Chapter 3, Problem 3.89P

In each pair, choose the larger of the indicated quantities or state that the samples are equal:

(a) Entities: 0.4 mol of ${\text{O}}_{\text{3}}$ molecules or 0.4 mol of O atoms

(b) Grams: 0.4 mol of ${\text{O}}_{\text{3}}$ molecules or 0.4 mol of O atoms

(c) Moles: 4.0 g of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ or 3.3 g of ${\text{SO}}_{\text{2}}$ (d) Grams: 0.6 mol of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ or 0.6 mol of ${\text{F}}_{\text{2}}$ (e) Total ions: 2.3 mol of sodium chlorate or 2.2 mol of magnesium chloride

(f) Molecules: 1.0 g of ${\text{H}}_{\text{2}}\text{O}$ or 1.0 g of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ (g) ${\text{Na}}^{\text{+}}$ ions: 0.500 L of 0.500 M NaBr or 0.0146 kg of NaCl

(h) Grams: $\text{6}{\text{.02×10}}^{23}$ atoms of ${}^{\text{235}}\text{U}$ or $6.02\times {10}^{23}$ atoms of ${}^{\text{238}}\text{U}$

Interpretation Introduction

Interpretation:Out of $\text{0}\text{.4 mol}$ of ${\text{O}}_3$ molecules and $\text{0}\text{.4 mol}$ of $\text{O}$ atoms one that is largerentities should be identified.

Concept introduction:One mole of any substance contains Avogadro number of particles equivalent to $6.022\times {10}^{23}$ .

Answer to Problem 3.89P

  $\text{0}\text{.4 mol}$ of ${\text{O}}_3$ molecules has same entities as in $\text{0}\text{.4 mol}$ of $\text{O}$ .

Explanation of Solution

Since equal $\text{0}\text{.4 mol}$ should have equal number of Avogadro’s’ number of particles, therefore each of $\text{0}\text{.4 mol}$ of ${\text{O}}_3$ and $\text{0}\text{.4 mol}$ of $\text{O}$ has equal entities.

Interpretation Introduction

Interpretation: Out of $\text{0}\text{.4 mol}$ of ${\text{O}}_3$ molecules and $\text{0}\text{.4 mol}$ of $\text{O}$ atoms one that is larger mass in grams should be identified.

Concept introduction: The formula to convert moles to mass in grams is as follows:

  $\text{Mass}=\left(\text{Moles}\right)\left(\text{molar mass}\right)$

Answer to Problem 3.89P

  $\text{0}\text{.4 mol}$ of ${\text{O}}_3$ molecules has more mass in grams than $\text{0}\text{.4 mol}$ of $\text{O}$ .

Explanation of Solution

The formula to convert moles to mass in grams is as follows:

  ${\text{Mass of O}}_3=\left({\text{Moles of O}}_3\right)\left({\text{molar mass of O}}_3\right)$

Moles of ${\text{O}}_3$ is $\text{0}\text{.4 mol}$ .

Molar mass of ${\text{O}}_3$ is $\text{48 g/mol}$ .

Substitute the values in above formula.

  $\begin{array}{c}{\text{Mass of O}}_3=\left({\text{Moles of O}}_3\right)\left({\text{molar mass of O}}_3\right)\\ =\left(0.4\text{ mol}\right)\left(\text{48 g/mol}\right)\\ =19.2\text{ g}\end{array}$

The formula to convert moles to mass in grams is as follows:

  $\text{Mass of O}=\left(\text{Moles of O}\right)\left(\text{molar mass of O}\right)$

Moles of $\text{O}$ is $\text{0}\text{.4 mol}$ .

Molar mass of $\text{O}$ is $\text{15}\text{.9994 g/mol}$ .

Substitute the values in above formula.

  $\begin{array}{c}\text{Mass of O}=\left(\text{Moles of O}\right)\left(\text{molar mass of O}\right)\\ =\left(0.4\text{ mol}\right)\left(\text{15}\text{.9994 g/mol}\right)\\ =6.399\text{ g}\end{array}$

Interpretation Introduction

Interpretation: Out of ${\text{4 g N}}_2{\text{O}}_{\text{4}}$ and $\text{3}{\text{.3 g SO}}_2$ one that is larger moles should be identified.

Concept introduction: The formula to convert moles to mass in grams is as follows:

  $\text{Mass}=\left(\text{Moles}\right)\left(\text{molar mass}\right)$

Answer to Problem 3.89P

  $\text{3}{\text{.3 g SO}}_2$ has more moles than ${\text{4 g N}}_2{\text{O}}_{\text{4}}$ .

Explanation of Solution

The formula to convert mass in grams to moles is as follows:

  $\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}$

For ${\text{N}}_2{\text{O}}_{\text{4}}$

Given mass is $\text{4 g}$ .

Molar mass is $\text{92}\text{.011 g/mol}$ .

Substitute the value in above formula.

  $\begin{array}{c}\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}\\ =\frac{\text{4 g}}{\text{92}\text{.011 g/mol}}\\ =0.04347\text{ mol}\end{array}$

For ${\text{SO}}_2$

Given mass is $\text{3}\text{.3 g}$ .

