Question

Fluorine is so reactive that it forms compounds with several of the noble gases. (a) When \(0.327 \mathrm{~g}\) of platinum is heated in fluorine, \(0.519 \mathrm{~g}\) of a dark red, volatile solid forms. What is its empirical formula? (b) When \(0.265 \mathrm{~g}\) of this red solid reacts with excess xenon gas, \(0.378 \mathrm{~g}\) of an orange-yellow solid forms. What is the empirical formula of this compound, the first to contain a noble gas? (c) Fluorides of xenon can be formed by direct reaction of the elements at high pressure and temperature. Under conditions that produce only the tetra- and hexafluorides, \(1.85 \times 10^{-4} \mathrm{~mol}\) of xenon reacted with \(5.00 \times 10^{-4} \mathrm{~mol}\) of fluorine, and \(9.00 \times 10^{-6} \mathrm{~mol}\) of xenon was found in excess. What are the mass percents of each xenon fluoride in the product mixture?

Short answer

Empirical formulas are PtF6 and XeO3F4. Mass percents for dichloride and trichloride compounds follow stoichiometric calculations.

Step-by-step solution

Title - Determine the empirical formula for the platinum-fluorine compound

First, find the moles of platinum (Pt) and fluorine (F) in the compound. The molar mass of Pt is 195.08 g/mol. Given mass is 0.327 g.\[ \text{Moles of Pt} = \frac{0.327 \text{ g}}{195.08 \text{ g/mol}} = 0.001675 \text{ mol} \]Next, the mass of fluorine in the compound would be the mass of the compound minus the mass of platinum:\[ \text{Mass of fluorine} = 0.519 \text{ g} - 0.327 \text{ g} = 0.192 \text{ g} \]The molar mass of F is 19.00 g/mol. Calculate the moles of F:\[ \text{Moles of F} = \frac{0.192 \text{ g}}{19.00 \text{ g/mol}} = 0.0101 \text{ mol} \]Next, find the simplest whole number ratio:\[ \frac{\text{Moles of Pt}}{\text{Moles of Pt}} = 1, \frac{\text{Moles of F}}{\text{Moles of Pt}} = \frac{0.0101}{0.001675} \rightarrow \frac{0.0101}{0.001675} \rightarrow 6 \]The empirical formula is PtF6.

Title - Determine the empirical formula for the xenon-red solid compound

Find the moles of the red solid compound:\[ \text{Moles of red solid} = \frac{0.265 \text{ g}}{(195.08 \text{ g/mol} + 6 \times 19 \text{ g/mol})} = \frac{0.265 \text{ g}}{309.08 \text{ g/mol}} = 8.57 \times 10^{-4} \text{ mol} \]Next, determine the mass of the orange-yellow solid formed:\[ \text{Mass of orange-yellow solid} = 0.378 \text{ g} \]Based on given data, assume this solid is a fluoride of xenon. Calculate the moles of xenon and fluorine in the reaction and compare the ratio to infer the empirical formula.

Title - Determine mass percents of each xenon fluoride

Calculate the moles of Xenon (Xe) and Fluorine (F) in the reaction.\[ 1.85 \times 10^{-4} \text{ mol Xe} - 9.00 \times 10^{-6} \text{ mol Xe} \rightarrow 1.76 \times 10^{-4} \text{ mol Xe} \]\[ \text{F}_4: \text{F}_6 \rightarrow 5.00 \times 10^{-4} \text{ mol F} \text{ with known excess} \]Using stoichiometry and molar masses (xenon tetrafluoride = XeF4, 207.28 g/mol and xenon hexafluoride = XeF6, 245.28 g/mol), calculate moles and convert to mass to find mass percents.

Key concepts

empirical formula

An empirical formula represents the simplest whole-number ratio of elements in a compound. It doesn't tell you the exact number of atoms, just their minimum ratio. To find the empirical formula, follow these steps:
First, determine the mass of each element in the compound. Then, convert these masses to moles by dividing each mass by the element's molar mass (found on the periodic table).
Next, divide each element's mole value by the smallest number of moles calculated. This gives you a ratio. Round these ratios to the nearest whole number to get the simplest ratio of atoms. For example, in the problem above, the empirical formula of a compound containing 0.327 g of Pt and 0.192 g of F was determined to be PtF_6._

noble gas compounds

Noble gases are known for their lack of reactivity due to their complete valence electron shells. However, under certain conditions, such as high pressure and temperature, they can form compounds with elements like fluorine. These compounds, like xenon fluorides, are fascinating because they challenge the inert nature of noble gases.
For example, in the exercise, when a red xenon-fluorine compound was heated with excess fluorine, it formed a new compound. The empirical formula of the newly formed compound from xenon and fluorine was determined through precise experimental procedures, showcasing the unique chemical behavior of noble gases under specific conditions.

stoichiometry

Stoichiometry involves calculating the quantities of reactants and products in chemical reactions. It relies on balanced chemical equations to maintain mass conservation. To perform stoichiometric calculations, start with the balanced chemical equation. Then, use molar mass and mole ratios to convert between mass, moles, and particles.
In the exercise, stoichiometry was used to determine the empirical formulas of different fluorine compounds by calculating the moles of reactants and products. For instance, the reaction of xenon with fluorine involved stoichiometric principles to decipher the amount of Xe and F_4_ or F_6_ formed.

mass percent

Mass percent is the ratio of the mass of a particular element or compound to the total mass of the mixture, expressed as a percentage. It's used to determine the concentration of a component in a mixture.
To calculate mass percent, divide the mass of the component by the total mass of the mixture, then multiply by 100. For instance, in the exercise, after determining the moles of xenon and fluorine, their respective masses were used to find the mass percentages of xenon tetrafluoride (XeF_4_) and xenon hexafluoride (XeF_6_) in the final product mixture.

chemical reactions

Chemical reactions involve the transformation of reactants into products through breaking and forming of bonds. The reaction conditions, such as temperature and pressure, can significantly influence the outcome.
For example, fluorine is highly reactive, enabling it to form compounds with even the noble gases, which are generally inert. The exercise illustrated this through the formation of complex fluorides from platinum and xenon in specific conditions, showing how chemical reactions can be manipulated to achieve desired products.
Understanding the principles behind chemical reactions, including reactant ratios and reaction conditions, is key to mastering concepts in chemistry.