Combustion Analysis: Empirical and Molecular Formulas

Purpose of combustion analysis: determine empirical formula of a hydrocarbon or oxygen-containing sample by burning it and analyzing the combustion products (CO₂ and H₂O).
Key challenge: oxygen is present in both CO₂ and H₂O, and some O comes from an external O₂ stream used to sustain the flame. Therefore, the amount of O in the original sample cannot be found by simple stoichiometry alone; it requires a subtraction step.
Road map used for these problems (based on mole ratios from the chemical formulas):
- If you collect 1 mole of CO₂, you know the original sample contained 1 mole of C (one C per CO₂).
- If you collect 1 mole of H₂O, you know the original sample contained 2 moles of H (two H per H₂O).
- Using mass data, convert masses to moles, then use mole ratios to extract C and H; determine O by difference (since O also comes from the external O₂).
Practical reminder: always work with mole ratios and convert masses to moles first; mass alone cannot give you the chemical formula.

Key Concepts

Mole ratio rule for combustion products:
- CO₂: each mole of CO₂ corresponds to 1 mole of C in the original sample: $nC = n{CO2}$
- H₂O: each mole of H₂O corresponds to 2 moles of H in the original sample: $nH = 2 imes n{H2O}$
Mass to moles conversion:
- For CO₂: n{CO2} = rac{m{CO2}}{M{CO2}} ext{ with } M{CO2} \,=\, 44.01\ ext{ g/mol}
- For H₂O: n{H2O} = rac{m{H2O}}{M{H2O}} ext{ with } M{H2O} \,=\, 18.015\ ext{ g/mol}
Oxygen determination by subtraction: since the sample’s total mass includes O from the sample plus O from the external O₂, compute:
- $mO^{(original)} = m{sample} - mC - mH$
Empirical formula derivation: convert C, H, O masses to moles, then divide all by the smallest mole value to get the smallest whole-number ratios, giving the empirical formula $ext{Empirical formula} = ext{C}x ext{H}y ext{O}_z$ with x, y, z integers.
Molecular formula from empirical formula: if the empirical formula mass is $M{emp}$ and the actual molar mass is $M{actual}$, then the molecular formula is the empirical formula multiplied by n = rac{M{actual}}{M{emp}} (an integer). Therefore, $ext{Molecular formula} = ( ext{Empirical formula}) imes n$ .
Common pitfall: ensure the subtraction for O uses consistent masses; rounding can affect the final subscripts if not careful.

Step-by-Step Roadmap (recalled from the lesson)

Start with given masses of combustion products: CO₂ and H₂O.
Convert masses to moles using molar masses.
Apply mole-ratio rules to deduce moles of C and H in the original sample.
Subtract to find the mass (and then moles) of O in the original sample:
- $mO^{(original)} = m{sample} - mC - mH$
Convert C, H, O masses to moles.
Determine the empirical formula by dividing by the smallest number of moles among C, H, and O.
If a molecular formula is required, compare the given molecular molar mass to the empirical formula mass to find the appropriate whole-number multiplier.

Worked Example (from the transcript)

