Problem 131 - Chemistry The Molecular Science

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 \(900{ }^{\circ} \mathrm{C}\) 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 oxide \(\mathrm{U}_{x} \mathrm{O}_{y}\) is obtained if \(\mathrm{UO}_{2} \mathrm{NO}_{3} \cdot n \mathrm{H}_{2} \mathrm{O}\) is heated to temperatures greater than \(800{ }^{\circ} \mathrm{C}\) in air. However, if you heat it gently, only the water of hydration is lost. If you have \(0.865 \mathrm{~g} \mathrm{UO}_{2} \mathrm{NO}_{3} \cdot n \mathrm{H}_{2} \mathrm{O}\) and obtain \(0.679 \mathrm{~g}\) \(\mathrm{UO}_{2} \mathrm{NO}_{3}\) on heating, how many molecules of water of hydration were there in each formula unit of the original compound?

Step-by-Step Solution

Answer

The empirical formula of the oxide is \(\mathrm{UO}_3\) and it is called Uranium(VI) oxide. Four water molecules are in the original hydrated compound.

Step 1: Calculate Moles of Uranium Metal

To find the moles of uranium, use the molar mass of uranium (U), which is approximately \( 238.03 \, \text{g/mol} \). The mass of uranium metal is given as \( 0.169 \, \text{g} \). \[\text{Moles of U} = \frac{0.169 \, \text{g}}{238.03 \, \text{g/mol}} \approx 0.00071 \, \text{mol}\]

Step 2: Calculate Moles of Oxide Formed

Since the uranium reacts with oxygen to form \( \mathrm{U}_x \mathrm{O}_y \), the total mass is given as \( 0.199 \, \text{g} \). To find the moles of oxide formed, simplify the mole ratio calculation:Mass of O in the compound: \[0.199 \, \text{g} - 0.169 \, \text{g} = 0.030 \, \text{g} \text{ of O}\]Moles of O:\[\frac{0.030 \, \text{g}}{16.00 \, \text{g/mol}} \approx 0.00188 \, \text{mol}\]

Step 3: Determine Empirical Formula of the Oxide

The mole ratio of uranium to oxygen determines the empirical formula. From Step 1, moles of U = 0.00071 mol and moles of O = 0.00188 mol. Divide both by the smallest number of moles:\[\text{U ratio} = \frac{0.00071}{0.00071} = 1\]\[\text{O ratio} = \frac{0.00188}{0.00071} \approx 2.65\]Since the O ratio rounds to 3, the empirical formula is \( \mathrm{U}\mathrm{O}_3 \). The name of this oxide is Uranium(VI) oxide.

Step 4: Moles of \( \mathrm{U} \mathrm{O}_3 \) Formed

Since each \( \mathrm{U} \mathrm{O}_3 \) molecule contains one uranium atom (moles of U = moles of \( \mathrm{U} \mathrm{O}_3 \)), moles of \( \mathrm{U}_3 \mathrm{O}_y \) = 0.00071 mol.

Step 5: Water of Hydration Calculation

For part (b), the initial mass of \( \mathrm{UO}_2\mathrm{NO}_3 \cdot n\mathrm{H}_2\mathrm{O} \) = 0.865 g and final mass of \( \mathrm{UO}_2\mathrm{NO}_3 \) = 0.679 g.The mass of water lost:\[0.865 \, \text{g} - 0.679 \, \text{g} = 0.186 \, \text{g}\text{ (H}_2\mathrm{O}\text{)}\]Moles of water lost:\[\frac{0.186 \, \text{g}}{18.02 \, \text{g/mol}} \approx 0.01032 \, \text{mol}\text{ (H}_2\mathrm{O}\text{)}\]Moles of \( \mathrm{UO}_2\mathrm{NO}_3 \) = \( \frac{0.679 \, \text{g}}{284.05 \, \text{g/mol}} \approx 0.00239 \, \text{mol} \).Water-to-compound mole ratio:\[\frac{0.01032 \, \text{mol} \text{ (H}_2\mathrm{O}\text{)}}{0.00239 \, \text{mol}\text{ (compound)}} \approx 4.32\]This rounds to 4, so \( n = 4 \) water molecules per formula unit.

Key Concepts

Empirical Formula

Understanding the empirical formula is crucial in chemistry because it helps us determine the simplest ratio of different atoms in a compound. In this exercise, we observe a reaction where uranium metal is heated to form an oxide, represented as \( \mathrm{U}_{x} \mathrm{O}_{y} \). To find the empirical formula, we calculate the moles of each element involved and determine their simplest whole number ratio.

