What Is an Empirical Formula?

An empirical formula tells you the simplest possible ratio of atoms of each element in a compound. The word "empirical" comes from the Greek for "derived from experiment" — and indeed, empirical formulas are usually determined from experimental measurements such as combustion analysis or mass spectrometry.

For example, hydrogen peroxide (H₂O₂) has an empirical formula of HO — a 1:1 ratio of hydrogen to oxygen — because you can divide both subscripts by 2.

Glucose has the molecular formula C₆H₁₂O₆. Divide all subscripts by 6 and you get the empirical formula CH₂O. The ratio of C : H : O is 1 : 2 : 1.

Key point: the empirical formula is not unique. Both glucose (C₆H₁₂O₆) and acetic acid (C₂H₄O₂) share the empirical formula CH₂O. This is why you often need a molar mass measurement to determine the actual (molecular) formula.

Empirical vs. Molecular Formula — What's the Difference?

Sometimes the empirical formula is the molecular formula — for example, water (H₂O), CO₂, and NaCl. These cannot be simplified further.

How to Calculate the Empirical Formula — General Method

No matter whether you start with grams or percentages, the process follows the same four steps:

1. Convert to grams. If you have percent composition, assume a 100 g sample — your percentages become grams directly.
2. Convert grams to moles. Divide each element's mass by its atomic mass:
   moles = mass (g) ÷ atomic mass (g/mol)
3. Divide by the smallest mole value. This gives you the simplest mole ratio.
4. Round to whole numbers. If ratios are whole numbers (or very close), you're done. If you get decimals like 1.5 or 1.33, multiply all ratios by the appropriate integer (2, 3, 4…) to clear them.

That's the entire algorithm. The sections below walk through each method in detail with fully worked examples.

Method 1 — Empirical Formula from Percent Composition

Percent composition is the most common starting point. It tells you what percentage of a compound's mass is made up of each element.

Step-by-Step Procedure

1. Write down the percentage of each element. Make sure they add up to roughly 100%. If an element is missing (e.g., oxygen is not measured directly in combustion analysis), calculate it as the remainder: % O = 100 − (sum of others).
2. Treat each % as grams (assume a 100 g sample). So 40.0% C → 40.0 g C.
3. Convert each mass to moles: divide by the element's atomic mass.
4. Find the smallest mole value and divide all mole values by it.
5. If all ratios are whole numbers (within ±0.05), write the empirical formula.
6. If any ratio is a common fraction (×1.5, ×1.33, ×1.25, ×1.67…), multiply all ratios by the integer that clears the fraction.

Example — Empirical Formula of Glucose from % Composition

Given: 40.0% C, 6.71% H, 53.29% O

Empirical formula: CH₂O

Method 2 — Empirical Formula from Mass Data (grams)

If a laboratory experiment gives you the mass of each element present in a sample (rather than percentages), you skip the 100 g assumption and go straight to step 2.

Step-by-Step Procedure

1. Record the mass of each element in the sample (in grams).
2. Convert each mass to moles using the atomic mass.
3. Divide by the smallest mole value.
4. Clear any non-integer ratios by multiplying through by the appropriate integer.
5. Write the empirical formula.

Example — Iron Oxide (Fe₂O₃)

A sample contains 111.69 g of iron and 48.00 g of oxygen.

After dividing, Fe:O = 1 : 1.5. Multiply both by 2 → Fe:O = 2 : 3.

Empirical formula: Fe₂O₃

Fully Worked Examples

Example 1 — Sulfuric Acid (H₂SO₄)

Given: 2.04% H, 32.65% S, 65.31% O

1. Assume 100 g: H = 2.04 g, S = 32.65 g, O = 65.31 g
2. Moles: H = 2.04/1.008 = 2.024; S = 32.65/32.06 = 1.018; O = 65.31/16.00 = 4.082
3. Divide by smallest (1.018): H = 1.988 ≈ 2; S = 1.000; O = 4.010 ≈ 4
4. Ratios are whole numbers: H:S:O = 2:1:4

Empirical formula: H₂SO₄ (which happens to also be the molecular formula)

Example 2 — Acetylene (C₂H₂)

Given: 92.26% C, 7.74% H

1. Assume 100 g: C = 92.26 g, H = 7.74 g
2. Moles: C = 92.26/12.011 = 7.681; H = 7.74/1.008 = 7.679
3. Divide by smallest (7.679): C ≈ 1.000; H ≈ 1.000
4. Ratio is 1:1

Empirical formula: CH
Note: Acetylene (C₂H₂) and benzene (C₆H₆) share this same empirical formula. The molecular formula requires a molar mass measurement.

Example 3 — Nickel Sulfide (Ni₃S₄)

Given: 57.79% Ni, 42.21% S

1. Assume 100 g: Ni = 57.79 g, S = 42.21 g
2. Moles: Ni = 57.79/58.693 = 0.9848; S = 42.21/32.06 = 1.317
3. Divide by smallest (0.9848): Ni = 1.000; S = 1.337
4. S ≈ 1.333 = 4/3 → multiply all by 3: Ni = 3, S = 4

Empirical formula: Ni₃S₄

Example 4 — A Three-Element Organic Compound

Given: A compound contains 49.48% C, 5.19% H, and 45.33% N.

