Chapter 3, Problem 3.121AP

(a)

Interpretation Introduction

Interpretation: The values of $a,b,c,d,x,y$ and $z$ in the formula of given compounds of uranium and the charge on the molecule ${\text{U}}_a{\text{O}}_b$ is to be determined.

Concept introduction: The compound ${\text{UO}}_x{\left({\text{NO}}_3\right)}_y{\left({\text{H}}_2\text{O}\right)}_x$ decomposes to ${\text{U}}_3{\text{O}}_8$ at higher temperature and decomposes to ${\text{UO}}_3$ at moderate or lower temperature.

To determine: The values of $a$ and $b$ in the compound ${\text{U}}_a{\text{O}}_b$ and the charge on $\text{U}$ in this molecule.

Answer to Problem 3.121AP

Solution

The value of $a$ is $1$ and $b$ is $2$. The charge on uranium in ${\text{UO}}_3$ is $+6$.

Explanation of Solution

Explanation

Given

The mass percent of uranium in ${\text{U}}_a{\text{O}}_b$ is $83.22\%$.

The mass percent of oxygen in ${\text{U}}_a{\text{O}}_b$ is $16.78\%$.

The number of moles of the given element is calculated by using the formula,

$\text{Number}\text{of}\text{moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}$ (1)

Substitute the values of mass and molar mass of uranium in the equation (1).

$\begin{array}{c}\text{Moles}\text{of}\text{U}=\frac{0.8322\text{g}}{238.0\text{g/mol}}\\ =0.00349\text{mol}\end{array}$

Substitute the values of mass and molar mass of oxygen in the equation (1).

$\begin{array}{c}\text{Moles}\text{of}\text{O}=\frac{0.1678\text{g}}{16.00\text{g/mol}}\\ =0.0105\text{mol}\end{array}$

The calculated values are divided by the smallest number of moles to determine the simplest whole number ratio of moles of each constituent,

For uranium $\left(\text{U}\right)$,

$\text{U}=\frac{0.00349\text{mol}}{0.00349\text{mol}}=1$

For oxygen $\left(\text{O}\right)$,

$\text{O}=\frac{0.0105\text{mol}}{0.00349\text{mol}}=3.0$

Therefore, the formula of given compound is ${\text{UO}}_3$.

The charge on the uranium in ${\text{UO}}_3$ is calculated by using the formula,

$\text{Charge}\text{on}\text{U}+3\times \text{charge}\text{on}\text{O}=\text{Charge}\text{on}\text{molecule}$

Substitute the values of charge on oxygen and molecule in the above equation.

$\begin{array}{c}\text{U}+3\times -2=0\\ \text{U}=+6\end{array}$

Thus, the value of $a$ is $1$ and $b$ is $2$ and the charge on uranium in ${\text{UO}}_3$ is $+6$.

(b)

Interpretation Introduction

To determine: The values of $a$ and $b$ in the compound ${\text{U}}_c{\text{O}}_d$ and the charge on $\text{U}$ in this molecule.

Answer to Problem 3.121AP

Solution

The value of $c$ is $3$ and $d$ is $8$. The charge on uranium in ${\text{U}}_3{\text{O}}_8$ is $+5.3$.

Explanation of Solution

Explanation

Given

The mass percent of uranium in ${\text{U}}_c{\text{O}}_d$ is $84.8\%$.

The mass percent of oxygen in ${\text{U}}_c{\text{O}}_d$ is $15.2\%$.

The number of moles of the given element is calculated by using the formula,

$\text{Number}\text{of}\text{moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}$ (1)

Substitute the values of mass and molar mass of uranium in the equation (1).

$\begin{array}{c}\text{Moles}\text{of}\text{U}=\frac{0.848\text{g}}{238.0\text{g/mol}}\\ =0.00356\text{mol}\end{array}$

Substitute the values of mass and molar mass of oxygen in the equation (1).

