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Wize University Chemistry Textbook > Stoichiometry

Combustion Reactions + Empirical and Molecular Formulae

Empirical and Molecular Formulae ProblemsPrevious Section
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Introduction to Combustion Reactions

Combustion reactions: A reaction between a hydrocarbon (CxHyOz) with oxygen to produce carbon dioxide (CO2) and water (H2O).

CxHyOz (g) + O2 (g)  →  CO2 (g) + H2O (g)C_xH_yO_z\ \left(g\right)\ +\ O_2\ \left(g\right)\ \ \rightarrow\ \ CO_2\ \left(g\right)\ +\ H_2O\ \left(g\right)


Combustion reactions are commonly used in elemental analysis to determine the empirical formula of a compound.

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Steps to Balance a combustion reaction:

  1. Balance the carbons
  2. Balance the hydrogens
  3. Balance the oxygens using ONLY the coefficient of elemental oxygen (O2)
  4. If the equation has any fractions (often times the coefficient in front of O2 is a fraction), multiply the ENTIRE equation by 2


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Example: Balancing the Following Reaction


C2H6O(g)   +   O2 (g)    →    CO2 (g)   +   H2O (g)C_2H_6O\left(g\right)\ \ \ +\ \ \ O_2\ \left(g\right)\ \ \ \ \rightarrow\ \ \ \ CO_2\ \left(g\right)\ \ \ +\ \ \ H_2O\ \left(g\right)




C2H6O(g)   +   3O2 (g)    →    2CO2 (g)   +   3H2O (g)C_2H_6O\left(g\right)\ \ \ +\ \ \ 3O_2\ \left(g\right)\ \ \ \ \rightarrow\ \ \ \ 2CO_2\ \left(g\right)\ \ \ +\ \ \ 3H_2O\ \left(g\right)

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Determining Empirical Formulae from Combustion Reactions

Steps to Calculate Empirical Formulae:

  1. Determine the moles of carbon dioxide, n(CO2). This will give you the moles of carbon since 1 n(CO2) = 1 n(C).
  2. Determine the moles of water, n(H2O). This will give you moles of hydrogen since 1 n(H2O) = 2 n(H).
  3. Calculate the mass of carbon and the mass of hydrogen using their molar masses.
  4. Calculate the mass of oxygen by subtracting the masses of carbon and hydrogen from the mass of the hydrocarbon
  5. Calculate the moles of oxygen using the mass of oxygen and its molar mass.
  6. With the moles of C, H, and O known, divide all 3 mole values by the LOWEST value. If there are any fractions, multiply by a common factor to get whole numbers.


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Example: Determine the Empirical Formula from a Combustion Reaction

A 55.4 g sample of a compound containing only C, H and O was burned in the air to produce 143.2 g of carbon dioxide and 68.4 g of water. Determine the empirical formula of this compound.


1) 143.2 g CO2 (g) ⋅ (1 mol CO244.009 g CO2) ⋅ (1 mol C1 mol CO2) = 3.254 mol C143.2\ g\ CO_2\ \left(g\right)\ \cdot\ \left(\frac{1\ mol\ CO_2}{44.009\ g\ CO_2}\right)\ \cdot\ \left(\frac{1\ mol\ C}{1\ mol\ CO_2}\right)\ =\ 3.254\ mol\ C


2) 68.4 g H2O ⋅ (1 mol H2O18.015 g H2O) ⋅ (2 mol H1 mol H2O) = 7.594 mol H68.4\ g\ H_2O\ \cdot\ \left(\frac{1\ mol\ H_2O}{18.015\ g\ H_2O}\right)\ \cdot\ \left(\frac{2\ mol\ H}{1\ mol\ H_2O}\right)\ =\ 7.594\ mol\ H


3) 3.254 mol C ⋅ (12.011 g C1 mol C) = 39.08 g C3.254\ mol\ C\ \cdot\ \left(\frac{12.011\ g\ C}{1\ mol\ C}\right)\ =\ 39.08\ g\ C

7.594 mol H ⋅ (1.008 g H1 mol H) = 7.655 g H7.594\ mol\ H\ \cdot\ \left(\frac{1.008\ g\ H}{1\ mol\ H}\right)\ =\ 7.655\ g\ H


4) Mass of O = Mass of compound − mass of C − mass of HMass\ of\ O\ =\ Mass\ of\ compound\ -\ mass\ of\ C\ -\ mass\ of\ H

Mass of O = 55.4 g compound − 39.08 g C − 7.655 g HMass\ of\ O\ =\ 55.4\ g\ compound\ -\ 39.08\ g\ C\ -\ 7.655\ g\ H

