Chapter 11, Problem 11.16P

(a)

Interpretation Introduction

Interpretation:

Empirical formula of compound that has $\text{Tl}$ and $\text{Br}$ 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 represents simplest positive integer ratio of atoms in the compound. It only gives proportions of elements in compound. Molecular formula consists of chemical symbols for the respective elements followed by numeric subscript that denotes the number of atom of each element present in molecule.

Answer to Problem 11.16P

Empirical formula of compound is $\text{TlBr}\text{.}$

Explanation of Solution

Mass % of $\text{Tl}$ is $71.89\%$ and $\text{Br}$ is $28.11\%$. Therefore, mass of $\text{Tl}$ is $71.89\text{ g}$ and $\text{Br}$ is $28.11\text{ g}$ in $100\text{ g}$ of compound.

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

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

Substitute $71.89\text{ g}$ for given mass of $\text{Tl}$ and $\text{204}\text{.3833 g/mol}$ for molar mass of $\text{Tl}$ in equation (1).

  $\begin{array}{c}\text{number of moles of Tl}=\frac{71.89\text{ g}}{\text{204}\text{.3833 g/mol}}\\ =\text{0}\text{.3517 mol}\end{array}$

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

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

Substitute $28.11\text{ g}$ for given mass of $\text{Br}$ and $\text{79}\text{.904 g/mol}$ for molar mass of $\text{Br}$ in equation (2).

  $\begin{array}{c}\text{number of moles of Br}=\frac{28.11\text{ g}}{\text{79}\text{.904 g/mol}}\\ =\text{0}\text{.3517 mol}\end{array}$

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

  $\text{Preliminary formula for compound}={\text{Ti}}_{\text{0}\text{.3517}}{\text{Br}}_{\text{0}\text{.3517 }}$

Each subscript of $\text{Tl}$ and $\text{Br}$ is divided by smallest value to determine empirical formula of compound. The smallest value is $\text{0}\text{.3517}$.

  $\begin{array}{c}\text{Empirical formula of compound}={\text{Tl}}_{\frac{\text{0}\text{.3517}}{\text{0}\text{.3517}}}{\text{Br}}_{\frac{\text{0}\text{.3517}}{\text{0}\text{.3517}}}\\ =\text{TlBr}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Empirical formula of compound that has $\text{Pb}$ and $\text{Cl}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 11.16P

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

Explanation of Solution

Mass % of $\text{Pb}$ is $\text{74}\text{.51 }\%$ and $\text{Cl}$ is $25.49\%$. Therefore, mass of $\text{Pb}$ is $\text{74}\text{.51 g}$ and $\text{Cl}$ is $25.49\text{ g}$ in $100\text{ g}$ of compound.

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

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

Substitute $\text{74}\text{.51 g}$ for given mass of $\text{Pb}$ and $\text{207}\text{.2 g/mol}$ for molar mass of $\text{Pb}$ in equation (3).

  $\begin{array}{c}\text{number of moles of Pb}=\frac{\text{74}\text{.51 g}}{\text{207}\text{.2 g/mol}}\\ =\text{0}\text{.3596 mol}\end{array}$

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

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

Substitute $25.49\text{ g}$ for given mass of $\text{Cl}$ and $\text{35}\text{.453 g/mol}$ for molar mass of $\text{Cl}$ in equation (4).

  $\begin{array}{c}\text{number of moles of Cl}=\frac{25.49\text{ g}}{\text{35}\text{.453 g/mol}}\\ =\text{0}\text{.7189 mol}\end{array}$

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

  $\text{Preliminary formula for compound}={\text{Pb}}_{\text{0}\text{.3596}}{\text{Cl}}_{0.7189}{}_{}$

Each subscript of $\text{Pb}$ and $\text{Cl}$ is divided by smallest value to determine empirical formula of compound. The smallest value is 0.3596.

