Chapter 8, Problem 85AP

Complete the following table.

l>

Mass of Sample	Moles of Sample	Molecules in Sample	Atoms in Sample
4.24 g

msp;  ${\text{C}}_6{\text{H}}_6$



0.224 mol

msp;  ${\text{H}}_2\text{O}$

msp;  $\text{2}\text{.71}\times {\text{10}}^{22}$ molecules

msp;  ${\text{CO}}_2$

1.26 mol HCl

td>

msp;  $\text{2}\text{.21}\times {\text{10}}^{24}$ molecules

msp;  ${\text{H}}_2\text{O}$
0.297 g

msp;  ${\text{CH}}_3\text{OH}$

Interpretation Introduction

Interpretation:

The given table should be completed.

Mass of sample	Moles of sample	Molecules in sample	Atoms in sample
4.24 g${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$
	0.224 mol${\text{H}}_{\text{2}}\text{O}$
		$2.71\times {10}^{22}$ molecules${\text{CO}}_{\text{2}}$
	1.26 mol$\text{HCl}$
		$4.21\times {10}^{24}$ molecules${\text{H}}_{\text{2}}\text{O}$
0.297 g${\text{CH}}_{\text{3}}\text{OH}$

Concept Introduction:

Number of moles is related to mass and molar mass as follows:

$\text{n}=\frac{\text{m}}{\text{M}}$

Here, m is mass and M is molar mass.

According to Avogadro’s law, 1 mole of a substance contains$6.023\times {10}^{23}$ atoms. This is known as Avogadro’s number and denoted by symbol${\text{N}}_{\text{A}}$.

Thus, number of molecules can be calculated from number of moles using the following conversion factor:

$\left(\frac{6.023\times {10}^{23}\text{ molecules}}{1\text{ mol}}\right)$.

Answer to Problem 85AP

Mass of sample	Moles of sample	Molecules in sample	Atoms in sample
4.24 g${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$	0.0543 mol${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$	$3.27\times {10}^{22}\text{ molecules}$ ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$	$3.924\times {10}^{23}\text{ atoms}$ ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$
4.032 g${\text{H}}_{\text{2}}\text{O}$	0.224 mol${\text{H}}_{\text{2}}\text{O}$	$3.27\times {10}^{22}\text{ molecules}$ ${\text{H}}_{\text{2}}\text{O}$	$3.924\times {10}^{23}\text{ atoms}$ ${\text{H}}_{\text{2}}\text{O}$
1.98 g${\text{CO}}_{\text{2}}$	$4.5\times {10}^{-2}$ mol${\text{CO}}_{\text{2}}$	$2.71\times {10}^{22}$ molecules${\text{CO}}_{\text{2}}$	$8.13\times {10}^{22}$ atoms${\text{CO}}_{\text{2}}$
45.94 g$\text{HCl}$	1.26 mol$\text{HCl}$	$7.58\times {10}^{23}$ molecules$\text{HCl}$	$15.18\times {10}^{23}$ atoms$\text{HCl}$
125.8 g${\text{H}}_{\text{2}}\text{O}$	6.98 mol${\text{H}}_{\text{2}}\text{O}$	$4.21\times {10}^{24}$ molecules${\text{H}}_{\text{2}}\text{O}$	$12.63\times {10}^{24}$ atoms${\text{H}}_{\text{2}}\text{O}$
0.297 g${\text{CH}}_{\text{3}}\text{OH}$	0.000927 mol${\text{CH}}_{\text{3}}\text{OH}$	$5.58\times {10}^{21}$ molecules${\text{CH}}_{\text{3}}\text{OH}$	$3.35\times {10}^{22}$ atoms${\text{CH}}_{\text{3}}\text{OH}$

Explanation of Solution

Mass of${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$ is 4.24 g and molar mass of${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$ is 78.11 g/mol thus, number of moles can be calculated as follows:

$\begin{array}{c}\text{n}=\frac{\text{4}\text{.24 g}}{\text{78}\text{.11 g/mol}}\\ =0.0543\text{ mol}\end{array}$

Since, 1 mol of a substance contains$6.023\times {10}^{23}$ molecules thus; number of molecules in 0.0543 mol can be calculated as follows:

