Problem 78 - Chemistry and Chemical Reactivity

Question

A xenon fluoride can be prepared by heating a mixture of Xe and \(\mathrm{F}_{2}\) gases to a high temperature in a pressure-proof container. Assume that xenon gas was added to a 0.25-L container until its pressure reached 0.12 atm at \(0.0^{\circ}\) 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 of the xenon fluoride is XeF2.

Step 1: Calculate Initial Amount of Xe

We begin by using the Ideal Gas Law, \( PV = nRT \), to calculate the moles of Xenon (Xe). Given: \( P_{\text{Xe}} = 0.12 \, \text{atm} \), \( V = 0.25 \, \text{L} \), \( T = 273 \text{K} \) (as 0°C). Use \( R = 0.0821 \, \text{atm} \, \cdot\, \text{L} / \text{mol} \, \cdot \, \text{K} \).Calculate: \[ n_{\text{Xe}} = \frac{P_{\text{Xe}} \cdot V}{R \cdot T} = \frac{0.12 \times 0.25}{0.0821 \times 273} \approx 0.00134 \, \text{mol} \]

Step 2: Total Initial Pressure and F2 Added

The fluorine was added until the total pressure reached 0.72 atm.The pressure of \(F_2\) added is the difference between total pressure and xenon pressure:\[ P_{\text{F}_2\text{,initial}} = 0.72 - 0.12 = 0.60 \, \text{atm} \]

Step 3: Calculate Initial Amount of F2

Using the Ideal Gas Law for the initial amount of \(F_2\):Given: \( P_{\text{F}_2\text{,initial}} = 0.60 \, \text{atm} \), \( V = 0.25 \, \text{L} \), \( R = 0.0821 \, \text{atm} \, \cdot\, \text{L} / \text{mol} \, \cdot \, \text{K} \), \( T = 273 \, \text{K} \).Calculate:\[ n_{\text{F}_2\text{,initial}} = \frac{P_{\text{F}_2\text{,initial}} \cdot V}{R \cdot T} = \frac{0.60 \times 0.25}{0.0821 \times 273} \approx 0.00671 \, \text{mol} \]

Step 4: Calculate Amount of F2 Remaining

After the reaction, the pressure of \(F_2\) is given as 0.36 atm.Calculate moles of \(F_2\) remaining:Given: \( P_{\text{F}_2\text{,final}} = 0.36 \, \text{atm} \), \( V = 0.25 \, \text{L} \), \( R = 0.0821 \, \text{atm} \, \cdot\, \text{L} / \text{mol} \, \cdot \, \text{K} \), \( T = 273 \, \text{K} \).Calculate:\[ n_{\text{F}_2\text{,final}} = \frac{P_{\text{F}_2\text{,final}} \cdot V}{R \cdot T} = \frac{0.36 \times 0.25}{0.0821 \times 273} \approx 0.00402 \, \text{mol} \]

Step 5: Calculate Amount of F2 Reacted

Calculate the moles of \(F_2\) that reacted during the formation of xenon fluoride:\[ n_{\text{F}_2\text{,reacted}} = n_{\text{F}_2\text{,initial}} - n_{\text{F}_2\text{,final}} = 0.00671 - 0.00402 = 0.00269 \, \text{mol} \]

Step 6: Determine Empirical Formula Ratio

The reaction indicates that all xenon is consumed, with 0.00134 moles of Xe reacting with the 0.00269 moles of \(F_2\).To find the empirical formula, calculate the mole ratio of \(F_2\) to Xe:\[ \text{Ratio of } F_2 \text{ to } Xe = \frac{0.00269}{0.00134} \approx 2 \]This ratio implies that each atom of xenon reacts with approximately 2 molecules of fluorine.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is an essential principle in chemistry and physics, representing the relationship between the pressure (P), volume (V), temperature (T), and amount of gas in moles (n). It is expressed with the formula \( PV = nRT \), where \( R \) is the ideal gas constant (approximately 0.0821 \( ext{atm} \, ext{L/mol} \, ext{K} \)).

In the context of determining an empirical formula for a xenon fluoride, the Ideal Gas Law helps to calculate the number of moles of gases involved in the reaction. By knowing the initial conditions, such as pressure, volume, and temperature, we can determine the number of moles of xenon and fluorine. This is crucial for finding the mole ratio needed for the empirical formula.

For instance, we observed that for xenon, with a pressure of 0.12 atm in a 0.25-L container at 0°C, the moles can be calculated as follows:- \( n_{\text{Xe}} = \frac{(0.12 \, \text{atm}) \times (0.25 \, \text{L})}{0.0821 \, \text{atm} \cdot \text{L/mol} \cdot \text{K} \times 273 \, \text{K}} \approx 0.00134 \, \text{mol} \)

The law is applicable for each step of the reaction process, including calculating the initial amount of \( F_2 \) and the amount remaining after the reaction.

Mole Ratio

Understanding the mole ratio is fundamental in determining the empirical formula of any compound formed from a chemical reaction. The mole ratio refers to the proportion of reactants that combine to form products, which is usually derived from the coefficients in a balanced chemical equation or from actual experimental data.

In our exercise, the mole ratio is critical to calculate the empirical formula of xenon fluoride. After determining the moles of both xenon and fluorine initially present and after the reaction, it was found that 0.00134 moles of xenon reacted completely with 0.00269 moles of fluorine.

This was calculated as:- \( \text{Ratio of } F_2 \text{ to } Xe = \frac{0.00269}{0.00134} \approx 2 \)

The mole ratio helps explain that for every 1 mole of xenon, approximately 2 moles of fluorine are required. Such ratios are pivotal in formulating the empirical formula, which in this case results in \( \text{XeF}_2 \), indicating xenon combines with fluorine in a 1:2 ratio. This ratio directly links to conditions under which the compound was synthesized.

Chemical Reactions

Chemical reactions involve transforming reactants into products, and they are governed by the conservation of mass where matter is neither created nor destroyed. Each reaction has a unique set of reactants and products that can be represented through a chemical equation.

In the provided exercise, a chemical reaction occurs between xenon and fluorine gas to form xenon fluoride. The reaction is presumed to follow the stoichiometry implied by the observed mole ratio of reactants. Initially, the xenon reacts completely, indicating it is the limiting reactant, meaning it determines the amount of product formed.

- Xenon is fully consumed, and some \( F_2 \) remains post-reaction.- By measuring the pressure change and using the Ideal Gas Law, we quantified the reactants and products.

The chemical reaction is valuable for illustrating principles like limiting reactants, stoichiometry, and gas laws. In our example, this understanding assists in computing the empirical formula by ensuring each atom pairs with the correct number of other atoms during the formation of the compound. These calculations help comprehend how reactions are quantified and balanced, further supporting the determination of empirical formulas in gas reactions.