Question

When organic compounds are burned, they react with oxygen to form \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). Write balanced equations for the combustion reactions involving the following compounds. (Hint: When balancing combustion reactions, begin by balancing the \(\mathrm{C}\) and \(\mathrm{H}\) atoms first, and balance the \(\mathrm{O}\) atoms last). (a) \(\mathrm{C}_{4} \mathrm{H}_{10}\) (butane, used in lighters) (b) \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}\) (ethanol, used in gasohol and as race car fuel) (c) \(\mathrm{C}_{8} \mathrm{H}_{18}\) (octane, a component of gasoline)

Short answer

(a) \(2\mathrm{C}_{4}\mathrm{H}_{10} + 13\mathrm{O}_{2} \rightarrow 8\mathrm{CO}_{2} + 10\mathrm{H}_{2}\mathrm{O}\); (b) \(\mathrm{C}_{2}\mathrm{H}_{6}\mathrm{O} + 3\mathrm{O}_{2} \rightarrow 2\mathrm{CO}_{2} + 3\mathrm{H}_{2}\mathrm{O}\); (c) \(2\mathrm{C}_{8}\mathrm{H}_{18} + 25\mathrm{O}_{2} \rightarrow 16\mathrm{CO}_{2} + 18\mathrm{H}_{2}\mathrm{O}\)."}

Step-by-step solution

Write the Unbalanced Chemical Equation for Butane

The combustion of butane \(\mathrm{C}_{4}\mathrm{H}_{10}\) in oxygen \(\mathrm{O}_{2}\) produces carbon dioxide \(\mathrm{CO}_{2}\) and water \(\mathrm{H}_{2}\mathrm{O}\). The unbalanced equation is: \[\mathrm{C}_{4}\mathrm{H}_{10} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O} \]

Balance Carbon Atoms in Butane

There are 4 carbon atoms in butane, so you need 4 \(\mathrm{CO}_{2}\) molecules to balance the carbon. The equation becomes: \[\mathrm{C}_{4}\mathrm{H}_{10} + \mathrm{O}_{2} \rightarrow 4\mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O} \]

Balance Hydrogen Atoms in Butane

There are 10 hydrogen atoms in butane, so balance with 5 \(\mathrm{H}_{2}\mathrm{O}\) molecules. The equation is now: \[\mathrm{C}_{4}\mathrm{H}_{10} + \mathrm{O}_{2} \rightarrow 4\mathrm{CO}_{2} + 5\mathrm{H}_{2}\mathrm{O} \]

Balance Oxygen Atoms in Butane

Count the oxygen atoms on the right: 8 from \(4\mathrm{CO}_{2}\) and 5 from \(5\mathrm{H}_{2}\mathrm{O}\), totaling 13. Thus, use \(\frac{13}{2}\) \(\mathrm{O}_{2}\) or 6.5 \(\mathrm{O}_{2}\). Multiply the entire equation by 2 to avoid fractions: \[2\mathrm{C}_{4}\mathrm{H}_{10} + 13\mathrm{O}_{2} \rightarrow 8\mathrm{CO}_{2} + 10\mathrm{H}_{2}\mathrm{O} \]

Write the Unbalanced Chemical Equation for Ethanol

For ethanol \(\mathrm{C}_{2}\mathrm{H}_{6}\mathrm{O}\), the initial equation is: \[\mathrm{C}_{2}\mathrm{H}_{6}\mathrm{O} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O} \]

Balance Carbon Atoms in Ethanol

With 2 carbon atoms in ethanol, balance with 2 \(\mathrm{CO}_{2}\) molecules: \[\mathrm{C}_{2}\mathrm{H}_{6}\mathrm{O} + \mathrm{O}_{2} \rightarrow 2\mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O} \]

Balance Hydrogen Atoms in Ethanol

There are 6 hydrogen atoms, so use 3 \(\mathrm{H}_{2}\mathrm{O}\) molecules: \[\mathrm{C}_{2}\mathrm{H}_{6}\mathrm{O} + \mathrm{O}_{2} \rightarrow 2\mathrm{CO}_{2} + 3\mathrm{H}_{2}\mathrm{O} \]

Balance Oxygen Atoms in Ethanol

There are 4 oxygen in \(2\mathrm{CO}_{2}\) and 3 in \(3\mathrm{H}_{2}\mathrm{O}\), totaling 7. With 1 in ethanol, you need 3 more from \(\mathrm{O}_{2}\), so use \(3\mathrm{O}_{2}\):\[\mathrm{C}_{2}\mathrm{H}_{6}\mathrm{O} + 3\mathrm{O}_{2} \rightarrow 2\mathrm{CO}_{2} + 3\mathrm{H}_{2}\mathrm{O} \]