$\begin{array}{c}\text{Moles}\text{of}\text{O}=\frac{0.152\text{g}}{16.00\text{g/mol}}\\ =0.0095\text{mol}\end{array}$

The calculated values are divided by the smallest number of moles to determine the simplest whole number ratio of moles of each constituent,

For uranium $\left(\text{U}\right)$,

$\text{U}=\frac{0.00356\text{mol}}{0.00356\text{mol}}=1$

For oxygen $\left(\text{O}\right)$,

$\text{O}=\frac{0.0095\text{mol}}{0.00356\text{mol}}=2.67$

The numbers obtained are multiplied with $3$ to obtain the simplest whole number ratio of the corresponding elements. Therefore, the formula of given compound is ${\text{U}}_3{\text{O}}_8$.

The charge on the uranium in ${\text{U}}_3{\text{O}}_8$ is calculated by using the formula,

$\text{3}\times \text{Charge}\text{on}\text{U}+8\times \text{charge}\text{on}\text{O}=\text{Charge}\text{on}\text{molecule}$

Substitute the values of charge on oxygen and molecule in the above equation.

$\begin{array}{c}\text{3U}+8\times -2=0\\ \text{U}=+5.3\end{array}$

Thus, the value of $c$ is $3$ and $d$ is $8$ and the charge on uranium in ${\text{U}}_3{\text{O}}_8$ is $+5.3$.

(c)

Interpretation Introduction

To determine: The values of $x,y$ and $z$ in the compound ${\text{UO}}_x{\left({\text{NO}}_3\right)}_y{\left({\text{H}}_2\text{O}\right)}_x$.

Answer to Problem 3.121AP

Solution

The value of $x$ is $2$, $y$ is $2$ and the value of $z$ is $6$ in the given compound.

Explanation of Solution

Explanation

Given

The compound ${\text{UO}}_x{\left({\text{NO}}_3\right)}_y{\left({\text{H}}_2\text{O}\right)}_x$ decomposes to ${\text{UO}}_x{\left({\text{NO}}_3\right)}_y$ on heating which at higher temperature converted into uranium oxide.

The mass of compound ${\text{UO}}_x{\left({\text{NO}}_3\right)}_y{\left({\text{H}}_2\text{O}\right)}_x$ is $1.328\text{g}$.

The mass of ${\text{UO}}_x{\left({\text{NO}}_3\right)}_y$ in the reaction is $1.042\text{g}$.

The mass of uranium oxide in the reaction is $0.742\text{g}$.

The reaction involving the decomposition of ${\text{UO}}_x{\left({\text{NO}}_3\right)}_y{\left({\text{H}}_2\text{O}\right)}_x$ into ${\text{UO}}_x{\left({\text{NO}}_3\right)}_y$ is,

${\text{UO}}_x{\left({\text{NO}}_3\right)}_y{\left({\text{H}}_2\text{O}\right)}_x\stackrel{\Delta }{\to }{\text{UO}}_x{\left({\text{NO}}_3\right)}_y+z{\text{H}}_2\text{O}$

The mass of water in the reaction is calculated by subtracting the mass of ${\text{UO}}_x{\left({\text{NO}}_3\right)}_y{\left({\text{H}}_2\text{O}\right)}_x$ and ${\text{UO}}_x{\left({\text{NO}}_3\right)}_y$ which is given as,

$\begin{array}{c}\text{Mass}\text{of}{\text{H}}_2\text{O}=1.328g-1.042g\\ =0.286\text{g}\end{array}$

The compound ${\text{UO}}_x{\left({\text{NO}}_3\right)}_y$ at higher temperature decomposes into the ${\text{U}}_3{\text{O}}_8$, ${\text{NO}}_2$, $\text{NO}$ and ${\text{O}}_2$ which is shown by the equation.

