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 \(0.048-\mathrm{g}\) 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 -L. flask at \(22^{\circ} \mathrm{C},\) what would its pressure be?

Step-by-Step Solution

Answer

Pressure of water vapor is 0.0197 atm, and oxygen pressure is 0.0269 atm.

Step 1: Calculate Moles of Octane

First, determine the number of moles of octane. The molar mass of octane (\(\mathrm{C}_8\mathrm{H}_{18}\)) is 114.22 g/mol. Use the formula:\[moles = \frac{\text{mass}}{\text{molar mass}}\]Substitute the given mass:\[moles = \frac{0.048 \text{ g}}{114.22 \text{ g/mol}} \approx 4.20 \times 10^{-4} \text{ mol}\]

Step 2: Calculate Moles of Water Produced

According to the balanced chemical equation, 2 moles of \(\mathrm{C}_8\mathrm{H}_{18}\) produce 18 moles of \(\mathrm{H}_2\mathrm{O}\). Use stoichiometry to find moles of \(\mathrm{H}_2\mathrm{O}\):\[moles \; \mathrm{H}_2\mathrm{O} = \left(4.20 \times 10^{-4} \right) \times \frac{18}{2} = 3.78 \times 10^{-3} \text{ mol}\]

Step 3: Calculate Pressure of Water Vapor

Use the ideal gas law to find the pressure of the water vapor. The ideal gas law equation is \( PV = nRT \). Remember, \( R = 0.0821 \text{ L atm/mol K} \), and temperature in Kelvin is \( 303.15 \text{ K} \) (i.e., \( 30^{\circ}\text{C} + 273.15 \)). Rearrange for \( P \):\[P = \frac{nRT}{V}\]Substitute the known values:\[P = \frac{(3.78 \times 10^{-3})(0.0821)(303.15)}{4.75} \approx 0.0197 \text{ atm}\]

Step 4: Calculate Pressure of Oxygen Needed

Next, find the moles of \(\mathrm{O}_2\) needed for complete combustion. From the stoichiometry, 2 moles of \(\mathrm{C}_8\mathrm{H}_{18}\) require 25 moles of \(\mathrm{O}_2\).\[moles \; \mathrm{O}_2 = \left(4.20 \times 10^{-4} \right) \times \frac{25}{2} = 5.25 \times 10^{-3} \text{ mol}\]Use the ideal gas law at 22°C, which is \( 295.15 \text{ K} \) for the temperature in Kelvin:\[P = \frac{nRT}{V}\]Substitute the values into the equation:\[P = \frac{(5.25 \times 10^{-3})(0.0821)(295.15)}{4.75} \approx 0.0269 \text{ atm}\]

Key Concepts

Ideal Gas Law

The ideal gas law is a cornerstone of chemistry when dealing with gases, linking pressure, volume, temperature, and number of moles in a simple equation. This law is expressed as \( PV = nRT \), where \( P \) is the pressure in atm, \( V \) is the volume in liters, \( n \) is the number of moles, \( R \) is the gas constant (0.0821 L atm/mol K), and \( T \) is the temperature in Kelvin.
To solve problems using the ideal gas law, remember:

• Convert temperature from Celsius to Kelvin by adding 273.15.
• Ensure that the volume is in liters and pressure is in atm to use \( R = 0.0821 \).
• Rearrange the formula to solve for the unknown variable, such as pressure or volume, depending on what you need.

In this exercise, we use it twice. First, to find the water vapor pressure at a given temperature in a flask. Secondly, to calculate the needed pressure of oxygen for the complete combustion process.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It is based on the balanced chemical equation, providing the mole ratios of substances involved, as seen in our combustion example of octane.

• The balanced equation: \( 2 \text{C}_8\text{H}_{18} + 25 \text{O}_2 \rightarrow 16 \text{CO}_2 + 18 \text{H}_2\text{O} \) gives the ratio of octane to oxygen and products.
• From this, the stoichiometric coefficients help determine how many moles of each substance are needed or produced.

For octane combustion, stoichiometry tells us that 2 moles of octane are needed to produce 18 moles of water. This relationship allows us to determine the moles of water produced for a given amount of octane burned.

Molar Mass

Molar mass is the mass of one mole of a substance, usually expressed in g/mol. To calculate the molar mass of a compound, you sum the atomic masses of all its atoms.
For octane \( \text{C}_8\text{H}_{18} \):

• Carbon (C) has an atomic mass of approximately 12.01 g/mol. So, for 8 atoms: \( 8 \times 12.01 = 96.08 \).
• Hydrogen (H) has an atomic mass of approximately 1.008 g/mol. For 18 atoms: \( 18 \times 1.008 = 18.144 \).
• Add these to get octane's molar mass: \( 96.08 + 18.144 = 114.22 \text{ g/mol} \).

Knowing the molar mass allows you to convert between grams and moles, an essential step when using the ideal gas law and stoichiometry.

Chemical Reaction

A chemical reaction involves the rearrangement of atoms to transform reactants into products. The combustion of octane is a classic example, where hydrocarbons react with oxygen to produce carbon dioxide and water.
In combustion:

• Reactants: Octane \( \text{C}_8\text{H}_{18} \) and Oxygen \( \text{O}_2 \).
• Products: Carbon dioxide \( \text{CO}_2 \) and Water \( \text{H}_2\text{O} \).
• The reaction must be balanced, meaning the number of each type of atom is the same on both sides of the equation.

This balance ensures the law of conservation of mass, allowing precise calculation of reactants needed and products formed. In our exercise, balanced equations allow us to predict the amount of water vapor pressure from the mass of octane burned.