Chapter 2, Problem 2.75P

(a)

Interpretation Introduction

Interpretation:Molar mass and number of $\text{H}$ atoms ${\text{C}}_6{\text{H}}_5\text{COONH}$ should be determined.

Concept introduction:Molecular mass of a chemical compound is defined as sum of masses of all atoms contained in a molecule.

The formula to calculate molecular mass of compound is as follows:

  $\text{Molecular mass of compound}=\left[\begin{array}{l}\Sigma \left(\text{number of atoms of element}\right)\\ \left(\text{atomic mass of element}\right)\end{array}\right]$

Explanation of Solution

A molecular is a representation of a molecule by chemical symbol to indicate different types of atoms and how each atom connected. Here the number of hydrogen atoms in ${\text{C}}_6{\text{H}}_5\text{COONH}$ is $9$ atoms.

Formula to calculate molar mass of ${\text{C}}_6{\text{H}}_5\text{COONH}$ is as follows:

  $\text{Molecular mass}=\left[\begin{array}{l}\left(7\right)\left(\text{atomic mass of C}\right)+\\ \left(9\right)\left(\text{atomic mass of H}\right)+\\ \left(2\right)\left(\text{atomic mass of O}\right)+\\ \left(1\right)\left(\text{atomic mass of N}\right)\end{array}\right]$

Atomic mass of $\text{C}$ is $12.01\text{ amu}$ .

Atomic mass of $\text{O}$ is $16\text{ amu}$ .

Atomic mass of $\text{N}$ is $14.01\text{ amu}$ .

Atomic mass of $\text{H}$ is $1.008\text{ amu}$ .

Substitute the values in the above formula.

  $\begin{array}{c}\text{Molecular mass}=\left[\begin{array}{l}\left(7\right)\left(\text{atomic mass of C}\right)+\\ \left(9\right)\left(\text{atomic mass of H}\right)+\\ \left(2\right)\left(\text{atomic mass of O}\right)+\\ \left(1\right)\left(\text{atomic mass of N}\right)\end{array}\right]\\ \text{Molecular mass}=\left[\begin{array}{l}\left(7\right)\left(12.01\text{ amu}\right)+\left(9\right)\left(1.008\text{ amu}\right)+\left(2\right)\left(16\text{ amu}\right)+\\ \left(1\right)\left(14.01\text{ amu}\right)\end{array}\right]\\ =84.07\text{ amu}+9.072\text{ amu}+\text{3}2\text{ amu}+14.01\text{ amu}\\ =139.15\text{ amu}\end{array}$

(b)

Interpretation Introduction

Interpretation: Molar mass and number of $\text{N}$ atoms ${\text{N}}_2{\text{H}}_6{\text{SO}}_4$ should be determined.

Concept introduction:Molecular mass of a chemical compound is defined as sum of masses of all atoms contained in a molecule.

The formula to calculate molecular mass of compound is as follows:

  $\text{Molecular mass of compound}=\left[\begin{array}{l}\Sigma \left(\text{number of atoms of element}\right)\\ \left(\text{atomic mass of element}\right)\end{array}\right]$

Explanation of Solution

A molecular is a representation of a molecule by chemical symbol to indicate different types of atoms and how each atom connected. Here the number of $\text{N}$ atoms in ${\text{N}}_2{\text{H}}_6{\text{SO}}_4$ is $2$ atoms.

Formula to calculate molar mass of ${\text{N}}_2{\text{H}}_6{\text{SO}}_4$ is as follows:

  $\text{Molecular mass}=\left[\begin{array}{l}\left(2\right)\left(\text{atomic mass of N}\right)+\\ \left(6\right)\left(\text{atomic mass of H}\right)+\\ \left(1\right)\left(\text{atomic mass of S}\right)+\\ \left(4\right)\left(\text{atomic mass of O}\right)\end{array}\right]$

Atomic mass of $\text{S}$ is $32.06\text{ amu}$ .

Atomic mass of $\text{O}$ is $16\text{ amu}$ .

