Chapter 3, Problem 64E

Bauxite, the principal ore used in the production of aluminum, has a molecular formula of Al₂O₃ · 2H₂O. The ·H₂O in the formula are called waters of hydration. Each formula unit of the compound contains two water molecules.

  a.    What is the molar mass of bauxite?

  b.    What is the mass of aluminum in 0.58 mole of bauxite?

  c.    How many atoms of aluminum are in 0.58 mole of bauxite?

  d.    What is the mass of 2.1 × 10²⁴ formula units of bauxite?

Interpretation Introduction

Interpretation:

The molar mass of bauxite, mass and number of aluminum atoms has to be determined.  Also, the mass of given formula units of Bauxite has to be determined.

Concept introduction:

Molar mass:

Molar mass of a substance is defined as the mass of the substance in grams of one mole of that substance.

The molar mass of any compound can be calculated by adding of atomic weight of individual atoms present in it.

The amount of substance containing $\text{12}\text{g}$ of pure carbon is called a mole.  One mole of atoms always contains $\text{6}\text{.022}\times {\text{10}}^{\text{23}}$ molecules.  The number of molecules in one mole is also called Avogadro’s number.

Answer to Problem 64E

The molar mass of bauxite is $\underset{\_}{\text{137}\text{.992}\text{g/mol}\text{.}}$

Explanation of Solution

The molar mass of Bauxite $\left({\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}{\text{.2H}}_{\text{2}}\text{O}\right)$

The atomic weight of Aluminum $\left(\text{Al}\right)$ is $\text{26}\text{.98}\text{g}/\text{mol}.$

The atomic weight of Oxygen $\left(\text{O}\right)$ is $\text{16}\text{.00}\text{g}/\text{mol}.$

The atomic weight of Hydrogen $\left(\text{H}\right)$ is $\text{1}\text{.008}\text{g}/\text{mol}.$

Calculate the molar mass of Bauxite

In Bauxite $\left({\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}{\text{.2H}}_{\text{2}}\text{O}\right)$, two Aluminum, five Oxygen and four Hydrogen atoms are present.

$\text{Molar}\text{mass}\text{of}\text{Bauxite}\text{=}\left[\begin{array}{c}\text{2×atomic}\text{weight}\text{of}\text{Al}\text{+ 5×atomic}\text{weight}\text{of}\text{O}\\ \text{+}\text{4×atomic}\text{weight}\text{of}\text{H}\end{array}\right]$

Substitute the value of atomic weight of $\text{Al,}\text{O}$ and $\text{H}$ in the above equation.

$\begin{array}{c}\text{Molar}\text{mass}\text{of}\text{Bauxite = }\left[\begin{array}{l}\text{2×}\left(\text{26}\text{.98}\text{g/mol}\right)\text{ + 5×}\left(\text{16}\text{.00}\text{g/mol}\right)\\ \text{ + 4×}\left(\text{1}\text{.008}\text{g/mol}\right)\end{array}\right]\\ \text{=}\underset{\_}{\text{137}\text{.992}\text{g/mol}}\text{.}\end{array}$

The molar mass of bauxite is $\underset{\_}{\text{137}\text{.992}\text{g/mol}\text{.}}$

Interpretation Introduction

Interpretation:

The molar mass of bauxite, mass and number of aluminum atoms has to be determined.  Also, the mass of given formula units of Bauxite has to be determined.

Concept introduction:

Molar mass:

Molar mass of a substance is defined as the mass of the substance in grams of one mole of that substance.

The molar mass of any compound can be calculated by adding of atomic weight of individual atoms present in it.

The amount of substance containing $\text{12}\text{g}$ of pure carbon is called a mole.  One mole of atoms always contains $\text{6}\text{.022}\times {\text{10}}^{\text{23}}$ molecules.  The number of molecules in one mole is also called Avogadro’s number.

