Question

A given sample of a xenon fluoride compound contains molecules of the type \(\mathrm{XeF}_{n},\) where \(n\) is some whole number. Given that \(9.03 \times 10^{20}\) molecules of \(\mathrm{XeF}_{n}\) weigh \(0.368 \mathrm{g},\) determine the value for \(n\) in the formula.

Short answer

Given the weight and the number of molecules of a xenon fluoride compound, we can find the value of \(n\) in the formula XeF_n by following these steps: 1. Calculate the moles of XeF_n: Moles = \(\frac{9.03 \times 10^{20} \textrm{ molecules}}{6.022 \times 10^{23} \textrm{ molecules/mol}}\) 2. Calculate the molar mass of XeF_n: Molar mass = \(\frac{0.368 \textrm{ g}}{\textrm{moles of XeF}_n}\) 3. Set up an equation to find \(n\): Molar mass of XeF_n = 131.29 g/mol + n * 19.00 g/mol 4. Rearrange the equation and solve for \(n\): n = \(\frac{\textrm{Calculated molar mass} - 131.29 \textrm{ g/mol}}{19.00 \textrm{ g/mol}}\) Plug in the calculated molar mass from step 2, solve for \(n\), and obtain the value of \(n\) in the molecular formula XeF_n.

Step-by-step solution

Calculate the moles of XeF_n

First, we need to calculate the number of moles of XeF_n molecules present in the given sample. We will use the following formula: Moles = (Number of molecules) / (Avogadro's number) Where Avogadro's number is \( 6.022 \times 10^{23} \) molecules/mol. Let's calculate the moles of XeF_n in the sample: Moles of XeF_n = \(\frac{9.03 \times 10^{20} \textrm{ molecules}}{6.022 \times 10^{23} \textrm{ molecules/mol}}\)

Calculate the molar mass of XeF_n

To calculate the molar mass of XeF_n, we can use the following formula: Molar mass = \(\frac{\textrm{mass}}{\textrm{moles}}\) Given that the mass of the sample is 0.368 g, let's calculate the molar mass of XeF_n: Molar mass of XeF_n = \(\frac{0.368 \textrm{ g}}{\textrm{moles of XeF}_n}\) (which we calculated in step 1)

Calculate the value of n

Now we can write the molar mass of XeF_n as the sum of the molar masses of xenon (Xe) and n times the molar mass of fluorine (F): Molar mass of XeF_n = Molar mass of Xe + n * Molar mass of F The molar masses of xenon and fluorine are 131.29 g/mol and 19.00 g/mol, respectively. Now we can set up an equation to solve for n: Molar mass of XeF_n = 131.29 g/mol + n * 19.00 g/mol Now, substitute the molar mass of XeF_n obtained in step 2 and solve for n: \(\textrm{Calculated molar mass} = 131.29 \textrm{ g/mol} + n * 19.00 \textrm{ g/mol}\) Rearrange the equation to solve for n: n = \(\frac{\textrm{Calculated molar mass} - 131.29 \textrm{ g/mol}}{19.00 \textrm{ g/mol}}\)

Determine the value of n

Now, plug the calculated molar mass of XeF_n from step 2 into the equation above and solve for n. The value of n should be a whole number. After solving for n, you will find the value of n in the molecular formula of the xenon fluoride compound, XeF_n.

Key concepts

Avogadro's Number

Understanding Avogadro's number is fundamental when dealing with quantities in chemistry. Defined as approximately \( 6.022 \times 10^{23} \) entities per mole, it's the bridge between the microscopic world of atoms and molecules and the macroscopic world that we can measure in the laboratory.

When a problem, like the xenon fluoride compound exercise, gives the number of molecules, we use Avogadro's number to convert this figure to moles—a standard unit in chemistry that corresponds to Avogadro's number of particles, whether they are atoms, ions, or molecules. This conversion is essential in stoichiometry, allowing us to relate quantities of reactants and products in chemical reactions. Simply put, Avogadro's number helps us quantify substances on a scale that's practical for scientific studies.

Molecular Formula

The molecular formula reveals the types and numbers of atoms present in a molecule. In the case of xenon fluoride \( \mathrm{XeF}_{n} \), 'Xe' stands for xenon, 'F' for fluorine, and 'n' represents the number of fluorine atoms bonded to a xenon atom.

To deduce the correct value of 'n,' we utilize the actual mass of a sample and combine it with Avogadro's number and the molar masses of the individual elements. By understanding the molecular formula, we gain insights into the compound's structure, properties, and how it reacts with other chemical species.

Stoichiometry

Stoichiometry is a section of chemistry that deals with the quantitative relationships between the substances involved in chemical reactions. It's predicated on the conservation of mass and the concept of the mole.

By using stoichiometry in the exercise, we determine the number of moles of \( \mathrm{XeF}_{n} \) and then calculate the compound’s molar mass to ultimately find 'n.' Stoichiometry is not just theoretical; it's what chemists use to predict the amounts of products formed in a reaction from a given amount of reactants and vice versa. It's the roadmap for understanding chemical equations and the very heartbeat of chemical analysis and synthesis.

Xenon Fluoride

Xenon fluoride compounds are an interesting subject as xenon is a noble gas, which traditionally doesn’t form many compounds. However, under the right conditions, xenon can react with fluorine to create various xenon fluoride compounds, such as \( \mathrm{XeF}_{2} \) , \( \mathrm{XeF}_{4} \) , and \( \mathrm{XeF}_{6} \).

The exercise at hand involves a xenon fluoride compound with an unknown number of fluorine atoms. By determining the correct 'n' in the molecular formula \( \mathrm{XeF}_{n} \), we learn not just about the compound's stoichiometry but also its possible physical and chemical properties, which are governed by the types and numbers of atoms present.