Atomic mass of $\text{N}$ is $14.01\text{ amu}$ .

Atomic mass of $\text{H}$ is $1.008\text{ amu}$ .

Substitute the values in the above equation.

  $\begin{array}{c}\text{Molecular mass}=\left[\begin{array}{l}\left(2\right)\left(\text{atomic mass of N}\right)+\\ \left(6\right)\left(\text{atomic mass of H}\right)+\\ \left(1\right)\left(\text{atomic mass of S}\right)+\\ \left(4\right)\left(\text{atomic mass of O}\right)\end{array}\right]\\ =\left[\begin{array}{l}\left(2\right)\left(14.01\text{ amu}\right)+\left(6\right)\left(1.008\text{ amu}\right)+\left(1\right)\left(32.06\text{ amu}\right)+\\ \left(4\right)\left(16\text{ amu}\right)\end{array}\right]\\ =28.02\text{ amu}+9.072\text{ amu}+32.06\text{ amu}+64\text{ amu}\\ =130.13\text{ amu}\end{array}$

(c)

Interpretation Introduction

Interpretation: The number of $\text{O}$ atoms and molar mass ${\text{Pb}}_4{\text{SO}}_4\left({\text{CO}}_3\right)\left(\text{OH}\right){}_2$ should be determined.

Concept introduction:Molecular mass of a chemical compound is defined as sum of masses of all atoms contained in a molecule.

The formula to calculate molecular mass of compound is as follows:

  $\text{Molecular mass of compound}=\left[\begin{array}{l}\Sigma \left(\text{number of atoms of element}\right)\\ \left(\text{atomic mass of element}\right)\end{array}\right]$

Explanation of Solution

A molecular is a representation of a molecule by chemical symbol to indicate different types of atoms and how each atom connected. The number of $\text{O}$ atoms in ${\text{Pb}}_4{\text{SO}}_4\left({\text{CO}}_3\right)\left(\text{OH}\right){}_2$ is $12$ atoms.

Formula to calculate molar mass of ${\text{Pb}}_4{\text{SO}}_4\left({\text{CO}}_3\right)\left(\text{OH}\right){}_2$ is as follow

  $\text{Molecular mass}=\left[\begin{array}{l}\left(4\right)\left(\text{atomic mass of Pb}\right)+\\ \left(1\right)\left(\text{atomic mass of S}\right)+\\ \left(2\right)\left(\text{atomic mass of C}\right)+\\ \left(12\right)\left(\text{atomic mass of O}\right)+\\ \left(2\right)\left(\text{atomic mass of H}\right)\end{array}\right]$

Atomic mass of $\text{Pb}$ is $207.2\text{ amu}$ .

Atomic mass of $\text{S}$ is $32.06\text{ amu}$ .

Atomic mass of $\text{O}$ is $16\text{ amu}$ .

Atomic mass of $\text{N}$ is $14.01\text{ amu}$ .

Atomic mass of $\text{H}$ is $1.008\text{ amu}$ .

Substitute the value in the above equation.

  $\begin{array}{c}\text{Molecular mass}=\left[\begin{array}{l}\left(4\right)\left(\text{atomic mass of Pb}\right)+\\ \left(1\right)\left(\text{atomic mass of S}\right)+\\ \left(2\right)\left(\text{atomic mass of C}\right)+\\ \left(12\right)\left(\text{atomic mass of O}\right)+\\ \left(2\right)\left(\text{atomic mass of H}\right)\end{array}\right]\\ =\left[\begin{array}{l}\left(4\right)\left(207.2\text{ amu}\right)+\left(1\right)\left(32.06\text{ amu}\right)+\left(2\right)\left(12.01\text{ amu}\right)+\\ \left(12\right)\left(16\text{ amu}\right)+\left(2\right)\left(1.008\text{ amu}\right)\end{array}\right]\\ =828.8\text{ amu}+32.06\text{ amu}+24.02\text{ amu}+192\text{ amu}+2.016\text{ amu}\\ =1078.9\text{ amu}\end{array}$