Chapter 16, Problem 114A

(a)

Interpretation Introduction

Interpretation : The mass of one mole of iron, copper, sulfur, and mercury is to be calculated.

Concept Introduction : Mole concept is the method used to define the amount of a compound and the value of 1 mole is $6.022\times {10}^{23}$ Avogadro number.

Answer to Problem 114A

Mass of 1 mole iron $\left(Fe\right)$ is $55.8\text{ g}$

Mass of 1 mole sulphur $\left(S\right)$ is $32.1\text{ g}$

Mass of 1 mole copper $\left(Cu\right)$ is $63.5\text{ g}$

Mass of 1 mole mercury $\left(Hg\right)$ is $200.6\text{ g}$

Explanation of Solution

Atomic weight of iron $\left(Fe\right)$ is $58.5\text{ g/mol}$

Atomic weight of sulphur $\left(S\right)$ is $32.1\text{ g/mol}$

Atomic weight of copper $\left(Cu\right)$ is $63.5\text{ g/mol}$

Atomic weight of mercury $\left(Hg\right)$ is $200.6\text{ g/mol}$

Moles of solute is the ratio of the mass of a compound to the atomic/molecular mass of that compound.

The expression for moles is as follows:

  $Moles=\frac{Mass}{Molarmass}$

So, 1 mole of any compound contains mass in grams equal to that of its atomic or molecular weight. Hence, the mass of 1 mole of the given compounds is as follows:

Mass of 1 mole iron $\left(Fe\right)$ is $55.8\text{ g}$

Mass of 1 mole sulphur $\left(S\right)$ is $32.1\text{ g}$

Mass of 1 mole copper $\left(Cu\right)$ is $63.5\text{ g}$

Mass of 1 mole mercury $\left(Hg\right)$ is $200.6\text{ g}$

(b)

Interpretation Introduction

Interpretation : The number of atoms of iron, sulfur, and mercury present is to be calculated.

Concept Introduction : Mole concept is the method used to define the amount of a compound and the value of 1 mole is $6.022\times {10}^{23}$ Avogadro number.

Answer to Problem 114A

One mole of iron, copper, sulfur, and mercury all contain $6.022\times {10}^{23}$ atoms.

Explanation of Solution

Mole concept is the method used to define the amount of a compound and the value of 1 mole is $6.022\times {10}^{23}$ Avogadro number. Hence, 1 mole of iron, copper, sulfur, and mercury all contain $6.022\times {10}^{23}$ atoms.

(c)

Interpretation Introduction

Interpretation : The number of moles contained by $25\text{ g}$ of iron, sulfur, and mercury is to be calculated.

Concept Introduction : Mole concept is the method used to define the amount of a compound and the value of 1 mole is $6.022\times {10}^{23}$ Avogadro number.

Answer to Problem 114A

Moles of iron $\left(Fe\right)$ present in $25\text{ g}$ is $0.448\text{ moles}$ .

Moles of sulfur $\left(S\right)$ present in $25\text{ g}$ is $0.779\text{ moles}$ .

Moles of copper $\left(Cu\right)$ present in $25\text{ g}$ is $0.394\text{ moles}$ .

Moles of mercury $\left(Hg\right)$ present in $25\text{ g}$ is $0.12\text{ moles}$

Explanation of Solution

Atomic weight of iron $\left(Fe\right)$ is $55.8\text{ g/mol}$

Atomic weight of iron $\left(S\right)$ is $32.1\text{ g/mol}$

Atomic weight of copper $\left(Cu\right)$ is $63.5\text{ g/mol}$

Atomic weight of iron $\left(Hg\right)$ is $200.6\text{ g/mol}$

Mass of iron, copper, sulfur, and mercury present is $25\text{ g}$ .

Moles of solute is the ratio of the mass of a compound to the atomic/molecular mass of that compound.

The expression for moles is as follows:

  $Moles=\frac{Mass}{Molarmass}$

Substitute $25\text{ g}$ as the mass of iron and $55.8\text{ g/mol}$ as the molar mass in the mole expression.

  $\begin{array}{c}Moles=\frac{25\text{ g}}{55.8\text{ g/mol}}\\ =0.448\text{ moles moles}\end{array}$

Substitute $25\text{ g}$ as the mass of iron and $32.1\text{ g/mol}$ as the molar mass in the mole expression.

  $\begin{array}{c}Moles=\frac{25\text{ g}}{32.1\text{ g/mol }}\\ =0.779\text{ moles}\end{array}$

Substitute, $25\text{ g}$ as the mass of iron and $63.5\text{ g/mol}$ as the molar mass in the mole expression.

  $\begin{array}{c}Moles=\frac{25\text{ g}}{32.1\text{ g/mol }}\\ =0.394\text{ moles}\end{array}$

Substitute, $25\text{ g}$ as the mass of iron and $200.6\text{ g/mol}$ as the molar mass in the mole expression.

  $\begin{array}{c}Moles=\frac{25\text{ g}}{200.6\text{ g/mol}}\\ =0.125\text{ moles}\end{array}$