Combustion Analysis Calculator

Determine the empirical formula of a C, H, O compound from combustion data. Enter the masses of sample, CO₂ produced, and H₂O produced.

Mass of the unknown organic compound sample

Mass of carbon dioxide produced from combustion

Mass of water produced from combustion

Mass of oxygen consumed (if compound contains oxygen)

Mass C = Mass CO₂ × (12.011/44.010), Mass H = Mass H₂O × (2.016/18.015), Mass O = Sample - C - H. Moles = mass/atomic mass. Divide by smallest to get ratios.
5.00 mg sample produces 15.42 mg CO₂ and 6.32 mg H₂O. Mass C = 15.42 × (12.011/44.010) = 4.21 mg. Mass H = 6.32 × (2.016/18.015) = 0.71 mg. Mass O = 5.00 - 4.21 - 0.71 = 0.08 mg. Ratios give empirical formula.

What is combustion analysis?

Combustion analysis (elemental analysis) burns an organic compound in excess oxygen to produce CO₂, H₂O, and possibly N₂. By measuring the masses of CO₂ and H₂O produced, we can determine the empirical formula: mass of C = mass(CO₂) × (12.01/44.01), mass of H = mass(H₂O) × (2.016/18.015).

How do you determine if oxygen is present?

Sum the masses of C and H found. If this is less than the original sample mass, the difference is oxygen: mass(O) = mass(sample) - mass(C) - mass(H). If the sum equals the sample mass, the compound contains only C, H (and possibly N, which requires separate analysis).

How do you find the empirical formula?

1) Calculate masses of C, H, O from combustion data, 2) Convert to moles (÷ atomic mass), 3) Divide all by the smallest mole value, 4) Multiply to get whole numbers. Example: C=4.36mg→0.363mmol, H=0.73mg→0.73mmol, ratio C:H = 1:2 → CH₂.

What is the difference between empirical and molecular formula?

Empirical formula is the simplest whole-number ratio of atoms (e.g., CH₂O for glucose). Molecular formula is the actual number of atoms (C₆H₁₂O₆ for glucose). To get molecular formula, divide the molar mass by the empirical formula mass: n = Molar Mass / Empirical Mass.