Chapter 2, Problem 131QRT

Uranium is used as a fuel, primarily in the form of uranium(IV) oxide, in nuclear power plants. This question considers some uranium chemistry.

1. (a) A small sample of uranium metal (0.169 g) is heated to 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?
2. (b) The oxide U_xO_y is obtained if UO₂NO₃ · n H₂O is heated to temperatures greater than 800 °C in air. However, if you heat it gently, only the water of hydration is lost. If you have 0.865 g UO₂NO₃ · n H₂O and obtain 0.679 g UO₂NO₃ on heating, how many molecules of water of hydration were there in each formula unit of the original compound?

Interpretation Introduction

Interpretation:

A small sample of uranium metal is heated to ${\text{900}}^{\text{o}}\text{C}$ in air to give $\text{0}\text{.199g}$ of a dark oxide, ${\text{U}}_{\text{x}}{\text{O}}_{\text{y}}$.  The number of moles of uranium metal has to be determined.  The empirical formula of the oxide ${\text{U}}_{\text{x}}{\text{O}}_{\text{y}}$, the name of the oxide and the number of moles of ${\text{U}}_{\text{x}}{\text{O}}_{\text{y}}$ has to be determined.

Explanation of Solution

    $\begin{array}{l}{\text{Molar of O in the sample = U}}_{\text{x}}{\text{O}}_{\text{y}}\text{compound mass-U mass}\\ \text{ 0}{\text{.199g U}}_{\text{x}}{\text{O}}_{\text{y}}\text{-0}\text{.169gU=0}\text{.030gO}\end{array}$

Find moles of uranium and oxygen in sample:

    $\begin{array}{l}\text{0}\text{.169g U×}\frac{\text{1mol U}}{\text{238}\text{.03g U}}\text{=7}{\text{.10×10}}^{\text{-4}}\text{mol U}\\ \text{0}\text{.030g O×}\frac{\text{1mol O}}{\text{15}\text{.9994g O}}\text{=1}{\text{.9×10}}^{\text{-3}}\text{mol O}\end{array}$

Set up mole ratio:

    $\text{7}{\text{.10 × 10}}^{\text{-4}}\text{mol U:1}{\text{.9 × 10}}^{\text{-3}}\text{mol O}$

Divide all the numbers in the ratio by the smallest number of moles, $\text{7}{\text{.10 × 10}}^{\text{-4}}\text{mol U}$ $\text{l U:2}\text{.6O}$.

2.6 is close to both 2.66 and 2.5, and since this number has an uncertainty of, either one is equally valid.  It is unclear whether we should round down or up to find the whole number ratio.  So, look at each case.

If the real ratio is $\text{2:2/3}$, we multiply by 3, and get $\text{3U:8O}$ and an empirical of ${\text{U}}_{\text{3}}{\text{O}}_{\text{8}}$.

If the real ratio is ,we multiply by 2, and get $\text{2U:5O}$ and an empirical formula of ${\text{U}}_{\text{3}}{\text{O}}_{\text{8}}$.

Of the two empirical formula of ${\text{U}}_{\text{3}}{\text{O}}_{\text{8}}$ and ${\text{U}}_2{\text{O}}_5$, the second formula makes more sense.  The common ion of oxygen is oxide ion, ${\text{O}}^{\text{2-}}$.  ${\text{U}}_{\text{3}}{\text{O}}_{\text{8}}$ looks like the sample combination of ${\text{U}}^{\text{5+}}{\text{and O}}^{\text{2-}}$.  The formula is either wrong, or contains some mixture of uranium oxides with different formulas, such as a mixture of ${\text{UO}}_3{\text{ and U}}_{\text{2}}{\text{O}}_5$.  With this insight, let’s presume that the ${\text{U}}_2{\text{O}}_5$ formula is the right formula.  ${\text{U}}_2{\text{O}}_5$ would be called uranium (V) oxide.

