Problem 72 - Chemistry and Chemical Reactivity

Question

A xenon fluoride can be prepared by heating a mixture of \(\mathrm{Xe}\) and \(\mathrm{F}_{2}\) gases to a high temperature in a pressureproof container. Assume that xenon gas was added to a \(0.25-\mathrm{L}\) container until its pressure reached \(0.12 \mathrm{atm}\) at \(0.0^{\circ} \mathrm{C} .\) Fluorine gas was then added until the total pressure reached 0.72 atm at \(0.0^{\circ} \mathrm{C}\). After the reaction was complete, the xenon was consumed completely, and the pressure of the \(\mathrm{F}_{2}\) remaining in the container was 0.36 atm at \(0.0^{\circ} \mathrm{C} .\) What is the empirical formula of the xenon fluoride?

Step-by-Step Solution

Answer

The empirical formula is \(\mathrm{XeF_2}\).

Step 1: Determine Moles of Xenon

First, calculate the moles of xenon gas (Xe) using the ideal gas law. The initial pressure of xenon is 0.12 atm in a 0.25 L container. Use the equation \( PV = nRT \) , where R = 0.0821 L·atm/mol·K and T = 273.15 K (0°C in Kelvin).\[ n = \frac{PV}{RT} = \frac{0.12 \times 0.25}{0.0821 \times 273.15} \approx 0.00134 \text{ moles of } \text{Xe} \]

Step 2: Determine Moles of Initial Fluorine

Next, calculate the moles of fluorine gas (F_2) before the reaction. The total pressure of the system is 0.72 atm, so the initial pressure of fluorine is 0.72 - 0.12 = 0.60 ext{ atm}. Use the ideal gas law:\[ n = \frac{PV}{RT} = \frac{0.60 \times 0.25}{0.0821 \times 273.15} \approx 0.00665 \text{ moles of } \text{F}_{2} \]

Step 3: Determine Moles of Remaining Fluorine

After the reaction, the pressure of fluorine is 0.36 atm. Calculate the remaining moles of F_2:\[ n = \frac{PV}{RT} = \frac{0.36 \times 0.25}{0.0821 \times 273.15} \approx 0.00399 \text{ moles of } \text{F}_{2} \]

Step 4: Calculate Moles of Fluorine Consumed

Subtract the moles of remaining fluorine from the initial moles to find the moles of F_2 consumed:\[\text{Moles of } F_2 \text{ consumed} = 0.00665 - 0.00399 = 0.00266 \text{ moles} \]

Step 5: Determine Ratio and Empirical Formula

Find the ratio of consumed fluorine to xenon, which is \frac{0.00266}{0.00134} = 1.99. This ratio is approximately 2:1, indicating two moles of fluorine react with one mole of xenon, suggesting the empirical formula is \text{\(XeF_2\)}.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is fundamental in the study of gases in chemistry. It relates the pressure, volume, temperature, and number of moles of a gas together in the equation: \( PV = nRT \). In this equation, \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant (0.0821 L·atm/mol·K), and \( T \) is the temperature in Kelvin.

To use this law effectively, you need to ensure that units match. Typically, pressure is measured in atmospheres (atm), volume in liters (L), and temperature must always be in Kelvin. To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature.

For this particular exercise, the Ideal Gas Law was employed to calculate the moles of xenon and fluorine gas. By rearranging the formula to \( n = \frac{PV}{RT} \), you can directly calculate the number of moles present when you know pressure, volume, and temperature. This lays the groundwork for further stoichiometric calculations needed to find the empirical formula.

Empirical Formula

An empirical formula is the simplest whole-number ratio of atoms in a compound. It is often determined by chemical analysis and calculations, as seen in this exercise involving xenon fluoride.

The calculated method for deriving an empirical formula first involves determining the moles of each element involved in the chemical reaction, using data such as mass, pressure, volume, and temperature. This was done by utilizing the Ideal Gas Law to calculate moles of xenon and fluorine gas before and after the reaction.

After determining the moles of consumed substances, the next step is to find their mole ratio. In our example, the exercise shows that the consumed moles of fluorine to xenon are in a 2:1 ratio. This calculation leads us directly to suggest the empirical formula \( XeF_2 \). This means that in the compound, xenon and fluorine combine in the simplest ratio of one atom of xenon to two atoms of fluorine.

Gas Reaction Calculations

Gas reaction calculations often involve using the stoichiometry of a reaction to relate quantities of reactants and products. In this exercise, the reaction between xenon and fluorine gases was considered to form a xenon fluoride compound.

The calculations began by using the Ideal Gas Law to determine the initial and remaining moles of fluorine gas. By subtracting the post-reaction moles from the initial moles, the moles of fluorine consumed in the reaction were found.

The step then follows to calculate the mole ratio between the reactants. In many reactions, this ratio provides insight into the empirical formula of the product. The consumed moles of fluorine divided by the moles of xenon gave a value close to 2, suggesting that for every mole of xenon, two moles of fluorine reacted. This highlights the importance and elegance of using stoichiometry in gas reactions to easily identify stoichiometric relationships and formulate chemical equations.