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A question gives you 40.92% carbon, 4.58% hydrogen and the rest oxygen, and asks for the formula. Most students can start. Many lose the mark at one step: they see a ratio of 1.33 and round it to 1. This post walks through the five steps NCERT uses, two complete worked examples (one from percentages, one from combustion data), and the slips examiners see every year.
Where is this in the syllabus?
Empirical and molecular formulae are in section 1.9 of NCERT Class 11 chemistry, Unit 1, Some Basic Concepts of Chemistry. The JEE (Main) 2026 syllabus lists "percentage composition, empirical and molecular formulae" in Unit 1 and, again in Unit 13, "calculations of empirical formulae and molecular formulae" from organic quantitative analysis. The NEET (UG) 2026 syllabus has the same two entries. The JEE (Advanced) 2026 brochure asks for "determination of empirical and molecular formulae of simple compounds by combustion method only". Syllabi change, so check the bulletin for your exam year.
What is the difference between the two formulas?
NCERT puts it in one line: an empirical formula is "the simplest whole number ratio of various atoms present in a compound", while the molecular formula "shows the exact number of different types of atoms present in a molecule".
Formaldehyde (CH2O), acetic acid (C2H4O2) and glucose (C6H12O6) all share the empirical formula CH2O, so all three are 40.0% carbon by mass. Percentage data alone cannot tell them apart. You need the molar mass as well.
What are the five steps?
Assume 100 g of compound, so each percentage becomes a mass in grams.
Convert to moles by dividing each mass by the atomic mass (16.00 for oxygen, not 32).
Divide by the smallest number of moles.
Clear decimals by multiplying. A ratio ending in about .5 means multiply everything by 2; .33 or .67 by 3; .25 or .75 by 4; .2 by 5.
Find n. Divide the molar mass by the empirical formula mass and multiply the subscripts by n.
Worked example 1: vitamin C from percentages
PubChem lists ascorbic acid as C6H8O6 with a molar mass of 176.12 g/mol. Working back from that formula gives 40.92% C, 4.58% H and 54.50% O, so here is the forward calculation a student would do.
Carbon: 40.92 / 12.01 = 3.407 mol
Hydrogen: 4.58 / 1.008 = 4.544 mol
Oxygen: 54.50 / 16.00 = 3.406 mol
Divide by the smallest, 3.406: C = 1.000, H = 1.334, O = 1.000.
The hydrogen ratio is 1⅓, not 1. Multiply all three by 3 to get C3H4O3.
Empirical formula mass = 3(12.01) + 4(1.008) + 3(16.00) = 88.06 g/mol. Then n = 176.12 / 88.06 = 2.00, so the molecular formula is C6H8O6, which matches PubChem.
If you had rounded 1.334 to 1, you would have written CHO, with a mass of 29.02 g/mol, and got n = 176.12 / 29.02 = 6.07. An n that is not a whole number is a warning that an earlier step went wrong.
Worked example 2: from combustion data
These are sample numbers chosen to show the method. A 0.250 g sample of a compound containing only C, H and O burns completely to give 0.3664 g of CO2 and 0.1500 g of H2O. Its molar mass is 60 g/mol.
Mass of carbon = 0.3664 × 12.01 / 44.01 = 0.1000 g
Mass of hydrogen = 0.1500 × 2.016 / 18.02 = 0.01679 g
Mass of oxygen, by difference = 0.250 − 0.1000 − 0.0168 = 0.1332 g
Moles: C = 0.1000 / 12.01 = 0.00833; H = 0.01679 / 1.008 = 0.01666; O = 0.1332 / 16.00 = 0.00833. The ratio is 1 : 2 : 1, so the empirical formula is CH2O (30.03 g/mol), n = 60 / 30.03 = 2.0, and the molecular formula is C2H4O2. That is the formula PubChem gives for acetic acid. The shortcut formulas in NCERT's organic chemistry unit, percentage of carbon = 12 × m2 × 100 / (44 × m), give the same result: about 40.0% carbon and 6.7% hydrogen.
Why do students find this harder than it looks?
Teachers Peter Nassiff and Wendy Czerwinski note that students often learn empirical formula calculations by rote, without knowing why the method works, and they built a paperclip activity to show that experiments can only ever give a ratio. Worked examples help, with a limit. In Sweller and Cooper's algebra experiments (1985), studying worked examples made later problems faster and less error-prone, but the gain was specific to problems with the same structure. So vary the type: percentages, combustion data and gas-volume data each test a different step.
Which mistakes cost the most marks?
Rounding 1.33 or 1.5 to a whole number. Multiply instead.
Rounding too early. Keep three or four significant figures until step 3.
Using 32 for oxygen. You are counting atoms, so use 16.00. NCERT's exercise 1.3 says "dioxygen", which tempts this slip.
Forgetting oxygen in combustion problems. It is never measured directly; find it by difference.
Treating the CO2 mass as carbon. Only 12.01 / 44.01 of it is carbon.
If your ratio keeps coming out as an odd decimal and you cannot see why, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer, so you can find the step where your numbers drifted.
A practice routine
Day 1: three percentage problems whose ratios end in .5, .33 and .25.
Day 2: two combustion problems, including NCERT's exercise 1.34 on the welding gas.
Day 3: connect to moles more broadly with our posts on mole concept errors and limiting reagent problems.
A week later: redo one problem of each type from memory, without notes.
Tonight, solve NCERT's Problem 1.2 without looking at the solution and check that you reach C2H4Cl2.