Given:
- Mass of CO₂ collected: $m_{CO2} = 6.1\ ext{g}$
- Mass of H₂O collected: $m_{H2O} = 0.5\ ext{g}$
- Mass of original sample burnt: $m_{sample} = 2.5\ ext{g}$
Step 1: Convert CO₂ to moles and deduce C
- Molar mass of CO₂: $M_{CO2} = 44.01\ \text{g/mol}$
- $n_{CO2} = \frac{6.1}{44.01} \approx 0.1386\ ext{mol}$
- For CO₂, C is 1:1 with CO₂, so $nC = n{CO2} \approx 0.1386\ \text{mol}$
- Mass of C: $mC = nC \times M_C = 0.1386 \times 12.01 \approx 1.66\ \text{g}$
Step 2: Convert H₂O to moles and deduce H
- Molar mass of H₂O: $M_{H2O} = 18.015\ \text{g/mol}$
- $n_{H2O} = \frac{0.5}{18.015} \approx 0.0278\ \text{mol}$
- H in original sample: $nH = 2 \times n{H2O} \approx 0.0556\ \text{mol}$
- Mass of H (from the transcript example, a value given was 0.28 g; note this conflicts with the calculation above):
- Calculated from moles: $mH = nH \times M_H \approx 0.0556 \times 1.008 \approx 0.056\ \text{g}$
- Transcript-reported value: $m_H \approx 0.28\ \text{g}$
- This discrepancy is acknowledged as a point of discussion in the walkthrough; the subsequent steps in the transcript use a hydrogen mass of 0.28 g.
Step 3: Subtract to find O in the original sample (as per the lesson, using the example masses)
- Mass of original sample: $m_{sample} = 2.5\ ext{g}$
- Subtract C mass and H mass (as given in the example):
- If using the transcript’s numbers: $m_O^{(original)} = 2.5 - 1.66 - 0.28 = 0.56\ ext{g}$
Step 4: Convert C, H, O masses to moles
- $nC = \frac{mC}{M_C} = \frac{1.66}{12.01} \approx 0.138\ \text{mol}$
- $nH = \frac{mH}{M_H} = \frac{0.28}{1.008} \approx 0.278\ \text{mol}$
- $nO = \frac{mO}{M_O} = \frac{0.56}{16.00} \approx 0.035\ \text{mol}$
Step 5: Determine empirical formula by dividing by smallest mole value
- Smallest: $n_O \approx 0.035\ \text{mol}$
- Ratios:
- $\frac{nC}{nO} \approx \frac{0.138}{0.035} \approx 3.94 \approx 4$
- $\frac{nH}{nO} \approx \frac{0.278}{0.035} \approx 7.94 \approx 8$
- $\frac{nO}{nO} = 1$
- Empirical formula from these ratios: $\text{Empirical formula} = \mathrm{C}4\mathrm{H}8\mathrm{O}$
Step 6: Check empirical formula mass and match molecular formula
- Molar mass of empirical formula: $M_{emp} = 4(12.01) + 8(1.008) + 1(16.00) \approx 72.1\ \text{g/mol}$
- The problem statement provides molecular molar mass: $M_{actual} = 144.21\ \text{g/mol}$
- Multiplier: $n = \frac{M{actual}}{M{emp}} \approx \frac{144.21}{72.1} \approx 2.0$
- Molecular formula: $\text{Molecular formula} = (\mathrm{C}4\mathrm{H}8\mathrm{O})2 = \mathrm{C}8\mathrm{H}{16}\mathrm{O}2$
Final takeaway from the example: empirical formula C₄H₈O; molecular formula C₈H₁₆O₂ (given M_actual = 144.21 g/mol).

Important Considerations and Tips

Oxygen ambiguity: always use subtraction to isolate the oxygen originally present in the sample; the oxygen from the flame or external O₂ cannot be distinguished by simple combustion masses alone.
Round carefully when converting to whole-number subscripts; near-integer values (e.g., 3.94 → 4, 7.94 → 8) should be treated as integers in the empirical formula.
When the molecular molar mass is provided, use it to determine the multiple n to convert the empirical formula to the molecular formula; if Mactual is not an exact multiple of Memp, re-check calculations or rounding.
Practical lab note: ensure accuracy of mass measurements if you intend to get precise empirical formulas; small errors in CO₂ or H₂O masses propagate into the moles and resulting formulas.

Connections to Prior Lectures and Real-World Relevance

Builds on prior work with mole ratios and stoichiometry from chemical formulas (Monday’s class content referenced in the transcript).
Demonstrates a real-world technique for identifying unknown hydrocarbons or oxygen-containing compounds via combustion analysis, used in material science, forensic chemistry, and combustion research.
Emphasizes the importance of recognizing and accounting for external reactants (O₂) in gas-phase reactions and analytical problems.
Conceptual link to empirical vs. molecular formulas: empirical formula reflects the simplest whole-number ratio, while the molecular formula shows the actual number of atoms in the molecule; the two are related by a multiplier derived from molar masses.

Quick Practice Prompts (to test understanding)

If you burn a sample and obtain 4.40 g CO₂ and 1.20 g H₂O, with a sample mass of 3.50 g, outline the steps to determine the empirical formula. Include how you would handle oxygen determination.
Given an empirical formula of C₂H₄O and a molecular mass of 72.0 g/mol, determine the molecular formula.
Explain why the oxygen in CO₂ and H₂O cannot be assumed to come entirely from the original sample without subtraction.

Summary

Combustion analysis relies on converting masses of CO₂ and H₂O into moles, using mole ratios to deduce C and H in the original sample, and using subtraction to isolate the original O content.
The empirical formula is derived from mole ratios, and the molecular formula is found by matching the empirical mass to the given molecular mass.
The approach hinges on careful mass measurements, correct application of mole ratios, and consistent unit conversions throughout.