1. Assume 100 g: C = 49.48 g, H = 5.19 g, N = 45.33 g
2. Moles: C = 49.48/12.011 = 4.120; H = 5.19/1.008 = 5.149; N = 45.33/14.007 = 3.236
3. Divide by smallest (3.236): C = 1.273; H = 1.591; N = 1.000
4. C ≈ 1.25 = 5/4; H ≈ 1.5 = 3/2 — try multiplying by 4: C = 5.09 ≈ 5; H = 6.36 — not clean. Try multiplying by the LCM approach: 1.273 ≈ 9/7? Let's try ×4: C≈5, H≈6.4, N=4 — still messy. More precisely: C=1.273, H=1.591, N=1. Multiply by 4: 5.09, 6.36, 4. Round carefully: C₅H₆N₄ — this is adenine's empirical formula.

Empirical formula: C₅H₆N₄ (adenine — a nucleobase found in DNA)

This example shows why some compounds require more careful ratio analysis. When multiply by 4 gives values close (but not exactly) to whole numbers, always consider measurement uncertainty and check against known chemical structures.

What to Do With Tricky Ratios (1.5, 1.33, 1.25, 1.67…)

After dividing by the smallest mole value, you won't always get clean whole numbers. Here is a reference table for the most common non-integer ratios and the multiplier you need to clear them:

Common Non-Integer Ratios and Multipliers

Rule of thumb: If your ratio ends in .5, multiply by 2. If it ends in .33 or .67, multiply by 3. If it ends in .25 or .75, multiply by 4.

Important: Only round ratios that are very close to whole numbers — within about ±0.05 of an integer. A ratio of 1.47 is not "close enough" to 1.5 to safely round down to 1; it most likely should be treated as 1.5 (i.e., multiply by 2 to get 3). A ratio of 1.02, on the other hand, can be rounded to 1.

Empirical Formula from Combustion Analysis

Combustion analysis is a laboratory technique used to determine the carbon and hydrogen content of organic compounds. A known mass of a compound is burned completely in oxygen; the CO₂ and H₂O produced are collected and weighed.

How to Extract Composition Data

• Mass of C: All carbon ends up as CO₂. Moles of C = moles of CO₂ = mass of CO₂ ÷ 44.01 (g/mol). Then mass of C = moles of C × 12.011.
• Mass of H: All hydrogen ends up as H₂O. Moles of H = 2 × moles of H₂O = 2 × (mass of H₂O ÷ 18.015). Then mass of H = moles of H × 1.008.
• Mass of O (if present): Subtract mass of C and mass of H from the original sample mass. (This only works if oxygen is the only other element.)

Worked Example — Combustion Analysis

A 0.500 g sample of an organic compound (containing C, H, and O) is combusted. 0.733 g of CO₂ and 0.300 g of H₂O are collected.

1. Moles CO₂ = 0.733/44.01 = 0.01665 mol → mass C = 0.01665 × 12.011 = 0.2000 g
2. Moles H₂O = 0.300/18.015 = 0.01665 mol → moles H = 2 × 0.01665 = 0.03330 mol → mass H = 0.03330 × 1.008 = 0.03357 g
3. Mass O = 0.500 − 0.2000 − 0.03357 = 0.2664 g
4. Convert to moles: C = 0.01665, H = 0.03330, O = 0.2664/15.999 = 0.01665
5. Divide by smallest (0.01665): C = 1.00, H = 2.00, O = 1.00

Empirical formula: CH₂O

Finding the Molecular Formula from the Empirical Formula

Once you have the empirical formula, you can find the molecular formula if you also know the compound's molar mass (usually from mass spectrometry or given in the problem).

Formula

n = Molar Mass of Compound ÷ Molar Mass of Empirical Formula

The molecular formula = (empirical formula) × n

Example

Empirical formula: CH₂O (molar mass of EF = 12.011 + 2×1.008 + 15.999 = 30.026 g/mol)
Given molar mass of compound: 180.16 g/mol

n = 180.16 ÷ 30.026 = 5.999 ≈ 6

Molecular formula = C₁×₆H₂×₆O₁×₆ = C₆H₁₂O₆ (glucose)

Another Example

Empirical formula: CH (molar mass = 13.019 g/mol)
Given molar mass: 78.11 g/mol (benzene)

n = 78.11 ÷ 13.019 = 6.00

Molecular formula: C₆H₆

Common Mistakes and How to Avoid Them

1. Rounding Too Early

Rounding mole values before you divide them leads to compounding errors. Always carry at least four significant figures through to the ratio step before you round.

Wrong: 3.330 mol rounded to 3, then 6.657 mol rounded to 7 → ratio 3:7 (incorrect)
Right: 3.330/3.330 = 1.000; 6.657/3.330 = 1.999 ≈ 2 → ratio 1:2 ✓

2. Rounding a Non-Integer Ratio to the Nearest Whole Number

If you get a ratio of 1.50, do not round it to 2. Multiply all ratios by 2 to get 3.