Answer to Problem 64E

The mass of aluminum is $\underset{\_}{\text{31}\text{.3}\text{g}\text{.}}$

Explanation of Solution

The mass of Aluminum $\left(\text{Al}\right)$ in $\text{0}\text{.58}\text{mol}$ of Bauxite.

The moles of Bauxite is $\text{0}\text{.58}\text{mol}\text{.}$

Since, one mole of ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}{\text{.2H}}_{\text{2}}\text{O}$ is equal to two moles of Aluminum, Therefore, $\text{0}\text{.58}\text{mol}$ of ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}{\text{.2H}}_{\text{2}}\text{O}$ is equal to,

$\text{2×}\text{0}\text{.58}\text{mol = 1}\text{.16}\text{mol}$ of Aluminum $\left(\text{Al}\right)$

The atomic mass of Aluminum $\left(\text{Al}\right)$ is $\text{26}\text{.98}\text{g}/\text{mol}$.

Calculate the mass of aluminum

The mass of Aluminum $\left(\text{Al}\right)$ in ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}{\text{.2H}}_{\text{2}}\text{O}$ is calculated as,

$\begin{array}{c}\text{Moles}\text{of}\text{Al = }\frac{\text{Mass}\text{of}\text{Al}}{\text{Atomic}\text{mass}\text{of}\text{Al}}\\ \text{Mass}\text{of}\text{Al = Moles}\text{of}\text{Al × Atomic}\text{mass}\text{of}\text{Al}\end{array}$

Substitute the values of moles and atomic mass of $\text{Al}$ in the above equation.

$\begin{array}{c}\text{Mass}\text{of}\text{Al = Moles}\text{of}\text{Al × Atomic}\text{mass}\text{of}\text{Al}\\ \text{= }\left(\text{1}\text{.16}\text{mol}\right)\left(\text{26}\text{.98}\text{g}/\text{mol}\right)\\ \text{= }\underset{\_}{\text{31}\text{.3}\text{g}}.\end{array}$

The mass of aluminum is $\underset{\_}{\text{31}\text{.3}\text{g}\text{.}}$

Interpretation Introduction

Interpretation:

The molar mass of bauxite, mass and number of aluminum atoms has to be determined.  Also, the mass of given formula units of Bauxite has to be determined.

Concept introduction:

Molar mass:

Molar mass of a substance is defined as the mass of the substance in grams of one mole of that substance.

The molar mass of any compound can be calculated by adding of atomic weight of individual atoms present in it.

The amount of substance containing $\text{12}\text{g}$ of pure carbon is called a mole.  One mole of atoms always contains $\text{6}\text{.022}\times {\text{10}}^{\text{23}}$ molecules.  The number of molecules in one mole is also called Avogadro’s number.

Answer to Problem 64E

The number of aluminum atoms is $\underset{\_}{\text{6}\text{.986}\text{×}{\text{10}}^{\text{23}}\text{atoms}\text{.}}$

Explanation of Solution

The number of Aluminum $\left(\text{Al}\right)$ atoms in $\text{0}\text{.58}\text{mol}$ of Bauxite.

$\text{0}\text{.58}\text{mol}$ of ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}{\text{.2H}}_{\text{2}}\text{O}$ is equal to $\text{1}\text{.16}\text{mol}$ of Aluminum $\left(\text{Al}\right)$.

The atomic weight of Aluminum $\left(\text{Al}\right)$ is $\text{26}\text{.98}\text{g}/\text{mol}$.

Calculate the number of Aluminum atoms in ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}{\text{.2H}}_{\text{2}}\text{O}$

$\text{Number}\text{of}\text{Al}\text{atoms = Moles}\text{of}\text{Al}\text{in}{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}{\text{.2H}}_{\text{2}}\text{O × 6}{\text{.022×10}}^{\text{23}}$

Substitute the value of number of moles of $\text{Al}$ in above equation.

