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How to Find Empirical Formulas: The Step-by-Step Method Every Student Needs

12 min read 27 March 2026

Key Takeaways

Contents

  1. What is an Empirical Formula?
  2. The 4-Step Method
  3. Three Worked Examples
  4. From Empirical to Molecular Formula
  5. Common Exam Mistakes
  6. Bonus: Hydrated Salts
  7. Frequently Asked Questions

Finding empirical formulas from percentage composition data is one of the most frequently examined skills in both GCSE and IB Chemistry. The good news is that it follows a fixed, repeatable method that works every single time. Once you learn the process, you can apply it to any question the exam throws at you.

This guide covers the complete method from first principles, walks through three fully worked examples of increasing difficulty, and highlights the specific mistakes that cost students marks every year. There is also a link to our free Formula Calculator so you can check your working instantly.

What is an Empirical Formula?

A chemical formula can be expressed in two ways, and understanding the difference is essential before you attempt any calculation.

Empirical Formula

The empirical formula is the simplest whole-number ratio of atoms of each element in a compound. It tells you the proportions, not the actual numbers. For example, the empirical formula of glucose is CH2O. This tells us there is one carbon atom for every two hydrogen atoms and one oxygen atom.

Molecular Formula

The molecular formula shows the actual number of atoms of each element in one molecule of the compound. For glucose, the molecular formula is C6H12O6. Notice that this is simply the empirical formula (CH2O) multiplied by 6.

To get from the empirical formula to the molecular formula, you need one additional piece of information: the relative formula mass (Mr) of the compound. This is often provided in the question, or you may be told it was determined by mass spectrometry.

The 4-Step Method

Every empirical formula question, regardless of the exam board or difficulty level, can be solved using the same four steps.

The Method (memorise this)

  1. Write down the percentage (or mass) of each element. If percentages do not add up to 100%, the missing portion is almost certainly oxygen.
  2. Divide each value by the Ar of that element. This converts the data into moles. Use the Ar values appropriate for your exam board.
  3. Divide every result by the smallest number. This gives you the simplest ratio.
  4. If any ratio is not a whole number, multiply all ratios by the smallest integer that makes them all whole. Common multipliers are x2 (for values ending in .5), x3 (for .33 or .67), and x4 (for .25 or .75).

That is the entire method. The rest is just practice.

Three Worked Examples

Example 1: A Simple Ratio (No Multiplier Needed)

Worked Example 1: Magnesium oxide

A compound contains 60.3% magnesium and 39.7% oxygen. Find its empirical formula.

Step 1:Mg = 60.3%, O = 39.7%. These add to 100%, so all elements are accounted for.
Step 2:Divide by Ar. Mg: 60.3 / 24 = 2.5125. O: 39.7 / 16 = 2.4813.
Step 3:Divide by the smallest (2.4813). Mg: 2.5125 / 2.4813 = 1.013. O: 2.4813 / 2.4813 = 1.000.
Step 4:Both values round to 1. No multiplier needed.
Empirical formula: MgO

Example 2: A Ratio That Needs Multiplying

Worked Example 2: Ethanoic acid

A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen. Find its empirical formula.

Step 1:C = 40.0%, H = 6.7%, O = 53.3%. Total = 100%.
Step 2:Divide by Ar. C: 40.0 / 12 = 3.333. H: 6.7 / 1 = 6.7. O: 53.3 / 16 = 3.331.
Step 3:Divide by the smallest (3.331). C: 1.001. H: 2.011. O: 1.000.
Step 4:Ratios round cleanly to 1 : 2 : 1. No multiplier needed.
Empirical formula: CH2O

Example 3: Missing Oxygen and a Multiplier

Worked Example 3: Iron oxide from experimental data

In a combustion analysis, a compound is found to contain 72.4% iron. The only other element present is oxygen. Given that the Mr of the compound is 231.5, find the molecular formula.