Wrong: 1:1.50 → rounded to 1:2 → wrong formula
Right: ×2 → 2:3 ✓

3. Forgetting to Account for Oxygen

In combustion analysis problems, oxygen is rarely measured directly. Always check: if percentages don't add up to 100%, the remainder is typically oxygen.

4. Using the Wrong Atomic Mass

Common pitfall: using 16 for sulfur instead of oxygen, or using 14 for carbon instead of nitrogen. Double-check your periodic table values.

5. Confusing Empirical and Molecular Formula

The empirical formula is the simplest ratio. CH₂O is the empirical formula of both glucose (C₆H₁₂O₆) and acetic acid (C₂H₄O₂). You need a molar mass to tell them apart.

6. Assuming the Percentages Always Add to Exactly 100%

Due to rounding in published data, percentages may total 99.8% or 100.2%. This is normal. Proceed as usual — the small discrepancy won't affect the result.

Quick-Reference Atomic Masses for Common Elements

Use these values when calculating empirical formulas by hand. Values are rounded to four significant figures.

Standard Atomic Masses — Common Elements

Frequently Asked Questions

What is the empirical formula of water?

Water's molecular formula is H₂O. The ratio of H to O is 2:1, and this cannot be simplified further, so the empirical formula is also H₂O.

Can two different compounds have the same empirical formula?

Yes — this is very common. Acetylene (C₂H₂) and benzene (C₆H₆) both have the empirical formula CH. Glucose (C₆H₁₂O₆) and acetic acid (C₂H₄O₂) both have the empirical formula CH₂O. This is why the empirical formula alone is not enough to identify a compound — you also need the molar mass.

What if percentages don't add up to 100%?

First, check whether there is a missing element (usually oxygen in organic compounds). If oxygen isn't listed, assume % O = 100 − (sum of other percentages) and include it in your calculation. If they're just slightly off (e.g., 99.7%), proceed normally — this is a measurement rounding issue.

How do I know when to multiply by 2, 3, or 4?

Look at all the decimal parts of your ratios. If any ratio ends in .5, multiply all by 2. If any ratio ends in .33 or .67, multiply all by 3. If any ends in .25 or .75, multiply all by 4. When in doubt, check the table in the tricky ratios section above.

What is the empirical formula of CO₂?

CO₂ (carbon dioxide) already represents the simplest ratio of C to O (1:2). It cannot be simplified, so the empirical formula is CO₂ — the same as its molecular formula.

Does the empirical formula tell you the structure of a molecule?

No. The empirical formula only tells you the ratio of atoms, not how they are arranged or bonded. Structural formulas and 3D models are needed for that information.

How do I find the empirical formula if only the masses of individual elements are given?

Convert each mass to moles using the element's atomic mass, then follow the same division and rounding steps. You do not need to assume a 100 g sample — that trick is only needed when you have percentages. See Method 2 above for a full worked example.

What is the empirical formula of sodium chloride (table salt)?

NaCl. Ionic compounds like salts don't have discrete molecules, so their "formula" is always the empirical (simplest) ratio. The ratio of Na⁺ to Cl⁻ ions in the crystal lattice is 1:1.

Why do we divide by the smallest number of moles?

We divide by the smallest value so that the element with the fewest moles gets a ratio of 1. This gives us the relative ratio of all elements compared to the least abundant element, which is the most reduced form of the ratio.

What's the difference between an empirical formula and a structural formula?

An empirical formula (e.g., CH₂O) shows the simplest atom ratio. A molecular formula (e.g., C₆H₁₂O₆) shows the actual number of atoms per molecule. A structural formula goes further and shows how the atoms are connected — for example, showing glucose as a ring structure with specific bonds between atoms.

Can the empirical formula have a subscript of 1?

Yes, but by convention, a subscript of 1 is never written explicitly. If your calculation gives C:H:N = 1:2:1, you write CH₂N, not C₁H₂N₁.

What if I'm working with a hydrate?

A hydrate is a compound with water molecules incorporated into its crystal structure. For example, copper(II) sulfate pentahydrate is CuSO₄·5H₂O. To find the empirical formula of the anhydrous compound, you need to know the number of water molecules per formula unit — usually determined by heating the hydrate to drive off the water and measuring the mass loss.

Summary — Empirical Formula Calculator Method at a Glance

Here is the complete four-step method in its simplest form, suitable for printing or quick reference:

1. Mass → Grams: If given percentages, assume 100 g (% = g). If given grams, use those directly.
2. Grams → Moles: Divide each element's mass by its atomic mass.
3. Divide by smallest: Divide all mole values by the smallest mole value to get ratios.
4. Clear decimals: If ratios are whole numbers (±0.05), write the formula. If not, multiply all by 2, 3, or 4 to get whole numbers.

Bonus step: To find the molecular formula from the empirical formula, divide the known molar mass of the compound by the molar mass of the empirical formula. That whole number (n) is the multiplier: Molecular Formula = (Empirical Formula) × n.