Chapter 8, Problem 18E

Interpretation Introduction

(a)

Interpretation:

The mass in grams for $\text{1}{\text{.21×10}}^{\text{24}}$ atoms of krypton is to be calculated.

Concept introduction:

One mole of atoms, molecules or particles is defined as a quantity whose mass is equivalent to its gram atomic mass or molecular mass. One mole of any substance possesses fixed Avogadro’s number, which is equivalent to $\text{6}{\text{.022×10}}^{\text{23}}$ atoms.

Answer to Problem 18E

The mass of $\text{1}{\text{.21×10}}^{\text{24}}$ atoms of krypton is $\text{168}\text{.350}\text{g}$.

Explanation of Solution

First the number of moles of krypton is calculated by the formula shown below.

$\text{Moles}=\frac{\text{Given}\text{atoms}\text{of}\text{Kr}}{\text{Avogadro's}\text{number}}$ … (1)

The atoms of krypton is $\text{1}\text{.21}\times {\text{10}}^{\text{24}}$.

Substitute the value of given atoms of krypton and the value of Avogadro’s number in equation (1) to calculate the number of moles of krypton.

$\begin{array}{rll}\text{Moles}& =& \frac{\text{1}\text{.21}\times {\text{10}}^{\text{24}}\text{atom}\text{Kr}}{\text{6}\text{.022}\times {\text{10}}^{\text{23}}\text{atom}\text{Kr}}\\ & =& 2.\text{009}\text{mol}\end{array}$

The mass in grams for $\text{1}{\text{.21×10}}^{\text{24}}$ atoms of krypton is calculated by the formula is shown below.

$\text{Mass}\text{of}\text{Kr}=\text{Moles}\text{of}\text{Kr}\times \text{Molar}\text{mass}\text{of}\text{Kr}$ … (2)

The molar mass of krypton is $83.798\text{g}{\text{mol}}^{-1}$.

Substitute the value of moles and molar mass of krypton in equation (2) to calculate the mass.

$\begin{array}{rll}\text{Mass}\text{of}\text{Kr}& =& \text{Moles}\text{of}\text{Kr}\times \text{Molar}\text{mass}\text{of}\text{Kr}\\ & =& \text{2}\text{.009}\text{mol}\times \text{83}\text{.798}\text{g}{\text{mol}}^{-1}\\ & =& \text{168}\text{.350}\text{g}\end{array}$

Therefore, the mass of $\text{1}{\text{.21×10}}^{\text{24}}$ atoms of krypton is $\text{168}\text{.350}\text{g}$.

Conclusion

The mass of $\text{1}\text{.21}\times {\text{10}}^{\text{24}}$ atoms of krypton is $\text{168}\text{.350}\text{g}$.

Interpretation Introduction

(b)

Interpretation:

The mass in grams for $\text{6}\text{.33}\times {\text{10}}^{\text{22}}$ molecules of dinitrogen oxide ${\text{N}}_{\text{2}}\text{O}$ is to be calculated.

Concept introduction:

One mole of atoms, molecules or particles is defined as a quantity whose mass is equivalent to its gram atomic mass or molecular mass. One mole of any substance possesses fixed Avogadro’s number, which is equivalent to $\text{6}{\text{.022×10}}^{\text{23}}$ atoms.

Answer to Problem 18E

The mass of $\text{6}\text{.33}\times {\text{10}}^{\text{22}}$ molecules of dinitrogen oxide ${\text{N}}_{\text{2}}\text{O}$ is $\text{4}\text{.62}\text{g}$.

Explanation of Solution

The mass of $\text{6}\text{.33}\times {\text{10}}^{\text{22}}$ molecules of dinitrogen oxide, ${\text{N}}_{\text{2}}\text{O}$ oxide, is calculated by the formula shown below.

$\text{Mass}=\frac{\text{Number}\text{of}\text{molecules}\text{of}{\text{N}}_{\text{2}}\text{O}}{\text{Avogadro's}\text{number}}\times \text{Molar}\text{mass}\text{of}{\text{N}}_{\text{2}}\text{O}$ … (3)

The molar mass of ${\text{N}}_{\text{2}}\text{O}$ is $44.013\text{g}{\text{mol}}^{-1}$.

Substitute the values of number of molecules of ${\text{N}}_{\text{2}}\text{O}$, Avogadro’s number and the molar mass in equation (3) to calculate the mass.

