Example 1 — Percent Composition
40.00% C, 6.71% H, and 53.29% O become 40.00 g, 6.71 g, and 53.29 g in a 100 g sample. The mole ratio reduces to 1:2:1, so the formula is CH₂O.
Try This ExampleEmpirical Formula Calculator
Find an empirical formula from element masses or percent composition, with mole ratios and step-by-step calculations. You can also calculate a molecular formula from an empirical formula and molar mass.
How to Use the Empirical Formula Calculator
Use the Empirical Formula tab for lab data or homework percent composition. Use the Molecular Formula tab when you already know an empirical formula and the compound molar mass.
Empirical formula tab
Choose Percent Composition or Mass Data, enter one row for each element, then calculate. When you enter percentages, the calculator assumes a 100 g sample; when you enter masses, it uses those gram amounts directly. The result shows grams, atomic masses, moles, mole ratios, and the final whole-number ratio.
Molecular formula tab
Enter the empirical formula, such as CH₂O, and the molecular molar mass in g/mol. The calculator finds the empirical formula mass, divides the two masses, rounds to the whole-number multiplier, and expands the formula.
What Is an Empirical Formula?
An empirical formula shows the simplest whole-number ratio of the elements in a compound.
In chemistry, the empirical formula is the reduced ratio behind the composition of a substance. If a compound contains carbon, hydrogen, and oxygen in a 1:2:1 ratio, its empirical formula is CH₂O. That does not automatically mean the molecule has only one carbon atom; it only states the simplest ratio.
Molecular formulas can be multiples of empirical formulas. Glucose is C₆H₁₂O₆, but all three subscripts reduce by 6, so the empirical formula is CH₂O. Some compounds, such as water, already sit in their simplest ratio, so H₂O is both the empirical and molecular formula.
How to Find an Empirical Formula
To calculate an empirical formula, convert the data into mole amounts, then reduce those moles to the smallest whole-number ratio.
The reason the calculation uses moles is that grams do not compare atoms directly. Twelve grams of carbon and twelve grams of hydrogen are not the same number of atoms, because carbon atoms are much heavier than hydrogen atoms. Dividing each mass by its atomic mass converts the data into mole amounts, which can be compared as element counts.
After every element is in moles, divide by the smallest mole value. That makes the least abundant element equal to 1 and expresses every other element relative to it. Only then should you decide whether the ratios are already whole numbers or whether they need a common multiplier such as ×2, ×3, ×4, or ×5.
Step 1Convert each element amount to moles
For mass data, divide grams by the element's atomic mass. For percent composition, first assume a 100 g sample so the percentages become grams.
Step 2Divide every mole value by the smallest
This normalizes the smallest element to about 1 and turns the other mole values into a ratio against it.
Step 3Check whether the ratios are whole numbers
Values such as 1.999 or 3.001 are rounding noise and can be treated as 2 or 3. Values such as 1.5 or 1.333 need another adjustment.
Step 4Multiply fractional ratios if necessary
Use the same multiplier for every element: 1.5 means ×2, 1.333 means ×3, 1.25 means ×4, and 1.20 means ×5.
Step 5Use the final ratios as subscripts
A 1:2:1 ratio for C, H, and O becomes CH₂O. A subscript of 1 is not written.
Ratio discipline
Do not round every ratio blindly. The difference between 1.50 and 2.00 changes the formula. A value like 1.50 needs a common multiplier before it becomes a formula subscript.
Empirical Formula from Percent Composition
Percent composition problems use one extra idea: assume a 100 g sample. Then each percentage becomes the same number of grams.
If a compound is 40.00% carbon, 6.71% hydrogen, and 53.29% oxygen, a 100 g sample contains 40.00 g C, 6.71 g H, and 53.29 g O. Those masses are then converted to moles with atomic masses from the periodic table.
The 100 g assumption does not change the ratio. It simply chooses a convenient sample size so percent values can be read as gram values. After the grams-to-moles step, the rest of the calculation is the same as any mass-data empirical formula problem.
| Element | Percent | Mass in 100 g | Moles | Ratio |
|---|---|---|---|---|
| C | 40.00% | 40.00 g | 3.330 mol | 1.000 |
| H | 6.71% | 6.71 g | 6.657 mol | 1.999 |
| O | 53.29% | 53.29 g | 3.331 mol | 1.000 |
The mole ratios round to 1:2:1, so the empirical formula is CH₂O. When you enter percentages, the calculator shows this same table and warns you if the percentages are not close to 100%.
