Combustion Analysis Calculator

What Is Combustion Analysis?

Combustion analysis is a quantitative laboratory technique used to determine the empirical formula of an unknown organic compound containing carbon, hydrogen, and optionally oxygen. The sample is burned completely in excess oxygen, and the masses of carbon dioxide (CO₂) and water (H₂O) produced are measured. From these combustion products, chemists calculate how much carbon and hydrogen were in the original sample, then find the oxygen content by difference.

The complete combustion reaction for a general organic compound is:

$$\text{C}_\alpha\text{H}_\beta\text{O}_\gamma + a\text{O}_2 \longrightarrow b\text{CO}_2 + c\text{H}_2\text{O}$$

All carbon from the sample ends up in CO₂, and all hydrogen ends up in H₂O. This makes combustion analysis a powerful method for identifying unknown organic substances in research and education.

How to Find the Empirical Formula

The calculation follows three main steps:

Calculate element masses from the combustion product masses.
Convert masses to moles using atomic molar masses.
Normalize mole ratios to the smallest whole-number subscripts.

The key mass relationships are:

$$m_C = m_{CO_2} \cdot \frac{M_C}{M_{CO_2}}$$

$$m_H = m_{H_2O} \cdot \frac{2M_H}{M_{H_2O}}$$

For C, H, O compounds, oxygen mass is found by difference:

$$m_O = m_{sample} - m_C - m_H$$

For hydrocarbons (compounds with only carbon and hydrogen), the sample mass is not required because no oxygen needs to be calculated by difference.

How to Find the Molecular Formula

Once you have the empirical formula, divide the compound's molar mass by the empirical formula mass to get an integer multiplier $n$:

$$n = \frac{\text{Molar mass}}{\text{Empirical formula mass}}$$

Multiply each subscript in the empirical formula by $n$ to obtain the molecular formula. For example, if the empirical formula is CH₂O with empirical mass 30.03 g/mol and the molar mass is 90.08 g/mol, then $n \approx 3$ and the molecular formula is C₃H₆O₃.

Worked Example: C, H, O Compound

A 12.915 g sample of an unknown compound produces 18.942 g CO₂ and 7.749 g H₂O upon combustion. The molar mass is 90.0779 g/mol. Using this calculator:

Carbon mass: $18.942 \times \frac{12.011}{44.010} = 5.169$ g
Hydrogen mass: $7.749 \times \frac{2 \times 1.008}{18.015} = 0.867$ g
Oxygen mass: $12.915 - 5.169 - 0.867 = 6.879$ g

Normalizing the mole ratios gives empirical formula CH₂O. With $n = 90.08 / 30.03 \approx 3$, the molecular formula is C₃H₆O₃ (glucose or a related isomer).

Related Chemistry Tools

For related calculations, try our Empirical Formula Calculator for direct mass-percentage input, Molar Mass Calculator for molecular weight lookups, and the Chemical Equation Balancer to balance combustion reactions.

Frequently Asked Questions

What is the difference between empirical and molecular formula?

The empirical formula shows the simplest whole-number ratio of atoms (for example CH₂O). The molecular formula shows the actual number of atoms per molecule (for example C₆H₁₂O₆). The molecular formula is always a whole-number multiple of the empirical formula.

Why is sample mass needed for C, H, O compounds but not hydrocarbons?

For compounds containing oxygen, the oxygen mass cannot be measured directly from combustion products. It is calculated by subtracting the carbon and hydrogen masses from the original sample mass. Hydrocarbons contain only carbon and hydrogen, so both elements are fully accounted for by the CO₂ and H₂O masses.

Do I need the molar mass to find the empirical formula?

No. The empirical formula depends only on the combustion product masses (and sample mass for C, H, O compounds). Molar mass is required only when you want to determine the molecular formula.

What assumptions does combustion analysis make?

The method assumes complete combustion: all carbon becomes CO₂, all hydrogen becomes H₂O, and no carbon remains as soot or other products. The sample must contain only C, H, and O (or only C and H for hydrocarbons). Compounds with nitrogen, sulfur, or halogens require modified analysis.

How are non-integer mole ratios handled?

Mole ratios are scaled to the nearest whole numbers. For example, a carbon-to-hydrogen ratio of 1:1.60 is converted to 5:8 by multiplying both values until they are close to integers. This calculator applies the same rounding logic used in standard organic chemistry textbooks.

Can two different compounds give the same combustion analysis result?

Yes. Isomers with the same molecular formula (such as glucose and fructose, both C₆H₁₂O₆) produce identical combustion data. Combustion analysis identifies the molecular formula but not the structural arrangement of atoms.

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