Chapter 11, Problem 11.15P

(a)

Interpretation Introduction

Interpretation:

Empirical formula of compound that has lithium and oxygen has to be determined.

Concept Introduction:

Mole is S.I. unit. The number of moles is calculated as ratio of mass of compound to molar mass of compound.

Molar mass is sum of the total mass in grams of all atoms that make up mole of particular molecule that is mass of 1 mole of compound. The S.I unit is $\text{g/mol}$.

The expression to relate number of moles, mass and molar mass of compound is as follows:

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

Empirical formula is simplest positive integer ratio of atoms in the compound. It only gives proportions of elements in compound.

Answer to Problem 11.15P

Empirical formula of compound is ${\text{Li}}_{\text{2}}\text{O}$.

Explanation of Solution

Mass % of $\text{Li}$ is $46.45\%$ and $\text{O}$ is $53.55\%$ in $100\%$ compound. Therefore the mass of compound to be $100\text{ g}$. Therefore, mass of $\text{Li}$ is $46.4\text{5 g}$ and $\text{O}$ is $53.55\text{ g}$ in $100\text{ g}$ of compound.

Mass of $\text{Li}$ is $46.4\text{5 g}$.

Molar mass of $\text{Li}$ is $\text{6}\text{.941 g/mol}$.

The formula to calculate number of moles of $\text{Li}$ is as follows:

  $\text{Number of moles of Li}=\frac{\text{Given mass of Li}}{\text{molar mass of Li}}$        (1)

Substitute $46.4\text{5 g}$ for given mass of $\text{Li}$ and $\text{6}\text{.941 g/mol}$ for molar mass of $\text{Li}$ in equation (1).

  $\begin{array}{c}\text{number of moles of Li}=\frac{46.45\text{ g}}{\text{6}\text{.941 g/mol}}\\ =\text{6}\text{.6921 mol}\end{array}$

Mass of $\text{O}$ is $53.55\text{ g}$.

Molar mass of $\text{O}$ is $\text{15}\text{.999 g/mol}$.

The formula to calculate number of moles of $\text{O}$ is as follows:

  $\text{Number of moles of O}=\frac{\text{Given mass of O}}{\text{molar mass of O}}$        (2)

Substitute $53.55\text{ g}$ for given mass of $\text{O}$ and $\text{15}\text{.999 g/mol}$ for molar mass of $\text{O}$ in equation (2).

  $\begin{array}{c}\text{number of moles of O}=\frac{53.55\text{ g}}{\text{15}\text{.999 g/mol}}\\ =\text{3}\text{.3470 mol}\end{array}$

Preliminary formula for compound is formed with moles of $\text{Li}$ and $\text{O}$ written in subscripts. Therefore it can be written as follows:

  $\text{Preliminary formula for compound}={\text{Li}}_{\text{6}\text{.6921}}{\text{O}}_{\text{3}\text{.3470 }}$

Each of subscript of $\text{Li}$ and $\text{O}$ is divided by smallest value to determine empirical formula of compound. The smallest value is $\text{3}\text{.3470}$.

  $\begin{array}{c}\text{Empirical formula of compound}={\text{Li}}_{\frac{\text{6}\text{.6921}}{\text{3}\text{.3470}}}{\text{O}}_{\frac{\text{3}\text{.3470}}{\text{3}\text{.3470}}}\\ ={\text{Li}}_{1.9994}\text{O}\\ \approx {\text{Li}}_2\text{O}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Empirical formula of compound that has $\text{59}\text{.78 }\%$ lithium and $40.22\%$ nitrogen has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 11.15P

Empirical formula of compound is ${\text{Li}}_3\text{N}$.

Explanation of Solution

Mass % of $\text{Li}$ is $\text{59}\text{.78 }\%$ and $\text{N}$ is $40.22\%$ in $100\%$ compound. Therefore the mass of compound is $100\text{ g}$, mass of $\text{Li}$ is $\text{59}\text{.78 g}$ and $\text{N}$ is $40.22\text{ g}$ in $100\text{ g}$ of compound.

Mass of $\text{Li}$ is $\text{59}\text{.78 g}$.

