Problem 34 - Chemistry and Chemical Reactivity

Question

The hydrocarbon octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) burns to give \(\mathrm{CO}_{2}\) and water vapor: $$ 2 \mathrm{C}_{8} \mathrm{H}_{18}(\mathrm{g})+25 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 16 \mathrm{CO}_{2}(\mathrm{g})+18 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$ If a 0.048 -g sample of octane burns completely in \(\mathrm{O}_{2},\) what will be the pressure of water vapor in a \(4.75-\mathrm{L}\) flask at \(30.0^{\circ} \mathrm{C} ?\) If the \(\mathrm{O}_{2}\) gas needed for complete combustion was contained in a \(4.75-\mathrm{L}\) flask at \(22^{\circ} \mathrm{C},\) what would its pressure be?

Step-by-Step Solution

Answer

Water vapor pressure is 0.0197 atm; \(\mathrm{O}_{2}\) pressure is 0.0267 atm in the flask.

Step 1: Calculate Moles of Octane

To find the moles of octane, we start by using the molar mass. The molar mass of octane, \( \text{C}_8\text{H}_{18} \), is calculated as follows: \( \text{Molar mass} = 8 \times 12.01 + 18 \times 1.01 = 114.22 \, \text{g/mol} \). Given that we have a 0.048 g sample, the number of moles \( n \) is given by \( n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.048 \, \text{g}}{114.22 \, \text{g/mol}} = 4.20 \times 10^{-4} \, \text{mol} \).

Step 2: Use Stoichiometry to Determine Moles of Water Vapor

From the balanced chemical equation, 2 moles of \( \text{C}_8\text{H}_{18} \) produce 18 moles of \( \text{H}_2\text{O} \). Thus, 1 mole of \( \text{C}_8\text{H}_{18} \) produces 9 moles of \( \text{H}_2\text{O} \). Therefore, \( 4.20 \times 10^{-4} \, \text{mol} \) of octane will produce \( 4.20 \times 10^{-4} \times 9 = 3.78 \times 10^{-3} \, \text{mol} \) of \( \text{H}_2\text{O} \).

Step 3: Calculate Pressure of Water Vapor Using Ideal Gas Law

We apply the Ideal Gas Law, \( PV = nRT \), to find the pressure of water vapor. Here, \( n = 3.78 \times 10^{-3} \, \text{mol} \), \( R = 0.0821 \, \text{L atm} / \text{mol K} \), and \( T = 30.0^{\circ} \text{C} = 303.15 \, \text{K} \). Solving for \( P \), \( P = \frac{nRT}{V} = \frac{3.78 \times 10^{-3} \times 0.0821 \times 303.15}{4.75} = 0.0197 \, \text{atm} \).

Step 4: Calculate Moles of \(\mathrm{O}_{2}\) Required for Combustion

From the stoichiometry of the equation, \( 2 \text{C}_8\text{H}_{18} \) requires 25 \( \text{O}_2 \). Thus, each mole of octane requires \( 12.5 \) moles of \( \text{O}_2 \). Therefore, \( 4.20 \times 10^{-4} \, \text{mol} \) of octane will require \( 4.20 \times 10^{-4} \times 12.5 = 5.25 \times 10^{-3} \, \text{mol} \) of \( \text{O}_2 \).

Step 5: Determine Pressure of \(\mathrm{O}_{2}\) Using Ideal Gas Law

Again, using the Ideal Gas Law formula \( PV = nRT \), where \( n = 5.25 \times 10^{-3} \, \text{mol} \), \( T = 22^{\circ} \text{C} = 295.15 \, \text{K} \). Calculating \( P \), we have \( P = \frac{5.25 \times 10^{-3} \times 0.0821 \times 295.15}{4.75} = 0.0267 \, \text{atm} \).

Key Concepts

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. In simpler terms, it helps us know how much of one substance will react with another in a balanced chemical equation. It uses the coefficients of a chemical equation to find relationships between different substances involved in the reaction.
For example, in the combustion of octane, the equation given is:

• \[2 \text{C}_8\text{H}_{18}(\text{g}) + 25 \text{O}_2(\text{g}) \rightarrow 16 \text{CO}_2(\text{g}) + 18 \text{H}_2\text{O}(\text{g})\]

This tells us that 2 moles of octane react with 25 moles of oxygen to produce 16 moles of carbon dioxide and 18 moles of water. By using stoichiometry, one can predict how many moles of water are formed when a certain amount of octane is burned, as was calculated in Step 2 of the solution. This relation is crucial for solving problems related to chemical reactions.

Molar Mass

Molar mass is the mass of one mole of a substance. It is expressed in grams per mole (g/mol) and is derived from the atomic masses of the elements in a compound. Understanding molar mass is essential for converting between the mass of a substance and the amount in moles.To find the molar mass of octane, we sum up the atomic masses of its constituent elements:

• Carbon (C): Atomic mass = 12.01 g/mol. Total for 8 carbons = \(8 \times 12.01 = 96.08 \text{ g/mol}\)
• Hydrogen (H): Atomic mass = 1.01 g/mol. Total for 18 hydrogens = \(18 \times 1.01 = 18.18 \text{ g/mol}\)
• Molar mass of octane = 114.22 g/mol

Using molar mass, we can find the moles of octane in a 0.048 g sample as shown in Step 1 of the solution. This step is key in calculating how much product is formed in the reaction.

Chemical Equations

Chemical equations represent the formulas of reactants and products in a reaction. They provide information not only about the reactants and products but also about the quantity of each substance involved, given by the coefficients in the balanced equation.In this reaction of octane combustion:

• The left side (reactants) consists of octane \((\text{C}_8\text{H}_{18})\) and oxygen \((\text{O}_2)\)
• The right side (products) consists of carbon dioxide \((\text{CO}_2)\) and water \((\text{H}_2\text{O})\)
• The equation is balanced to reflect that matter is neither created nor destroyed (law of conservation of mass), showing equal numbers of each type of atom on either side.

Balancing the equation is fundamental to using stoichiometry, as it ensures that calculations of moles and masses during the reaction are accurate.

Combustion Reactions

Combustion reactions are chemical reactions where a substance combines with oxygen and releases energy in the form of heat and light. It is a type of oxidation-reduction reaction, and hydrocarbons like octane commonly serve as fuel. A complete combustion reaction of a hydrocarbon results in carbon dioxide and water as products, as seen in the combustion of octane:

• This specific reaction contributes to its applications in engines and heating.
• Complete combustion ensures maximum energy release and involves oxygen as a necessary reactant to convert all fuel into products.
• In the reaction provided, all of the octane reacts completely due to sufficient oxygen provided according to the stoichiometry, resulting in the formation of water vapor, and it's helpful in calculating the pressure it creates in a given flask volume.

Having a clear understanding of combustion reactions allows individuals to understand energy balances and emissions in various chemical and industrial processes.