$\begin{array}{rll}\text{Mass}& =& \frac{\text{6}\text{.33}\times {\text{10}}^{\text{22}}}{\text{6}\text{.022}\times {\text{10}}^{\text{23}}}\times \text{44}\text{.02}\text{g}\\ & =& 0.105\times \text{44}\text{.02}\text{g}\\ & =& \text{4}\text{.62}\text{g}\end{array}$

Therefore, the mass of $\text{6}\text{.33}\times {\text{10}}^{\text{22}}$ molecules of dinitrogen oxide, ${\text{N}}_{\text{2}}\text{O}$ is $\text{4}\text{.62}\text{g}$

Conclusion

The mass of $\text{6}\text{.33}\times {\text{10}}^{\text{22}}$ molecules of dinitrogen oxide, ${\text{N}}_{\text{2}}\text{O}$ is $\text{4}\text{.62}\text{g}$.

Interpretation Introduction

(c)

Interpretation:

The mass in grams for $\text{4}\text{.17}\times {\text{10}}^{\text{21}}$ formula units of magnesium perchlorate $\text{Mg}{\left({\text{ClO}}_{\text{4}}\right)}_{\text{2}}$ is to be calculated.

Concept introduction:

One mole of atoms, molecules or particles is defined as a quantity whose mass is equivalent to its gram atomic mass or molecular mass. One mole of any substance possesses fixed Avogadro’s number, which is equivalent to $\text{6}\text{.022}\times {\text{10}}^{\text{23}}$ atoms.

Answer to Problem 18E

The mass of $\text{4}\text{.17}\times {\text{10}}^{\text{21}}$ formula units of magnesium perchlorate $\text{Mg}{\left({\text{ClO}}_{\text{4}}\right)}_{\text{2}}$ is $\text{1}\text{.545}\text{g}$.

Explanation of Solution

The moles of magnesium perchlorate, $\text{Mg}{\left({\text{ClO}}_{\text{4}}\right)}_{\text{2}}$ is calculated by the formula shown below.

$\text{Moles}=\frac{\text{Given}\text{formula}\text{unit}\text{of}\text{Mg}{\left({\text{ClO}}_{\text{4}}\right)}_{\text{2}}}{\text{Avogadro's}\text{number}}$ … (4)

The formula unit of magnesium perchlorate is $\text{4}{\text{.17×10}}^{\text{21}}$.

Substitute the value of the formula unit of magnesium perchlorate and the value of Avogadro’s number in equation (4) to calculate the moles of magnesium perchlorate.

$\begin{array}{rll}\text{Moles}& =& \frac{\text{4}{\text{.17×10}}^{\text{21}}}{\text{6}{\text{.022×10}}^{\text{23}}}\\ & =& 6.925\times {10}^{-3}\text{mol}\end{array}$

The mass in grams for $\text{4}{\text{.17×10}}^{\text{21}}$ formula units of magnesium perchlorate $\text{Mg}{\left({\text{ClO}}_{\text{4}}\right)}_{\text{2}}$ is calculated by the formula shown below.

$\text{Mass}\text{of}\text{Mg}{\left({\text{ClO}}_{\text{4}}\right)}_{\text{2}}=\text{Moles}\text{of}\text{Mg}{\left({\text{ClO}}_{\text{4}}\right)}_{\text{2}}\times \text{Molar}\text{mass}\text{of}\text{Mg}{\left({\text{ClO}}_{\text{4}}\right)}_{\text{2}}$ … (5)

The molar mass of $\text{Mg}{\left({\text{ClO}}_{\text{4}}\right)}_{\text{2}}$ is $\text{223}\text{.206}\text{g}{\text{mol}}^{-1}$

Substitute the value of moles of $\text{Mg}{\left({\text{ClO}}_{\text{4}}\right)}_{\text{2}}$ and the molar mass of $\text{Mg}{\left({\text{ClO}}_{\text{4}}\right)}_{\text{2}}$ in equation (5) to calculate the mass.

$\begin{array}{rll}\text{Mass}\text{of}\text{Mg}{\left({\text{ClO}}_{\text{4}}\right)}_{\text{2}}& =& 6.925\times {10}^{-3}\text{mol}\times \text{223}\text{.206}\text{g}{\text{mol}}^{-1}\\ & =& \text{1}\text{.545}\text{g}\end{array}$

Therefore, the mass of $\text{4}{\text{.17×10}}^{\text{21}}$ formula units of magnesium perchlorate $\text{Mg}{\left({\text{ClO}}_{\text{4}}\right)}_{\text{2}}$ is $\text{1}\text{.545}\text{g}$.

Conclusion

The mass of $\text{4}{\text{.17×10}}^{\text{21}}$ formula units of magnesium perchlorate, $\text{Mg}{\left({\text{ClO}}_{\text{4}}\right)}_{\text{2}}$ is $\text{1}\text{.545}\text{g}$.