Free Combustion Analysis Calculator

Combustion analysis is a foundational technique in organic chemistry used to determine the empirical formula of unknown compounds containing carbon, hydrogen, and often oxygen. The Combustion Analysis Calculator presented here is a free online tool that automates the entire workflow: from entering combustion data—such as sample mass, the masses of $\mathrm{CO_2}$ and $\mathrm{H_2O}$ produced, and optionally the molar mass—it directly outputs both the empirical and molecular formulas. This combustion analysis calculator is designed for students, educators, and laboratory professionals who need quick, reliable formula determination.

Understanding Combustion Analysis

In a standard combustion analysis, a precisely weighed organic sample is burned completely in an excess of oxygen. All carbon in the sample is converted to carbon dioxide, and all hydrogen to water vapor. The produced $\mathrm{CO_2}$ and $\mathrm{H_2O}$ are collected and weighed. From these masses, the amounts of carbon and hydrogen originally present can be calculated. If the compound contains oxygen, its mass is obtained by subtracting the carbon and hydrogen masses from the original sample mass.

Deriving the Empirical Formula – Step by Step

The process of extracting the empirical formula from combustion data can be broken into three clear steps:

1. Calculate the mass of each element.
   For carbon:

   $$m_{\mathrm{C}} = m_{\mathrm{CO_2}} \times \frac{M_{\mathrm{C}}}{M_{\mathrm{CO_2}}}$$

   For hydrogen:

   $$m_{\mathrm{H}} = m_{\mathrm{H_2O}} \times \frac{2\,M_{\mathrm{H}}}{M_{\mathrm{H_2O}}}$$

   For oxygen (when present):

   $$m_{\mathrm{O}} = m_{\mathrm{sample}} - m_{\mathrm{C}} - m_{\mathrm{H}}$$

   The molar masses used are $M_{\mathrm{C}} = 12.0107\ \mathrm{g/mol}$, $M_{\mathrm{H}} = 1.00794\ \mathrm{g/mol}$, $M_{\mathrm{CO_2}} = 44.0095\ \mathrm{g/mol}$, $M_{\mathrm{H_2O}} = 18.0153\ \mathrm{g/mol}$, and $M_{\mathrm{O}} = 15.999\ \mathrm{g/mol}$.

2. Convert the masses to moles.

   $$n_{\mathrm{C}} = \frac{m_{\mathrm{C}}}{M_{\mathrm{C}}},\quad n_{\mathrm{H}} = \frac{m_{\mathrm{H}}}{M_{\mathrm{H}}},\quad n_{\mathrm{O}} = \frac{m_{\mathrm{O}}}{M_{\mathrm{O}}}$$

3. Find the simplest whole‑number ratio. Divide each mole amount by the smallest among them. If the resulting values are not integers, multiply them by a common factor to clear any fractional parts (for example, 1.5 becomes 3, or 1.6 ≈ 8/5 leads to multiplication by 5). The final integers become the subscripts in the empirical formula.

From Empirical to Molecular Formula

The empirical formula shows the simplest ratio of atoms, but the molecular formula reveals the actual number of each atom in a molecule. To move from one to the other, two pieces of information are needed: the empirical formula and the compound’s molar mass.

• Compute the empirical‑formula molar mass (EFM): $$\mathrm{EFM} = (\text{subscript}_\mathrm{C} \times M_{\mathrm{C}}) + (\text{subscript}_\mathrm{H} \times M_{\mathrm{H}}) + (\text{subscript}_\mathrm{O} \times M_{\mathrm{O}})$$
• Determine the factor $n$: $$n = \frac{M_{\mathrm{molecular}}}{\mathrm{EFM}}$$
• Multiply each subscript in the empirical formula by $n$ to obtain the molecular formula.

If the molar mass has not been entered, the tool will still output the empirical formula; the molecular formula appears only when the molar mass is supplied.

How to Use the Combustion Analysis Calculator

Using this free online combustion analysis calculator requires only a few inputs. Start by selecting the substance type—either a C‑H‑O compound or a hydrocarbon (C and H only). Then provide the measured data:

• Sample mass (required for C‑H‑O compounds; for hydrocarbons this field is optional because oxygen is absent).
• Mass of $\mathrm{CO_2}$ and mass of $\mathrm{H_2O}$ from the combustion analysis.
• Molar mass of the compound if you also want the molecular formula.