$\begin{array}{c}\text{Number}\text{of}\text{Al}\text{atoms = Moles}\text{of}\text{Al}\text{in}{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}{\text{.2H}}_{\text{2}}\text{O × 6}{\text{.022×10}}^{\text{23}}\\ \text{= 1}\text{.16×6}{\text{.022×10}}^{\text{23}}\\ \text{=}\underset{\_}{\text{ 6}{\text{.986×10}}^{\text{23}}\text{atoms}}.\end{array}$

The number of aluminum atoms is $\underset{\_}{\text{6}\text{.986}\text{×}{\text{10}}^{\text{23}}\text{atoms}\text{.}}$

Interpretation Introduction

Interpretation:

The molar mass of bauxite, mass and number of aluminum atoms has to be determined.  Also, the mass of given formula units of Bauxite has to be determined.

Concept introduction:

Molar mass:

Molar mass of a substance is defined as the mass of the substance in grams of one mole of that substance.

The molar mass of any compound can be calculated by adding of atomic weight of individual atoms present in it.

The amount of substance containing $\text{12}\text{g}$ of pure carbon is called a mole.  One mole of atoms always contains $\text{6}\text{.022}\times {\text{10}}^{\text{23}}$ molecules.  The number of molecules in one mole is also called Avogadro’s number.

Answer to Problem 64E

The mass of bauxite in $\text{2}\text{.1}\text{×}{\text{10}}^{\text{24}}$ formula unit is $\underset{\_}{\text{481}\text{.2}\text{g}\text{.}}$

Explanation of Solution

The mass of $\text{2}\text{.1}\times {\text{10}}^{\text{24}}$ formula units of Bauxite.

The formula units of Bauxite is $\text{2}\text{.1}\times {\text{10}}^{\text{24}}$.

The molar mass of Bauxite is $\text{137}\text{.99}\text{g}$.

Calculate the mass of Bauxite in $\text{2}\text{.1}\text{×}{\text{10}}^{\text{24}}$ formula units,

$\begin{array}{c}\text{Mass}\text{of}\text{Bauxite = Formula}\text{units}\text{of}\text{Bauxite × }\left(\frac{\text{1}\text{mol}{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}{\text{.2H}}_{\text{2}}\text{O}}{\text{6}{\text{.022×10}}^{\text{23}}\text{formula}\text{units}}\right)\\ \text{ ×}\left(\frac{\text{Molar}\text{mass}\text{of}\text{Bauxite}}{\text{1}\text{mol}{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}{\text{.2H}}_{\text{2}}\text{O}}\right)\end{array}$

Substitute the values of molar mass and formula unit of Bauxite in above equation.

$\begin{array}{c}\text{Mass}\text{of}\text{Bauxite = Formula}\text{units}\text{of}\text{Bauxite × }\left(\frac{\text{1}\text{mol}{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}{\text{.2H}}_{\text{2}}\text{O}}{\text{6}{\text{.022×10}}^{\text{23}}\text{formula}\text{units}}\right)\\ \text{ × }\left(\frac{\text{Molar}\text{mass}\text{of}\text{Bauxite}}{\text{1}\text{mol}{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}{\text{.2H}}_{\text{2}}\text{O}}\right)\\ \text{= 2}{\text{.1×10}}^{\text{24}}\text{formula units × }\left(\frac{\text{1}\text{mol}{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}{\text{.2H}}_{\text{2}}\text{O}}{\text{6}{\text{.022×10}}^{\text{23}}\text{formula}\text{units}}\right)\\ \text{ ×}\left(\frac{\text{137}\text{.99}\text{g}}{\text{1}\text{mol}{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}{\text{.2H}}_{\text{2}}\text{O}}\right)\\ \text{=}\underset{\_}{\text{481}\text{.2}\text{g}}.\end{array}$

The mass of bauxite in $\text{2}\text{.1}\text{×}{\text{10}}^{\text{24}}$ formula unit is $\underset{\_}{\text{481}\text{.2}\text{g}\text{.}}$