Question

A compound containing xenon and fluorine was prepared by shining sunlight on a mixture of Xe \((0.526 \mathrm{g})\) and excess \(\mathrm{F}_{2}\) gas. If you isolate \(0.678 \mathrm{g}\) of the new compound, what is its empirical formula?

Short answer

The empirical formula is \( \text{XeF}_2 \).

Step-by-step solution

Calculate moles of Xe

First, find the moles of xenon in the compound. The molar mass of xenon (Xe) is approximately 131.29 g/mol. Use the formula: \[ ext{Moles of Xe} = \frac{0.526 ext{ g}}{131.29 ext{ g/mol}} \approx 0.00401 ext{ moles} \]

Calculate moles of the compound

Find the total mass of the isolated compound and determine the moles of the compound formed. The mass of the compound is 0.678 g. We use it to find the difference in mass from xenon to get the fluorine contribution.

Determine mass of fluorine

Calculate the mass of fluorine that combined with xenon to form the compound by subtracting the mass of xenon from the total mass of the compound:\[ ext{Mass of F} = 0.678 ext{ g} - 0.526 ext{ g} = 0.152 ext{ g} \]

Calculate moles of F

Determine the number of moles of fluorine (F). The molar mass of fluorine (F) is approximately 19.00 g/mol, but keep in mind that fluorine exists as diatomic molecules (F extsubscript{2}), so we consider it as 38.00 g/mol.\[ ext{Moles of F} = \frac{0.152 ext{ g}}{19.00 ext{ g/mol}} \approx 0.00800 ext{ moles of F}\]

Determine mole ratio

Determine the simplest whole-number mole ratio of xenon to fluorine. This can be done by dividing the moles of each element by the smallest number of moles calculated above:\[ ext{Ratio of Xe:} \frac{0.00401}{0.00401} \approx 1 \]\[ ext{Ratio of F:} \frac{0.00800}{0.00401} \approx 2 \]

Write the empirical formula

Using the ratio from Step 5, write the empirical formula for the compound. Since the mole ratio is approximately 1:2, the empirical formula is \( \text{XeF}_2 \).

Key concepts

Mole Calculation

Understanding mole calculation is an essential aspect of chemical analysis. It refers to determining the number of moles in a given mass of substance, which is fundamental in understanding chemical reactions and formulas. To calculate moles, you need to know the mass of the substance and its molar mass (the mass of one mole of a substance). This is given by the formula:

• Moles = \( \frac{\text{mass of substance (g)}}{\text{molar mass (g/mol)}} \)

In our exercise, we calculated moles of xenon ( \( \text{Xe} \)) in the compound by using its molar mass, approximately 131.29 g/mol. With 0.526 g of xenon in our sample, we found that it equates to approximately 0.00401 moles of Xe. It’s important to follow this approach to ensure our chemical equations balance correctly.

Mass Conversion

Mass conversion is the process of determining the mass components of a compound from given data. It allows you to bridge the gap between different measurements and understand the composition of compounds. In chemistry, it often involves converting between mass and moles.
In our exercise, understanding mass conversion helped us find how much fluorine combined with xenon by using the difference in mass.

• The compound’s total mass is 0.678 g.
• Mass of xenon used is 0.526 g.
• By subtraction, the mass of fluorine is 0.152 g.

This step is crucial, as it tells us what proportion of the compound each element constitutes.

Chemical Compound Analysis

Chemical compound analysis involves identifying the elemental composition and structure of a given compound. It’s crucial in determining empirical formulas, which depict the simplest ratio of elements within a compound.
By analyzing our xenon-fluorine compound, we derived essential data: the mass of xenon and fluorine contributed individually to the overall mass. From this, chemistry enthusiasts can calculate the mass and moles of each component, leading to understanding the compound’s structure. Ultimately, through accurate calculations, you'll determine the empirical formula, showing the mole ratio of the compound.

Stoichiometry

Stoichiometry is the field of chemistry that deals with the quantitative relationships of the elements and compounds involved in chemical reactions. It uses the balanced chemical equation to determine the relationships among reactants and products.
Through stoichiometry, in our exercise, we identified the mole ratio of xenon to fluorine by comparing their moles. We used the smallest number of moles calculated to form a proportional relationship:

• Xenon: \( \frac{0.00401 \text{ mol}}{0.00401 \text{ mol}} = 1 \)
• Fluorine: \( \frac{0.00800 \text{ mol}}{0.00401 \text{ mol}} \approx 2 \)

This process, known as determining stoichiometric coefficients, led us to conclude that the empirical formula of the compound is \( \text{XeF}_2 \), indicating 1 xenon atom for every 2 fluorine atoms.