Problem 115 - Chemistry and Chemical Reactivity

Question

Uranium is used as a fuel, primarily in the form of uranium(IV) oxide, in nuclear power plants. This question considers some uranium chemistry. (a) A small sample of uranium metal \((0.169 \mathrm{g})\) is heated to between 800 and \(900^{\circ} \mathrm{C}\) in air to give \(0.199 \mathrm{g}\) of a dark green oxide, \(\mathrm{U}_{x} \mathrm{O}_{y}\). How many moles of uranium metal were used? What is the empirical formula of the oxide, \(\mathrm{U}_{x} \mathrm{O}_{y} ?\) What is the name of the oxide? How many moles of \(\mathrm{U}_{x} \mathrm{O}_{y}\) must have been obtained? (b) The naturally occurring isotopes of uranium are \(^{234} \mathrm{U}\) " \(^{235} \mathrm{U},\) and \(^{238} \mathrm{U} .\) Knowing that uranium's atomic weight is \(238.02 \mathrm{g} / \mathrm{mol},\) which isotope must be the most abundant? (c) If the hydrated compound \(\mathrm{UO}_{2}\left(\mathrm{NO}_{3}\right)_{2} \cdot z \mathrm{H}_{2} \mathrm{O}\) is heated gently, the water of hydration is lost. If you have \(0.865 \mathrm{g}\) of the hydrated compound and obtain \(0.679 \mathrm{g}\) of \(\mathrm{UO}_{2}\left(\mathrm{NO}_{3}\right)_{2}\) on heating, how many molecules of water of hydration are in each formula unit of the original compound? (The oxide \(\mathrm{U}_{x} \mathrm{O}_{y}\) is obtained if the hydrate is heated to temperatures over \(800^{\circ} \mathrm{C}\) in the air.)

Step-by-Step Solution

Answer

(a) Empirical formula: \(UO_3\), oxide name: uranium trioxide, moles of oxide: 0.000710 mol. (b) Most abundant isotope: \(^{238}U\). (c) Water molecules per unit: 6.

Step 1: Calculate Moles of Uranium Metal

First, we need to calculate the number of moles of uranium metal used. Given the mass is 0.169 g and the atomic weight of uranium is approximately 238.02 g/mol, the number of moles of uranium can be calculated using the formula:\[\text{Moles of U} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.169 \, \text{g}}{238.02 \, \text{g/mol}}\]Thus, compute the number of moles:\[\text{Moles of U} = 0.000710 \text{ mol}\]

Step 2: Determine Moles of Oxygen in the Oxide

To determine the moles of oxygen in the uranium oxide, find the mass of oxygen in the oxide. Subtract the initial mass of uranium from the final mass of the oxide:\[\text{Mass of O} = 0.199 \, \text{g (oxide)} - 0.169 \, \text{g (U)} = 0.030 \, \text{g}\]Now, calculate the number of moles of oxygen using the molar mass of oxygen (16.00 g/mol):\[\text{Moles of O} = \frac{0.030 \, \text{g}}{16.00 \, \text{g/mol}} = 0.001875 \, \text{mol}\]

Step 3: Calculate Empirical Formula of the Oxide

The ratio of moles of uranium to moles of oxygen is given by the moles calculated in Steps 1 and 2. Divide each by the smallest number to simplify the ratio:\[\text{U:O} = \frac{0.000710}{0.000710} : \frac{0.001875}{0.000710} = 1 : 2.64 \approx 1 : 3\]Thus, the empirical formula is approximately \(UO_3\).

Step 4: Identify the Name of the Oxide

Since the empirical formula is approximately \(UO_3\), the compound is often referred to as uranium trioxide.

Step 5: Determine Moles of Oxide Formed

Since the entire oxide is formed from the transformation of uranium, and no uranium is lost, the moles of uranium in \(UxOy\) equates to the original moles of U. Thus, the number of moles of \(UxOy\) is:\[\text{Moles of } UxOy = 0.000710 \, \text{mol}\]

Step 6: Identify the Most Abundant Uranium Isotope

The atomic weight of uranium is closer to 238, indicating that \(^{238}U\) is the most abundant isotope, as it contributes more to the average atomic weight of uranium.