First, we calculate the moles of uranium metal used. Given that the mass is \( 0.169 \text{ g} \) and the molar mass of uranium is approximately \( 238.03 \text{ g/mol} \), we use the formula:

• \( \text{Moles of U} = \frac{0.169}{238.03} \approx 0.00071 \text{ mol} \)

Next, we find the moles of oxygen. The total mass of the oxide is \( 0.199 \text{ g} \), and the mass of oxygen is the difference:

• \( 0.199 \text{ g} - 0.169 \text{ g} = 0.030 \text{ g} \)

Convert this mass to moles:

• \( \frac{0.030 \text{ g}}{16.00 \text{ g/mol}} \approx 0.00188 \text{ mol} \)

To find the empirical formula, compare the moles of uranium and oxygen by dividing both by the smallest value:

• U to O ratio is: \( \frac{0.00071}{0.00071} = 1 \) and \( \frac{0.00188}{0.00071} \approx 2.65 \)

When we round to whole numbers, the O ratio becomes 3, resulting in an empirical formula of \( \mathrm{UO}_3 \). This compound is known as Uranium(VI) oxide.

Moles Calculation

Calculating moles is a fundamental process in chemistry that allows us to relate mass to the number of particles. In this exercise, we start by calculating the number of moles of the uranium sample and the oxide it forms.

To begin, we use the mass and atomic weight of uranium to find the moles of uranium metal. Using the formula:

• \( \text{Moles of U} = \frac{0.169 \text{ g}}{238.03 \text{ g/mol}} \approx 0.00071 \text{ mol} \)

For the oxide, we focus on the mass change due to oxygen. The oxygen mass involved is calculated by subtracting the original uranium mass from the oxide's total mass:

• \( \text{Mass of O} = 0.199 \text{ g} - 0.169 \text{ g} = 0.030 \text{ g} \)

Convert the oxygen mass to moles:

• \( \text{Moles of O} = \frac{0.030 \text{ g}}{16.00 \text{ g/mol}} \approx 0.00188 \text{ mol} \)

The empirical formula development hinges on these calculations. Once we establish the moles of uranium and oxygen, we determine their ratio to find the simplest whole-number ratio. This ensures the empirical formula reflects the atomic proportions in the compound. In the end, moles help us connect mass with molecular amounts in chemical reactions efficiently.

Water of Hydration

Water of hydration refers to water molecules that are chemically bound within a crystalline structure. When dealing with compounds like \( \mathrm{UO}_{2} \mathrm{NO}_{3} \cdot n \mathrm{H}_{2} \mathrm{O} \), it's essential to figure out how many water molecules are part of the crystal lattice, as this affects the compound's overall properties.

In this problem, \( \mathrm{UO}_{2} \mathrm{NO}_{3} \cdot n \mathrm{H}_{2} \mathrm{O} \) is heated, losing water and forming \( \mathrm{UO}_{2} \mathrm{NO}_{3} \). The initial mass is \( 0.865 \text{ g} \), and post-heating, it's \( 0.679 \text{ g} \). First, calculate the mass of water lost:

• \( \text{Mass of } \mathrm{H}_{2} \mathrm{O} = 0.865 \text{ g} - 0.679 \text{ g} = 0.186 \text{ g} \)

Convert the mass of water to moles:

• \( \text{Moles of } \mathrm{H}_{2} \mathrm{O} = \frac{0.186 \text{ g}}{18.02 \text{ g/mol}} \approx 0.01032 \text{ mol} \)

Next, find the moles of \( \mathrm{UO}_{2} \mathrm{NO}_{3} \):

• \( \text{Moles of } \mathrm{UO}_{2} \mathrm{NO}_{3} = \frac{0.679 \text{ g}}{284.05 \text{ g/mol}} \approx 0.00239 \text{ mol} \)

To find \( n \), the number of water molecules per formula unit, we calculate the ratio:

• \( \text{Water-to-compound mole ratio} = \frac{0.01032}{0.00239} \approx 4.32 \)

This ratio, rounded to the nearest whole number, gives \( n = 4 \). Therefore, each formula unit of \( \mathrm{UO}_{2} \mathrm{NO}_{3} \cdot n \mathrm{H}_{2} \mathrm{O} \) initially contained four water molecules, which is significant for understanding the compound's behavior and properties.