Chapter 2, Problem 156GQ

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 °C in air to give 0.199 g of a dark green oxide, U_xO_y How many moles of uranium metal were used? What is the empirical formula of the oxide, U_xO_y? What is the name of the oxide? How many moles of U_xO_y must have been obtained?

(b) The naturally occurring isotopes of uranium are ²³⁴U, ²³⁵U, and ²³⁸U. Knowing that uranium’s atomic weight is 238.02 g/mol, which isotope must be the most abundant?

(c) If the hydrated compound UO₂(NO₃)₂ · z H₂O is heated gently, the water of hydration is lost. If you have 0.865 g of the hydrated compound and obtain 0.679 g of UO₂(NO₃)₂ on heating, how many waters of hydration are in each formula unit of the original compound? (The oxide U_x, O_y is obtained if the hydrate is heated to temperatures over 800°C in the air.)

Interpretation Introduction

Interpretation: The moles of uranium, empirical formula of oxide $U_x{O}_y$, name of $U_x{O}_y$ and the moles obtained should be determined under given conditions.

Concept introduction:

• Empirical formula of a compound represents the smallest whole number relative ratio of elements in that compound.
• Equation for number moles from mass and molar mass,

  $Numberofmoles=\frac{Massingrams}{Molarmass}$

• Mass percent of elements of a compound is the ratio of mass of element to the mass of whole compound and multiplied with hundred.
• Isotopes are atoms of the same element having the same number of protons but different number of neutrons.
• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mole $\left(\text{g/mol}\right)$.

Answer to Problem 156GQ

Empirical formula of the compound ${\text{U}}_{\text{2}}{\text{O}}_{\text{5}}$

Name: $\text{Uranium}\left(\text{V}\right)\text{oxide}$

Amount of uranium metal used is $\text{7}\text{.1}{\text{×10}}^{\text{-4}}\text{mol}$

Explanation of Solution

Here,

A small sample of uranium metal (0.169 g) is heated to between $800and900{}^{0}C$ in air to give $0.199g$ of product $U_x{O}_y$

Then the amount of uranium used in this process can be calculated as follows,

  $\begin{array}{l}\text{Mass}\text{of}\text{uranium}\text{metal}\text{=}\text{0}\text{.169}\text{g}\\ \text{Molar}\text{mass}\text{of}\text{uranium}\text{=}\text{238}\text{.02}\text{g/mol}\\ \\ \text{Amount}\text{of}\text{uranium}\text{=}\frac{\text{0}\text{.169}\text{g}}{\text{238}\text{.02}\text{g/mol}}\\ \\ \text{=}\text{7}\text{.1}{\text{×10}}^{\text{-4}}\text{mol}\end{array}$

  $\begin{array}{l}Massofoxygenreacted=0.199g-0.169g\\ =0.030g\\ \\ Molesofoxygenreacted=\frac{0.030g}{16g/mol}=1.875\times {10}^{-3}mole\end{array}$

The mole ratio between the elements is,

$Ti:O=7.1\times {10}^{-4}:1.875\times {10}^{-3}$

Dividing the every element’s number of moles by the smallest number of mole.

  $\begin{array}{l}\text{U}\text{:}\text{O}\text{=}\frac{\text{7}\text{.1}{\text{×10}}^{\text{-4}}}{\text{7}\text{.1}{\text{×10}}^{\text{-4}}}\text{:}\frac{\text{1}{\text{.875×10}}^{\text{-3}}}{\text{7}\text{.1}{\text{×10}}^{\text{-4}}}\\ \\ \text{=}\text{1:}\text{2}\text{.5}\end{array}$

Empirical formula of a compound represents the smallest whole number relative ratio of elements in that compound.

To get the whole number ration between these elements multiplying the above mentioned ratio by 2.

  $\begin{array}{l}\text{U:}\text{O}\text{=}\text{2×}\left(\text{1:}\text{2}\text{.5}\right)\\ \\ \text{=}\text{2}\text{:}\text{5}\Rightarrow \text{2U}\text{:}\text{5}\text{O}\end{array}$

Thus the empirical formula of the compound is ${\text{U}}_{\text{2}}{\text{O}}_{\text{5}}$

This compound can be named as $Uranium\left(V\right)oxide$

Interpretation Introduction

Interpretation: The most abundant isotope from naturally occurring isotopes should be identified using given atomic weight of uranium.

Concept introduction:

• Empirical formula of a compound represents the smallest whole number relative ratio of elements in that compound.
• Equation for number moles from mass and molar mass,

  $Numberofmoles=\frac{Massingrams}{Molarmass}$

• Mass percent of elements of a compound is the ratio of mass of element to the mass of whole compound and multiplied with hundred.
• Isotopes are atoms of the same element having the same number of protons but different number of neutrons.
• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mole $\left(\text{g/mol}\right)$.

Answer to Problem 156GQ

${}^{238}U$ is the most abundant isotope of uranium.

Explanation of Solution

Given

Three isotopes of Uranium are:

            $\begin{array}{l}{}^{\text{234}}\text{U}\\ {}^{\text{235}}\text{U}\\ {}^{\text{238}}\text{U}\end{array}$

Among these isotopes uranium – 238 ${}^{\text{238}}\text{U}$ is the most abundant isotope.  The actual atomic mass of uranium is 238.  Mass of  ${}^{\text{238}}\text{U}$ isotope closes to the actual mass of uranium atom. So there is a possibility to be more stable also.

Therefore the most abundant isotope of uranium atom is uranium – 238, ${}^{\text{238}}\text{U}$.

Interpretation Introduction

Interpretation: The waters of hydration in each formula unit of original given compound should be identified by using given data.

Concept introduction:

• Empirical formula of a compound represents the smallest whole number relative ratio of elements in that compound.
• Equation for number moles from mass and molar mass,

  $Numberofmoles=\frac{Massingrams}{Molarmass}$

• Mass percent of elements of a compound is the ratio of mass of element to the mass of whole compound and multiplied with hundred.
• Isotopes are atoms of the same element having the same number of protons but different number of neutrons.
• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mole $\left(\text{g/mol}\right)$

Answer to Problem 156GQ

Value of z in the $U{O}_2{\left(N{O}_3\right)}_2.z{H}_{2}O$ compound is 6

Explanation of Solution

When $0.865g$ of hydrated compound ($U{O}_2{\left(N{O}_3\right)}_2.z{H}_{2}O$ ) is heated, $0.679g$ of anhydrate compound produced.

  $Massofwaterlost=0.865g-0.679g=0.186g$

Then $\text{Amount of water =}\frac{\text{0}\text{.186}\text{g}}{\text{18}\text{.01}\text{g/mol}}\text{=}\text{0}\text{.01032}\text{mol}$

  $\begin{array}{l}AmountofU{O}_2{\left(N{O}_3\right)}_2=\frac{0.679gU{O}_2{\left(N{O}_3\right)}_2}{394.0375gU{O}_2{\left(N{O}_3\right)}_2/mol}\\ \\ =0.00172mol\end{array}$

Formula unit z in the compound is,

  $\begin{array}{l}\frac{0.010547mol{H}_{2}O}{0.00172molU{O}_2{\left(N{O}_3\right)}_2}=6.13\\ \end{array}$

For convenience this value rounded to the nearest whole number to find z,

Thus,

Value of z in the $U{O}_2{\left(N{O}_3\right)}_2.z{H}_{2}O$ compound is 6