# Empirical Formula Calculator

Free empirical formula calculator. Determine the simplest formula from percent composition data. Step-by-step calculation with element ratios.

## What this calculates

Find the empirical formula (simplest whole-number ratio of atoms) from percent composition data. Enter the mass percentages for up to three elements and the calculator will determine the empirical formula and its molar mass.

## Inputs

- **Element 1** — options: Carbon (C) - 12.011, Hydrogen (H) - 1.008, Oxygen (O) - 15.999, Nitrogen (N) - 14.007, Sulfur (S) - 32.06, Chlorine (Cl) - 35.45, Sodium (Na) - 22.990, Iron (Fe) - 55.845, Calcium (Ca) - 40.078, Phosphorus (P) - 30.974 — Select the first element.
- **Element 1: Mass %** (%) — min 0, max 100 — Mass percent of the first element.
- **Element 2** — options: Carbon (C) - 12.011, Hydrogen (H) - 1.008, Oxygen (O) - 15.999, Nitrogen (N) - 14.007, Sulfur (S) - 32.06, Chlorine (Cl) - 35.45, Sodium (Na) - 22.990, Iron (Fe) - 55.845, Calcium (Ca) - 40.078, Phosphorus (P) - 30.974 — Select the second element.
- **Element 2: Mass %** (%) — min 0, max 100 — Mass percent of the second element.
- **Element 3** — options: None (2 elements only), Carbon (C) - 12.011, Hydrogen (H) - 1.008, Oxygen (O) - 15.999, Nitrogen (N) - 14.007, Sulfur (S) - 32.06, Chlorine (Cl) - 35.45, Sodium (Na) - 22.990, Iron (Fe) - 55.845, Calcium (Ca) - 40.078, Phosphorus (P) - 30.974 — Select the third element or 'None'.
- **Element 3: Mass %** (%) — min 0, max 100 — Mass percent of the third element (0 if none).

## Outputs

- **Empirical Formula** — formatted as text — The simplest whole-number ratio of elements.
- **Empirical Formula Mass** (g/mol) — Molar mass of the empirical formula.
- **Calculation Steps** — formatted as text — Detailed step-by-step calculation.

## Frequently Asked Questions

**Q: What is an empirical formula?**

A: An empirical formula shows the simplest whole-number ratio of atoms in a compound. For example, glucose has a molecular formula of C₆H₁₂O₆ but an empirical formula of CH₂O (ratio 1:2:1).

**Q: How do I find the molecular formula from the empirical formula?**

A: Divide the known molecular mass by the empirical formula mass to get a whole number n. Multiply all subscripts in the empirical formula by n. For glucose: molecular mass (180.16) / empirical mass of CH₂O (30.026) = 6, so the molecular formula is C₆H₁₂O₆.

**Q: What if my percentages do not add up to 100%?**

A: If the percentages do not sum to 100%, the missing percentage often belongs to oxygen (in combustion analysis). For 3 elements, make sure all percentages sum to approximately 100% for accurate results.

**Q: Why do I assume a 100 g sample?**

A: Assuming 100 g makes the math simple: percentages directly become grams. 40% carbon in 100 g means 40 g of carbon. Then divide by atomic mass to get moles.

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Source: https://vastcalc.com/calculators/chemistry/empirical-formula
Category: Chemistry
Last updated: 2026-04-21