Convert mass percent composition or mass data into empirical formulas. Use measured molar mass to determine molecular formulas. See step-by-step mole calculations, ratios, and subscript determination.
Enter the percent composition or mass data for your compound to determine its empirical and molecular formulas.
Start with percent composition data from combustion analysis or elemental analysis. Assume you have exactly 100 grams of compound—this converts percentages directly into grams. If your unknown is 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen, you're working with 40.0 g C, 6.7 g H, and 53.3 g O.
Convert each mass to moles using atomic weights. Carbon: 40.0 g ÷ 12.011 g/mol = 3.33 mol. Hydrogen: 6.7 g ÷ 1.008 g/mol = 6.65 mol. Oxygen: 53.3 g ÷ 15.999 g/mol = 3.33 mol. Keep at least three significant figures through these intermediate steps to avoid rounding errors in your final answer.
Watch units throughout. Grams cancel with grams in the denominator of atomic weight, leaving moles. If you accidentally divide by atomic number instead of atomic weight, your mole counts will be completely wrong. Carbon's atomic number is 6, but its atomic weight is 12.011—those give very different results.
Divide every mole value by the smallest mole value in your set. This normalizes one element to 1.00 while preserving the ratios among all elements. If your moles are 3.33 C, 6.65 H, and 3.33 O, divide by 3.33 to get 1.00 C : 2.00 H : 1.00 O.
Round ratios to integers only when they're within about ±0.1 of a whole number. A ratio of 1.97 rounds to 2. A ratio of 1.05 rounds to 1. But 1.50 or 1.33 should not be rounded directly—these indicate you need multiplication to reach whole numbers.
The empirical formula from our example is CH2O. This represents the simplest whole-number ratio of atoms in the compound. Many different molecules can share the same empirical formula—formaldehyde (CH2O), acetic acid (C2H4O2), and glucose (C6H12O6) all reduce to CH2O.
The empirical formula alone doesn't tell you which compound you have. You need the actual molar mass, typically from mass spectrometry, freezing point depression, or vapor density measurements. Compare this measured value to the empirical formula mass.
Calculate the empirical formula mass first. CH2O weighs 12.011 + 2(1.008) + 15.999 = 30.03 g/mol. If your mass spec shows a molecular ion at m/z = 180, divide to find the multiplier: 180 ÷ 30.03 = 5.99, which rounds to 6. The molecular formula is C6H12O6.
The multiplier must be a positive integer. If your calculation gives 2.7 or 4.3, something is wrong—either the measured molar mass has error, the percent composition is off, or you made an arithmetic mistake. Recheck your inputs before accepting a non-integer result.
Core equation: Molecular formula = n × Empirical formula, where n = (measured molar mass) / (empirical formula mass). This relationship connects experimental mass data to actual molecular structure.
Converting % to moles:
moles = (percent / 100) × 100 g / atomic weight
Simplified: moles = percent / atomic weight
Finding mole ratios:
ratio_i = moles_i / min(all moles)
Scaling to molecular:
n = M_measured / M_empirical (round to integer)
The 100 g assumption in the first step is just for convenience—it makes percentages equal grams numerically. You could use any sample size; the ratios would be the same since percent composition is scale-invariant.
Problem: Combustion analysis of an unknown organic compound gives 85.6% carbon and 14.4% hydrogen. Mass spectrometry shows the molecular ion at m/z = 56. Find the molecular formula.
Step 1: Convert to moles
C: 85.6 / 12.011 = 7.13 mol
H: 14.4 / 1.008 = 14.29 mol
Step 2: Find mole ratios
C: 7.13 / 7.13 = 1.00
H: 14.29 / 7.13 = 2.00
Empirical formula: CH2
Step 3: Calculate empirical mass
M = 12.011 + 2(1.008) = 14.03 g/mol
Step 4: Find multiplier
n = 56 / 14.03 = 3.99 ≈ 4
Step 5: Write molecular formula
Molecular formula: C4H8
C4H8 could be several isomers: 1-butene, 2-butene, isobutylene, or cyclobutane. The formula alone doesn't distinguish them—you'd need NMR or IR spectroscopy for structural identification.
Ratio ≈ 1.5: Multiply all ratios by 2. The pattern 1:1.5:2 becomes 2:3:4. This commonly appears in compounds like Fe2O3 (iron oxide).
Ratio ≈ 1.33: Multiply all ratios by 3. The pattern 1:1.33 becomes 3:4. You see this in molecules where one element has 4/3 the count of another.
Ratio ≈ 1.25: Multiply all ratios by 4. The pattern 1:1.25 becomes 4:5. Less common, but it appears in some phosphate and sulfate compounds.
Ratio ≈ 1.67: Multiply all ratios by 3. The pattern 1:1.67 becomes 3:5. This one tricks students because 1.67 looks like it might round to 2.
If none of these multipliers work and you still have decimals, check your original calculations. Combustion analysis data typically has ±0.3% error, which can shift ratios just enough to confuse the pattern. Try recalculating with slight adjustments to see if a clean formula emerges.
• Isomers indistinguishable: Many compounds share the same molecular formula. C2H6O could be ethanol or dimethyl ether—spectroscopy is needed to tell them apart.
• Experimental uncertainty: Combustion analysis has inherent error. Small deviations in percent composition can shift mole ratios enough to complicate formula determination.
• Discrete molecules only: This method works for molecular compounds. Ionic compounds, polymers, and network solids don't have true molecular formulas in the same sense.