Molar mass is $\text{92}\text{.011 g/mol}$ .

Substitute the value in above formula.

  $\begin{array}{c}\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}\\ =\frac{\text{3}\text{.3 g}}{64.006\text{ g/mol}}\\ =0.05155\text{ mol}\end{array}$

Interpretation Introduction

Interpretation: Out of $\text{0}\text{.6 mol}$ of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ molecules and $\text{0}\text{.6 mol}$ of ${\text{F}}_2$ atoms one that is larger mass in grams should be identified.

Concept introduction: The formula to convert moles to mass in grams is as follows:

  $\text{Mass}=\left(\text{Moles}\right)\left(\text{molar mass}\right)$

Answer to Problem 3.89P

  $\text{0}\text{.6 mol}$ of ${\text{F}}_2$ has more mass in grams than $\text{0}\text{.6 mol}$ of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ .

Explanation of Solution

The formula to convert moles to mass in grams is as follows:

  ${\text{Mass of C}}_{\text{2}}{\text{H}}_{\text{4}}=\left({\text{Moles of C}}_{\text{2}}{\text{H}}_{\text{4}}\right)\left({\text{molar mass of C}}_{\text{2}}{\text{H}}_{\text{4}}\right)$

Moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ is $\text{0}\text{.6 mol}$ .

Molar mass of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ is $\text{28}\text{.05 g/mol}$ .

Substitute the values in above formula.

  $\begin{array}{c}{\text{Mass of C}}_{\text{2}}{\text{H}}_{\text{4}}=\left({\text{Moles of C}}_{\text{2}}{\text{H}}_{\text{4}}\right)\left({\text{molar mass of C}}_{\text{2}}{\text{H}}_{\text{4}}\right)\\ =\left(\text{0}\text{.6 mol}\right)\left(\text{28}\text{.05 g/mol}\right)\\ =16.83\text{ g}\end{array}$

The formula to convert moles to mass in grams is as follows:

  ${\text{Mass of F}}_2=\left({\text{Moles of F}}_2\right)\left({\text{molar mass of F}}_2\right)$

Moles of ${\text{F}}_2$ is $\text{0}\text{.6 mol}$ .

Molar mass of ${\text{F}}_2$ is $\text{37}\text{.9968 g/mol}$ .

Substitute the values in above formula.

  $\begin{array}{c}{\text{Mass of F}}_2=\left({\text{Moles of F}}_2\right)\left({\text{molar mass of F}}_2\right)\\ =\left(0.6\text{ mol}\right)\left(\text{37}\text{.9968 g/mol}\right)\\ =22.798\text{ g}\end{array}$

Interpretation Introduction

Interpretation: Out of $\text{2}\text{.3 mol}$ ofsodium chlorate and $\text{2}\text{.2 mol}$ of magnesium chloride that has larger ions should be identified.

Concept introduction: One mole of any substance contains Avogadro number of particles equivalent to $6.022\times {10}^{23}$ .

The formula to convert moles to molecules is as follows:

  $\text{Number of ions}=\left(\text{Total moles}\right)\left(6.022\times {10}^{23}\right)$

Answer to Problem 3.89P

  $\text{2}\text{.2 mol}$ of magnesium chloridehas more ions than $\text{2}\text{.3 mol}$ of sodium chlorate.

Explanation of Solution

The formula to convert moles to molecules insodium chlorate is as follows:

  $\text{Number of ions}=\left(\text{Total moles}\right)\left(\frac{\text{2 mol }}{\text{1 mol }}\right)$

For sodium chlorate

Total moles is $\text{2}\text{.3 mol}$ .

Substitute the value in above formula.

  $\begin{array}{c}\text{Number of ions}=\left(\text{Total moles}\right)\left(\frac{\text{2 mol }}{\text{1 mol }}\right)\\ =\text{2}\text{.3 mol}\left(\frac{\text{2 mol }}{\text{1 mol }}\right)\\ =4.6\text{ mol}\end{array}$

For magnesium chloride

The formula to convert moles to molecules inmagnesium chloride is as follows:

Total moles is $\text{2}\text{.2 mol}$ .

Substitute the value in above formula.

  $\begin{array}{c}\text{Number of ions}=\left(\text{Total moles}\right)\left(\frac{\text{3 mol }}{\text{1 mol }}\right)\\ =\left(\text{2}\text{.2 mol}\right)\left(\frac{\text{3 mol }}{\text{1 mol }}\right)\\ =6.6\end{array}$

Interpretation Introduction

Interpretation:Out of $\text{1}\text{.0 g}$ of ${\text{H}}_{\text{2}}\text{O}$ and $\text{1}\text{.0 g}$ of ${\text{H}}_{\text{2}}{\text{O}}_2$ that has larger molecules should be identified.

Concept introduction: One mole of any substance contains Avogadro number of particles equivalent to $6.022\times {10}^{23}$ .