Write the Unbalanced Chemical Equation for Octane

For octane \(\mathrm{C}_{8}\mathrm{H}_{18}\), the equation starts as: \[\mathrm{C}_{8}\mathrm{H}_{18} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O} \]

Balance Carbon Atoms in Octane

With 8 carbon atoms, use 8 \(\mathrm{CO}_{2}\) molecules: \[\mathrm{C}_{8}\mathrm{H}_{18} + \mathrm{O}_{2} \rightarrow 8\mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O} \]

Balance Hydrogen Atoms in Octane

There are 18 hydrogen atoms, so use 9 \(\mathrm{H}_{2}\mathrm{O}\) molecules: \[\mathrm{C}_{8}\mathrm{H}_{18} + \mathrm{O}_{2} \rightarrow 8\mathrm{CO}_{2} + 9\mathrm{H}_{2}\mathrm{O} \]

Balance Oxygen Atoms in Octane

There are 16 oxygen from \(8\mathrm{CO}_{2}\) and 9 from \(9\mathrm{H}_{2}\mathrm{O}\), totaling 25. Thus, use \(\frac{25}{2}\) \(\mathrm{O}_{2}\) or 12.5 \(\mathrm{O}_{2}\). Multiply the entire equation by 2: \[2\mathrm{C}_{8}\mathrm{H}_{18} + 25\mathrm{O}_{2} \rightarrow 16\mathrm{CO}_{2} + 18\mathrm{H}_{2}\mathrm{O} \]

Key concepts

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry, especially important in reactions like combustion. When we write a chemical equation, it represents a chemical reaction where reactants turn into products. The law of conservation of mass requires that the mass of reactants equals the mass of products. Therefore, we must have the same number of each type of atom on both sides of the equation. If you start with 4 carbon atoms in the reactants, you should also have 4 in the products.

To balance equations correctly, follow these steps:

• First, write the unbalanced equation.
• Adjust the coefficients (the numbers in front of chemicals) to balance the equation.
• Start with one element and proceed to others systematically, typically leaving hydrogen and oxygen for last.
• Finally, verify that you have the same number of each atom on both sides.

This methodical approach is particularly helpful in complex reactions such as those found in organic chemistry.

Organic Chemistry

Organic chemistry involves studying compounds mainly composed of carbon and hydrogen atoms. These can include other elements like oxygen, nitrogen, sulfur, etc. Organic molecules are the building blocks of life, making up everything from fuels like butane and octane to substances like ethanol, which can be found in beverages and fuels alike.

Understanding organic chemical reactions, such as combustion, is crucial. Combustion involves an organic compound (like hydrocarbons) reacting with oxygen to form carbon dioxide and water as products. Familiar organic compounds include:

• Butane ( \( \mathrm{C}_{4}\mathrm{H}_{10} \)), found in lighters.
• Ethanol ( \( \mathrm{C}_{2}\mathrm{H}_{6}\mathrm{O} \)), used as a solvent and in fuel blends.
• Octane ( \( \mathrm{C}_{8}\mathrm{H}_{18} \)), a major gasoline component.

These reactions not only show the variety of organic compounds but also highlight the significance of combustion reactions in everyday life.

Stoichiometry

Stoichiometry is the part of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. Understanding stoichiometry is key to predicting yields and proportions of products formed or reactants consumed. In a combustion reaction, stoichiometry lets us determine how much oxygen is needed to completely burn a given amount of fuel.

Take, for example, the combustion of butane:

• The balanced stoichiometric equation is: \[ 2\mathrm{C}_{4}\mathrm{H}_{10} + 13\mathrm{O}_{2} \rightarrow 8\mathrm{CO}_{2} + 10\mathrm{H}_{2}\mathrm{O} \]
• For every molecule of butane, you need a certain number of oxygen molecules for complete combustion.
• By using coefficients, stoichiometry allows precise calculations.

With stoichiometry, one can smoothly carry out chemical calculations, ensuring enough supplies of reactants, and reducing waste in industrial processes.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They outline the reactants transforming into products and follow strict conventions. Components in a chemical equation include reactants, products, coefficients, and sometimes conditions like temperature and catalysts.

Steps to write and understand chemical equations:

• Identify the reactants and write them on the left of the arrow.
• Write the products on the right side of the arrow.
• Add coefficients to balance the equation, which indicates the proportions of each substance involved.

Correctly writing and balancing chemical equations ensures the reaction is accurately represented, which is vital not only in academic settings but also in real-world applications like industry and research.