    $\begin{array}{l}{\text{Molar mass of U}}_{\text{2}}{\text{O}}_{\text{5}}\text{=2(238}\text{.03g/molU)+5(15}\text{.9994g/molO)=556}{\text{.06g/molU}}_{\text{2}}{\text{O}}_{\text{5}}\\ \text{0}{\text{.199gU}}_{\text{2}}{\text{O}}_{\text{5}}\text{×}\frac{{\text{1molU}}_{\text{2}}{\text{O}}_{\text{5}}}{\text{556}{\text{.06gU}}_{\text{2}}{\text{O}}_{\text{5}}}\text{=3}{\text{.58×10}}^{\text{-4}}{\text{molU}}_{\text{2}}{\text{O}}_{\text{5}}\end{array}$

Interpretation Introduction

Interpretation:

The oxide ${\text{U}}_{\text{x}}{\text{O}}_{\text{y}}$ is obtained if ${\text{UO}}_{\text{2}}{\text{NO}}_{\text{3}}{\text{.nH}}_{\text{2}}\text{O}$ is heated to temperatures greater than ${\text{800}}^{\text{o}}\text{C}$ in air.  If $\text{0}\text{.865g}$ ${\text{UO}}_{\text{2}}{\text{NO}}_{\text{3}}{\text{.nH}}_{\text{2}}\text{O}$ and obtain $\text{0}\text{.679g}$ ${\text{UO}}_{\text{2}}{\text{NO}}_{\text{3}}$ on heating, the number of molecules of water of hydration were there in each formula unit of the original compound has to be calculated.

Explanation of Solution

    ${\text{Mass of H}}_{\text{2}}\text{O=0}{\text{.865gUO}}_{\text{2}}{\text{(NO}}_{\text{3}}\text{)}{\text{.nH}}_{\text{2}}\text{O-0}{\text{.679g UO}}_{\text{2}}{\text{(NO}}_{\text{3}}\text{)=0}{\text{.186g H}}_{\text{2}}\text{O}$.

    ${\text{Mass of H}}_{\text{2}}\text{O = 2(1}\text{.0079g/mol H)+15}\text{.9994g/mol O = 18}{\text{.0152g/mol H}}_{\text{2}}\text{O}$.

${\text{Molar mass of UO}}_{\text{2}}{\text{(NO}}_{\text{3}}\text{) = 238}\text{.03g/mol U+5(15}\text{.9994g/mol O) + 14}\text{.0067 g/mol N = 332}{\text{.03g/mol UO}}_{\text{2}}{\text{(NO}}_{\text{3}}\text{)}$

  ${\text{Mol of H}}_{\text{2}}\text{O =0}{\text{.186g H}}_{\text{2}}\text{O×}\frac{{\text{1mol H}}_{\text{2}}\text{O}}{\text{18}{\text{.0152g H}}_{\text{2}}\text{O}}\text{=0}{\text{.0103mol H}}_{\text{2}}\text{O}$.

$\begin{array}{l}{\text{Moles of UO}}_{\text{2}}{\text{(NO}}_{\text{3}}\text{)}{\text{.nH}}_{\text{2}}\text{O in sample:}\\ \text{0}{\text{.679g UO}}_{\text{2}}{\text{(NO}}_{\text{3}}\text{)×}\frac{{\text{1mol UO}}_{\text{2}}{\text{(NO}}_{\text{3}}\text{)}}{\text{332}{\text{.03gUO}}_{\text{2}}{\text{(NO}}_{\text{3}}\text{)}}\text{ × }\frac{{\text{1mol UO}}_{\text{2}}{\text{(NO}}_{\text{3}}\text{)}{\text{.nH}}_{\text{2}}\text{O}}{{\text{1mol UO}}_{\text{2}}{\text{(NO}}_{\text{3}}\text{)}}\text{=0}{\text{.00204molUO}}_{\text{2}}{\text{(NO}}_{\text{3}}\text{)}{\text{.nH}}_{\text{2}}\text{O}\end{array}$

Mole ratio:

    $\text{0}{\text{.0103 mol H}}_{\text{2}}\text{O : 0}{\text{.00204 mol UO}}_{\text{2}}{\text{(NO}}_{\text{3}}\text{)}{\text{.nH}}_{\text{2}}\text{O}$.

Divide by the smallest number of moles, 0.00204 mol, and round to whole numbers:

    ${\text{5 H}}_{\text{2}}{\text{O : 1UO}}_{\text{2}}{\text{(NO}}_{\text{3}}\text{)}{\text{.nH}}_{\text{2}}\text{O}$.

That means $\text{n=5}$, the formula is ${\text{UO}}_{\text{2}}{\text{(NO}}_{\text{3}}\text{)}{\text{.5H}}_{\text{2}}\text{O}$, and there are 5 molecules of water of hydration in the original hydrated compound.