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4.5: Classifying Chemical Reactions - Combustion Reactions

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    Learning Objectives
     
    • Define combustion reactions
    • Identify when a combustion reaction will occur
    • Write a balanced chemical equation that describes what happens when a combustion reaction occurs
    • Solve stoichometry problems with combustion reactions

    Humans interact with one another in various and complex ways, and we classify these interactions according to common patterns of behavior. When two humans exchange information, we say they are communicating. When they exchange blows with their fists or feet, we say they are fighting. Faced with a wide range of varied interactions between chemical substances, scientists have likewise found it convenient (or even necessary) to classify chemical interactions by identifying common patterns of reactivity. This module will provide an introduction to combustion reactions, one of the most prevalent types of chemical reactions.

    Combustion Reactions

    Combustion reactions are another common class of reactions.  These reactions provide the energy to drive your car, heat your home, and cook your food.  Anytime something burns, it is a combustion reaction.  These reactions are critical to our economy but they are also responsible for emitting 35 gigatons of CO2 into the atmosphere in 2020.  In a combustion reaction a fuel, like methane, reacts with oxygen to produce carbon dioxide and water.  Almost all combustion reactions require additional oxygen, which comes from the atmosphere, with the reactants.  In a combustion reaction all the carbon in the reactants is converted into carbon dioxide and all the hydrogen in the reactants is converted into water.

    Example \(\PageIndex{1}\): Writing Equations for Combustion Reactions

    Write balanced chemical equations for the combustion reaction described here:

    1. the combustion of methane, CH4
    2. the combustion of ethane, C2H6

    Solution

    (a) In addition to methane, CH4, the combustion reaction also requires oxygen from the atmosphere.  The products of the combustion reaction will be H2O and CO2. 

    \[\ce{CH4}(g)+\ce{2O2}(g)\rightarrow \ce{CO2}(g)+\ce{2H2O}(l) \nonumber \]

     

    (b) In addition to ethane, C2H6, the combustion reaction also requires oxygen from the atmosphere.  The products of the combustion reaction will be H2O and CO2. 

    \[\ce{2C2H6}(g)+\ce{7O2}(g)\rightarrow \ce{4CO2}(g)+\ce{6H2O}(l) \nonumber \]

    Notice that balancing the reaction requires doubling the number of ethane molecules so that the coefficient for oxygen is a whole number.

    Exercise \(\PageIndex{1}\)

    Write the balanced equation for the combustion of methanol, CH3OH

    Answer

    \[\ce{2CH3OH}(l)+\ce{3O2}(g)\rightarrow \ce{2CO2}(g)+\ce{4H2O}(l) \nonumber \]

    Combustion in Biological Systems

    Similar reactions also occur in biological systems.  The overall chemical reaction that occurs in aerobic respiration is basically the same as the combustion of glucose.  In biological systems these reactions are broken down into small steps to control the release of energy. The overall reaction for the combustion of glucose is:

    \[\ce{C6H12O6}(s)+\ce{6O2}(g)\rightarrow \ce{6CO2}(g)+\ce{6H2O}(l) \nonumber \]

    Combustion Analysis

    The elemental composition of hydrocarbons and related compounds may be determined via a gravimetric method known as combustion analysis. In a combustion analysis, a weighed sample of the compound is heated to a high temperature under a stream of oxygen gas, resulting in its complete combustion to yield gaseous products of known identities. The complete combustion of hydrocarbons, for example, will yield carbon dioxide and water as the only products. The gaseous combustion products are swept through separate, preweighed collection devices containing compounds that selectively absorb each product (Figure \(\PageIndex{1}\)). The mass increase of each device corresponds to the mass of the absorbed product and may be used in an appropriate stoichiometric calculation to derive the mass of the relevant element.

    This diagram shows an arrow pointing from O subscript 2 into a tube that leads into a vessel containing a red material, labeled “Sample.” This vessel is inside a blue container with a red inner lining which is labeled “Furnace.” An arrow points from the tube to the right into the vessel above the red sample material. An arrow leads out of this vessel through a tube into a second vessel outside the furnace. An line points from this tube to a label above the diagram that reads “C O subscript 2, H subscript 2 O, O subscript 2, and other gases.” Many small green spheres are visible in the second vessel which is labeled below, “H subscript 2 O absorber such as M g ( C l O subscript 4 ) subscript 2.” An arrow points to the right through the vessel, and another arrow points right heading out of the vessel through a tube into a third vessel. The third vessel contains many small blue spheres. It is labeled “C O subscript 2 absorber such as N a O H.” An arrow points right through this vessel, and a final arrow points out of a tube at the right end of the vessel. Outside the end of this tube at the end of the arrow is the label, “O subscript 2 and other gases.”
    Figure \(\PageIndex{1}\): This schematic diagram illustrates the basic components of a combustion analysis device for determining the carbon and hydrogen content of a sample.
    Example \(\PageIndex{2}\): Combustion Analysis

    Polyethylene is a hydrocarbon polymer used to produce food-storage bags and many other flexible plastic items. A combustion analysis of a 0.00126-g sample of polyethylene yields 0.00394 g of CO2 and 0.00161 g of H2O. What is the empirical formula of polyethylene?

