Empirical Formula from Combustion Analysis (C, H, O)
This empirical formula from combustion analysis method is used when a compound contains only
carbon, hydrogen, and oxygen. The sample is burned in excess oxygen so that all carbon becomes CO₂ and all
hydrogen becomes H₂O. Measuring the masses of CO₂ and H₂O lets the amounts of C and H be determined directly.
Oxygen in the original compound is then found by mass difference. Converting masses to moles and
reducing to the simplest whole-number ratio gives the empirical formula.
What combustion analysis measures
In a complete combustion experiment, the products are trapped and weighed:
CO₂ tells how much carbon was present, and H₂O tells how much hydrogen was present.
Oxygen in the unknown is not measured as a product here; it is inferred as whatever sample mass remains after
subtracting the masses of C and H.
Workflow (how the calculation proceeds)
- Convert measured CO₂ mass to moles of CO₂, then set moles of C equal to moles of CO₂.
- Convert measured H₂O mass to moles of H₂O, then set moles of H equal to 2 times moles of H₂O.
- Convert moles of C and H to their masses; subtract from the sample mass to get oxygen mass by difference.
- Convert oxygen mass to moles of O.
- Divide all mole amounts by the smallest to obtain a mole ratio; scale to whole numbers for subscripts.
Key relations (C, H, O only)
\[
\begin{aligned}
n_{\mathrm{CO_2}} &= \frac{m_{\mathrm{CO_2}}}{M_{\mathrm{CO_2}}}, &\quad n_{\mathrm{C}} &= n_{\mathrm{CO_2}} \\[0.35em]
n_{\mathrm{H_2O}} &= \frac{m_{\mathrm{H_2O}}}{M_{\mathrm{H_2O}}}, &\quad n_{\mathrm{H}} &= 2\cdot n_{\mathrm{H_2O}} \\[0.35em]
m_{\mathrm{C}} &= n_{\mathrm{C}}\cdot A_{\mathrm{C}}, &\quad m_{\mathrm{H}} &= n_{\mathrm{H}}\cdot A_{\mathrm{H}} \\[0.35em]
m_{\mathrm{O}} &= m_{\text{sample}} - m_{\mathrm{C}} - m_{\mathrm{H}}, &\quad n_{\mathrm{O}} &= \frac{m_{\mathrm{O}}}{A_{\mathrm{O}}}
\end{aligned}
\]
After finding \(n_{\mathrm{C}}, n_{\mathrm{H}}, n_{\mathrm{O}}\), the empirical subscripts are obtained from the reduced ratio:
\[
\text{ratio}_i = \frac{n_i}{\min(n_{\mathrm{C}}, n_{\mathrm{H}}, n_{\mathrm{O}})} \quad \text{then scale to integers.}
\]
Choosing whole-number subscripts (common scaling rules)
Experimental data rarely lands exactly on integers because of rounding and measurement uncertainty, especially
because oxygen is determined by difference. If ratios come out close to familiar fractions, multiply all ratios by
a small integer:
| Ratio looks like |
Closest fraction |
Multiply all ratios by |
| 1.50 |
\(\frac{3}{2}\) |
2 |
| 1.33 |
\(\frac{4}{3}\) |
3 |
| 1.25 |
\(\frac{5}{4}\) |
4 |
| 1.20 |
\(\frac{6}{5}\) |
5 |
| 1.67 |
\(\frac{5}{3}\) |
3 |
A good check is that the scaled subscripts should be small and chemically reasonable. If scaling produces very
large numbers, re-check units, significant figures, and whether the sample truly contains only C, H, and O.
Worked numbers (same as the example button)
Sample = 0.2000 g, CO₂ = 0.2998 g, H₂O = 0.0819 g.
Use \(M_{\mathrm{CO_2}}=44.009\ \mathrm{g\cdot mol^{-1}}\), \(M_{\mathrm{H_2O}}=18.015\ \mathrm{g\cdot mol^{-1}}\),
and atomic masses \(A_{\mathrm{C}}=12.011\), \(A_{\mathrm{H}}=1.008\), \(A_{\mathrm{O}}=15.999\ \mathrm{g\cdot mol^{-1}}\).
Step 1. Convert CO₂ to moles, then infer moles and mass of carbon.
\[
\begin{aligned}
n_{\mathrm{CO_2}} &= \frac{0.2998}{44.009} = 0.006812\ \mathrm{mol} \\
n_{\mathrm{C}} &= n_{\mathrm{CO_2}} = 0.006812\ \mathrm{mol} \\
m_{\mathrm{C}} &= n_{\mathrm{C}}\cdot A_{\mathrm{C}}
= 0.006812\cdot 12.011
= 0.08182\ \mathrm{g}
\end{aligned}
\]
Step 2. Convert H₂O to moles, then infer moles and mass of hydrogen.
\[
\begin{aligned}
n_{\mathrm{H_2O}} &= \frac{0.0819}{18.015} = 0.004545\ \mathrm{mol} \\
n_{\mathrm{H}} &= 2\cdot n_{\mathrm{H_2O}}
= 2\cdot 0.004545
= 0.009090\ \mathrm{mol} \\
m_{\mathrm{H}} &= n_{\mathrm{H}}\cdot A_{\mathrm{H}}
= 0.009090\cdot 1.008
= 0.00916\ \mathrm{g}
\end{aligned}
\]
Step 3. Find oxygen by difference and convert to moles of oxygen atoms.
\[
\begin{aligned}
m_{\mathrm{O}} &= 0.2000 - 0.08182 - 0.00916
= 0.1090\ \mathrm{g} \\
n_{\mathrm{O}} &= \frac{0.1090}{15.999}
= 0.006813\ \mathrm{mol}
\end{aligned}
\]
Step 4. Reduce to the simplest whole-number ratio.
\[
\begin{aligned}
\text{C ratio} &= \frac{0.006812}{0.006812} = 1.00 \\
\text{H ratio} &= \frac{0.009090}{0.006812} = 1.33 \\
\text{O ratio} &= \frac{0.006813}{0.006812} = 1.00
\end{aligned}
\]
The ratio is approximately C : H : O = 1.00 : 1.33 : 1.00. Since 1.33 is close to \(\frac{4}{3}\), multiply all
ratios by 3 to get integers: 3 : 4 : 3. Therefore, the empirical formula is
C3H4O3.
Quality checks and common pitfalls
- Complete combustion: If combustion is incomplete (soot/CO forms), calculated carbon will be too low.
- Water capture: If some H₂O escapes the trap, hydrogen will be underestimated.
- Oxygen by difference amplifies error: small mass errors in CO₂ or H₂O can noticeably affect \(m_{\mathrm{O}}\).
- Units: masses must be in grams to match molar masses in \(\mathrm{g\cdot mol^{-1}}\).
- Scope: This procedure assumes only C, H, and O. If other elements exist (N, halogens, S, metals), additional analysis is required.
From empirical to molecular formula (if molar mass is known)
If the compound’s molar mass \(M\) is available, the molecular formula is an integer multiple of the empirical formula.
Compute the factor:
\[
k = \frac{M}{M_{\text{empirical}}}
\]
Round \(k\) to the nearest whole number and multiply every empirical subscript by \(k\).