${\text{UO}}_x{\left({\text{NO}}_3\right)}_y\stackrel{\Delta }{\to }{\text{U}}_3{\text{O}}_8+{\text{NO}}_2+\text{NO}+{\text{O}}_2$

The mass of nitrogen oxides and oxygen in the reaction is calculated by subtracting the mass of ${\text{UO}}_x{\left({\text{NO}}_3\right)}_y$ and ${\text{U}}_3{\text{O}}_8$ which is given as,

$\begin{array}{c}\text{Mass}\text{of}\text{N}\text{and}\text{O}=1.042g-0.742g\\ =0.300\text{g}\end{array}$

Since, the nitrogen and oxygen are present in $2:5$ amount in the side products of the reaction. Therefore, the mass of nitrogen and oxygen is calculated as,

For nitrogen

$\begin{array}{c}\text{Mass}\text{of}\text{N}=\frac{2}{7}\times 0.300\text{g}\\ =0.0857\text{g}\end{array}$

For oxygen

$\begin{array}{c}\text{Mass}\text{of}\text{O}=\frac{5}{7}\times 0.300\text{g}\\ =0.214\text{g}\end{array}$

The mass of uranium and oxygen in ${\text{U}}_3{\text{O}}_8$ is calculated by multiplying their mass percent with the mass given in the reaction.

For uranium

$\begin{array}{c}\text{Mass}\text{of}\text{U}=84.8\%\times 0.742\text{g}\\ =0.629\text{g}\end{array}$

For oxygen

$\begin{array}{c}\text{Mass}\text{of}\text{O}=15.2\%\times 0.742\text{g}\\ =0.113\text{g}\end{array}$

The total mass of oxygen in the given compound is,

$\begin{array}{c}\text{Total}\text{mass}\text{of}\text{O}=0.113\text{g}+0.214\text{g}\\ =0.327\text{g}\end{array}$

The number of moles of the given element is calculated by using the formula,

$\text{Number}\text{of}\text{moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}$ (1)

Substitute the values of mass and molar mass of uranium in the equation (1).

$\begin{array}{c}\text{Moles}\text{of}\text{U}=\frac{0.629\text{g}}{238.0\text{g/mol}}\\ =0.00264\text{mol}\end{array}$

Substitute the values of mass and molar mass of oxygen in the equation (1).

$\begin{array}{c}\text{Moles}\text{of}\text{O}=\frac{0.327\text{g}}{16.00\text{g/mol}}\\ =0.0204\text{mol}\end{array}$

Substitute the values of mass and molar mass of nitrogen in the equation (1).

$\begin{array}{c}\text{Moles}\text{of}\text{N}=\frac{0.0857\text{g}}{14.0\text{g/mol}}\\ =0.0061\text{mol}\end{array}$

Substitute the values of mass and molar mass of water in the equation (1).

$\begin{array}{c}\text{Moles}\text{of}{\text{H}}_2\text{O}=\frac{0.286\text{g}}{18.01\text{g/mol}}\\ =0.0159\text{mol}\end{array}$

The calculated values are divided by the smallest number of moles to determine the simplest whole number ratio of moles of each constituent,

For uranium $\left(\text{U}\right)$,

$\text{U}=\frac{0.00264\text{mol}}{0.00264\text{mol}}=1$

For oxygen $\left(\text{O}\right)$,

$\text{O}=\frac{0.0204\text{mol}}{0.00264\text{mol}}=8.0$

For nitrogen $\left(\text{N}\right)$,

$\text{N}=\frac{0.0061\text{mol}}{0.00264\text{mol}}=2.0$

For water $\left({\text{H}}_2\text{O}\right)$,

$\text{O}=\frac{0.0159\text{mol}}{0.00264\text{mol}}=6.0$

Therefore, the formula of given compound is ${\text{UO}}_2{\left({\text{NO}}_3\right)}_2{\left({\text{H}}_2\text{O}\right)}_6$.

Conclusion

The value of $a$ is $1$ and $b$ is $2$. The charge on uranium in ${\text{UO}}_3$ is $+6$.

The value of $c$ is $3$ and $d$ is $8$. The charge on uranium in ${\text{U}}_3{\text{O}}_8$ is $+5.3$.

The value of $x$ is $2$, $y$ is $2$ and the value of $z$ is $6$ in the given compound