Step 7: Calculate Moles of Water Lost on Heating

Determine the mass of water lost by subtracting the mass of the dehydrated compound from the hydrated compound:\[\text{Mass of } H_2O = 0.865 \, \text{g} - 0.679 \, \text{g} = 0.186 \, \text{g}\]Calculate moles of water using its molar mass (18.02 g/mol):\[\text{Moles of } H_2O = \frac{0.186 \, \text{g}}{18.02 \, \text{g/mol}} = 0.01032 \, \text{mol}\]

Step 8: Determine Number of Water Molecules (z)

The moles of the dehydrated subpart \(UO_2(NO_3)_2\) is found using its mass and molar mass (394.05 g/mol):\[\text{Moles of } UO_2(NO_3)_2 = \frac{0.679 \, \text{g}}{394.05 \, \text{g/mol}} = 0.001723 \, \text{mol}\]Calculate the number of water molecules (z) per formula unit by dividing the moles of water by the moles of \(UO_2(NO_3)_2\):\[z = \frac{0.01032 \, \text{mol of } H_2O}{0.001723 \, \text{mol of } UO_2(NO_3)_2} \approx 6\]

Step 9: Conclusion

Therefore, the empirical formula of the oxide is \(UO_3\), the most abundant isotope is \(^{238}U\), and there are approximately 6 water molecules of hydration per formula unit of the original compound.

Key Concepts

Moles Calculation

In chemistry, calculating moles is a fundamental concept that helps us understand the quantity of a substance in chemical reactions. The mole is a unit used to measure the amount of a substance and is defined as the amount of any chemical substance that contains the same number of entities as there are atoms in 12 grams of pure carbon-12. This number is Avogadro's number, approximately \(6.022 \times 10^{23}\). To calculate the number of moles, the formula used is: - \(\text{Moles} = \frac{\text{mass}}{\text{molar mass}}\) For example, to calculate the moles of uranium metal used in the original problem, we use its given mass \(0.169 \, \text{g}\) and its molar mass \(238.02 \, \text{g/mol}\). By dividing the mass by its molar mass, we calculate the moles of uranium: - \(\text{Moles of U} = \frac{0.169 \, \text{g}}{238.02 \, \text{g/mol}} = 0.000710 \, \text{mol}\) Understanding how to calculate moles is crucial in stoichiometry, which involves using relationships between reactants and products in a chemical reaction.

Empirical Formula

An empirical formula represents the simplest whole-number ratio of elements in a compound, giving insight into the kinds, not the amounts, of atoms in a molecule. It’s an essential tool in chemistry for identifying unknown compounds. To determine the empirical formula from the given data, you need the moles of each element. In the problem, we had to find the empirical formula of the dark green oxide formed from uranium. Here’s how you do it: - First, calculate the moles of uranium (already calculated previously as 0.000710 mol). - Next, determine the moles of oxygen by subtracting the mass of uranium used from the total mass of the oxide, and then divide by the molar mass of oxygen. The calculation showed that the ratio of uranium to oxygen is approximately 1:3, yielding an empirical formula of \(UO_3\). This highlights the importance of simplifying mole ratios to get a clear understanding of the compound’s composition.

Isotopes

Isotopes are variants of a particular chemical element that have different numbers of neutrons, and thus different mass numbers, but the same number of protons. This means they retain the chemical properties of the element but differ in atomic mass and physical properties. In the case of uranium, the naturally occurring isotopes are \(^{234} \text{U}\), \(^{235} \text{U}\), and \(^{238} \text{U}\). Understanding isotopes is vital in nuclear chemistry because different isotopes have different stabilities and nuclear properties. In the exercise, we deduced that the isotope \(^{238} \text{U}\) is the most abundant, as its atomic mass of 238 contributes the most to the average atomic weight of uranium, which is \(238.02 \, \text{g/mol}\). This insight is crucial for applications involving isotopes, such as nuclear reactors where specific isotopes, like \(^{235} \text{U}\), are used as fuel.

Hydration Water Molecules

Compounds that contain water molecules within their crystalline structure are known as hydrates. These water molecules, referred to as water of hydration, play a significant role in the formation and stability of the crystal lattice. When hydrates are gently heated, the water is "driven off," resulting in an anhydrous compound with a lower mass. In the example problem, we began with \(0.865 \, \text{g}\) of the hydrate and ended with \(0.679 \, \text{g}\) of the anhydrous compound \(UO_2(NO_3)_2\). To find the number of hydration water molecules (z) per formula unit, calculate the moles of water lost and divide by the moles of the dehydrated compound. This process revealed approximately 6 molecules of water per unit, indicating the original compound's composition. Understanding hydration is essential, especially in solid state chemistry, as it affects the properties and usability of chemical compounds.