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

A 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) $\mathrm{A}$ 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 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 $800^{\circ} \mathrm{C}$ in the air.)

A 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) $\mathrm{A}$ 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
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 $800^{\circ} \mathrm{C}$ in the air.)

Key Concepts

Stoichiometric Calculations

Stoichiometric calculations involve using the known molar masses and masses of substances to convert between mass and moles. This is fundamental to quantify reactants and products in chemical reactions, determine empirical formulas, and calculate reaction yields, which is essential in many areas of chemistry including analytical and industrial processes.

Empirical Formula Determination

Empirical formula determination is the process of finding the simplest whole-number ratio of atoms in a compound based on experimental mass data. This method helps reveal the basic composition of a compound and is a key technique in analytical chemistry for characterizing unknown substances.

Nomenclature of Inorganic Compounds

The systematic nomenclature of inorganic compounds involves naming them based on their constituents and the oxidation states of the elements present. This ensures clear communication in the scientific community and accurate interpretation of the compound’s structure and properties.

Isotopic Abundance and Atomic Weight

Isotopic abundance refers to the relative amounts of different isotopes of an element as they occur naturally, which influences the element's average atomic weight. Understanding isotopic distribution is important in fields like nuclear chemistry and radiochemistry, where the behavior and applications of different isotopes can vary significantly.

Hydration and Waters of Hydration

Hydrates are compounds that include water molecules within their crystalline structure, known as waters of hydration. Determining the number of water molecules per formula unit involves analyzing the mass difference before and after heating, a common technique in the characterization of hydrated compounds and important in understanding their physical and chemical properties.

Transcript

00:01 All right, so the first step here, the first thing they're asking is to find is the moles of uranium.

00:07 So we'll use this equation here, most equal mass developed atomic wave, and we can get some moles of uranium.

00:19 It's going to be 0 .169 to by 238 .102, which is 7 .1 .169, which is 7 .1 .69, which is 7 .1.

00:32 Times 10 to the part of negative 4 moles.

00:42 Next, they're asking us to find the empirical formula.

00:48 So we're gonna make a quick table with the mass, the molds, and the subscript.

01:01 So the mass is 0 .169.

01:04 And they give us that the total mass of the oxide is 0 .199.

01:08 So i'm gonna subtract the mass of uranium from that get the mass of oxide.

01:15 In order to find the moles, we do the same thing.

01:17 We divide by the motor mass.

01:19 We get 7 .1 times 10 to the power of negative 4.

01:24 And here i get the oxygen i divide by 16 to find the moles.

01:30 And i get 1 .875 times 10 to the power of negative 3.

01:41 Next, we are going to divide both these molds by the smallest number, which happens to be 7 .1 times 10 to the power of negative 4.

01:50 And when i first divided them, i got 1 and 2 .6 .4.

01:55 Now, that has to be a whole number.

01:57 So i will multiply both of them by 3 to get 3 and roughly 8.

02:07 So the formula, the empirical formula of this compound is u308.

02:13 You're asking for the name.

02:14 It is tri -uranium, octa, oxide.

02:28 So they're also asking for the moles of this so we quickly find the molar mass of this which is going to be three times 238 .102 plus 8 times 16 which turns out to be 842 .106 grams per north.

02:53 Divide that by not to find the moles we do 0 .199 divide by 8 .192 .8 .5 .5.

03:01 Divide by 8 .5 .5.

03:02 142 .06 we get 2 .36...

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