Mass of O = 8.666 g OMass\ of\ O\ =\ 8.666\ g\ O


5) 8.666 g O ⋅ (1 mol O15.999 g O) = 0.5417 mol O8.666\ g\ O\ \cdot\ \left(\frac{1\ mol\ O}{15.999\ g\ O}\right)\ =\ 0.5417\ mol\ O


6) Number of C = 3.254 mol C0.5417 mol = 6.01 C ∼ 6 CNumber\ of\ C\ =\ \frac{3.254\ mol\ C}{0.5417\ mol}\ =\ 6.01\ C\ \sim\ 6\ C

Number of H = 7.594 mol H0.5417 mol = 14.02 H ∼ 14 HNumber\ of\ H\ =\ \frac{7.594\ mol\ H}{0.5417\ mol}\ =\ 14.02\ H\ \sim\ 14\ H

Number of O = 0.5417 mol O0.5417 mol = 1 ONumber\ of\ O\ =\ \frac{0.5417\ mol\ O}{0.5417\ mol}\ =\ 1\ O

∴ Empirical Formula = C6H14O\therefore\ Empirical\ Formula\ =\ C_6H_{14}O

Practice: Combustion Analysis Calculation

When 2.11 g of an unknown hydrocarbon (compound made up of only carbon and hydrogen atoms) was burned in excess oxygen as part of a combustion analysis. In the process 6.32 g of CO2 and 3.45 g of H2O are produced. What is the empirical formula of the hydrocarbon?
Extra Practice
Empirical formula: Combustion reaction
When 2.11 g of an unknown hydrocarbon (compound made up of only carbon and hydrogen atoms) was burned in excess oxygen as part of a combustion analysis. In the process 6.32 g of CO2 and 3.45 g of H2O are produced. What is the mass percent of carbon? What is the empirical formula of the hydrocarbon?
Empirical Formula
A 1.00 g sample of a species containing only C, H, and O was completely combusted in pure oxygen, and 1.5 g CO2 and 0.41g H2O were obtained. What is the empirical formula of the compound?
Combustion and Empirical Formula
A hydrocarbon (0.875 g) containing only C, H, and O yielded 2.21 g CO2 and 0.387 g H2O after complete combustion.
What is the empirical formula?
Combustion, Stoichiometry
What mass of KClO3 is required to produce proper amount of O2 needed to achieve complete combustion of 78.88 g CH4?
2 KClO3 → 2 KCl + 3 O2
Combustion Analysis Calculation
When 2.11 g of an unknown hydrocarbon (compound made up of only carbon and hydrogen atoms) was burned in excess oxygen as part of a combustion analysis. In the process 6.32 g of CO2 and 3.45 g of H2O are produced. What is the empirical formula of the hydrocarbon?
Stoichiometry 2
An unknown hydrocarbon contains 40.0% C and 6.66% H by weight.
a. What is the empirical formula of the compound?
b. What is the molecular formula if molecular weight of the compound is 180 gmol-1?
Gas stoichiometry and Empirical Formula
Complete combustion of a hydrocarbon yields 1.5 L CO2 and 2 L H2O at 298 K and 101 kPa. Whats the empirical formula of the compound?
Empirical Formula
A 1.00 g sample of a species containing only C, H, and O was completely combusted in pure oxygen, and 1.5 g CO2 and 0.41g H2O were obtained. What is the empirical formula of the compound? (enter answer: C_H_O_ For example C2H2O5)
Empirical Formulae
This question focuses on material from 1.3. Empirical Formulae.
Empirical Formulae
3.0 g of an organic compound, CxHyOz, is combusted in air to produce 7.75 g of carbon dioxide and 3.70 g of water vapour. Determine the empirical formula of this compound.
Empirical Formulae
An organic compound, CxHyClz, was subject to elemental analysis, and the compound was found to contain 64.6% carbon by mass and 11.5% hydrogen by mass. Determine the empirical formula of this compound.
Mass Percentage Practice
0.2976g of an unknown compound gives 0.3147g of CO2 and 0.3026g of H2O on combustion. The only other element known to be present is N. What is the empirical formula?
Empirical Formulae
5. 5.0 g of an organic compound, CxHyOz, is combusted in air to produce 11.87 g of carbon dioxide and 6.08 g of water vapour. Determine the empirical formula of this compound.
Empirical formula: Combustion reaction
When 2.11 g of an unknown hydrocarbon (compound made up of only carbon and hydrogen atoms) was burned in excess oxygen as part of a combustion analysis. In the process 6.32 g of CO2 and 3.45 g of H2O are produced. What is the empirical formula of the hydrocarbon?
Combustion Analysis
For a hydrocarbon (a molecule made up of only C and H atoms) combustion analysis allows us to determine the: