Empirical Formula Calculator
Find the empirical formula from percent composition, or determine the molecular formula from the empirical formula and molar mass. Also calculates percent composition from any chemical formula.
How to Find the Empirical Formula
Given the percent composition of a compound (the mass percentage of each element), follow these four steps:
- Assume a 100 g sample. This converts percentages directly to grams — a 40% carbon sample becomes 40 g of carbon.
- Convert grams to moles. Divide each mass by the atomic mass of the element: n = m ÷ MW.
- Divide by the smallest moles value. This gives the simplest ratio between the elements.
- Round to whole numbers. If the ratios are close to integers (e.g. 1.00 or 1.99), round them. If they are fractional (e.g. 1.5, 1.33), multiply all ratios by 2, 3, or 4 until they become integers.
Example: Find the empirical formula of a compound containing 40.0% C, 6.7% H, and 53.3% O.
- Assume 100 g → 40.0 g C, 6.7 g H, 53.3 g O
- Moles: C = 40.0 ÷ 12.011 = 3.33; H = 6.7 ÷ 1.008 = 6.65; O = 53.3 ÷ 15.999 = 3.33
- Divide by smallest (3.33): C = 1.00; H = 2.00; O = 1.00
- All integers → empirical formula = CH₂O
On percentages summing to 100%: in theory the percent composition must add to exactly 100%, but in practice experimental values may not sum to exactly 100% due to rounding (each percentage is usually reported to 1–2 decimals) or trace elements not included in the analysis. A total of 99.5–100.5% is normal; larger deviations suggest measurement error or a missing element.
Empirical Formula vs Molecular Formula
| Aspect | Empirical Formula | Molecular Formula |
|---|---|---|
| Definition | Simplest whole-number ratio of atoms | Actual number of atoms in one molecule |
| Example | CH₂O | C₆H₁₂O₆ (glucose) |
| Unique per compound? | No — many compounds share it | Yes — unique per compound |
| How to find | From percent composition | Empirical × (MW ÷ empirical MW) |
| Data needed | Percent composition only | Percent composition + molar mass |
Percent Composition Formula
The mass percentage of an element in a compound is:
Example — H₂O (water):
- H: 2 × 1.008 = 2.016 g/mol → (2.016 ÷ 18.015) × 100% = 11.19% H
- O: 1 × 15.999 = 15.999 g/mol → (15.999 ÷ 18.015) × 100% = 88.81% O
- Total: 100% ✓
Worked Examples
Example 1: Ethanol (C₂H₆O)
Given composition: 52.14% C, 13.13% H, 34.73% O. Known MW: 46.07 g/mol.
- Moles: C = 52.14 ÷ 12.011 = 4.341; H = 13.13 ÷ 1.008 = 13.026; O = 34.73 ÷ 15.999 = 2.171
- Divide by smallest (2.171): C = 2.00; H = 6.00; O = 1.00
- Empirical formula: C₂H₆O (already integer ratios)
- Empirical MW: (2×12.011) + (6×1.008) + 15.999 = 46.068 g/mol
- Multiplier: 46.07 ÷ 46.068 ≈ 1 → molecular formula = C₂H₆O (same as empirical)
Example 2: Iron(III) Oxide (Fe₂O₃)
Given composition: 69.94% Fe, 30.06% O.
- Moles: Fe = 69.94 ÷ 55.845 = 1.252; O = 30.06 ÷ 15.999 = 1.879
- Divide by smallest (1.252): Fe = 1.000; O = 1.500
- Multiply all by 2 to clear the half: Fe = 2; O = 3
- Empirical formula: Fe₂O₃
Example 3: Glucose (C₆H₁₂O₆)
Given composition: 40.0% C, 6.7% H, 53.3% O. Known MW: 180.16 g/mol.
- Moles: C = 3.33; H = 6.65; O = 3.33
- Divide by smallest: C = 1; H = 2; O = 1 → empirical CH₂O
- Empirical MW: 12.011 + 2×1.008 + 15.999 = 30.026 g/mol
- Multiplier: 180.16 ÷ 30.026 ≈ 6
- Molecular formula: CH₂O × 6 = C₆H₁₂O₆ (glucose)
Frequently Asked Questions
What is an empirical formula?
The empirical formula is the simplest whole-number ratio of atoms in a compound. For example, the empirical formula of glucose (C₆H₁₂O₆) is CH₂O, because the ratio C:H:O is 1:2:1.
Different compounds can share the same empirical formula — CH₂O is also the empirical formula of formaldehyde and acetic acid.
How do you find the empirical formula from percent composition?
Four steps:
- Assume a 100 g sample so percent = grams.
- Divide each element's mass by its atomic mass to get moles.
- Divide all mole values by the smallest to get the simplest ratio.
- Multiply by a common factor (2, 3, 4) if needed to make all ratios whole numbers.
What is the difference between empirical and molecular formulas?
Empirical formula — the simplest whole-number ratio of atoms (CH₂O).
Molecular formula — the actual number of atoms in one molecule (C₆H₁₂O₆ for glucose).
The molecular formula is always a whole-number multiple of the empirical formula.
How do you find the molecular formula from the empirical formula?
Divide the known molar mass of the compound by the molar mass of the empirical formula. Round the result to the nearest whole number — this is the multiplier. Multiply every subscript in the empirical formula by this multiplier.
Example: Empirical CH₂O has MW 30.026. A compound with MW 180.16 has multiplier 180.16 ÷ 30.026 ≈ 6, so molecular formula = (CH₂O)₆ = C₆H₁₂O₆.
What is percent composition?
Percent composition is the mass percentage of each element in a compound. It is calculated as:
mass % = (atoms × atomic mass ÷ molar mass) × 100%
Percent composition is used to identify unknown compounds by comparing experimental analysis to known values.
Can two different compounds have the same empirical formula?
Yes. CH₂O is the empirical formula for three very different compounds:
- Formaldehyde — CH₂O (MW 30)
- Acetic acid — C₂H₄O₂ (MW 60)
- Glucose — C₆H₁₂O₆ (MW 180)
All three have a 1:2:1 C:H:O ratio, but with different molecular formulas and very different properties.
Related Chemistry Tools
- Molar Mass Calculator — find the molar mass of any compound
- Grams to Moles Calculator — convert mass to moles using molecular weight
- Moles to Grams Calculator
- Equation Balancer — balance chemical equations
- Molarity Calculator
- All Chemistry Calculators
All calculations on CoolConversion use atomic weights from the IUPAC 2021 standard, ensuring accuracy consistent with internationally recognised standards.
