Chapter 3, Problem 34E

Interpretation Introduction

Interpretation: The given table is to be completed.

Concept introduction: The amount corresponding to any substance or compound can be conveniently evaluated through mole concept. This theory associates the substance’s moles (amount) and its respective molar mass. Mathematically, this relationship is depicted by the formula, $\text{Moles}=\text{Given}\text{mass}/\text{Molar}\text{mass}$ . The Avogadro’s number interconverts substance’s moles to its molecules from where its respective atoms can be interpreted.

Answer to Problem 34E

The completetable is as follows.

Mass of Sample	Moles of Sample	Molecules in Sample	Total Atoms in Sample
$4.24\text{g}$${\text{C}}_6{\text{H}}_6$	$0.0543\text{mol}$${\text{C}}_6{\text{H}}_6$	$3.27\times {10}^{22}$ molecules ${\text{C}}_6{\text{H}}_6$	$3.92\times {10}^{23}$ total atoms in ${\text{C}}_6{\text{H}}_6$ sample
$4.035\text{g}$${\text{H}}_2\text{O}$	$0.224\text{mol}$${\text{H}}_2\text{O}$	$1.35\times {10}^{23}$ molecules ${\text{H}}_2\text{O}$	$4.05\times {10}^{23}$ total atoms in ${\text{H}}_2\text{O}$ sample
$1.98\text{g}$${\text{CO}}_2$	$0.045\text{mol}$${\text{CO}}_2$	$2.71\times {10}^{22}$ molecules ${\text{CO}}_2$	$8.13\times {10}^{22}$ total atoms in ${\text{CO}}_2$ sample
$0.297\text{g}$${\text{CH}}_3\text{OH}$	$9.264\times {10}^{-3}\text{mol}$${\text{CH}}_3\text{OH}$	$5.58\times {10}^{21}$ molecules ${\text{CH}}_3\text{OH}$	$3.35\times {10}^{22}$ total atoms in ${\text{CH}}_3\text{OH}$ sample

Explanation of Solution

The given mass of benzene $\left({\text{C}}_6{\text{H}}_6\right)$ is $4.24\text{g}$ .

The number of moles of ${\text{C}}_6{\text{H}}_6$ can be calculated as follows.

  $\text{Moles}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$  (1)

The molar mass of ${\text{C}}_6{\text{H}}_6$ is $78.11\text{g/mol}$ .

Substitute the values in equation (1).

  $\begin{array}{c}\text{Moles}\text{of}{\text{C}}_6{\text{H}}_6=\frac{4.24\text{g}}{78.11\text{g/mol}}\\ =0.0543\text{mol}\end{array}$

The number of molecules can be calculated as follows:

  $\text{Number}\text{of}\text{molecules}=\text{Moles}\times \text{Avogadro's}\text{number}$ (2)

The value of Avogadro’s number is $6.023\times {10}^{23}{\text{mol}}^{-1}$ .

Substitute the values in equation (2).

  $\text{Number}\text{of}\text{molecules}=\text{Moles}\times 6.023\times {10}^{23}{\text{mol}}^{-1}$(3)

Substitute the respective values in equation (3).

  $\begin{array}{c}\text{Number}\text{of}{\text{C}}_6{\text{H}}_6\text{molecules}=0.0543\text{mol}\times 6.023\times {10}^{23}{\text{mol}}^{-1}\\ =3.27\times {10}^{22}\end{array}$

The number of atoms can be calculated as follows.

  $\text{Total}\text{atoms}=\text{Atoms}\text{per}\text{molecule}\times \text{Total}\text{molecules}$ (4)

The total number of atoms in ${\text{C}}_6{\text{H}}_6$ is $12$ .

Substitute the respective values in equation (4).

  $\begin{array}{c}\text{Total}\text{atoms}\text{in}{\text{C}}_6{\text{H}}_6=\text{12}\times 3.27\times {10}^{22}\\ =3.92\times {10}^{23}\end{array}$

The given moles of water $\left({\text{H}}_2\text{O}\right)$ are $0.224\text{mol}$ .

The molar mass of water is $18.015\text{g/mol}$ .

The mass of water can be calculated using equation (1) as follows.

  $\begin{array}{c}0.224\text{mol}=\frac{\text{Mass}\text{of}{\text{H}}_2\text{O}}{18.015\text{g/mol}}\\ \text{Mass}\text{of}{\text{H}}_2\text{O}=0.224\text{mol}\times 18.015\text{g/mol}\\ \text{Mass}\text{of}{\text{H}}_2\text{O}=4.035\text{g}\end{array}$

The number of molecules of ${\text{H}}_2\text{O}$ can be calculated using equation (3) as follows.

  $\begin{array}{c}\text{Number}\text{of}{\text{H}}_2\text{O}\text{molecules}=0.224\text{mol}\times 6.023\times {10}^{23}{\text{mol}}^{-1}\\ =1.35\times {10}^{23}\end{array}$

The number of atoms in ${\text{H}}_2\text{O}$ molecule are $3$ .

The total number of atoms in ${\text{H}}_2\text{O}$ can be calculated using equation (4) as follows.

  $\begin{array}{c}\text{Total}\text{atoms}\text{in}{\text{H}}_2\text{O}=3\times 1.35\times {10}^{23}\\ =4.05\times {10}^{23}\end{array}$

The given number of molecules of ${\text{CO}}_2$ are $2.71\times {10}^{22}$ .

