Combustion Analysis

Core Concept

Combustion analysis is a technique used primarily to determine the percentages of carbon and hydrogen (and sometimes other elements) in an organic compound.

The sample is burned in an oxygen-rich environment, producing CO₂ and H₂O. By measuring the mass or volume of CO₂ and H₂O formed, one can calculate the amount of carbon and hydrogen originally present. This information is then used to derive empirical or molecular formula.

Practice Tips

All Carbon Ends Up in CO₂: Convert the mass of CO₂ to the mass of carbon to find the amount of carbon in the original compound.
All Hydrogen Ends Up in H₂O: Convert the mass of H₂O to the mass of hydrogen to find the amount of hydrogen.
Use the Sample Mass to Find Oxygen (if applicable): For compounds with oxygen, subtract the mass of carbon and hydrogen from the total mass to find the oxygen content.
Always Simplify Ratios: Divide by the smallest number of moles to get the simplest whole-number ratio.

Test Yourself

Assorted Multiple Choice

A 0.500 gram sample of a pure hydrocarbon is burned completely in a combustion analysis apparatus. The process yields 1.57 grams of carbon dioxide and 0.642 grams of water. Based on these measurements, which of the following steps is required to determine the mass of carbon present in the original sample?

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Empirical Formula Determination for Hydrocarbons (C and H only)

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Problem Solving

Steps in Combustion Analysis

Burn the Sample in Excess Oxygen:
- The compound combusts completely in an oxygen-rich environment, producing CO₂ and H₂O.
- The mass of CO₂ and H₂O produced is measured.
Determine the Mass of Carbon and Hydrogen:
- All carbon in the original compound ends up as carbon in CO₂, and all hydrogen ends up as hydrogen in H₂O.
- Calculate the mass of carbon from CO₂ and the mass of hydrogen from H₂O.
Calculate the Moles of Carbon and Hydrogen:
- Use the masses of carbon and hydrogen to find their moles in the sample.
Determine the Amount of Oxygen (If Present):
- If the compound also contains oxygen, subtract the mass of carbon and hydrogen from the total sample mass to find the mass of oxygen.
- Calculate the moles of oxygen from this mass.
Find the Empirical Formula:
- Divide the moles of each element by the smallest number of moles.
- This gives the simplest whole-number ratio, which is the empirical formula of the compound.

  
    Combustion Analysis Table
    
        Step
        Example Application (0.500 g sample produces 1.467 g CO₂ and 0.600 g H₂O)
    
        1. Burn the Sample in Excess Oxygen
        Combust the 0.500 g sample in oxygen, which yields 1.467 g of CO₂ and 0.600 g of H₂O.
    
        2. Determine the Mass of Carbon from CO₂
        
            - Molar mass of CO₂ = 44.01 g/mol.

            - Moles of CO₂ = \( \frac{1.467 \, \text{g}}{44.01 \, \text{g/mol}} = 0.0333 \, \text{mol} \).

            - Since each mole of CO₂ has 1 mole of C, moles of C = 0.0333 mol.

            - Mass of C = \( 0.0333 \, \text{mol} \times 12.01 \, \text{g/mol} = 0.400 \, \text{g} \).
        
        3. Determine the Mass of Hydrogen from H₂O
        
            - Molar mass of H₂O = 18.02 g/mol.

            - Moles of H₂O = \( \frac{0.600 \, \text{g}}{18.02 \, \text{g/mol}} = 0.0333 \, \text{mol} \).

            - Each mole of H₂O has 2 moles of H, so moles of H = \( 0.0333 \times 2 = 0.0666 \, \text{mol} \).

            - Mass of H = \( 0.0666 \, \text{mol} \times 1.01 \, \text{g/mol} = 0.0673 \, \text{g} \).
        
        4. Determine Mass of Oxygen (if present)
        Since the compound is a hydrocarbon (only C and H), we skip this step. If oxygen were present, we would subtract the masses of C and H from the initial sample mass to find the mass of O.
    
        5. Calculate Moles and Find Empirical Formula
        
            - Moles of C = 0.0333 mol; Moles of H = 0.0666 mol.

            - Divide by the smallest number of moles (0.0333) to get the ratio:

                - Carbon: \( 0.0333 / 0.0333 = 1 \)

                - Hydrogen: \( 0.0666 / 0.0333 = 2 \)

            - The empirical formula is CH₂.

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