Problem 156 - 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 g) is heated to between 800 and \(900^{\circ} \mathrm{C}\) in air to give 0.199 g of a dark green oxide, Ux \(\mathrm{O}_{y}\) 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 \(^{238}\) 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 waters 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 \(\left.800^{\circ} \mathrm{C} \text { in the air. }\right)\)

Step-by-Step Solution

Answer

(a) The empirical formula is UO₃; the oxide is uranium trioxide. (b) ^238U is the most abundant isotope. (c) There are 6 waters of hydration (z=6).

Step 1: Calculating Moles of Uranium Metal

To find the moles of uranium, we use the equation \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). The mass of uranium metal given is 0.169 g. Using uranium's atomic weight (molar mass) of 238.02 g/mol, the moles of uranium is calculated as follows: \[ \text{Moles of } \text{U} = \frac{0.169 \text{ g}}{238.02 \text{ g/mol}} \approx 0.000710 \text{ mol} \]

Step 2: Determining Empirical Formula of the Oxide

The oxide has a mass of 0.199 g. Uranium's moles are already calculated as 0.000710 mol. To find the moles of oxygen, we subtract the mass of uranium from the total mass of the oxide: \[ \text{Mass of Oxygen} = 0.199 \text{ g} - 0.169 \text{ g} = 0.03 \text{ g} \]The molar mass of oxygen (O) is 16.00 g/mol, so the moles of oxygen in the oxide is:\[ \text{Moles of O} = \frac{0.03 \text{ g}}{16.00 \text{ g/mol}} \approx 0.001875 \text{ mol} \]To find the empirical formula, we need the mole ratio of uranium to oxygen. Dividing by the smallest number of moles gives:\[ \text{U:O ratio} = \frac{0.000710}{0.000710} : \frac{0.001875}{0.000710} \approx 1:2.64 \]Rounding gives a ratio of 1:3, so the empirical formula is \( \text{UO}_3 \).

Step 3: Naming the Oxide

The oxide with the empirical formula \( \text{UO}_3 \) is named "uranium trioxide."

Step 4: Calculating Moles of Uranium Oxide Produced

To find out how many moles of \( \text{UO}_3 \) are formed, divide the mass of the oxide by its molar mass. The molar mass of \( \text{UO}_3 \) is calculated as:\[ 238.02 + (3 \times 16.00) = 286.02 \text{ g/mol} \]Using the mass of the oxide:\[ \text{Moles of } \text{UO}_3 = \frac{0.199 \text{ g}}{286.02 \text{ g/mol}} \approx 0.000696 \text{ mol} \]

Step 5: Identifying Most Abundant Isotope

Since the atomic weight is close to the mass number of \({}^{238}\text{U} \), \({}^{238}\text{U} \) is the most abundant isotope.

Step 6: Finding Waters of Hydration

The hydrated compound was 0.865 g, which reduced to 0.679 g of anhydrous \( \text{UO}_2\left(\text{NO}_3\right)_2 \). The lost mass is due to water:\[ \text{Mass of water} = 0.865 \text{ g} - 0.679 \text{ g} = 0.186 \text{ g} \]The molar mass of water (\( \text{H}_2\text{O} \)) is 18.02 g/mol, so the moles of water:\[ \text{Moles of water} = \frac{0.186 \text{ g}}{18.02 \text{ g/mol}} \approx 0.01032 \text{ mol} \]The moles of the compound can be determined with:\[ \text{Moles of } \text{UO}_2\left(\text{NO}_3\right)_2 = \frac{0.679 \text{ g}}{238.02 + 2(62.00)} \approx 0.00169 \text{ mol} \]The ratio of water to the compound is:\[ z \approx \frac{0.01032}{0.00169} \approx 6.11 \]Rounding to the nearest whole number, \( z = 6 \). Therefore, there are 6 waters of hydration.

Key Concepts

Nuclear Fuel

Uranium's role as a nuclear fuel is crucial to the functioning of nuclear power plants. This element, primarily used in the form of uranium(IV) oxide, undergoes a nuclear reaction known as fission. In nuclear fission, uranium atoms split into smaller parts, releasing a tremendous amount of energy, which is utilized to generate electricity. While there are different types of nuclear fuel, uranium's unique properties make it a popular choice.

Some key advantages of using uranium include:

• High energy yield: Uranium is highly efficient, producing more energy per gram compared to traditional fossil fuels.
• Long life span: Nuclear plants using uranium can operate for long periods before requiring refueling.
• Reduced greenhouse gases: Unlike fossil fuels, nuclear fission does not produce carbon dioxide during electricity generation.

Understanding uranium's role as a nuclear fuel can help students appreciate its importance in today's energy landscape.

Empirical Formula

In chemistry, understanding an empirical formula is vital, as it reveals the simplest ratio of elements in a compound. For example, in the problem we discussed, UO e is determined as the empirical formula of the oxide. The empirical formula represents the lowest whole number that describes the proportion of uranium to oxygen in this compound.

To determine an empirical formula:

• Find the moles of each element present in a combustion or reaction process.
• Divide the amount of moles of each element by the smallest number obtained.
• Express the ratio in its simplest form, rounding to the nearest whole number if needed.

By understanding the concept of an empirical formula, students can predict the composition and proportions of elements in chemical reactions.

Isotopes of Uranium

Uranium has several isotopes, each with the same number of protons but a different number of neutrons. The naturally occurring isotopes include \(^{234} ext{U}\), \(^{235} ext{U}\), and \(^{238} ext{U}\). Among these, \(^{238} ext{U}\) is the most abundant isotope, accounting for more than 99% of natural uranium.

Understanding isotopes is crucial because:

• Different isotopes have different nuclear stability, which affects their radioactivity and suitability as nuclear fuel.
• \(^{235} ext{U}\) is particularly important for nuclear power, as it is fissile – meaning it can sustain a nuclear chain reaction.
• Enrichment processes increase \(^{235} ext{U}\) concentration to make it viable for use in nuclear reactors.

Studying isotopes provides insight into the nuances of nuclear chemistry and the versatility of uranium.

Hydrated Compounds in Chemistry

Hydrated compounds contain water molecules within their crystal structure, an essential concept in understanding how chemicals behave. In the exercise, we explored the hydrated compound UO e is determined as the empirical formula of the oxide. The empirical formula represents the lowest whole number that describes the proportion of uranium to oxygen in this compound. The process of gently heating the compound to remove water is called dehydration.

Key points about hydrated compounds:

• They are important in various chemical reactions and processes, where water molecules adjust the compound's physical and chemical properties.
• The amount of water in a hydrated compound can affect its weight, and can be determined by calculating the water of hydration.
• Hydrated compounds often change color or texture when heated, aiding identification and categorization in labs.

By understanding hydrated compounds, students can better grasp the behavior of compounds in different environmental conditions.