Skip to main content
Chemistry LibreTexts

7.3: Composition, Decomposition, and Combustion Reactions

  • Page ID
    212496
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)
    Learning Objectives
    • Recognize composition, decomposition, and combustion reactions.
    • Predict the products of a combustion reaction.

    Three classifications of chemical reactions will be reviewed in this section. Predicting the products in some of them may be difficult, but the reactions are still easy to recognize.

    A composition reaction (sometimes also called a combination reaction or a synthesis reaction) produces a single substance from multiple reactants. A single substance as a product is the key characteristic of the composition reaction. There may be a coefficient other than one for the substance, but if the reaction has only a single substance as a product, it can be called a composition reaction. In the reaction

    \[\ce{2H_2(g) + O_2(g) → 2H_2O(ℓ)}\nonumber \]

    water is produced from hydrogen and oxygen. Although there are two molecules of water being produced, there is only one substance—water—as a product. So this is a composition reaction.

    A decomposition reaction starts from a single substance and produces more than one substance; that is, it decomposes. The key characteristics of a decomposition reaction are: one substance as a reactant and more than one substance as the products. For example, in the decomposition of sodium hydrogen carbonate (also known as sodium bicarbonate):

    \[\ce{2NaHCO_3(s) → Na_2CO_3(s) + CO_2(g) + H_2O(ℓ) }\nonumber \]

    sodium carbonate, carbon dioxide, and water are produced from the single substance sodium hydrogen carbonate.

    Composition and decomposition reactions are difficult to predict; however, they should be easy to recognize.

    Example \(\PageIndex{1}\): Identifying Reactions

    Identify each equation as a composition reaction, a decomposition reaction, or neither.

    1. \(\ce{Fe2O3 + 3SO3 → Fe2(SO4)3}\)
    2. \(\ce{NaCl + AgNO3 → AgCl + NaNO3}\)
    3. \(\ce{(NH4)2Cr2O7 → Cr2O3 + 4H2O + N2}\)
    Solution
    1. In this equation, two substances combine to make a single substance. This is a composition reaction.
    2. Two different substances react to make two new substances. This does not fit the definition of either a composition reaction or a decomposition reaction, so it is neither. In fact, you may recognize this as a double-replacement reaction.
    3. A single substance reacts to make multiple substances. This is a decomposition reaction.
    Exercise \(\PageIndex{1}\)

    Identify the equation as a composition reaction, a decomposition reaction, or neither.

    \[\ce{C3H8 → C3H4 + 2H2} \nonumber \]

    Answer

    decomposition

    A combustion reaction occurs when a reactant combines with oxygen, many times from the atmosphere, to produce oxides of all other elements as products; any nitrogen in the reactant is converted to elemental nitrogen, N2. Many reactants, called fuels, contain mostly carbon and hydrogen atoms, reacting with oxygen to produce CO2 and H2O. For example, the balanced chemical equation for the combustion of methane, CH4, is as follows:

    \[\ce{CH_4 + 2O_2 → CO_2 + 2H_2O}\nonumber \]

    Kerosene can be approximated with the formula \(\ce{C_{12}H_{26}}\), and its combustion equation is:

    \[\ce{2C_{12}H_{26} + 37O_2 → 24CO-2 + 26H_2O}\nonumber\]

    Sometimes fuels contain oxygen atoms, which must be counted when balancing the chemical equation. One common fuel is ethanol, \(\ce{C_2H_5OH}\), whose combustion equation is:

    \[\ce{C_2H_5OH + 3O_2 → 2CO_2 + 3H_2O}\nonumber \]

    If nitrogen is present in the original fuel, it is converted to \(\ce{N_2}\), not to a nitrogen-oxygen compound. Thus, for the combustion of the fuel dinitroethylene, whose formula is \(\ce{C_2H_2N_2O_4}\), we have:

    \[\ce{2C_2H_2N_2O_4 + O_2 → 4CO_2 + 2H_2O + 2N_2}\nonumber \]

    Example \(\PageIndex{2}\): Combustion Reactions

    Complete and balance each combustion equation.

    1. the combustion of propane (\(\ce{C3H8}\))
    2. the combustion of ammonia (\(\ce{NH3}\))

    Solution

    1. The products of the reaction are CO2 and H2O, so our unbalanced equation is \[\ce{C3H8 + O2 → CO2 + H2O} \nonumber \]

      Balancing (and you may have to go back and forth a few times to balance this), we get \[\ce{C3H8 + 5O2 → 3CO2 + 4H2O} \nonumber \]

    2. The nitrogen atoms in ammonia will react to make N2, while the hydrogen atoms will react with O2 to make H2O: \[\ce{NH3 + O2 → N2 + H2O} \nonumber \]

      To balance this equation without fractions (which is the convention), we get \[\ce{4NH3 + 3O2 → 2N2 + 6H2O} \nonumber \]

    Exercise \(\PageIndex{2}\)

    Complete and balance the combustion equation for cyclopropanol (\ce{C3H6O}\)).

    Answer

    \[\ce{C_3H_6O + 4O_2 → 3CO_2 + 3H_2O}\nonumber \]

    Propane_tank_20lb.jpg
    Figure \(\PageIndex{1}\) A 20 lb (9.1 kg) steel propane cylinder.

    Key Takeaways

    • A composition reaction produces a single substance from multiple reactants.
    • A decomposition reaction produces multiple products from a single reactant.
    • Combustion reactions are the combination of some compound with oxygen to make oxides of the other elements as products (although nitrogen atoms react to make N2).

    7.3: Composition, Decomposition, and Combustion Reactions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.