Enter the percentage by mass (or actual mass) and the element symbol for up to 5 elements. If percentages do not add up to 100%, oxygen will be calculated by difference. Optionally enter Mr to find the molecular formula.
An empirical formula gives the simplest whole number ratio of atoms of each element in a compound. It represents the minimum repeating unit of the compound's composition.
The empirical formula does not necessarily reflect the actual structure or number of atoms in a molecule — it only shows the ratio. For instance, both ethanoic acid (CH₃COOH) and glucose (C₆H₁₂O₆) share the same empirical formula CH₂O, even though they are completely different compounds.
Empirical formulas are the starting point for identifying unknown compounds. Analytical techniques such as combustion analysis and mass spectrometry provide data that chemists use to determine empirical — and ultimately molecular — formulas of new substances.
| Compound | Molecular Formula | Empirical Formula |
|---|---|---|
| Glucose | C₆H₁₂O₆ | CH₂O |
| Benzene | C₆H₆ | CH |
| Ethane | C₂H₆ | CH₃ |
| Hydrogen peroxide | H₂O₂ | HO |
| Ethyne | C₂H₂ | CH |
| Water | H₂O | H₂O |
| Sodium chloride | NaCl (ionic) | NaCl |
| Butene | C₄H₈ | CH₂ |
Note: The molecular formula is always a whole number multiple of the empirical formula. For ionic compounds such as NaCl, the empirical formula is the formula unit.
Step 1: Assume 100 g of compound (so percentages become masses in grams)
C = 40 g | H = 6.7 g | O = 53.3 g
Step 2: Divide by relative atomic mass (Ar)
Moles C = 40 ÷ 12 = 3.33 | Moles H = 6.7 ÷ 1 = 6.7 | Moles O = 53.3 ÷ 16 = 3.33
Step 3: Divide by the smallest value (3.33)
C : H : O = 3.33/3.33 : 6.7/3.33 : 3.33/3.33 = 1 : 2.01 : 1 ≈ 1 : 2 : 1
Step 4: Write empirical formula: CH₂O
Step 5 (molecular formula): If Mr = 180, empirical formula mass = 12+2+16 = 30. Multiplier = 180÷30 = 6. Molecular formula = C₆H₁₂O₆
| Decimal Ratio | Multiply by | Reason |
|---|---|---|
| x.5 (e.g. 1 : 1.5) | 2 | 1.5 × 2 = 3 (whole number) |
| x.33 (e.g. 1 : 1.33) | 3 | 1.33 × 3 ≈ 4 |
| x.25 (e.g. 1 : 1.25) | 4 | 1.25 × 4 = 5 |
| x.67 (e.g. 1 : 1.67) | 3 | 1.67 × 3 = 5 |
Combustion analysis is used to determine the empirical formula of organic compounds. The compound is burned completely in excess oxygen, and the products are absorbed and weighed.
0.300 g organic compound burned. CO₂ = 0.440 g, H₂O = 0.180 g.
Mass C = 0.440 × (12/44) = 0.120 g
Mass H = 0.180 × (2/18) = 0.020 g
Mass O = 0.300 − 0.120 − 0.020 = 0.160 g
Moles: C=0.010, H=0.020, O=0.010 → Ratio 1:2:1 → CH₂O
Hydrated salts contain water of crystallisation in their crystal lattice. The formula is written as the anhydrous salt followed by a dot and then xH₂O, where x is the number of water molecules per formula unit.
Given: 6.25 g hydrated salt heated to give 4.00 g anhydrous CuSO₄
Mass H₂O lost = 6.25 − 4.00 = 2.25 g
n(CuSO₄) = 4.00 / 159.5 = 0.02508 mol
n(H₂O) = 2.25 / 18 = 0.1250 mol
Ratio = 0.1250 / 0.02508 = 4.98 ≈ 5
Therefore: CuSO₄·5H₂O
An empirical formula shows the simplest whole number ratio of atoms of each element present in a compound. It does not necessarily show the actual number of atoms in a molecule. For example, glucose (C₆H₁₂O₆) has the empirical formula CH₂O because the ratio of C:H:O is 1:2:1.
Divide each percentage by the relative atomic mass (Ar) of the element to convert to moles. Then divide all values by the smallest to get the simplest ratio. If the ratios contain common decimal fractions such as .5 or .33, multiply through by 2 or 3 respectively to obtain whole numbers.
The empirical formula is the simplest whole number ratio of atoms. The molecular formula shows the actual number of each type of atom in one molecule. The molecular formula is always a whole number multiple of the empirical formula. You need the relative molecular mass (Mr) to find the molecular formula from the empirical formula.
Calculate the empirical formula mass (Mr of one unit of the empirical formula by adding up the atomic masses). Divide the actual Mr of the compound by the empirical formula mass to find the multiplier. Then multiply each subscript in the empirical formula by this whole number to get the molecular formula.
Combustion analysis is a method to determine the empirical formula of an organic compound by completely burning it in excess oxygen and measuring the masses of CO₂ and H₂O produced. All carbon appears in CO₂ (multiply mass of CO₂ by 12/44 to get mass of C) and all hydrogen appears in H₂O (multiply mass of H₂O by 2/18). Oxygen content is found by subtracting the masses of C and H from the total sample mass.
Water of crystallisation is water molecules incorporated into the crystal lattice of a hydrated salt, such as CuSO₄·5H₂O (hydrated copper(II) sulfate, which is blue). When gently heated, the water is driven off leaving white anhydrous CuSO₄. The number of water molecules per formula unit (x) is found by comparing the masses before and after heating.
Common empirical formulas include: CH₂O (glucose, acetic acid), CH (benzene, ethyne), C₂H₃ (vinyl chloride), CH₂ (alkenes and cycloalkanes), CH₄ (methane — also its molecular formula), and HO (hydrogen peroxide). Recognising these empirical formulas helps to quickly identify compound families and predict molecular formulas when given the Mr.