AP Chemistry: Combustion Analysis

Combustion analysis is a cornerstone experimental technique in chemistry for determining the empirical and molecular formulas of unknown organic compounds. This procedure transforms the abstract concept of molecular composition into measurable quantities—the masses of carbon dioxide and water produced—providing a direct link between laboratory data and chemical identity. Mastering this skill is essential for AP Chemistry success, has applications in forensic science and environmental monitoring, and builds foundational logic used in medical diagnostics like metabolic studies.

The Core Principle: Conservation of Mass and Atoms

The entire logic of combustion analysis rests on two unwavering chemical laws: the Law of Conservation of Mass and the Law of Conservation of Atoms. During complete combustion, an organic compound containing carbon, hydrogen, and possibly oxygen reacts with excess oxygen gas ($O_2$). The carbon in the original compound is converted exclusively into carbon dioxide ($C{O}_2$), and the hydrogen is converted exclusively into water ($H_{2}O$). Every atom of carbon in the original sample ends up in a $C{O}_2$ molecule, and every atom of hydrogen ends up in an $H_{2}O$ molecule.

This means we can work backwards. If you know the mass of $C{O}_2$ produced, you can calculate the mass and then moles of carbon that originated from your sample. Similarly, from the mass of $H_{2}O$ produced, you can find the mass and moles of hydrogen from your sample. The power of this technique is that it allows you to "see" the invisible atoms in your original compound by analyzing the products they form.

The Step-by-Step Calculation (For C/H Compounds)

Let's walk through the standard calculation for a compound containing only carbon and hydrogen. Imagine you combust 0.250 g of an unknown hydrocarbon and obtain 0.784 g of $C{O}_2$ and 0.214 g of $H_{2}O$.

Step 1: Find the mass and moles of Carbon.

• All the carbon is in the $C{O}_2$. The mass percent of C in $C{O}_2$ is:

\text{%C in } CO_2 = \frac{\text{molar mass of C}}{\text{molar mass of } CO_2} = \frac{12.01 \text{ g/mol}}{44.01 \text{ g/mol}} \approx 0.2729

• Therefore, mass of C from sample = mass of $C{O}_2$ × (mass fraction of C) = $0.784\text{ g}\times 0.2729\approx 0.214\text{ g C}$.
• Moles of C = $0.214\text{ g}/12.01\text{ g/mol}\approx 0.0178\text{ mol}$.

Step 2: Find the mass and moles of Hydrogen.

• All the hydrogen is in the $H_{2}O$. The mass percent of H in $H_{2}O$ is:

\text{%H in } H_2O = \frac{2 \times \text{molar mass of H}}{\text{molar mass of } H_2O} = \frac{2.016 \text{ g/mol}}{18.016 \text{ g/mol}} \approx 0.1119

• Mass of H from sample = mass of $H_{2}O$ × (mass fraction of H) = $0.214\text{ g}\times 0.1119\approx 0.0239\text{ g H}$.
• Moles of H = $0.0239\text{ g}/1.008\text{ g/mol}\approx 0.0237\text{ mol}$.

Step 3: Determine the Empirical Formula.

• Find the simplest whole number mole ratio: C : H = $0.0178:0.0237$.
• Divide both by the smallest number (0.0178):

C: $0.0178/0.0178=1$ H: $0.0237/0.0178\approx 1.33$

• A ratio of 1:1.33 is not whole. Multiply by the smallest integer to get whole numbers. Multiplying by 3 gives C: 3, H: 4.
• The empirical formula is therefore $C_3{H}_4$.

Accounting for Oxygen: The Crucial Extra Step

Most organic compounds of interest, from ethanol to glucose, contain oxygen. This adds a critical layer to the analysis because the mass of oxygen in the products ($C{O}_2$ and $H_{2}O$) comes from both the original compound and the excess $O_2$ gas used for combustion. You cannot directly trace oxygen atoms from products back to the sample.