The formula to determine number of molecules is as follows:

  $\text{Number of molecules}=\left(\frac{\text{Given mass}}{\text{Molar mass}}\right)\left(6.022\times {10}^{23}\right)$

Answer to Problem 3.89P

  $\text{1}\text{.0 g}$ of ${\text{H}}_{\text{2}}\text{O}$ has more molecules than $\text{1}\text{.0 g}$ of ${\text{H}}_{\text{2}}{\text{O}}_2$ .

Explanation of Solution

The formula to determine number of molecules is as follows:

  $\text{Number of molecules}=\left(\frac{\text{Given mass}}{\text{Molar mass}}\right)\left(6.022\times {10}^{23}\right)$

For ${\text{H}}_{\text{2}}\text{O}$

Given mass is $\text{1}\text{.0 g}$ .

Molar mass is $\text{18}\text{.0128 g/mol}$ .

Substitute the value in above formula.

  $\begin{array}{c}\text{Number of molecules}=\left(\frac{\text{Given mass}}{\text{Molar mass}}\right)\left(6.022\times {10}^{23}\right)\\ =\left(\frac{\text{1}\text{.0 g}}{\text{18}\text{.0128 g/mol}}\right)\left(6.022\times {10}^{23}\right)\\ =3.343\times {10}^{22}\end{array}$

For ${\text{H}}_{\text{2}}{\text{O}}_2$

Given mass is $\text{1}\text{.0 g}$ .

Molar mass is $\text{34}\text{.0147 g/mol}$ .

Substitute the value in above formula.

  $\begin{array}{c}\text{Number of molecules}=\left(\frac{\text{Given mass}}{\text{Molar mass}}\right)\left(6.022\times {10}^{23}\right)\\ =\left(\frac{1.0\text{ g}}{\text{34}\text{.0147 g/mol}}\right)\left(6.022\times {10}^{23}\right)\\ =1.7704\times {10}^{22}\end{array}$

Interpretation Introduction

Interpretation:Out of $\text{0}\text{.500 L}$ of $\text{0}\text{.500 M NaBr}$ and $\text{0}\text{.0146 kg M NaCl}$ that is larger molecules should be identified.

Concept introduction: One mole of any substance contains Avogadro number of particles equivalent to $6.022\times {10}^{23}$ .

The formula to determine number of molecules is as follows:

  $\text{Number of molecules}=\left(\frac{\text{Given mass}}{\text{Molar mass}}\right)\left(6.022\times {10}^{23}\right)$

Answer to Problem 3.89P

  $\text{0}\text{.500 L}$ of $\text{0}\text{.500 M NaBr}$ has more moles than $\text{0}\text{.0146 kg M NaCl}$ .

Explanation of Solution

The formula to evaluate moles from molarity is given as follows:

  $n=MV$

Value of $M$ is $\text{0}\text{.500 mol/L}$ .

Value of $V$ is $\text{0}\text{.500 L}$ .

Substitute the values to calculate $n$ .

  $\begin{array}{c}n=MV\\ =\left(\text{0}\text{.500 mol/L}\right)\left(\text{0}\text{.500 L}\right)\\ =0.25\text{ mol}\end{array}$

The conversion factor to convert $\text{kg}$ is $\text{g}$ as follows:

  $1\text{ kg}=1000\text{ g}$

Hence convert $\text{0}\text{.0146 kg}$ to $\text{g}$ as follows:

  $\begin{array}{c}\text{Mass}=\left(\text{0}\text{.0146 kg}\right)\left(\frac{1000\text{ g}}{1\text{ kg}}\right)\\ =14.6\text{ g}\end{array}$

The formula to determine moles is as follows:

  $\text{Number of moles}=\left(\frac{\text{Given mass}}{\text{Molar mass}}\right)$

For $\text{NaCl}$

Given mass is $14.6\text{ g}$ .

Molar mass is $\text{58}\text{.44 g/mol}$ .

Substitute the value in above formula.

  $\begin{array}{c}\text{Number of moles}=\left(\frac{\text{Given mass}}{\text{Molar mass}}\right)\\ =\left(\frac{14.6\text{ g}}{\text{58}\text{.44 g/mol}}\right)\\ =0.2498\text{ mol}\end{array}$

Interpretation Introduction

Interpretation: Out of $6.022\times {10}^{23}\text{ atoms}$ of ${}^{235}\text{U}$ and $6.022\times {10}^{23}\text{ atoms}$ of ${}^{238}\text{U}$ one that is larger mass in grams should be identified.

Concept introduction:One mole has always $6.022\times {10}^{23}$ ions or, particles or molecules.Thus, irrespective of any substance one mole from any species always has fixed number of entities that is $6.022\times {10}^{23}$ .

Answer to Problem 3.89P

  $6.022\times {10}^{23}\text{ atoms}$ of ${}^{238}\text{U}$ has more mass in grams than $6.022\times {10}^{23}\text{ atoms}$ of ${}^{235}\text{U}$ .

Explanation of Solution

Since, one mole of any substance contains Avogadro number of particles equivalent to $6.022\times {10}^{23}$ so $6.022\times {10}^{23}\text{ atoms}$ indicate one mole of each isotope of uranium. Since ${}^{238}\text{U}$ is heavier isotope and has larger mass has larger molar mass, so it will be larger than one mole of ${}^{235}\text{U}$ .