Step 1:Fe = 72.4%. Oxygen = 100 - 72.4 = 27.6%.
Step 2:Divide by Ar. Fe: 72.4 / 56 = 1.293. O: 27.6 / 16 = 1.725.
Step 3:Divide by the smallest (1.293). Fe: 1.000. O: 1.334.
Step 4:1.334 is close to 4/3, so multiply all by 3. Fe: 3. O: 4.
Empirical formula: Fe3O4
Step 5:Empirical Mr = (3 x 56) + (4 x 16) = 232. Multiplier = 231.5 / 232 = 1. Molecular formula is the same as the empirical formula.
Molecular formula: Fe3O4

From Empirical to Molecular Formula

Once you have the empirical formula, getting the molecular formula requires just two more steps:

  1. Calculate the Mr of the empirical formula. Add up the Ar values for all atoms in the empirical formula.
  2. Find the multiplier. Divide the given Mr of the compound by the empirical Mr. This gives you a whole number (the multiplier). Multiply every subscript in the empirical formula by this number.

Worked Example 4: Glucose (empirical to molecular)

A compound has the empirical formula CH2O and an Mr of 180. Find its molecular formula.

Step 1:Empirical Mr = 12 + (2 x 1) + 16 = 30.
Step 2:Multiplier = 180 / 30 = 6.
Step 3:Multiply all subscripts by 6. C1x6H2x6O1x6.
Molecular formula: C6H12O6

Common Exam Mistakes

These are the specific errors that examiners see every year. Each one is entirely preventable.

Forgetting the Missing Oxygen

If the question gives you percentage data that does not add up to 100%, the remaining percentage is almost always oxygen. Students who do not check the total will miss an element entirely and get the wrong formula.

Using Wrong Ar Values

GCSE exams use whole-number Ar values (H = 1, C = 12, O = 16), except for Cl = 35.5 and Cu = 63.5. IB exams use values from Section 7 of the data booklet (e.g. H = 1.01, C = 12.01, O = 16.00). Using IB precision in a GCSE exam, or vice versa, can produce different answers.

Rounding Too Early

If you round intermediate values to 1 decimal place during the calculation, the errors compound. Keep all decimal places in your calculator until the final step. Only round the subscripts in the empirical formula, and only when you are confident they are close to whole numbers.

Confusing Empirical and Molecular Formulas

Many students write the empirical formula when the question asks for the molecular formula, or the other way around. Read the question carefully. If it asks for the molecular formula, you need the Mr as well.

Bonus: Hydrated Salts

A related exam question type asks you to find the value of x in a hydrated salt formula like CuSO4·xH2O. The method is straightforward:

  1. Calculate the Mr of the anhydrous (dry) salt.
  2. Calculate the Mr of the hydrated salt (either given or calculated from masses before and after heating).
  3. Subtract: mass of water = hydrated Mr - anhydrous Mr.
  4. Divide by 18 (the Mr of H2O) to find x.

Our Formula Calculator includes a dedicated Hydrated Salts tab that does this calculation for you, with full working shown.

Frequently Asked Questions

What is the difference between an empirical formula and a molecular formula?

The empirical formula shows the simplest whole-number ratio of atoms in a compound. The molecular formula shows the actual number of atoms in one molecule. For example, glucose has an empirical formula of CH2O and a molecular formula of C6H12O6. They both describe the same compound, but the molecular formula contains six times as many of each atom.

Why do I need the Mr to find the molecular formula?

The empirical formula only gives you the ratio. Multiple compounds can share the same empirical formula: CH2O could be formaldehyde (Mr = 30), acetic acid (Mr = 60), or glucose (Mr = 180). The Mr tells you which one it actually is.

What should I do if my ratios do not give whole numbers?

If dividing by the smallest gives you a value ending in .5, multiply all ratios by 2. If you get a value near .33 or .67, multiply by 3. If the value is near .25 or .75, multiply by 4. The key is to find the smallest integer that converts all ratios to whole numbers. Our Formula Calculator tests multipliers 1 through 5 automatically.

Which Ar values should I use in the exam?

For GCSE exams (AQA, Edexcel, OCR), use the whole-number values from the periodic table provided in the exam, with the exceptions of Cl (35.5) and Cu (63.5). For IB Diploma exams, use the 2 decimal place values from Section 7 of the IB Data Booklet. Our calculator has a toggle to switch between GCSE and IB precision.

Try the Formula Calculator

Enter your percentage data and get the empirical formula instantly, with full step-by-step working.

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