The solution is to use the Law of Conservation of Mass for the original sample. The procedure is:

1. Calculate the masses of Carbon and Hydrogen from the $C{O}_2$ and $H_{2}O$ data as before.
2. Sum these masses. If they equal the mass of the original sample, the compound contains only C and H.
3. If the sum is less than the original sample mass, the difference is the mass of Oxygen in the original compound.

$$\text{mass of O in sample}=\text{mass of sample}-(\text{mass of C}+\text{mass of H})$$

Example with Oxygen: Combust 0.500 g of a compound containing C, H, and O. You get 0.733 g $C{O}_2$ and 0.300 g $H_{2}O$.

• Mass of C = $0.733\times (12.01/44.01)\approx 0.200\text{ g}$. Moles C ≈ 0.01666 mol.
• Mass of H = $0.300\times (2.016/18.016)\approx 0.0336\text{ g}$. Moles H ≈ 0.03333 mol.
• Mass of C + H = 0.200 + 0.0336 = 0.2336 g.
• Mass of original sample was 0.500 g. Therefore, mass of O = $0.500-0.2336=0.2664\text{ g}$.
• Moles of O = $0.2664\text{ g}/16.00\text{ g/mol}\approx 0.01665\text{ mol}$.
• Mole Ratio C : H : O = $0.01666:0.03333:0.01665$. Dividing by the smallest (0.01665):

C: 1.00, H: 2.00, O: 1.00.

• Empirical formula is $C{H}_{2}O$.

From Empirical to Molecular Formula

The empirical formula gives the simplest whole-number ratio of atoms. The molecular formula gives the actual number of atoms in a molecule. To find it, you need one more piece of data: the molar mass of the compound (often provided or determined via a separate experiment like mass spectrometry or freezing point depression).

The process is:

1. Calculate the molar mass of the empirical formula.
2. Divide the given molar mass of the compound by the empirical formula mass.

$$n=\frac{\text{molar mass of compound}}{\text{molar mass of empirical formula}}$$

1. Multiply all subscripts in the empirical formula by this integer n.

For our $C{H}_{2}O$ example, the empirical formula mass is ~30.03 g/mol. If the problem states the molar mass is approximately 180 g/mol, then $n=180/30\approx 6$. The molecular formula is $(C{H}_{2}O{)}_6$, or $C_6{H}_{12}{O}_6$ (glucose).

Common Pitfalls

1. Forgetting to Double-Check for Oxygen: The most common error is assuming a compound contains only C and H. Always sum the calculated masses of C and H and compare to the initial sample mass. If the masses don't add up, the missing mass is almost certainly oxygen (assuming complete combustion of a C, H, O compound).

1. Incorrect Molar Masses and Ratios: Using atomic masses like 12, 1, and 16 is often acceptable for AP exams, but be consistent. The more critical mistake is rounding intermediate mole values too aggressively. Carry at least 4 significant figures through mole calculations until you determine the ratio. Premature rounding of 1.33 to 1.3 can derail finding the correct multiplier (3).

1. Confusing Empirical and Molecular Formulas: Remember that the combustion data alone typically gives only the empirical formula. You cannot determine the molecular formula without an independent measure of molar mass. Presenting the empirical formula as the final answer when the problem asks for the molecular formula (and provides molar mass) will cost you full credit.

1. Misinterpreting the Water Data: The hydrogen from your compound ends up in water ($H_{2}O$). Each mole of $H_{2}O$ contains two moles of H atoms. Failing to multiply by two when calculating moles of H is a fatal error. Always use: Moles H = 2 × (mass $H_{2}O$ / molar mass $H_{2}O$).

Summary

• Combustion analysis leverages the conservation of atoms to determine the elemental composition of an organic compound from the masses of $C{O}_2$ and $H_{2}O$ produced.
• The core calculation involves finding the mass and moles of carbon from the $C{O}_2$ data and hydrogen from the $H_{2}O$ data.
• For compounds containing oxygen, the mass of oxygen is found by subtracting the masses of C and H from the original sample mass, a step that must never be overlooked.
• The empirical formula is derived from the simplest whole-number mole ratio of C, H, and O (if present).
• The molecular formula is found by comparing the compound's known molar mass to the mass of the empirical formula unit.
• Success depends on meticulous tracking of significant figures, correct use of molar masses, and a clear, step-by-step logical approach to the arithmetic.