Chapter 1, Problem 12Q

Calculate the atomic percentages for each of the following compounds.

1. a. HfO₂
2. b. BeCl₂
3. c. Ti(OH)₄
4. d. FeO
5. e. SiO₂
6. f. B(OH)₃

Interpretation Introduction

Interpretation:

Atomic percentages of each atoms in given compound has to be calculated.

Concept introduction:

Atomic percentage:

Percentage of one type of atom that present in a molecule is known as atomic percentage.

    $\text{Atomic}\text{\%}\text{=}\frac{\text{Total mass}\text{of}\text{atom}}{\text{Total}\text{mass}\text{of}\text{compound}}\text{×100}$

Explanation of Solution

Given: ${\text{HfO}}_{\text{2}}$

Molar mass of ${\text{HfO}}_{\text{2}}$= $\text{210}\text{.49 g/mol}$

Atomic percentages are,

  $\begin{array}{l}\text{\%}\text{Hafnium}\text{=}\frac{\text{178}\text{.49}}{\text{210}\text{.49}}\text{×100}\text{=}\text{84}\text{.80}\text{\%}\\ \text{\%}\text{oxygen}\text{=}\frac{\text{2×16}}{\text{210}\text{.49}}\text{×100}\text{=}\text{15}\text{.20}\text{\%}\end{array}$

Interpretation Introduction

Interpretation:

Atomic percentages of each atoms in given compound has to be calculated.

Concept introduction:

Atomic percentage:

Percentage of one type of atom that present in a molecule is known as atomic percentage.

    $\text{Atomic}\text{\%}\text{=}\frac{\text{Total mass}\text{of}\text{atom}}{\text{Total}\text{mass}\text{of}\text{compound}}\text{×100}$

Explanation of Solution

Given: ${\text{BeCl}}_{\text{2}}$

Molar mass of ${\text{BeCl}}_{\text{2}}$= $\text{79}\text{.92}\text{g/mol}$

Atomic percentages are,

  $\begin{array}{l}\text{\%}\text{Be=}\frac{9.01}{\text{79}\text{.92}}\text{×100}\text{=}11.27\text{\%}\\ \text{\%}\text{Cl}\text{=}\frac{\text{2×}35.45}{\text{79}\text{.92}}\text{×100}\text{=}88.71\text{\%}\end{array}$

Interpretation Introduction

Interpretation:

Atomic percentages of each atoms in given compound has to be calculated.

Concept introduction:

Atomic percentage:

Percentage of one type of atom that present in a molecule is known as atomic percentage.

    $\text{Atomic}\text{\%}\text{=}\frac{\text{Total mass}\text{of}\text{atom}}{\text{Total}\text{mass}\text{of}\text{compound}}\text{×100}$

Explanation of Solution

Given: ${\text{Ti(OH)}}_4$

Molar mass of ${\text{Ti(OH)}}_4$= $119.927g/mol$

Atomic percentages are,

  $\begin{array}{l}\text{\%}\text{Ti}\text{=}\frac{\text{47}\text{.87}}{\text{119}\text{.927}}\text{×100}\text{=}\text{39}\text{.91}\text{\%}\\ \text{\%}\text{O}\text{=}\frac{\text{4×16}}{\text{119}\text{.927}}\text{×100}\text{=}\text{53}\text{.37}\text{\%}\\ \text{\%}\text{H}\text{=}\frac{\text{4×1}\text{.008}}{\text{119}\text{.927}}\text{×100}\text{=}\text{3}\text{.36}\text{\%}\end{array}$

Interpretation Introduction

Interpretation:

Atomic percentages of each atoms in given compound has to be calculated.

Concept introduction:

Atomic percentage:

Percentage of one type of atom that present in a molecule is known as atomic percentage.

    $\text{Atomic}\text{\%}\text{=}\frac{\text{Total mass}\text{of}\text{atom}}{\text{Total}\text{mass}\text{of}\text{compound}}\text{×100}$

Explanation of Solution

Given: $\text{FeO}$

Molar mass of $\text{FeO}\text{=}$$\text{71}\text{.844 g/mol}$

Atomic percentages are,

  $\begin{array}{l}\text{\%}\text{Fe=}\frac{\text{55}\text{.84}}{\text{71}\text{.844}}\text{×100}\text{=}\text{77}\text{.73}\text{\%}\\ \text{\%}\text{O}\text{=}\frac{\text{16}}{\text{71}\text{.844}}\text{×100}\text{=}\text{22}\text{.27\%}\end{array}$

Interpretation Introduction

Interpretation:

Atomic percentages of each atoms in given compound has to be calculated.

Concept introduction:

Atomic percentage:

Percentage of one type of atom that present in a molecule is known as atomic percentage.

    $\text{Atomic}\text{\%}\text{=}\frac{\text{Total mass}\text{of}\text{atom}}{\text{Total}\text{mass}\text{of}\text{compound}}\text{×100}$

Explanation of Solution

Given: ${\text{SiO}}_{\text{2}}$

Molar mass of ${\text{SiO}}_{\text{2}}$= $\text{60}\text{.08 g/mol}$

Atomic percentages are,

  $\begin{array}{l}\text{\%}\text{Si}\text{=}\frac{\text{28}\text{.08}}{\text{60}\text{.08}}\text{×100}\text{=}\text{46}\text{.74}\text{\%}\\ \text{\%}\text{O}\text{=}\frac{\text{2×16}}{\text{60}\text{.08}}\text{×100}\text{=}\text{53}\text{.26}\text{\%}\end{array}$

Interpretation Introduction

Interpretation:

Atomic percentages of each atoms in given compound has to be calculated.

Concept introduction:

Atomic percentage:

Percentage of one type of atom that present in a molecule is known as atomic percentage.

    $\text{Atomic}\text{\%}\text{=}\frac{\text{Total mass}\text{of}\text{atom}}{\text{Total}\text{mass}\text{of}\text{compound}}\text{×100}$

Explanation of Solution

Given: ${\text{B(OH)}}_{\text{3}}$

Molar mass of ${\text{B(OH)}}_{\text{3}}$= $\text{61}\text{.83 g/mol}$

Atomic percentages are,

  $\begin{array}{l}\text{\%}\text{B}\text{=}\frac{\text{10}\text{.811}}{\text{61}\text{.83}}\text{×100}\text{=}\text{17}\text{.485\%}\\ \text{\%}\text{O }\text{=}\frac{\text{3×16}}{\text{61}\text{.83}}\text{×100}\text{=}\text{77}\text{.63\%}\\ \text{\%}\text{H }\text{=}\frac{\text{3×1}\text{.008}}{\text{61}\text{.83}}\text{×100}\text{=}\text{4}\text{.89\%}\end{array}$