Chapter 8, Problem 84AP

Complete the following table.

l>

Mass of Sample	Moles of Sample	Atoms in Sample
5.00 g AI

td>


0.00250 mol Fe

td>

msp;  $2.6\times {10}^{24}$ atoms Cu 0.00250 g Mg

td>

msp;  $2.7\times {10}^{-3}$ mol Na

msp;  $1.00\times {10}^4$ atoms U

Interpretation Introduction

Interpretation:

The given table should be completed.

Mass of sample	Moles of sample	Atoms in sample
5.00 g Al
	0.00250 mol Fe
		$2.6\times {10}^{24}$ atoms Cu
0.00250 g Mg
	$2.7\times {10}^{-3}\text{ mol}$ Na
		$1.00\times {10}^4$ atoms U

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 84AP

Mass of sample	Moles of sample	Atoms in sample
5.00 g Al	0.185 mol Al	$1.12\times {10}^{23}\text{ atoms}$ Al
0.14 g Fe	0.00250 mol Fe	$1.50\times {10}^{21}\text{ atoms}$ Fe
273.75 g Cu	4.3 mol Cu	$2.6\times {10}^{24}$ atoms Cu
0.00250 g Mg	0.00010 mol Mg	$6.023\times {10}^{19}\text{ atoms}$ Mg
0.0621 g Na	$2.7\times {10}^{-3}\text{ mol}$ Na	$1.63\times {10}^{21}\text{ atoms}$ Na
$\text{3}\text{.95}\times {\text{10}}^{-17}\text{ g}$ U	$1.66\times {10}^{-19}\text{ mol}$ U	$1.00\times {10}^4$ atoms U

Explanation of Solution

Mass of aluminium is 5 g and molar mass of Al is 26.98 g/mol thus, number of moles can be calculated as follows:

$\begin{array}{c}\text{n}=\frac{5\text{ g}}{26.98\text{ g/mol}}\\ =0.185\text{ mol}\end{array}$

Since, 1 mol of a substance contains$6.023\times {10}^{23}$ atoms thus; number of atoms in 0.185 mol can be calculated as follows:

$\begin{array}{c}\text{N}=0.185\text{ mol}\left(\frac{6.023\times {10}^{23}\text{ atoms}}{1\text{ mol}}\right)\\ =1.12\times {10}^{23}\text{ atoms}\end{array}$

Number of moles of iron is 0.00250 mol and molar mass of Fe is 55.845 g/mol thus, mass can be calculated as follows:

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

Putting the values,

$\begin{array}{c}\text{m}=0.00250\text{ mol}\times 55.845\text{ g/mol}\\ =\text{0}\text{.14 g}\end{array}$

Since, 1 mol of a substance contains$6.023\times {10}^{23}$ atoms thus; number of atoms in 0.00250 mol can be calculated as follows:

$\begin{array}{c}\text{N}=0.00250\text{ mol}\left(\frac{6.023\times {10}^{23}\text{ atoms}}{1\text{ mol}}\right)\\ =1.50\times {10}^{21}\text{ atoms}\end{array}$

Number of atoms of copper is$2.6\times {10}^{24}\text{ atoms}$

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

$\begin{array}{c}\text{n}=2.6\times {10}^{24}\text{ atoms}\left(\frac{1\text{ mol}}{6.023\times {10}^{23}\text{ atoms}}\right)\\ =4.3\text{ mol}\end{array}$

Molar mass of copper is 63.546 g/mol thus, mass of copper is:

$\begin{array}{c}\text{m}=\text{n}\times \text{M}\\ =4.3\text{ mol}\times \text{63}\text{.54 g/mol}\\ \text{=273}\text{.25 g}\end{array}$

Mass of magnesium is 0.00250 g and molar mass of Mg is 24.305 g/mol thus, number of moles can be calculated as follows:

$\begin{array}{c}\text{n}=\frac{\text{0}\text{.00250 g}}{24.305\text{ g/mol}}\\ =0.00010\text{ mol}\end{array}$

Since, 1 mol of a substance contains$6.023\times {10}^{23}$ atoms thus; number of atoms in 0.00010 mol can be calculated as follows:

$\begin{array}{c}\text{N}=0.00010\text{ mol}\left(\frac{6.023\times {10}^{23}\text{ atoms}}{1\text{ mol}}\right)\\ =6.023\times {10}^{19}\text{ atoms}\end{array}$

Number of moles of sodium is$2.7\times {10}^{-3}$ mol and molar mass of Na is 23 g/mol thus, mass can be calculated as follows:

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

Putting the values,

$\begin{array}{c}\text{m}=2.7\times {10}^3\text{ mol}\times 23\text{ g/mol}\\ =\text{0}\text{.0621g}\end{array}$

Since, 1 mol of a substance contains$6.023\times {10}^{23}$ atoms thus; number of atoms in$2.7\times {10}^{-3}$ mol can be calculated as follows:

$\begin{array}{c}\text{N}=2.7\times {10}^{-3}\text{ mol}\left(\frac{6.023\times {10}^{23}\text{ atoms}}{1\text{ mol}}\right)\\ =1.63\times {10}^{21}\text{ atoms}\end{array}$

Number of atoms of uranium is$1.0\times {10}^4\text{ atoms}$

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

$\begin{array}{c}\text{n}=1.0\times {10}^4\text{ atoms}\left(\frac{1\text{ mol}}{6.023\times {10}^{23}\text{ atoms}}\right)\\ =1.66\times {10}^{-19}\text{ mol}\end{array}$

Molar mass of uranium is 238.023 g/mol thus, mass of uranium is:

$\begin{array}{c}\text{m}=\text{n}\times \text{M}\\ =1.66\times {10}^{-19}\text{ mol}\times 238.023\text{ g/mol}\\ \text{=3}\text{.95}\times {\text{10}}^{-17}\text{ g}\end{array}$.

Conclusion

The complete table is as follows:

Mass of sample	Moles of sample	Atoms in sample
5.00 g Al	0.185 mol Al	$1.12\times {10}^{23}\text{ atoms}$ Al
0.14 g Fe	0.00250 mol Fe	$1.50\times {10}^{21}\text{ atoms}$ Fe
273.75 g Cu	4.3 mol Cu	$2.6\times {10}^{24}$ atoms Cu
0.00250 g Mg	0.00010 mol Mg	$6.023\times {10}^{19}\text{ atoms}$ Mg
0.0621 g Na	$2.7\times {10}^{-3}\text{ mol}$ Na	$1.63\times {10}^{21}\text{ atoms}$ Na
$\text{3}\text{.95}\times {\text{10}}^{-17}\text{ g}$ U	$1.66\times {10}^{-19}\text{ mol}$ U	$1.00\times {10}^4$ atoms U