$\begin{array}{c}{\text{N}}_{\text{molecules}}=0.0543\text{ mol}\left(\frac{6.023\times {10}^{23}\text{ molecules}}{1\text{ mol}}\right)\\ =3.27\times {10}^{22}\text{ molecules}\end{array}$

Since, 1 molecule of${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$ contains 12 atoms thus, number of atoms in$3.27\times {10}^{22}\text{ molecules}$ of${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$ will be:

$\begin{array}{c}{\text{N}}_{\text{atoms}}=12\left(3.27\times {10}^{22}\right)\\ =3.924\times {10}^{23}\text{ atoms}\end{array}$

Number of moles of${\text{H}}_{\text{2}}\text{O}$ is 0.244 mol and molar mass of${\text{H}}_{\text{2}}\text{O}$ is 18 g/mol thus, mass can be calculated as follows:

$\text{m}=\text{n}\times \text{M}$

Putting the values,

$\begin{array}{c}\text{m}=0.224\text{ mol}\times 18\text{ g/mol}\\ =\text{4}\text{.032 g}\end{array}$

Since, 1 mol of a substance contains$6.023\times {10}^{23}$ molecules thus; number of molecules in 0.224 mol can be calculated as follows:

$\begin{array}{c}{\text{N}}_{\text{molecules}}=0.224\text{ mol}\left(\frac{6.023\times {10}^{23}\text{ molecules}}{1\text{ mol}}\right)\\ =1.35\times {10}^{23}\text{ molecules}\end{array}$

Since, 1 mole of water molecule contains 3 atoms thus, number of atoms will be:

$\begin{array}{c}{\text{N}}_{\text{atoms}}=3\left(1.35\times {10}^{23}\right)\\ =4.05\times {10}^{23}\text{ atoms}\end{array}$

Number of molecules of${\text{CO}}_{\text{2}}$ is$2.71\times {10}^{22}\text{ molecules}$

Since, 1 mol of a substance contains$6.023\times {10}^{23}$ molecules thus; number of moles can be calculated as follows:

$\begin{array}{c}\text{n}=2.71\times {10}^{22}\text{ molecules}\left(\frac{1\text{ mol}}{6.023\times {10}^{23}\text{ molecules}}\right)\\ =4.5\times {10}^{-2}\text{ mol}\end{array}$

Molar mass of carbon dioxide is 44.01 g/mol thus, mass of carbon dioxide is:

$\begin{array}{c}\text{m}=\text{n}\times \text{M}\\ =4.5\times {10}^{-2}\text{ mol}\times 44.01\text{ g/mol}\\ \text{=1}\text{.98 g}\end{array}$

Since, 1 molecule of carbon dioxide contains 3 atoms thus, number of atoms can be calculated as follows:

$\begin{array}{c}{\text{N}}_{\text{atoms}}=3\times 2.71\times {10}^{22}\\ =8.13\times {10}^{22}\text{ atoms}\end{array}$

Number of moles of HCl is 1.26 mol and molar mass of HCl is 36.46 g/mol thus, mass can be calculated as follows:

$\text{m}=\text{n}\times \text{M}$

Putting the values,

$\begin{array}{c}\text{m}=1.26\text{ mol}\times 36.46\text{ g/mol}\\ =\text{45}\text{.94 g}\end{array}$

Since, 1 mol of a substance contains$6.023\times {10}^{23}$ molecules thus; number of molecules in 1.26 mol can be calculated as follows:

$\begin{array}{c}{\text{N}}_{\text{molecules}}=1.26\text{ mol}\left(\frac{6.023\times {10}^{23}\text{ molecules}}{1\text{ mol}}\right)\\ =7.58\times {10}^{23}\text{ molecules}\end{array}$

Since, 1 mol of HCl contains 2 atoms thus, number of atoms can be calculated as follows:

$\begin{array}{c}{\text{N}}_{\text{atoms}}=2\times 7.58\times {10}^{23}\\ \text{=15}\text{.18}\times {\text{10}}^{23}\text{ atoms}\end{array}$

