Chapter 6, Problem 14A

(a)

Interpretation Introduction

Interpretation:

The number of moles of the sample needs to be determined.

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

Answer to Problem 14A

Moles of Au = 0.133 mol

Explanation of Solution

symbol of gold = Au

mass of Au = 26.2 g

molar mass of Au = 196.966 gmol-1

The calculation of moles is shown below:

  $\begin{array}{c}\text{moles}\text{of}\text{Au}\text{=}\frac{\text{mass}\text{of}\text{Au}}{\text{molar}\text{mass}\text{of}\text{Au}}\\ \text{=}\frac{\text{26}\text{.2}\text{g}}{\text{196}\text{.966}{\text{gmol}}^{\text{-1}}}\\ \text{=}\text{0}\text{.133}\text{mol}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The number of moles of the sample needs to be determined.

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

Answer to Problem 14A

Moles of Ca = 1.035 mol

Explanation of Solution

symbol of calcium = Ca

mass of Ca = 41.5 g

molar mass of Ca = 40.078 gmol-1

The calculation of moles is shown below:

  $\begin{array}{c}\text{moles}\text{of}\text{Ca}\text{=}\frac{\text{mass}\text{of}\text{Ca}}{\text{molar}\text{mass}\text{of}\text{Ca}}\\ \text{=}\frac{\text{41}\text{.5}\text{g}}{\text{40}\text{.078}{\text{gmol}}^{\text{-1}}}\\ \text{=}\text{1}\text{.035}\text{mol}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The number of moles of the sample needs to be determined.

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

Answer to Problem 14A

Moles of Ba = 0.0024 mol

Explanation of Solution

symbol of Barium = Ba

mass of Ba = 335 mg

molar mass of Ba = 137.327 gmol-1

The calculation of moles is shown below:

  $\begin{array}{c}\text{moles}\text{of}\text{Ba}\text{=}\frac{\text{mass}\text{of}\text{Ba}}{\text{molar}\text{mass}\text{of}\text{Ba}}\\ \text{=}\frac{\text{0}\text{.335}\text{g}}{\text{137}\text{.327}{\text{gmol}}^{\text{-1}}}\\ \text{=}\text{0}\text{.0024}\text{mol}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The number of moles of the sample needs to be determined.

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

Answer to Problem 14A

Moles of Pd = 1.33×10²¹ mol

Explanation of Solution

symbol of palladium = Pd

mass of Pd = 1.42 ×10²³ g

molar mass of Pd = 106.42 gmol-1

The calculation of moles is shown below:

  $\begin{array}{c}\text{moles}\text{of}\text{Pd}\text{=}\frac{\text{mass}\text{of}\text{Pd}}{\text{molar}\text{mass}\text{of}\text{Pd}}\\ \text{=}\frac{\text{1}\text{.42}\text{×}{\text{10}}^{\text{23}}\text{g}}{\text{106}\text{.42}{\text{gmol}}^{\text{-1}}}\\ \text{=}\text{1}\text{.33}\text{×}{\text{10}}^{\text{21}}\text{mol}\end{array}$

(e)

Interpretation Introduction

Interpretation:

The number of moles of the sample needs to be determined.

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

Answer to Problem 14A

Moles of Ni = 5.196×10²⁰ mol

Explanation of Solution

symbol of nickel = Ni

mass of Ni = 3.05×10²⁵ mg

molar mass of Ni = 58.6934 gmol-1

The calculation of moles is shown below:

  $\begin{array}{c}\text{moles}\text{of}\text{Ni}\text{=}\frac{\text{mass}\text{of}\text{Ni}}{\text{molar}\text{mass}\text{of}\text{Ni}}\\ \text{=}\frac{\text{3}\text{.05}\text{×}{\text{10}}^{\text{22}}\text{g}}{\text{58}\text{.6934}{\text{gmol}}^{\text{-1}}}\\ \text{=}\text{5}\text{.196}\text{×}{\text{10}}^{\text{20}}\text{mol}\end{array}$

(f)

Interpretation Introduction

Interpretation:

The number of moles of the sample needs to be determined.

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

Answer to Problem 14A

Moles of Fe = 8.122 mol

Explanation of Solution

symbol of iron = Fe

mass of Fe = 1.00 lb

molar mass of Fe = 55.845 gmol-1

1 lb = 453.592 g

The calculation of moles is shown below:

  $\begin{array}{c}\text{moles}\text{of}\text{Fe}\text{=}\frac{\text{mass}\text{of}\text{Fe}}{\text{molar}\text{mass}\text{of}\text{Fe}}\\ \text{=}\frac{\text{453}\text{.592}\text{g}}{\text{55}\text{.845}{\text{gmol}}^{\text{-1}}}\\ \text{=}\text{8}\text{.122mol}\end{array}$

(g)

Interpretation Introduction

Interpretation:

The number of moles of the sample needs to be determined.

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

Answer to Problem 14A

Moles of C = 0.99 mol

Explanation of Solution

symbol of Carbon = C

mass of C = 12.01 g

molar mass of C = 12.0107 gmol-1

The calculation of moles is shown below:

  $\begin{array}{c}\text{moles}\text{of}\text{C}\text{=}\frac{\text{mass}\text{of}\text{C}}{\text{molar}\text{mass}\text{of}\text{C}}\\ \text{=}\frac{\text{12}\text{.01}\text{g}}{\text{12}\text{.0107}{\text{gmol}}^{\text{-1}}}\\ \text{=}\text{0}\text{.99}\text{mol}\end{array}$