Empirical Formula Calculator

Convert percent composition or mass data into the simplest whole‑number chemical formula. See every step, adjust rounding, and optionally get the molecular formula from molar mass.

Empirical formula calculator

Element data

Element Percent (%) Atomic mass (g/mol) Actions

Controls how close a ratio must be to 1, 1.5, 2, 2.5, 3, etc. to be treated as that value.

If provided, the tool will compute the molecular formula as a multiple of the empirical formula.

What is an empirical formula?

The empirical formula of a compound shows the simplest whole‑number ratio of atoms of each element present. It does not tell you the exact number of atoms in a single molecule, only their relative proportions.

  • Glucose: molecular formula = C6H12O6, empirical formula = CH2O
  • Hydrogen peroxide: molecular formula = H2O2, empirical formula = HO
  • Water: molecular and empirical formula are both H2O

How this empirical formula calculator works

The calculator follows the same steps you would use by hand in general chemistry:

  1. Convert each element’s percent or mass to moles.
  2. Divide all mole values by the smallest mole value to get a mole ratio.
  3. Recognize ratios that are close to simple fractions (1, 1.5, 2, 2.5, 3, etc.).
  4. Multiply all ratios by a small integer to obtain whole‑number subscripts.
  5. Optionally use the compound’s molar mass to find the molecular formula.

Key idea

If the mole ratios are \(n_1 : n_2 : \dots : n_k\), then the empirical formula is \[ \text{E.F.} = \mathrm{E}_1^{a_1}\mathrm{E}_2^{a_2}\dots \mathrm{E}_k^{a_k} \] where each \(a_i\) is the smallest whole number proportional to \(n_i\).

Step‑by‑step example (percent composition)

Suppose a compound is 40.0% C, 6.7% H, and 53.3% O by mass. Find its empirical formula.

  1. Assume 100 g of substance so that:
    • C: 40.0 g
    • H: 6.7 g
    • O: 53.3 g
  2. Convert grams to moles (using approximate atomic masses):
    • \(\text{mol C} = 40.0 \div 12.01 \approx 3.33\)
    • \(\text{mol H} = 6.7 \div 1.008 \approx 6.65\)
    • \(\text{mol O} = 53.3 \div 16.00 \approx 3.33\)
  3. Divide by the smallest mole value (≈ 3.33):
    • C: \(3.33 / 3.33 \approx 1.00\)
    • H: \(6.65 / 3.33 \approx 2.00\)
    • O: \(3.33 / 3.33 \approx 1.00\)
  4. Ratios are already whole numbers: 1 : 2 : 1
  5. Write the empirical formula: CH2O

From empirical formula to molecular formula

If you also know the compound’s molar mass (from experiment), you can find the molecular formula.

  1. Compute the empirical formula mass, \(M_{\text{emp}}\).
  2. Compute the ratio \(n = \dfrac{M_{\text{molar}}}{M_{\text{emp}}}\).
  3. If \(n\) is a whole number (or very close), multiply all empirical subscripts by \(n\).

For example, CH2O has \(M_{\text{emp}} \approx 30.03\ \text{g/mol}\). If the measured molar mass is 180.2 g/mol:

\[ n = \frac{180.2}{30.03} \approx 6 \quad\Rightarrow\quad \text{molecular formula} = \mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 \]

Common pitfalls and tips

  • Percent doesn’t sum to exactly 100%: small rounding errors are normal; the calculator normalizes the values.
  • Ratios like 1.33 or 1.67: these often correspond to 4/3 or 5/3; multiplying all ratios by 3 usually fixes them.
  • Hydrates (e.g., CuSO4·5H2O): treat water as another “element” (H and O) or calculate moles of anhydrous salt and water separately.
  • Significant figures: the empirical formula itself is exact integers, but your mole ratios depend on the precision of your experimental data.

FAQ

Can two different compounds have the same empirical formula?

Yes. Many compounds share the same empirical formula but differ in molecular formula and structure. For example, glucose, fructose, and acetic acid all reduce to CH2O empirically.

Why do we assume a 100 g sample for percent composition?

Assuming 100 g makes the math easy: each percent becomes that many grams. Any other sample size would give the same mole ratios after normalization.

What if my ratios never become nice whole numbers?

Try adjusting the rounding sensitivity and check that you used correct atomic masses and accurate experimental data. If large multipliers (like ×7 or ×8) are required, the data may be inconsistent or too imprecise.