Problem 107 - Chemistry Principles and Practice

Question

A The compound \(\mathrm{ClF}_{3}\) reacts with solid uranium to produce the volatile compounds \(\mathrm{UF}_{6}\) and ClF. \(\mathrm{ClF}_{3}\) can be used to separate uranium from plutonium, because plutonium reacts with \(\mathrm{ClF}_{3}\) to form the nonvolatile compounds \(\mathrm{PuF}_{4}\) and ClF. (a) Write equations for these two reactions. (b) Draw the Lewis structure of \(\mathrm{ClF}_{3}\). Is there anything unusual about the Lewis structure of \(\mathrm{ClF}_{3} ?\) (c) If a mixture of uranium and plutonium reacts with excess \(\mathrm{ClF}_{3}\) to produce \(43.5 \mathrm{~g} \mathrm{UF}_{6}\) and \(22.1 \mathrm{~g} \mathrm{PuF}_{4}\), what were the masses of the uranium and plutonium in the starting mixture? Use \(244 \mathrm{~g} / \mathrm{mol}\) for the molar mass of \(\mathrm{Pu}\).

Step-by-Step Solution

Answer

29.4 g of uranium, 16.3 g of plutonium.

Step 1: Write Chemical Equations

For the reaction of \( \mathrm{ClF}_{3} \) with solid uranium, the equation is:\[ \mathrm{U} + 3\mathrm{ClF}_{3} \rightarrow \mathrm{UF}_{6} + 3\mathrm{ClF} \]For the reaction of \( \mathrm{ClF}_{3} \) with plutonium, the equation is:\[ \mathrm{Pu} + 4\mathrm{ClF}_{3} \rightarrow \mathrm{PuF}_{4} + 4\mathrm{ClF} \]

Step 2: Draw the Lewis Structure of \(\mathrm{ClF}_3\)

The Lewis structure of \(\mathrm{ClF}_3\) shows chlorine in the center with three fluorine atoms bonded to it. Chlorine has two nonbonding electron pairs. The molecule has a T-shaped structure due to the repulsion between lone pairs, which is unusual for standard covalent compounds.

Step 3: Calculate Molar Masses

The molar mass of \( \mathrm{UF}_6 \) is calculated by adding the atomic mass of uranium (\(238.03 \mathrm{~g/mol}\)) and six fluorine atoms (\(6 \times 18.9984 \mathrm{~g/mol}\)) yielding \(352.02 \mathrm{~g/mol}.\)The molar mass of \( \mathrm{PuF}_4 \) is calculated by adding the atomic mass of plutonium (\(244 \mathrm{~g/mol}\)) and four fluorine atoms (\(4 \times 18.9984 \mathrm{~g/mol}\)) yielding \(331.99 \mathrm{~g/mol}.\)

Step 4: Calculate Moles of \(\mathrm{UF}_6\) and \(\mathrm{PuF}_4\)

Given 43.5 g of \( \mathrm{UF}_6 \) and 22.1 g of \( \mathrm{PuF}_4 \), calculate the moles:\[ \text{Moles of } \mathrm{UF}_6 = \frac{43.5 \text{ g}}{352.02 \text{ g/mol}} = 0.1235 \text{ mol} \]\[ \text{Moles of } \mathrm{PuF}_4 = \frac{22.1 \text{ g}}{331.99 \text{ g/mol}} = 0.0666 \text{ mol} \]

Step 5: Calculate Mass of Uranium and Plutonium

Using balanced equations, 1 mole of \(\mathrm{UF}_6\) is produced from 1 mole of uranium, meaning 0.1235 mol of uranium. Since the molar mass of uranium is \(238.03 \mathrm{~g/mol}\), the mass of uranium is:\[ \text{Mass of Uranium} = 0.1235 \text{ mol} \times 238.03 \text{ g/mol} = 29.4 \text{ g} \]Similarly, using \(\mathrm{PuF}_4\), 1 mole is produced per mole of plutonium, leading to 0.0666 mol of plutonium. With \(244 \mathrm{~g/mol}\) as the molar mass, the mass of plutonium is:\[ \text{Mass of Plutonium} = 0.0666 \text{ mol} \times 244 \text{ g/mol} = 16.3 \text{ g} \]

Key Concepts

Lewis Structures

Lewis structures are diagrams that represent the bonds and lone pairs of electrons in a molecule. To draw the Lewis structure for a molecule, you must:

• Count the total number of valence electrons from all the atoms involved.
• Arrange the atoms to show specific connections — often, a single atom acts as the central atom.
• Distribute the electrons around the atoms, starting with bonds and then lone pairs, to satisfy the octet rule wherever possible (though there are exceptions).

For \(\mathrm{ClF}_3\), chlorine is the central atom bonded with three fluorine atoms. It has a total of 28 valence electrons. This bond configuration is notable for having a T-shaped geometry due to two lone pairs on the central chlorine atom. This is unusual because it deviates from the standard octet structure, as chlorine can exceed an octet, thus accommodating more than eight valence electrons.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It is based on the mole ratio from the balanced chemical equation. Here's how it's used:

• **Mole Ratio:** From the balanced equation, determine the mole ratio between reactants and products. For example, in the \(\mathrm{U} + 3\mathrm{ClF}_3 \rightarrow \mathrm{UF}_6 + 3\mathrm{ClF}\), 1 mole of uranium reacts with 3 moles of \(\mathrm{ClF}_3\) to produce 1 mole of \(\mathrm{UF}_6\).
• **Convert to Moles:** Use the given mass of a product or reactant to find moles, using molar mass as a conversion factor.
• **Calculate Masses:** Once moles are known, convert back to grams if needed, to find the required mass of other substances participating in the reaction.

Stoichiometry allows the prediction of amounts of substances consumed and produced, crucial in determining how much uranium reacted in the \(\mathrm{UF}_6\) production.

Molecular Geometry

Molecular geometry is the three-dimensional arrangement of atoms in a molecule, influencing properties like polarity and reactivity. Common shapes include linear, tetrahedral, and trigonal planar.

• **VSEPR Theory:** This theory (Valence Shell Electron Pair Repulsion) is used to predict molecular shapes based on repulsion between electron pairs. \(\mathrm{ClF}_3\) is T-shaped, with two lone pairs causing the bonded atoms to repel each other.
• **Shape Effect:** The T-shaped arrangement of \(\mathrm{ClF}_3\) affects factors such as bond angles and molecular polarity, impacting how it interacts in chemical processes.

Understanding molecular geometry helps explain the stability and reactivity of molecules in a given environment.

Nuclear Chemistry

Nuclear chemistry focuses on the reactions and properties of atomic nuclei, often involving changes in an atom's nucleus resulting in a different element or isotope.

• **Reactivity Involving Nuclear Processes:** While not directly applicable in the given reaction, understanding nuclear characteristics is essential, especially when dealing with uranium and plutonium.
• **Isotopes and Stability:** Uranium and plutonium are vital in nuclear chemistry as they can exist in different isotopic forms and exhibit nuclear behavior under specific conditions – critical for energy generation and separation of nuclear materials.

This field's principles are crucial when considering the separation of nuclear materials like uranium from plutonium using chemical reactions such as those involving \(\mathrm{ClF}_3\).