If a compound contains only carbon, hydrogen and oxygen, find its oxygen mass by subtracting the masses of carbon and hydrogen from the original sample mass. During complete combustion, oxygen from the surroundings can enter the products. The oxygen in the products therefore cannot simply be assigned to the sample.
SEAB’s 2027 SEC G3 Chemistry syllabus (K324) includes finding empirical and molecular formulae from relevant data. The example below uses carbon dioxide and water measurements to recover the simplest whole-number ratio.
- Use carbon dioxide to find moles of carbon atoms.
- Use water to find moles of hydrogen atoms.
- Find sample oxygen by mass difference before comparing moles.
Worked example | A 4.60 g sample
A compound containing only carbon, hydrogen and oxygen is burned completely. A 4.60 g sample produces 8.80 g of carbon dioxide and 5.40 g of water. Use relative atomic masses C = 12, H = 1 and O = 16.
Carbon dioxide has molar mass 44 g/mol. The 8.80 g corresponds to 0.200 mol of CO₂ and therefore 0.200 mol of carbon atoms. The sample contains 0.200 × 12 = 2.40 g of carbon.
Water has molar mass 18 g/mol. The 5.40 g corresponds to 0.300 mol of H₂O. There are two hydrogen atoms per water molecule, so the sample contains 0.600 mol of hydrogen atoms, with mass 0.600 g.
The oxygen mass in the sample is 4.60 − 2.40 − 0.600 = 1.60 g, or 0.100 mol of oxygen atoms. The mole ratio is 0.200:0.600:0.100, which simplifies to 2:6:1. The empirical formula is C₂H₆O.
Check the source of each number
The mass difference only works as written because the question gives exactly three elements. If another element is present, you need to account for its mass before assigning the remainder to oxygen.
Check the units on the mole ratio. The masses 2.40:0.600:1.60 are mass ratios. Formula subscripts compare numbers of atoms, so first divide each mass by its element’s molar mass.
Try a second set of combustion data
A 3.00 g sample containing only C, H and O produces 4.40 g of CO₂ and 1.80 g of water. The sample contains 1.20 g of carbon and 0.200 g of hydrogen. The remaining 1.60 g is oxygen.
Those masses correspond to 0.100 mol C, 0.200 mol H and 0.100 mol O. The simplest ratio is 1:2:1, giving CH₂O. A molecular formula would require further information such as the relative molecular mass.
Frequently asked questions
Why do I multiply the water moles by two?
Each water molecule contains two hydrogen atoms.
Can the products weigh more than the sample?
Yes. Oxygen from the surroundings can be added during combustion.
Official information
SEAB’s 2027 SEC G3 Chemistry syllabus (K324), Chemical Calculations. Checked 9 November 2026.