    Solution

    The primary assumption in this exercise is that all the carbon in the sample combusted is converted to carbon dioxide, and all the hydrogen in the sample is converted to water:

    \[\mathrm{C_xH_y}(s)+\ce{excess\: O2}(g)\rightarrow x\ce{CO2}(g)+ \dfrac{y}{2} \ce{H2O}(g) \nonumber \]

    Note that a balanced equation is not necessary for the task at hand. To derive the empirical formula of the compound, only the subscripts x and y are needed.

    First, calculate the molar amounts of carbon and hydrogen in the sample, using the provided masses of the carbon dioxide and water, respectively. With these molar amounts, the empirical formula for the compound may be written as described in the previous chapter of this text. An outline of this approach is given in the following flow chart:

    This figure shows two flowcharts. The first row is a single flow chart. In this row, a rectangle at the left is shaded yellow and is labeled, “Mass of C O subscript 2.” This rectangle is followed by an arrow pointing right to a second rectangle. The arrow is labeled, “Molar mass.” The second rectangle is shaded pink and is labeled, “Moles of C O subscript 2.” This rectangle is followed by an arrow pointing right to a third rectangle. The arrow is labeled, “Stoichiometric factor.” The third rectangle is shaded pink and is labeled, “Moles of C.” This rectangle is followed by an arrow labeled “Molar mass” which points right to a fourth rectangle. The fourth rectangle is shaded yellow and is labeled “Mass of C.” Below, is a second flowchart. It begins with a yellow shaded rectangle on the left which is labeled, “Mass of H subscript 2 O.” This rectangle is followed by an arrow labeled, “Molar mass,” which points right to a second rectangle. The second rectangle is shaded pink and is labeled, “Moles of H subscript 2 O.” This rectangle is followed by an arrow pointing right to a third rectangle. The arrow is labeled, “Stoichiometric factor.” The third rectangle is shaded pink and is labeled “Moles of H.” This rectangle is followed to the right by an arrow labeled, “Molar mass,” which points to a fourth rectangle. The fourth rectangle is shaded yellow and is labeled “Mass of H.” An arrow labeled, “Sample mass” points down beneath this rectangle to a green shaded rectangle. This rectangle is labeled, “Percent composition.” An arrow extends beneath the pink rectangle labeled, “Moles of H,” to a green shaded rectangle labeled, “C to H mole ratio.” Beneath this rectangle, an arrow extends to a second green shaded rectangle which is labeled, “Empirical formula.”

    \[\mathrm{mol\: C=0.00394\:g\: CO_2\times\dfrac{1\:mol\: CO_2}{44.01\: g/mol}\times\dfrac{1\:mol\: C}{1\:mol\: CO_2}=8.95\times10^{-5}\:mol\: C} \nonumber \]

    \[\mathrm{mol\: H=0.00161\:g\: H_2O\times\dfrac{1\:mol\: H_2O}{18.02\:g/mol}\times\dfrac{2\:mol\: H}{1\:mol\: H_2O}=1.79\times10^{-4}\:mol\: H} \nonumber \]

    The empirical formula for the compound is then derived by identifying the smallest whole-number multiples for these molar amounts. The H-to-C molar ratio is

    \[\mathrm{\dfrac{mol\: H}{mol\: C}=\dfrac{1.79\times10^{-4}\:mol\: H}{8.95\times10^{-5}\:mol\: C}=\dfrac{2\:mol\: H}{1\:mol\: C}} \nonumber \]

    and the empirical formula for polyethylene is CH2.

     

    Exercise \(\PageIndex{2}\)

    A 0.00215-g sample of polystyrene, a polymer composed of carbon and hydrogen, produced 0.00726 g of CO2 and 0.00148 g of H2O in a combustion analysis. What is the empirical formula for polystyrene?

    Answer

    CH

    Summary

    Chemical reactions are classified according to similar patterns of behavior. A large number of important reactions are included in three categories: precipitation, acid-base, and combustion. Precipitation reactions involve the formation of one or more insoluble products. Acid-base reactions involve the transfer of hydrogen ions between reactants.

    Glossary

    combustion analysis
    gravimetric technique used to determine the elemental composition of a compound via the collection and weighing of its gaseous combustion products
     
    combustion reaction
    vigorous redox reaction producing significant amounts of energy in the form of heat and, sometimes, light

    Contributors and Attributions

    This page titled 4.5: Classifying Chemical Reactions - Combustion Reactions is shared under a CC BY license and was authored, remixed, and/or curated by Scott Van Bramer.