After clicking calculate, the tool displays the empirical formula, the empirical molar mass, and (if the molar mass was provided) the molecular formula. An optional setting lets you view the individual masses of C, H, and O in the original sample.

Worked Example 1 – C, H, O Compound

Consider an unknown C‑H‑O substance with the following combustion data:

• Sample mass = 12.915 g
• $\mathrm{CO_2}$ produced = 18.942 g
• $\mathrm{H_2O}$ produced = 7.749 g
• Molar mass = 90.0779 g/mol

Calculate element masses

$$m_{\mathrm{C}} = 18.942 \times \frac{12.0107}{44.0095} \approx 5.171\ \mathrm{g}$$ $$m_{\mathrm{H}} = 7.749 \times \frac{2 \times 1.00794}{18.0153} \approx 0.867\ \mathrm{g}$$ $$m_{\mathrm{O}} = 12.915 - 5.171 - 0.867 = 6.877\ \mathrm{g}$$

Moles

$$n_{\mathrm{C}} = \frac{5.171}{12.0107} = 0.4306\ \mathrm{mol},\quad n_{\mathrm{H}} = \frac{0.867}{1.00794} = 0.860\ \mathrm{mol},\quad n_{\mathrm{O}} = \frac{6.877}{15.999} = 0.4298\ \mathrm{mol}$$

Ratio – Divide by the smallest (0.4298):
C : H : O = 1.00 : 2.00 : 1.00 → Empirical formula: $\mathrm{CH_2O}$.

Molecular formula:
Empirical molar mass: $12.0107 + 2 \times 1.00794 + 15.999 = 30.0256\ \mathrm{g/mol}$
Factor $n = 90.0779 / 30.0256 \approx 3$
Therefore, molecular formula: $\mathrm{C_3H_6O_3}$ (i.e., $(\mathrm{CH_2O})_3$).

Worked Example 2 – Hydrocarbon

Now suppose a hydrocarbon (only C and H) yields:

• Sample mass = 12.501 g (not needed for C and H masses, but can be used as a cross‑check)
• $\mathrm{CO_2}$ = 33.057 g
• $\mathrm{H_2O}$ = 10.816 g
• Molar mass = 204.35 g/mol

Masses

$$m_{\mathrm{C}} = 33.057 \times \frac{12.0107}{44.0095} \approx 9.022\ \mathrm{g}$$ $$m_{\mathrm{H}} = 10.816 \times \frac{2 \times 1.00794}{18.0153} \approx 1.210\ \mathrm{g}$$

Moles

$$n_{\mathrm{C}} = \frac{9.022}{12.0107} = 0.751\ \mathrm{mol},\quad n_{\mathrm{H}} = \frac{1.210}{1.00794} = 1.201\ \mathrm{mol}$$

Ratio – Divide by the smaller value (0.751):
C : H = 1.00 : 1.599

Since 1.599 is not an integer, we express it as a fraction: $1.599 \approx \frac{8}{5}$. Multiply both terms by 5:
C : H = 5 : 8 → Empirical formula: $\mathrm{C_5H_8}$.

Molecular formula:
Empirical molar mass: $5 \times 12.0107 + 8 \times 1.00794 = 68.118\ \mathrm{g/mol}$
Factor $n = 204.35 / 68.118 \approx 3$
Hence molecular formula: $\mathrm{C_{15}H_{24}}$ (i.e., $(\mathrm{C_5H_8})_3$).

Handling Non‑Integer Ratios

As the hydrocarbon example demonstrates, the mole ratio does not always come out as a clean whole number. In such cases the decimal is converted to a fraction (e.g., 1.5 = 3/2, 1.599 ≈ 8/5) and all subscripts are multiplied by the denominator to produce the smallest possible integers. The calculator performs this conversion automatically, but understanding the logic helps you verify the output.

This free combustion analysis calculator online thus provides a complete solution for deducing formulas from combustion data—whether you are learning the manual method or need fast, accurate results for everyday laboratory work.