  $\begin{array}{c}\text{Empirical formula of compound}={\text{Pb}}_{\frac{\text{0}\text{.3596}}{0.3596}}{\text{Cl}}_{\frac{0.7189}{0.3596}}{}_{}\\ ={\text{PbCl}}_2\end{array}$

(c)

Interpretation Introduction

Interpretation:

Empirical formula of compound that has $\text{H}$ and $\text{N}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 11.16P

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

Explanation of Solution

Mass % of $\text{H}$ is $\text{17}\text{.76 }\%$ and $\text{N}$ is $82.24\%$. Therefore, mass of $\text{H}$ is $\text{17}\text{.76 g}$ and $\text{N}$ is $82.24\text{ g}$ in $100\text{ g}$ of compound.

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

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

Substitute $\text{17}\text{.76 g}$ for given mass of $\text{H}$ and $\text{1}\text{.00784 g/mol}$ for molar mass of $\text{H}$ in equation (5).

  $\begin{array}{c}\text{number of moles of H}=\frac{\text{17}\text{.76 g}}{\text{1}\text{.00784 g/mol}}\\ =\text{17}\text{.6218 mol}\end{array}$

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 $82.24\text{ g}$ for given mass of $\text{N}$ and $\text{14}\text{.0067 g/mol}$ for molar mass of $\text{N}$ in equation (6).

  $\begin{array}{c}\text{number of moles of N}=\frac{82.24\text{ g}}{\text{14}\text{.0067 g/mol}}\\ =\text{5}\text{.8714 mol}\end{array}$

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

  $\text{Preliminary formula for compound}={\text{H}}_{\text{17}\text{.6218}}{\text{N}}_{\text{5}\text{.7814 }}$

Each of subscript of $\text{H}$ and $\text{N}$ is divided by smallest value to determine empirical formula of compound. The smallest value is 5.8714.

  $\begin{array}{c}\text{Empirical formula of compound}={\text{H}}_{\frac{\text{17}\text{.6218}}{\text{5}\text{.8714}}}{\text{N}}_{\frac{\text{5}\text{.8714}}{\text{5}\text{.8714}}}\\ ={\text{H}}_{3.0012}\text{N}\\ \approx {\text{NH}}_{\text{3}}\end{array}$

(d)

Interpretation Introduction

Interpretation:

Empirical formula of compound that has $\text{Mg}$ and $\text{N}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 11.16P

Empirical formula of compound is ${\text{Mg}}_3{\text{N}}_2$.

Explanation of Solution

Mass % of $\text{Mg}$ is $\text{72}\text{.24 }\%$ and $\text{N}$ is $27.76\%$. Therefore, mass of $\text{Mg}$ is $\text{72}\text{.24 g}$ and $\text{N}$ is $27.76\text{ g}$ in $100\text{ g}$ of compound.

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

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

Substitute $\text{72}\text{.24 g}$ for given mass of $\text{Mg}$ and $\text{40}\text{.078 g/mol}$ for molar mass of $\text{Mg}$ in equation (7).

  $\begin{array}{c}\text{number of moles of Mg}=\frac{\text{72}\text{.24 g}}{\text{24}\text{.305 g/mol}}\\ =\text{2}\text{.9722 mol}\end{array}$

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

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

Substitute $27.76\text{ g}$ for given mass of $\text{N}$ and $\text{14}\text{.0067 g/mol}$ for molar mass of $\text{N}$ in equation (8).

  $\begin{array}{c}\text{number of moles of N}=\frac{27.76\text{ g}}{\text{14}\text{.0067 g/mol}}\\ =\text{1}\text{.9817 mol}\end{array}$

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

  $\text{Preliminary formula for compound}={\text{Mg}}_{\text{2}\text{.9722}}{\text{N}}_{\text{1}\text{.9817 }}$

Each of subscript of $\text{Mg}$ and $\text{N}$ is divided by smallest value to determine empirical formula of compound. The smallest value is 1.98197.

  $\begin{array}{c}\text{Empirical formula of compound}={\text{Mg}}_{\frac{\text{2}\text{.9722}}{1.9817}}{\text{N}}_{\frac{\text{1}\text{.9817}}{1.9817}}\\ ={\text{Mg}}_{\text{1}\text{.499}}\text{N}\\ \approx {\text{Mg}}_{1.50}\text{N}\end{array}$

Empirical formula will always be in simplest whole number. To get the simplest whole number; multiply the empirical formula by 2. Therefore, empirical formula of compound is ${\text{Mg}}_{\text{3}}{\text{N}}_{\text{2}}$.