Empirical & Molecular Formula Calculator
Convert percent composition, gram composition, or an empirical formula into the chemical formula you need.
What can you enter?
This calculator accepts the formats students usually see in homework, lab reports, and composition-analysis questions. Use element symbols, not full element names.
| Goal | Input example | What the tool does |
|---|---|---|
| Percent composition | C:40.0,H:6.7,O:53.3 | Treats the data as a 100 g sample, then finds CH2O. |
| Mass composition | Mg:2.43,O:1.60 | Converts grams to moles, then finds MgO. |
| Molecular formula | CH2O and 180.16 g/mol | Finds the multiplier 6 and gives C6H12O6. |
| Flexible separators | Fe 69.94 O 30.06 | Reads spaces, commas, colons, or equals signs between symbols and values. |
Formula path used
Worked example: glucose
Input: C:40.0,H:6.7,O:53.3 with molar mass 180.16 g/mol.
- Assume 100 g, so the sample has 40.0 g C, 6.7 g H, and 53.3 g O.
- Convert to moles: C = 40.0 / 12.011, H = 6.7 / 1.008, O = 53.3 / 15.999.
- Divide by the smallest mole value to get a ratio near 1 : 2 : 1.
- The empirical formula is CH2O. Its empirical mass is about 30.03 g/mol.
- 180.16 / 30.03 is about 6, so the molecular formula is C6H12O6.
Built-in example data
| Example | Composition data | Known molar mass | Expected result |
|---|---|---|---|
| Glucose | C 40.0%, H 6.7%, O 53.3% | 180.16 g/mol | CH2O -> C6H12O6 |
| Iron(III) oxide | Fe 69.94%, O 30.06% | Not needed | Fe2O3 |
| Caffeine | C 49.48%, H 5.19%, N 28.85%, O 16.48% | 194.19 g/mol | C4H5N2O -> C8H10N4O2 |
| Vitamin C | C 40.92%, H 4.58%, O 54.50% | 176.12 g/mol | C3H4O3 -> C6H8O6 |
| Magnesium oxide | Mg 2.43 g, O 1.60 g | Not needed | MgO |
Why this calculator is useful
Empirical formula problems are simple in idea but easy to misread because the answer depends on ratios, not direct masses. This tool keeps the mole conversion table visible, shows the whole-number ratio, and checks the molecular multiplier when molar mass is available.
| Where it helps | Why it matters |
|---|---|
| General chemistry homework | Shows every conversion step instead of only the final formula. |
| Combustion analysis | Turns measured C, H, O, or N data into atom ratios. |
| Lab report checks | Helps compare experimental composition with the expected compound formula. |
| Exam revision | Practices percent-to-moles and ratio rounding in one workflow. |
| Molecular formula questions | Connects empirical mass to the measured molar mass multiplier. |
How to read the output
The result card highlights the empirical formula first because that is the lowest whole-number ratio. The mole-ratio view shows how far each element is from the smallest mole value. If a known compound molar mass is entered, the formula flow shows how the empirical formula becomes the molecular formula.
For example, CH2O and C6H12O6 have the same empirical ratio, but C6H12O6 is the actual molecular formula for glucose because its molar mass is six times the empirical formula mass.
Common mistakes to avoid
- Using full element names like carbon instead of symbols like C.
- Dividing percentages directly by each other before converting to moles.
- Rounding 1.5 to 2 instead of multiplying every ratio by 2.
- Entering a molecular mass before checking the empirical formula mass.
- Forgetting that percent composition means a 100 g sample assumption.
Rounding and result checking
Ratios close to whole numbers, such as 1.99 or 3.02, are normally rounded to 2 or 3. Ratios near common fractions need a multiplier: 1.5 uses x2, 1.33 uses x3, 1.25 uses x4, and 1.67 uses x3. If the molecular multiplier is not close to a whole number, recheck the composition data, atomic masses, molar mass, and significant figures.
Related Chemistry Tools
Empirical & Molecular Formula Calculator FAQs
How to calculate empirical formula?
The empirical formula gives the simplest whole-number ratio of atoms in a compound. Follow these five steps: (1) Take the percentage of each element (treat as grams); (2) divide by the atomic mass to get moles; (3) divide every result by the smallest number of moles to get a ratio; (4) if any ratio is fractional, multiply all by the same small integer to get whole numbers; (5) write the formula. For example, a compound with 40% C, 6.7% H, 53.3% O gives the empirical formula CH2O. Steps: % → mass (g) → moles (÷ atomic mass) → ratio (÷ smallest) → whole numbers
How to calculate the molecular formula from the empirical formula?
The molecular formula is always a whole-number multiple of the empirical formula: Molecular formula = n × Empirical formula, where n = (Molecular mass) / (Empirical formula mass). For instance, the empirical formula of glucose is CH2O (mass 30 g/mol). Its molecular mass is 180 g/mol, so n = 180/30 = 6. Molecular formula = (CH2O)6 = C6H12O6. Always need molecular mass (from experiments like vapour density or mass spectrometry) to find the molecular formula. n = Mmolecular / Mempirical
How to calculate empirical formula from percent composition?
Assume you have 100 g of the compound, so each percentage directly becomes mass in grams. Then divide each mass by the corresponding atomic mass to get moles. Divide each mole value by the smallest one to get a ratio close to whole numbers. Worked example: a hydrocarbon contains 75% C and 25% H. C: 75/12 = 6.25 mol; H: 25/1 = 25 mol. Divide by 6.25 → 1 : 4. So the empirical formula is CH4, which is methane.
How to calculate the empirical formula of a compound?
If you are given experimental data — masses of elements, or percentages, or mass of compound and CO2/H2O produced on combustion — convert all information to moles of each element. Then find the simplest mole ratio. Example: 5.6 g of iron combines with 2.4 g of oxygen. Moles: Fe = 5.6/56 = 0.1; O = 2.4/16 = 0.15. Ratio Fe:O = 0.1 : 0.15 = 2 : 3. Empirical formula = Fe2O3, the formula of haematite (rust).
How to calculate molar mass from empirical formula?
This question can mean two things. (a) Empirical-formula mass — simply add the atomic masses of the atoms shown in the empirical formula. For CH2O: 12 + 2(1) + 16 = 30 g/mol. (b) Molar (molecular) mass — you need additional information such as vapour density, freezing-point depression, or the molecular formula. Once you know n = Mmolecular/Mempirical, Mmolecular = n × Mempirical.