Problem 7 - General Chemistry: Principles and Modern Applications

Question

Write balanced equations to represent the complete combustion of each of the following in excess oxygen: (a) butane, \(\mathrm{C}_{4} \mathrm{H}_{10} ;\) (b) isopropyl alcohol, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3} ;\) (c) lactic acid, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\)

Step-by-Step Solution

Answer

The balanced equations for the complete combustion are: (a) \(\mathrm{C}_{4} \mathrm{H}_{10} + 13/2 O_{2} \rightarrow 4 CO_{2} + 5 H_{2}O\) (b) \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3} + 4O_{2} \rightarrow 3CO_{2} + 4H_{2}O\) (c) \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH} + 3O_{2} \rightarrow 3CO_{2} + 4H_{2}O\)

Step 1: Combustion of Butane

Butane, \(\mathrm{C}_{4} \mathrm{H}_{10}\), reacts with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). The unbalanced equation is \( \mathrm{C}_{4} \mathrm{H}_{10} + O_{2} \rightarrow CO_{2} + H_{2}O\). To balance the equation, start by balancing the carbons, followed by the hydrogen and lastly the oxygen, providing the balanced equation \( \mathrm{C}_{4} \mathrm{H}_{10} + 13/2 O_{2} \rightarrow 4 CO_{2} + 5 H_{2}O\)

Step 2: Combustion of Isopropyl Alcohol

Isopropyl alcohol, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3}\), reacts with oxygen (O2) to also produce carbon dioxide (CO2) and water (H2O). The unbalanced equation is \( \mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3} + O_{2} \rightarrow CO_{2} + H_{2}O\). Again, balance the carbons first, then hydrogens, and finally the oxygens, to get the balanced equation \( \mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3} + 4O_{2} \rightarrow 3CO_{2} + 4H_{2}O\)

Step 3: Combustion of Lactic Acid

Lactic acid, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\), undergoes combustion with oxygen (O2) to yield carbon dioxide (CO2) and water (H2O). The unbalanced equation is \( \mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH} + O_{2} \rightarrow CO_{2} + H_{2}O\). Begin by balancing the carbons, then hydrogens, and finally the oxygens, resulting in the balanced equation \( \mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH} + 3O_{2} \rightarrow 3CO_{2} + 4H_{2}O\)

Key Concepts

Combustion Reactions

Combustion reactions are a major type of chemical reaction where a substance combines with oxygen to release heat and form products that usually include carbon dioxide and water. These reactions are crucial in organic chemistry, playing a key role in both understanding energy release from materials and creating new chemicals from burning hydrocarbons. To understand combustion reactions better, consider the burning of butane. Butane (\[\mathrm{C}_{4} \mathrm{H}_{10}\]) combusts in the presence of oxygen to produce carbon dioxide (\[\mathrm{CO}_{2}\]) and water (\[\mathrm{H}_{2}\mathrm{O}\]).
Combustion reactions are typically exothermic, meaning they release energy as heat. This process is not only important in laboratory settings but is also fundamental to many everyday processes, such as powering car engines or heating homes.

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry that ensures the law of conservation of mass is obeyed. This law states that matter is neither created nor destroyed in a chemical reaction. Thus, the number of atoms of each element must be the same on both sides of a chemical equation.

The steps for balancing include:

• Start by writing the unbalanced equation.
• Count the number of atoms for each element on both sides.
• Use coefficients to equalize the number of atoms for each element in the reactants and products.
• Check your work to ensure the entire equation is balanced.

When balancing the combustion of butane, begin by balancing the carbon atoms, then the hydrogen atoms, and finally the oxygen atoms. This chain of balancing ensures that the equation reflects the exact stoichiometry of the reactants and products.

Organic Chemistry

Organic chemistry is the branch of chemistry dealing with the structure, properties, and reactions of organic compounds and materials—primarily composed of carbon and hydrogen atoms. These compounds are the basis of all life on earth as they make up the cells and structures of organisms. In the combustion reactions covered, organic compounds like butane, isopropyl alcohol, and lactic acid provide a great example of organic chemistry in action.

Hydrocarbons like butane fuel combustion reactions by reacting with oxygen to produce energy. Moreover, functional groups such as the hydroxyl group in isopropyl alcohol or the carboxyl group in lactic acid dramatically affect the properties and reactivity of these organic substances. Understanding how these groups interact during chemical reactions is key in various applications from pharmaceuticals to energy production.

Stoichiometry

Stoichiometry is a section of chemistry that involves calculating the relative quantities of reactants and products in chemical reactions. It applies the principles of maths to chemical equations, utilizing the idea of moles—a measure of chemical quantity.

In stoichiometry, you typically:

• Identify the balanced chemical equation for the reaction.
• Determine the moles of a reactant or product from a given quantity.
• Utilize the mole ratio from the balanced equation to find the corresponding number of moles for other reactants or products.
• Convert the moles back to grams if needed.

For example, in the combustion of isopropyl alcohol, \[\mathrm{CH}_{3} \mathrm{CH} (\mathrm{OH}) \mathrm{CH}_{3} + 4 \mathrm{O}_{2} \rightarrow 3 \mathrm{CO}_{2} + 4 \mathrm{H}_{2} \mathrm{O}\], stoichiometry tells you how different quantities of reactants relate to quantities of products. Mastering stoichiometry is essential for accurate predictions and calculations in any chemical process.