The number of moles of ${\text{CO}}_2$ can be calculated using equation (3) as follows.

  $\begin{array}{c}2.71\times {10}^{22}=\text{Moles}\text{of}{\text{CO}}_2\times 6.023\times {10}^{23}{\text{mol}}^{-1}\\ \text{Moles}\text{of}{\text{CO}}_2=\frac{2.71\times {10}^{22}}{6.023\times {10}^{23}{\text{mol}}^{-1}}\\ \text{Moles}\text{of}{\text{CO}}_2=0.045\text{mol}\end{array}$

The molar mass of ${\text{CO}}_2$ is $44.01\text{g/mol}$ .

The mass of ${\text{CO}}_2$ can be calculated using equation (1) as follows.

  $\begin{array}{c}0.045\text{mol}=\frac{\text{Mass}\text{of}{\text{CO}}_2}{44.01\text{g/mol}}\\ \text{Mass}\text{of}{\text{CO}}_2=0.045\text{mol}\times 44.01\text{g/mol}\\ \text{Mass}\text{of}{\text{CO}}_2=1.98\text{g}\end{array}$

The number of atoms in ${\text{CO}}_2$ molecules are $3$ .

The total atoms in ${\text{CO}}_2$ can be calculated using equation (4) as follows.

  $\begin{array}{c}\text{Total}\text{atoms}\text{in}{\text{CO}}_2=3\times 2.71\times {10}^{22}\\ =8.13\times {10}^{22}\end{array}$

The given total number of atoms in ${\text{CH}}_3\text{OH}$ is $3.35\times {10}^{22}$ atoms.

The number of atoms in ${\text{CH}}_3\text{OH}$ molecule are $6$ .

The total number of molecules in ${\text{CH}}_3\text{OH}$ can be calculated using equation (4) as follows.

  $\begin{array}{c}3.35\times {10}^{22}\text{atoms}=6\text{atoms}\times \text{Total}\text{molecules}\text{in}{\text{CH}}_3\text{OH}\\ \text{Total}\text{molecules}\text{in}{\text{CH}}_3\text{OH}=\frac{3.35\times {10}^{22}\text{atoms}}{6\text{atoms}}\\ \text{Total}\text{molecules}\text{in}{\text{CH}}_3\text{OH}=5.58\times {10}^{21}\end{array}$

The number of moles of ${\text{CH}}_3\text{OH}$ can be calculated using equation (3) as follows.

  $\begin{array}{c}5.58\times {10}^{21}=\text{Moles}\text{of}{\text{CH}}_3\text{OH}\times 6.023\times {10}^{23}{\text{mol}}^{-1}\\ \text{Moles}\text{of}{\text{CH}}_3\text{OH}=\frac{5.58\times {10}^{21}}{6.023\times {10}^{23}{\text{mol}}^{-1}}\\ \text{Moles}\text{of}{\text{CH}}_3\text{OH}=9.264\times {10}^{-3}\text{mol}\end{array}$

The molar mass of ${\text{CH}}_3\text{OH}$ is $32.04\text{g/mol}$ .

The mass of ${\text{CH}}_3\text{OH}$ can be calculated using equation (1) as follows.

  $\begin{array}{c}9.264\times {10}^{-3}\text{mol}=\frac{\text{Mass}\text{of}{\text{CH}}_3\text{OH}}{32.04\text{g/mol}}\\ \text{Mass}\text{of}{\text{CH}}_3\text{OH}=9.264\times {10}^{-3}\text{mol}\times 32.04\text{g/mol}\\ \text{Mass}\text{of}{\text{CH}}_3\text{OH}=0.297\text{g}\end{array}$

Therefore, the complete table is as follows.

Mass of Sample	Moles of Sample	Molecules in Sample	Total Atoms in Sample
$4.24\text{g}$${\text{C}}_6{\text{H}}_6$	$0.0543\text{mol}$${\text{C}}_6{\text{H}}_6$	$3.27\times {10}^{22}$ molecules ${\text{C}}_6{\text{H}}_6$	$3.92\times {10}^{23}$ total atoms in ${\text{C}}_6{\text{H}}_6$ sample
$4.035\text{g}$${\text{H}}_2\text{O}$	$0.224\text{mol}$${\text{H}}_2\text{O}$	$1.35\times {10}^{23}$ molecules ${\text{H}}_2\text{O}$	$4.05\times {10}^{23}$ total atoms in ${\text{H}}_2\text{O}$ sample
$1.98\text{g}$${\text{CO}}_2$	$0.045\text{mol}$${\text{CO}}_2$	$2.71\times {10}^{22}$ molecules ${\text{CO}}_2$	$8.13\times {10}^{22}$ total atoms in ${\text{CO}}_2$ sample
$0.297\text{g}$${\text{CH}}_3\text{OH}$	$9.264\times {10}^{-3}\text{mol}$${\text{CH}}_3\text{OH}$	$5.58\times {10}^{21}$ molecules ${\text{CH}}_3\text{OH}$	$3.35\times {10}^{22}$ total atoms in ${\text{CH}}_3\text{OH}$ sample

Conclusion

The complete table for the given amount of substances is given above.