Combustion analysis and lab data give the percentage of each element in a compound. Enter the elements and their percentages to find the simplest whole-number formula, and the molecular formula if you know the molar mass.
Enter the element symbols.
Enter their percentages or grams.
Optionally enter the molar mass.
Steps: % ÷ atomic mass, ÷ smallest, × whole number
Molecular: n = molar mass ÷ empirical formula mass
An empirical formula gives the simplest whole-number ratio of atoms in a compound. This calculator takes the element symbols and either their percentages by mass or their masses in grams, converts each to moles, finds the simplest ratio, and multiplies up when the ratio has halves or thirds. If you also give the compound's molar mass, it scales the empirical formula to the molecular formula.
Chemists meet this problem whenever they analyse an unknown substance: combustion analysis or elemental analysis reports only percentages, and the formula must be worked out from them. It is a standard Class 11 chemistry topic and a favourite in board exams, JEE and NEET. Glucose, for example, analyses as 40% carbon, 6.7% hydrogen and 53.3% oxygen, which gives CH₂O, and its molar mass of 180 g/mol then reveals C₆H₁₂O₆.
1. Assume a 100 g sample, so each percentage becomes a mass in grams. If you have masses already, use them directly.
2. Divide each mass by the element's atomic mass to get moles.
3. Divide every mole value by the smallest one, so the smallest becomes 1.
4. If any ratio is close to a fraction like 1.5, 1.33 or 1.25, multiply all ratios by 2, 3 or 4 respectively to reach whole numbers.
5. Round to whole numbers and write the empirical formula with these subscripts.
6. For the molecular formula, compute n = molar mass ÷ empirical formula mass, round to a whole number, and multiply every subscript by n.
Percentages by mass do not compare atoms directly, because atoms of different elements have different masses. Iron oxide can be 70% iron by mass and still contain more oxygen atoms than iron atoms, since each iron atom weighs about 3.5 times an oxygen atom. Dividing by atomic mass converts grams into moles, which are proportional to numbers of atoms. Only then does the ratio reflect how atoms combine, which is what a formula expresses.
Experimental data carry error, so ratios rarely come out as exact whole numbers. A ratio of 1.98 is clearly 2, but 1.5 is not a rounding problem; it means the true ratio is 3 : 2. The calculator tries multipliers from 1 to 8 and picks the first that brings every ratio within 0.1 of a whole number. Rounding 1.5 up to 2, or 1.33 down to 1, is the most common way students get the wrong formula.
Many different compounds share an empirical formula. Formaldehyde, acetic acid and glucose are all CH₂O in simplest ratio, with molar masses of about 30, 60 and 180 g/mol. The molecular formula is a whole-number multiple of the empirical one, and the multiple comes from dividing the measured molar mass by the empirical formula mass. For ionic solids like NaCl or Fe₂O₃ there are no discrete molecules, so the empirical formula is the formula used.
Karan's lab analysis of a reddish-brown iron ore shows 69.94% iron and 30.06% oxygen by mass, and the reference molar mass given is 159.69 g/mol.
Moles = amount ÷ atomic mass: Fe: 69.94 ÷ 55.845 = 1.2524 O: 30.06 ÷ 15.999 = 1.8789
Divide by the smallest: Fe 1, O 1.5
Multiply by 2 to reach whole numbers: Fe 2, O 3 Empirical formula: Fe2O3
Molecular formula: n = 159.69 ÷ 159.687 ≈ 1 → Fe2O3
Answer: Empirical formula Fe2O3; Molecular formula Fe2O3; Empirical formula mass 159.687 g/mol
Dividing percentages directly by each other without converting to moles.
Rounding ratios like 1.5 or 1.33 to the nearest whole number instead of multiplying up.
Using wrong capitalisation, such as CO for cobalt, or entering the percentages in a different order from the symbols.
Rounding too early in the mole step, which can push a ratio away from a clean value.
Taking the molecular formula multiplier from a rough molar mass without rounding it to a whole number.
Class 11 chemistry problems on percentage composition and formulas.
Interpreting combustion and elemental analysis reports in organic chemistry labs.
Identifying minerals and ores from their composition.
Checking the purity or identity of a synthesised compound.
JEE and NEET practice on the mole concept.
What if my ratio is 2.5?
Multiply all ratios by 2 to get whole numbers.
Do percentages need to add to 100?
Not exactly; grams work too, since only the ratios matter.