Empirical Formula Calculator
This empirical formula calculator finds the empirical formula and molecular formula from percent composition or element masses step by step. Whether you call it an emp formula or an empirical formula, this tool handles all cases — including multi-element compounds and rounding edge cases.
Enter a chemical formula to check whether it is already in empirical (simplest) form.
What is an Empirical Formula?
An empirical formula is the simplest whole number ratio of atoms of each element in a compound. It does not necessarily represent the actual number of atoms in one molecule — it only shows the ratio. For example, glucose has a molecular formula of C₆H₁₂O₆, but its empirical formula is CH₂O, because the simplest ratio of C:H:O is 1:2:1.
A molecular formula shows the exact number of atoms of each element in one molecule of a compound. Benzene (C₆H₆) has a molecular formula of C₆H₆ but an empirical formula of CH, because the ratio of C:H is 1:1. Water (H₂O) is special — its empirical formula and molecular formula are the same, because the ratio 2:1 for H:O is already in its simplest form.
Key Formula: Molecular Formula = Empirical Formula × n, where n = Molar Mass of Compound ÷ Empirical Formula Mass
Empirical Formula vs Molecular Formula
Understanding the difference between empirical and molecular formula is one of the most important concepts in introductory chemistry. The difference between empirical formula and molecular formula comes down to one thing: simplification. An empirical formula is the simplified ratio; a molecular formula is the full count.
| Feature | Empirical Formula | Molecular Formula |
|---|---|---|
| Definition | Simplest whole number ratio of atoms | Actual number of atoms per molecule |
| Example (glucose) | CH₂O | C₆H₁₂O₆ |
| Example (benzene) | CH | C₆H₆ |
| Example (water) | H₂O | H₂O (same) |
| How to find | From % composition / masses | Empirical formula × n |
| Molar mass | Empirical formula mass | Actual molar mass |
| Can they be the same? | Yes — when n = 1 (e.g. H₂O, NaCl, CO₂) | |
The empirical and molecular formula are the same when n = 1, meaning the empirical formula mass equals the actual molar mass. Common examples include water (H₂O, M = 18), CO₂ (M = 44), and NaCl (M = 58.44). To convert from empirical to molecular, divide the given molar mass by the empirical formula mass to find n, then multiply each subscript by n.
Chemists determine empirical formulas from combustion analysis data — burning a compound and measuring the masses of CO₂ and H₂O produced. From these masses, they calculate the moles and hence the percentage of C, H, and O in the compound. This is the standard method to determine empirical formula for organic compounds in the laboratory.
How to Work Out Empirical Formula — Step by Step
Example 1: From Percentage Composition
Compound contains 40% C, 6.67% H, 53.33% O.
Full Worked Solution:
- Step 1: Assume 100g → C = 40g, H = 6.67g, O = 53.33g
- Step 2: Moles: C = 40/12.011 = 3.33 mol, H = 6.67/1.008 = 6.62 mol, O = 53.33/15.999 = 3.33 mol
- Step 3: Divide by smallest (3.33): C = 1.00, H = 1.99 ≈ 2, O = 1.00
- Step 4: All whole numbers → Empirical Formula = CH₂O
Example 2: From Mass Data
Sample contains 2.4g C, 0.4g H, 3.2g O.
- Step 1: Moles: C = 2.4/12.011 = 0.2 mol, H = 0.4/1.008 = 0.397 mol, O = 3.2/15.999 = 0.2 mol
- Step 2: Divide by smallest (0.2): C = 1, H = 1.99 ≈ 2, O = 1
- Result: Empirical Formula = CH₂O
How to Find Molecular Formula from Empirical Formula
Example 1: Glucose
Empirical: CH₂O | Molar Mass: 180 g/mol
EF mass = 12.011+2(1.008)+15.999 = 30.026 g/mol
n = 180 / 30.026 = 5.99 ≈ 6
Molecular formula = C₆H₁₂O₆ (Glucose) ✓
Example 2: Benzene
Empirical: CH | Molar Mass: 78 g/mol
EF mass = 12.011+1.008 = 13.019 g/mol
n = 78 / 13.019 = 5.99 ≈ 6
Molecular formula = C₆H₆ (Benzene) ✓
Example 3: Nitrogen Dioxide Dimer
Empirical: NO₂ | Molar Mass: 92 g/mol
EF mass = 14.007+2(15.999) = 46.005 g/mol
n = 92 / 46.005 = 2.0
Molecular formula = N₂O₄ ✓
Worked Examples — Common Search Questions
1 — How do you find the empirical formula from percentages?
