Empirical Formula Calculator
Determine the simplest whole-number ratio of elements in any chemical compound. Enter each element's symbol and its mass percentage to calculate the empirical formula instantly with a full step-by-step breakdown.
Empirical Formula Calculator
Element Percentages
Enter each element and its mass percentage in the compound.
Enter element symbols and their mass percentages to find the empirical formula.
Understanding Empirical Formulas
An empirical formula expresses the simplest whole-number ratio of atoms of each element present in a compound. Unlike the molecular formula, which reveals the exact count of every atom in a single molecule, the empirical formula strips that information down to its most reduced form. Chemists rely on empirical formulas when analyzing unknown substances because the ratio can be determined directly from experimental data such as percent composition or combustion analysis results.
Empirical Formula vs. Molecular Formula
Consider glucose: its molecular formula is C₆H₁₂O₆, meaning each molecule contains 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms. However, the ratio 6:12:6 simplifies to 1:2:1, so the empirical formula is CH₂O. Formaldehyde (CH₂O) and acetic acid (C₂H₄O₂) share the same empirical formula yet have very different molecular formulas, molar masses, and chemical properties. To move from an empirical formula to a molecular formula you need one additional piece of information: the compound's molar mass.
Steps to Find the Empirical Formula from Percent Composition
- Assume a 100 g sample so that every percentage converts directly to grams.
- Convert grams to moles by dividing the mass of each element by its atomic mass (from the periodic table).
- Divide by the smallest mole value to obtain the mole ratio for each element.
- Round or multiply to reach whole-number subscripts. If a ratio is close to 0.5, multiply everything by 2; if close to 0.33, multiply by 3; and so on.
Combustion Analysis
Combustion analysis is the standard laboratory technique for determining the empirical formula of organic compounds containing carbon, hydrogen, and often oxygen. A weighed sample is burned completely in excess oxygen. The carbon dioxide produced is trapped and weighed to find the mass of carbon, while the water produced is trapped to find the mass of hydrogen. If oxygen is also present in the compound, its mass is calculated by subtracting the combined mass of carbon and hydrogen from the original sample mass. Once the mass of each element is known, the mole-ratio method described above is applied.
Empirical vs. Molecular Formula Reference Table
The table below shows common compounds, their empirical formulas, molecular formulas, and molar masses. Notice how different substances can share the same empirical formula.
| Compound | Empirical Formula | Molecular Formula | Molar Mass (g/mol) | Multiplier (n) |
|---|---|---|---|---|
| Formaldehyde | CH₂O | CH₂O | 30.03 | 1 |
| Acetic Acid | CH₂O | C₂H₄O₂ | 60.05 | 2 |
| Glucose | CH₂O | C₆H₁₂O₆ | 180.16 | 6 |
| Benzene | CH | C₆H₆ | 78.11 | 6 |
| Acetylene | CH | C₂H₂ | 26.04 | 2 |
| Hydrogen Peroxide | HO | H₂O₂ | 34.01 | 2 |
| Water | H₂O | H₂O | 18.02 | 1 |
| Ribose | CH₂O | C₅H₁₀O₅ | 150.13 | 5 |
How to Find the Empirical Formula -- Worked Examples
Example 1: Compound with 40.0% C, 6.7% H, and 53.3% O
Find the empirical formula of a compound whose percent composition is 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen.
- Assume 100 g: 40.0 g C, 6.7 g H, 53.3 g O.
- Convert to moles: C = 40.0 / 12.01 = 3.33 mol; H = 6.7 / 1.008 = 6.65 mol; O = 53.3 / 16.00 = 3.33 mol.
- Divide by smallest (3.33): C = 1.00; H = 2.00; O = 1.00.
- Result: The empirical formula is CH₂O (the same empirical formula shared by formaldehyde, acetic acid, and glucose).
Example 2: Compound with 92.3% C and 7.7% H
Determine the empirical formula when a hydrocarbon is 92.3% carbon and 7.7% hydrogen by mass.
- Assume 100 g: 92.3 g C, 7.7 g H.
- Convert to moles: C = 92.3 / 12.01 = 7.69 mol; H = 7.7 / 1.008 = 7.64 mol.
- Divide by smallest (7.64): C = 1.007 ≈ 1; H = 1.000.
- Result: The empirical formula is CH. This corresponds to benzene (C₆H₆) or acetylene (C₂H₂) depending on the molar mass.
Example 3: Combustion Analysis -- 0.255 g compound yields 0.561 g CO₂ and 0.306 g H₂O
A 0.255 g sample of a compound containing C, H, and O is burned. The combustion produces 0.561 g of CO₂ and 0.306 g of H₂O. Find the empirical formula.
- Find moles of CO₂ and H₂O: CO₂ = 0.561 / 44.01 = 0.01275 mol; H₂O = 0.306 / 18.02 = 0.01699 mol.
- Moles of C and H: C = 0.01275 mol (one C per CO₂); H = 2 x 0.01699 = 0.03398 mol (two H per H₂O).
- Mass of C and H: C = 0.01275 x 12.01 = 0.1531 g; H = 0.03398 x 1.008 = 0.03425 g.
- Mass of O by difference: O = 0.255 - 0.1531 - 0.03425 = 0.0677 g; moles O = 0.0677 / 16.00 = 0.00423 mol.
- Divide by smallest (0.00423): C = 3.02 ≈ 3; H = 8.03 ≈ 8; O = 1.00.
- Result: The empirical formula is C₃H₈O (which could be propanol or isopropanol depending on structure).
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