Empirical Formula vs Molecular Formula
The empirical formula is the reduced ratio; the molecular formula is the actual atom count in one molecule.
| Point | Empirical Formula | Molecular Formula |
|---|---|---|
| Meaning | Simplest whole-number ratio | Actual numbers of atoms |
| Example | CH₂O | C₆H₁₂O₆ |
| Relationship | Reduced form | Integer multiple |
A molecular formula can equal its empirical formula, but it cannot be a non-integer multiple of it. That is why the molecular-formula calculator checks the mass ratio before expanding the subscripts.
How to Find the Molecular Formula from an Empirical Formula
Use the empirical formula mass and the compound molar mass to find the multiplier between the reduced ratio and the real formula.
Multiplier
n = molecular molar mass / empirical formula mass
Molecular Formula = (Empirical Formula)n
- 1.Calculate the empirical formula mass.
- 2.Divide the compound molar mass by the empirical formula mass.
- 3.Find the nearest whole-number multiplier n.
- 4.Multiply every empirical-formula subscript by n.
Worked example
Glucose has empirical formula CH₂O and a molecular molar mass of about 180.16 g/mol. The empirical formula mass is 12.011 + 2(1.008) + 15.999 = 30.026 g/mol. Divide the molecular mass by that empirical mass:
180.16 / 30.026 ≈ 6
(CH₂O)₆ = C₆H₁₂O₆
Because the multiplier is a whole number, multiply every subscript in CH₂O by 6: C becomes C₆, H₂ becomes H₁₂, and O becomes O₆.
Empirical Formula Examples
These three examples cover the cases most students miss: percent composition, direct mass data, and a fractional mole ratio.
Example 2 — Mass Data
12.01 g C, 2.016 g H, and 16.00 g O convert to about 1.000, 2.000, and 1.000 mol. The ratio is already whole-number, giving CH₂O.
Try This ExampleExample 3 — Fractional Ratio
A 1:1.5 ratio cannot be written as subscripts. Multiply every ratio by 2 to get 2:3, so a C:H example becomes C₂H₃.
Try This ExampleRelated Chemistry Calculators
Empirical formulas sit between elemental data, molar mass, and later stoichiometry work.
- Molar Mass CalculatorFind formula mass, atomic-weight sums, and percent composition.
- Grams to Moles CalculatorConvert a measured sample mass into moles before ratio work.
- Stoichiometry CalculatorUse mole ratios from a balanced equation after formulas are known.
- Chemical Equation BalancerBalance formulas and coefficients before reaction calculations.
Frequently Asked Questions
What is an empirical formula?
An empirical formula shows the simplest whole-number ratio of the elements in a compound. CH₂O says carbon, hydrogen, and oxygen appear in a 1:2:1 ratio; it does not say how many atoms are in one molecule.
How do you calculate an empirical formula?
Convert each element amount to moles, divide every mole value by the smallest mole value, adjust fractional ratios to whole numbers, and use those whole numbers as formula subscripts.
How do you find an empirical formula from percent composition?
Assume a 100 g sample. Each percentage then becomes the same number of grams, so 40.00% C becomes 40.00 g C. Convert those masses to moles and continue with the mole-ratio steps.
Why do you assume 100 g for percent composition?
Percent means parts per 100. A 100 g sample turns percentages into gram amounts without changing the element ratios, which makes the calculation direct and easy to check.
What do you do when the mole ratio is not a whole number?
Multiply all mole ratios by a small whole number. Common cases are 1.5 × 2, 1.333 × 3, 1.25 × 4, 1.667 × 3, and 1.20 × 5. The calculator checks these adjustments before writing the formula.
What is the difference between an empirical formula and a molecular formula?
The empirical formula is the reduced whole-number ratio. The molecular formula gives the actual atom counts in a molecule. Glucose has empirical formula CH₂O and molecular formula C₆H₁₂O₆.
How do you calculate a molecular formula from an empirical formula?
Find the empirical formula mass, divide the compound molar mass by that mass, then multiply every empirical-formula subscript by the whole-number multiplier.