Finding Empirical Formula from Percent Composition- Chemistry Guide

What Is an Empirical Formula?

The empirical formula shows the simplest whole-number ratio of atoms in a compound. That's it. Nothing fancy.

For example:

• Glucose (C₆H₁₂O₆) has an empirical formula of CH₂O
• Benzene (C₆H₆) has an empirical formula of CH
• Water is H₂O — the empirical formula is the same as the molecular formula

The empirical formula doesn't tell you how many atoms are actually in a molecule. It only tells you the ratio.

What Is Percent Composition?

Percent composition tells you the mass percentage of each element in a compound. If a compound is 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass, those numbers come from the formula itself.

Percent composition data is what you start with when finding an empirical formula. You'll usually get it from combustion analysis or similar experimental methods.

The Method: Step by Step

Here's the four-step process. Master this and you can solve any empirical formula problem.

Step 1: Convert Percentages to Grams

Assume you have 100 grams of the compound. This makes the math trivial — the percentages become the gram values directly.

If something is 40% carbon, you have 40 grams of carbon. 6.7% hydrogen means 6.7 grams of hydrogen. No conversion needed.

Step 2: Convert Grams to Moles

Divide each mass by the element's atomic mass.

• Carbon: 40g ÷ 12.01 g/mol = 3.33 mol
• Hydrogen: 6.7g ÷ 1.008 g/mol = 6.65 mol
• Oxygen: 53.3g ÷ 16.00 g/mol = 3.33 mol

Step 3: Find the Mole Ratio

Divide all mole values by the smallest number of moles you calculated.

In this case, the smallest is 3.33. So:

• Carbon: 3.33 ÷ 3.33 = 1
• Hydrogen: 6.65 ÷ 3.33 = 2
• Oxygen: 3.33 ÷ 3.33 = 1

Step 4: Write the Empirical Formula

Use the ratios as subscripts. The mole ratios are 1 : 2 : 1.

Empirical formula: CH₂O

Worked Example

Problem: A compound contains 72.0% iron and 28.0% oxygen by mass. Find the empirical formula.

Step 1: 100 g sample → 72.0 g Fe, 28.0 g O

Step 2: Convert to moles

• Fe: 72.0g ÷ 55.85 g/mol = 1.29 mol
• O: 28.0g ÷ 16.00 g/mol = 1.75 mol

Step 3: Divide by smallest value (1.29)

• Fe: 1.29 ÷ 1.29 = 1
• O: 1.75 ÷ 1.29 = 1.36

1.36 is not a whole number. You need to multiply to clear decimals.

Multiply both by 2 → Fe: 2, O: 2.72 ≈ 11/4 = 2.75

Try multiplying by 4 → Fe: 4, O: 5.44 ≈ 5.44

That still doesn't work cleanly. Let's check: 1.36 × 3 = 4.08 ≈ 4

Multiply by 3 instead → Fe: 3, O: 4.08 ≈ 4

Close enough for whole numbers. The empirical formula is Fe₃O₄. This is magnetite.

Quick Reference Table

Common Mistakes to Avoid

• Forgetting to divide by the smallest mole value. Some students divide by something random. Don't. Always the smallest.
• Rounding too early. Keep extra decimal places until the end.
• Panicking when you get decimals. Multiply by a small integer (2, 3, 4) to clear them. That's normal.
• Using wrong atomic masses. Memorize the common ones: C=12, H=1, O=16, N=14. For anything else, use the periodic table.

Practice Tips

Work through at least five problems before calling it done. The calculation itself isn't hard — the trap is rushing and making arithmetic errors.

When you get a decimal like 1.33 or 1.67, multiply everything by 3. These are fractions: 1.33 ≈ 4/3, 1.67 ≈ 5/3.

When you get 1.2, 1.4, or 1.5, multiply by 5 to clear the decimal.

The goal is whole numbers. If you can't get them with simple multiplication, check your arithmetic.