Molar mass of $\text{Li}$ is $\text{6}\text{.941 g/mol}$.

The formula to calculate number of moles of $\text{Li}$ is as follows:

  $\text{Number of moles of Li}=\frac{\text{Given mass of Li}}{\text{molar mass of Li}}$        (3)

Substitute $46.4\text{5 g}$ for given mass of $\text{Li}$ and $\text{6}\text{.941 g/mol}$ for molar mass of $\text{Li}$ in equation (3).

  $\begin{array}{c}\text{number of moles of Li}=\frac{\text{59}\text{.78 g}}{\text{6}\text{.941 g/mol}}\\ =\text{8}\text{.6125 mol}\end{array}$

Mass of $\text{N}$ is $40.22\text{ g}$.

Molar mass of $\text{N}$ is $\text{14}\text{.0067 g/mol}$.

The formula to calculate number of moles of $\text{N}$ is as follows:

  $\text{Number of moles of N}=\frac{\text{Given mass of N}}{\text{molar mass of N}}$        (4)

Substitute $40.22\text{ g}$ for given mass of $\text{N}$ and $\text{14}\text{.0067 g/mol}$ for molar mass of $\text{N}$ in equation (4).

  $\begin{array}{c}\text{number of moles of N}=\frac{40.22\text{ g}}{\text{14}\text{.0067 g/mol}}\\ =\text{2}\text{.8714 mol}\end{array}$

Preliminary formula of compound is formed with moles of $\text{Li}$ and $\text{N}$ written in subscripts. Therefore it can be written as follows:

  $\text{Preliminary formula for compound}={\text{Li}}_{\text{8}\text{.6125}}{\text{N}}_{\text{2}\text{.8714 }}$

Each of subscript of $\text{Li}$ and $\text{N}$ is divided by smallest value to determine empirical formula of compound. The smallest value is $\text{2}\text{.8714}$.

  $\begin{array}{c}\text{Empirical formula of compound}={\text{Li}}_{\frac{\text{8}\text{.6125}}{\text{2}\text{.8714}}}{\text{N}}_{\frac{\text{2}\text{.8714}}{\text{2}\text{.8714}}}\\ ={\text{Li}}_{2.9994}\text{N}\\ \approx {\text{Li}}_3\text{N}\end{array}$

(c)

Interpretation Introduction

Interpretation:

Empirical formula of compound that has $\text{14}\text{.17 }\%$ lithium and $85.83\%$ nitrogen has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 11.15P

Empirical formula of compound is ${\text{LiN}}_3$.

Explanation of Solution

Mass % of $\text{Li}$ is $\text{14}\text{.17 }\%$ and $\text{N}$ is $85.83\%$ in $100\%$ compound. Therefore, the mass of compound is $100\text{ g}$,  mass of $\text{Li}$ is $\text{14}\text{.17 g}$ and $\text{N}$ is $85.83\text{ g}$ in $100\text{ g}$ of compound.

Mass of $\text{Li}$ is $\text{14}\text{.17 g}$.

Molar mass of $\text{Li}$ is $\text{6}\text{.941 g/mol}$.

The formula to calculate number of moles of $\text{Li}$ is as follows:

  $\text{Number of moles of Li}=\frac{\text{Given mass of Li}}{\text{molar mass of Li}}$        (5)

Substitute $\text{14}\text{.17 g}$ for given mass of $\text{Li}$ and $\text{6}\text{.941 g/mol}$ for molar mass of $\text{Li}$ in equation (5).

  $\begin{array}{c}\text{number of moles of Li}=\frac{\text{14}\text{.17 g}}{\text{6}\text{.941 g/mol}}\\ =\text{2}\text{.0414 mol}\end{array}$

Mass of $\text{N}$ is $85.83\text{ g}$.

Molar mass of $\text{N}$ is $\text{14}\text{.0067 g/mol}$.

The formula to calculate number of moles of $\text{N}$ is as follows:

  $\text{Number of moles of N}=\frac{\text{Given mass of N}}{\text{molar mass of N}}$        (6)

Substitute $85.83\text{ g}$ for given mass of $\text{N}$ and $\text{14}\text{.0067 g/mol}$ for molar mass of $\text{N}$ in equation (6).

