Uranium is used as a fuel, primarily in the form of...

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.)

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.)

Key Concepts

Isotopic Abundance

Isotopic abundance refers to the relative amounts of different isotopes of an element that occur naturally. The average atomic weight of an element represents a weighted average based on these abundances, and understanding this concept helps in determining which isotope is most common in a natural sample.

Empirical Formula Determination

Empirical formula determination is the process of finding the simplest whole-number ratio of elements in a compound. This concept is crucial when dealing with experimental data where the mass composition of elements is known, helping chemists deduce the basic formula of the compound.

Hydrates and Water of Hydration

Hydrates are compounds that include a specific number of water molecules as part of their crystalline structure. The water of hydration can be removed by heating, which is a process used to study the composition of the compound. Determining the number of water molecules per formula unit involves comparing the mass loss upon heating to the molar masses of the components.

Mole Concept

The mole concept is a fundamental principle in chemistry that relates the mass of a substance to the number of particles present using Avogadro’s number. It allows chemists to convert between the mass of an element or compound and the number of moles, which is essential for quantitative chemical analysis and stoichiometric calculations.

Stoichiometry

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It uses molar ratios derived from a balanced chemical equation to predict yields, determine limiting reagents, and calculate amounts of substances consumed or produced in chemical reactions.

Transcript

00:01 This question relates to uranium.

00:06 First, we are given that 0 .169 grams of uranium reacted with oxygen and this reaction produced a uranium oxide of unknown identity.

00:31 This reaction produced 0 .199 grams of this uranium oxide.

00:39 So the first question we need to address is the number of moles of uranium that reacted.

00:48 The moles of uranium equals the mass of uranium that reacted 0 .169 grams divided by the atomic weight of uranium, which is found in the periodic table.

01:05 Uranium is a transition middle.

01:08 It's specifically an actinite.

01:11 And is found in the actanai series.

01:14 The atomic weight is 238 .03 grams per mole.

01:26 So the moles of iranian is 0 .169 divided by 238 .03 and this is 0 .0071 moles.

01:46 Next, we need to find empirical formula.

01:49 To find empirical formula, we start with with the grants of each of the elements that compose the compound, that are part of the compound.

02:00 We have uranium, 0 .169 grams, and oxygen.

02:10 That we don't have the grams of oxygen in the compound, but we can quickly calculate it by subtracting the mass of uranium from the mass of the compound obtained, that is 0 .199.

02:26 Minus 0 .169 grams.

02:36 So the mass of oxygen is 0 .03 grams.

02:47 Now we need to converge these grams to molds.

02:51 For uranium, we already did it, and we obtain 0 .0071.

03:02 For oxygen, we divide by the atomic mass of oxygen, which is 16 grams per.

03:10 And we get 0 .001875.

03:24 To find the empirical formula, we need to divide by the smallest number, by the smallest subscript to set one of the two elements as a whole number in this case is uranian which is 0 .0071.

03:45 Then uranium becomes one the subscript of uranium becomes one that of oxygen will be 2 .64 one which is obviously not a whole number and therefore we need to multiply by this by the smallest a whole number after one which is two and then start trying until we get two whole numbers here so if we multiply by two uranian subscript is 2, and oxygen subscript will be 5 .28.

04:34 This is obviously not a whole number, therefore 2 is not right.

04:38 We next multiply by 3, and we get approximately 8.

04:53 So the formula of this oxide is uranium 3, oxygen 8.

05:05 Next, we need to name this compound.

05:10 And this part of the question is a little bit tricky.

05:15 This is the formula of the compound, the empirical formula of the compound, and the formula of the compound, because this is an ionic compound.

05:27 What we know, what we can deduct from this formula is that the amount of negative charges is 16.

05:39 And we know that because this is an oxide.

05:42 And in the oxide, always oxygen is negative 2.

05:47 Therefore, we have 16 negative charges.

05:52 So it's 16 negative charges.

05:55 Therefore, we need to have 16 positive charges.

06:00 Okay? to have 16 positive charges, each uranium should be 16 divided by 3, which is the substrate of uranium.

06:12 This gives us 5 .3 as positive charges.

06:19 As you know, this is not possible.

06:24 There is no catalan with a non -hold number positive charges.

06:31 The positive charges of the catallions are positive 1, positive 2, positive 3, etc.

06:39 This implies that these iranian oxide has contained uraniums with more than one positive charge and indeed the uranium in this uranium oxide is known to have two positive charges in a fixed proportion it contains uranium five and uranium six okay in a in a fixed proportion and that explains why the number of oppositive charges here is not a whole number.

07:22 Therefore, the name of this compound is uranium 5, 6 oxide.

07:43 Next, we need to calculate the most of this compound that were obtained.

07:50 Remember that the moles, i'm going to write this up here, of this compound will be the mass divided by divided by the molar mass of the compound...

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