Number of molecules of${\text{H}}_{\text{2}}^{}\text{O}$ is$4.21\times {10}^{24}\text{ atoms}$

Since, 1 mol of a substance contains$6.023\times {10}^{23}$ molecules thus; number of moles can be calculated as follows:

$\begin{array}{c}\text{n}=4.21\times {10}^{24}\text{ molecules}\left(\frac{1\text{ mol}}{6.023\times {10}^{23}\text{ molecules}}\right)\\ =6.98\text{ mol}\end{array}$

Molar mass of water is 18 g/mol thus, mass of water is:

$\begin{array}{c}\text{m}=\text{n}\times \text{M}\\ =6.98\text{ mol}\times 18\text{ g/mol}\\ \text{=}125.8\text{ g}\end{array}$

Since, 1 molecule of water contains 3 atoms thus, number of atoms can be calculated as follows;

$\begin{array}{c}{\text{N}}_{\text{atoms}}=3\times 4.21\times {10}^{24}\text{ atoms}\\ \text{=12}\text{.63}\times {\text{10}}^{24}\text{ atoms}\end{array}$

Mass of${\text{CH}}_{\text{3}}\text{OH}$ is 0.297 g and molar mass is 32.04 g/mol thus, number of moles can be calculated as follows:

$\begin{array}{c}\text{n}=\frac{\text{m}}{\text{M}}\\ =\frac{0.297\text{ g}}{32.04\text{ g/mol}}\\ =0.000927\text{ mol}\end{array}$

Since, 1 mol of a substance contains$6.023\times {10}^{23}$ molecules thus, number of molecules can be calculated as follows:

$\begin{array}{c}{\text{N}}_{\text{molecules}}=0.000927\text{ mol}\left(\frac{6.023\times {10}^{23}\text{ molecules}}{1\text{ mol}}\right)\\ =5.58\times {10}^{21}\text{ molecules}\end{array}$

Since, 1 mol of${\text{CH}}_{\text{3}}\text{OH}$ contains 6 atoms thus, number of atoms can be calculated as follows:

$\begin{array}{c}{\text{N}}_{\text{atoms}}=6\times 5.58\times {10}^{21}\\ \text{=3}\text{.35}\times {\text{10}}^{22}\text{ atoms}\end{array}$.

Conclusion

The complete table is as follows:

Mass of sample	Moles of sample	Molecules in sample	Atoms in sample
4.24 g${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$	0.0543 mol${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$	$3.27\times {10}^{22}\text{ molecules}$ ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$	$3.924\times {10}^{23}\text{ atoms}$ ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$
4.032 g${\text{H}}_{\text{2}}\text{O}$	0.224 mol${\text{H}}_{\text{2}}\text{O}$	$3.27\times {10}^{22}\text{ molecules}$ ${\text{H}}_{\text{2}}\text{O}$	$3.924\times {10}^{23}\text{ atoms}$ ${\text{H}}_{\text{2}}\text{O}$
1.98 g${\text{CO}}_{\text{2}}$	$4.5\times {10}^{-2}$ mol${\text{CO}}_{\text{2}}$	$2.71\times {10}^{22}$ molecules${\text{CO}}_{\text{2}}$	$8.13\times {10}^{22}$ atoms${\text{CO}}_{\text{2}}$
45.94 g$\text{HCl}$	1.26 mol$\text{HCl}$	$7.58\times {10}^{23}$ molecules$\text{HCl}$	$15.18\times {10}^{23}$ atoms$\text{HCl}$
125.8 g${\text{H}}_{\text{2}}\text{O}$	6.98 mol${\text{H}}_{\text{2}}\text{O}$	$4.21\times {10}^{24}$ molecules${\text{H}}_{\text{2}}\text{O}$	$12.63\times {10}^{24}$ atoms${\text{H}}_{\text{2}}\text{O}$
0.297 g${\text{CH}}_{\text{3}}\text{OH}$	0.000927 mol${\text{CH}}_{\text{3}}\text{OH}$	$5.58\times {10}^{21}$ molecules${\text{CH}}_{\text{3}}\text{OH}$	$3.35\times {10}^{22}$ atoms${\text{CH}}_{\text{3}}\text{OH}$