  $\begin{array}{c}\text{number of moles of N}=\frac{85.83\text{ g}}{\text{14}\text{.0067 g/mol}}\\ =\text{6}\text{.1277 mol}\end{array}$

Preliminary formula for compound is formed with moles of $\text{Li}$ and $\text{N}$ written in subscripts. Therefore it can be written as follows:

  $\text{Preliminary formula for compound}={\text{Li}}_{\text{2}\text{.0414}}{\text{N}}_{\text{6}\text{.1277 }}$

Each of subscript of $\text{Li}$ and $\text{N}$ is divided by smallest value to determine empirical formula of compound. The smallest value is $\text{2}\text{.0414}$.

  $\begin{array}{c}\text{Empirical formula of compound}={\text{Li}}_{\frac{\text{2}\text{.0414}}{\text{2}\text{.0414}}}{\text{N}}_{\frac{\text{6}\text{.1277}}{\text{2}\text{.0414}}}\\ ={\text{LiN}}_{3.0017}\\ \approx {\text{LiN}}_3\end{array}$

(d)

Interpretation Introduction

Interpretation:

Empirical formula of compound that has calcium and chloride has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 11.15P

Empirical formula of compound is ${\text{CaCl}}_2$.

Explanation of Solution

Mass % of $\text{Ca}$ is $\text{36}\text{.11 }\%$ and $\text{Cl}$ is $63.89\%$ in $100\%$ compound. Therefore, the mass of compound is $100\text{ g}$, mass of $\text{Ca}$ is $\text{36}\text{.11 g}$ and $\text{Cl}$ is $63.89\text{ g}$ in $100\text{ g}$ of compound.

Mass of $\text{Ca}$ is $\text{36}\text{.11 g}$.

Molar mass of $\text{Ca}$ is $\text{40}\text{.078 g/mol}$.

The formula to calculate number of moles of $\text{Ca}$ is as follows:

  $\text{Number of moles of Ca}=\frac{\text{Given mass of Ca}}{\text{molar mass of Ca}}$        (7)

Substitute $\text{36}\text{.11 g}$ for given mass of $\text{Ca}$ and $\text{40}\text{.078 g/mol}$ for molar mass of $\text{Ca}$ in equation (7).

  $\begin{array}{c}\text{number of moles of Ca}=\frac{\text{36}\text{.11 g}}{\text{40}\text{.078 g/mol}}\\ =\text{0}\text{.9009 mol}\end{array}$

Mass of $\text{Cl}$ is $63.89\text{ g}$.

Molar mass of $\text{Cl}$ is $\text{35}\text{.453 g/mol}$.

The formula to calculate number of moles of $\text{Cl}$ is as follows:

  $\text{Number of moles of Cl}=\frac{\text{Given mass of Cl}}{\text{molar mass of Cl}}$        (8)

Substitute $63.89\text{ g}$ for given mass of $\text{Cl}$ and $\text{35}\text{.453 g/mol}$ for molar mass of $\text{Cl}$ in equation (8).

  $\begin{array}{c}\text{number of moles of Cl}=\frac{63.89\text{ g}}{\text{35}\text{.453 g/mol}}\\ =\text{1}\text{.8021 mol}\end{array}$

Preliminary formula for compound is formed with moles of $\text{Ca}$ and $\text{Cl}$ written in subscripts. Therefore it can be written as follows:

  $\text{Preliminary formula for compound}={\text{Ca}}_{\text{0}\text{.9009}}{\text{Cl}}_{\text{1}\text{.8021 }}$

Each of subscript of $\text{Ca}$ and $\text{Cl}$ is divided by smallest value to determine empirical formula of compound. The smallest value is $\text{0}\text{.9009}$$\text{2}\text{.0414}$.

  $\begin{array}{c}\text{Empirical formula of compound}={\text{Ca}}_{\frac{\text{0}\text{.9009}}{\text{0}\text{.9009}}}{\text{Cl}}_{\frac{\text{1}\text{.8021}}{\text{0}\text{.9009}}}\\ ={\text{CaCl}}_{2.0003}\\ \approx {\text{CaCl}}_2\end{array}$