (iron(III) oxide, or rust).

Here are the common mistakes students make when solving this classic empirical formula problem, along with how to avoid each.

1. Using the Wrong Atomic Masses

The Mistake:

Using atomic masses like $Fe = 55$ or $O = 16.0$ when the problem expects precise values (e.g., $Fe = 55.85$, $O = 16.00$). This shifts the mole ratio and can lead to a wrong formula.

How to Avoid:

Always use the standard atomic masses given in your textbook or exam data booklet. For this problem:

• Iron: $55.85 \, \text{g/mol}$
• Oxygen: $16.00 \, \text{g/mol}$

2. Confusing "Dioxygen" with "Oxygen Atom"

The Mistake:

Treating "30.1% dioxygen" as 30.1% oxygen atoms. Dioxygen ($O_2$) means the mass given is already for $O_2$ molecules, but in the oxide, oxygen is present as atoms.

How to Avoid:

Remember: dioxygen = $O_2$ . The percentage is the mass of oxygen atoms (since the oxide contains O atoms, not $O_2$ molecules). So you directly use 30.1 g of oxygen atoms per 100 g of compound.

3. Dividing by Atomic Mass Instead of Molar Mass

The Mistake:

Dividing the mass of oxygen by $32.00 \, \text{g/mol}$ (molar mass of $O_2$) instead of $16.00 \, \text{g/mol}$ (atomic mass of O).

How to Avoid:

Always convert mass of element to moles of atoms, not molecules.

Correct step:

• Moles of Fe = $\frac{69.9}{55.85}$
• Moles of O = $\frac{30.1}{16.00}$

4. Rounding Mole Ratios Too Early

The Mistake:

Rounding 1.25 to 1 or 1.3 before dividing by the smallest number, leading to a wrong ratio like $FeO$ instead of $Fe_2O_3$.

How to Avoid:

Keep at least 3 decimal places until the final step. Only round after dividing by the smallest mole value and checking if the ratio is close to a whole number.

Example:

• Moles of Fe = $69.9 / 55.85 = 1.251$
• Moles of O = $30.1 / 16.00 = 1.881$
• Divide by smallest (1.251):
  • Fe: $1.251 / 1.251 = 1.00$
  • O: $1.881 / 1.251 = 1.504$
• Multiply by 2 to get whole numbers: $Fe_2O_3$

5. Forgetting to Multiply to Get Whole Numbers

The Mistake:

Stopping at $Fe_{1}O_{1.5}$ and writing the formula as $FeO_{1.5}$.

How to Avoid:

Empirical formulas must have whole-number subscripts. If a ratio ends in .5, .33, or .25, multiply all subscripts by the smallest integer that clears the decimal:

• $1.5 \rightarrow$ multiply by 2
• $1.33 \rightarrow$ multiply by 3
• $1.25 \rightarrow$ multiply by 4

6. Writing the Formula Backwards

The Mistake:

Writing $O_3Fe_2$ instead of $Fe_2O_3$.

How to Avoid:

In empirical formulas, write the metal first, then the non-metal. Standard convention: $Fe_2O_3$, not $O_3Fe_2$.

Quick Checklist to Avoid All Mistakes

Final correct answer: $\boxed{Fe_2O_3}$