Treat each percentage as grams in a 100g sample. Then convert to moles by dividing each mass by the molar mass. Divide all mole values by the smallest mole value. Round the resulting ratios to the nearest whole number (handling .5, .33, .25 cases). Write the empirical formula using those whole numbers as subscripts. For 75% C, 25% H: C = 75/12.011 = 6.24 mol, H = 25/1.008 = 24.8 mol. Ratio H/C = 3.98 ≈ 4. Empirical formula: CH₄
2 — How to find empirical formula from percent composition step by step
Step 1: Write each % as a mass in a 100g sample. Step 2: Divide each mass by its atomic mass to get moles. Step 3: Divide all mole values by the smallest. Step 4: Round to whole numbers using multipliers if needed (.5 → ×2, .33 → ×3, .25 → ×4). Step 5: Write the formula with these integers as subscripts.
3 — How to determine empirical formula from percent composition
The standard method is to assume a 100g sample, convert each percentage to grams, convert grams to moles, and find the simplest whole number ratio. For a compound that is 27.3% C and 72.7% O: C = 27.3/12.011 = 2.27 mol, O = 72.7/15.999 = 4.54 mol. Ratio = 1:2. Empirical formula: CO₂
4 — How to calculate empirical formula from percent composition
The calculation always follows: mass (g) → moles (÷ molar mass) → ratio (÷ smallest) → whole numbers → formula. For a compound of 52.14% C, 13.13% H, 34.73% O: C=4.34, H=13.02, O=2.17 mol. Divide by 2.17: C=2, H=6, O=1. Empirical formula: C₂H₆O (ethanol unit).
5 — How to find molecular formula from percent composition
First find the empirical formula from the % data. Then calculate the empirical formula mass. Divide the compound's given molar mass by the empirical formula mass to get n. Multiply each subscript by n to get the molecular formula. This is the two-step process combined in Tool 3 above.
6 — How to find empirical and molecular formula together
Use Tool 3 (Full % to Molecular) on this page. Enter each element's percentage and the compound's molar mass. The calculator computes the empirical formula first, then uses n = molar mass ÷ EF mass to derive the molecular formula in one sequence.
7 — What is the empirical formula of a compound with 75% C and 25% H?
Moles: C = 75/12.011 = 6.244, H = 25/1.008 = 24.802. Divide by 6.244: C = 1, H = 3.97 ≈ 4. Empirical formula: CH₄ (methane unit). This could be methane itself if n=1, or a higher hydrocarbon if n>1.
8 — How to work out empirical formula when percentages do not add to 100%
If only C and H percentages are given (e.g. 85.6% C, 14.4% H = 100% ✓), verify they sum to 100%. If a third element is missing, calculate it as: remaining % = 100 − sum of given percentages. For example, if C=40%, H=6.67%, then O = 100−40−6.67 = 53.33%.
9 — How to handle decimal ratios in empirical formula calculation
After dividing by the smallest mole value, round only if the result is within 0.05 of a whole number. If the ratio ends in .5 (e.g. 1.5), multiply all ratios by 2. If it ends in .33 or .67, multiply all by 3. If it ends in .25 or .75, multiply all by 4. For C:H = 1:1.5 → multiply by 2 → C₂H₃.
10 — What is a hydrocarbon empirical formula?
A hydrocarbon empirical formula contains only carbon and hydrogen in their simplest ratio. Methane (CH₄) is both empirical and molecular. Ethylene (C₂H₄) has empirical formula CH₂. Acetylene (C₂H₂) has empirical formula CH. Benzene (C₆H₆) has empirical formula CH. The calculator handles all hydrocarbon formulas — just enter C and H percentages with their atomic masses.
Empirical Formula Examples Table
| Compound | Molecular Formula | Empirical Formula | n value |
|---|---|---|---|
| Water | H₂O | H₂O | 1 |
| Glucose | C₆H₁₂O₆ | CH₂O | 6 |
| Benzene | C₆H₆ | CH | 6 |
| Hydrogen peroxide | H₂O₂ | HO | 2 |
| Ethylene | C₂H₄ | CH₂ | 2 |
| Acetylene | C₂H₂ | CH | 2 |
| Sucrose | C₁₂H₂₂O₁₁ | C₁₂H₂₂O₁₁ | 1 |
| Phosphorus pentoxide | P₄O₁₀ | P₂O₅ | 2 |