Chapter 3, Problem 3.28P

(a)

Interpretation Introduction

Interpretation:Molecular formula of compound that has empirical formula as $\text{CH}$ should be written.

Concept introduction:An empirical formula depicts the ratio of simplified whole-number of atoms contained in a molecule. The molecular formula depicts exact number of each atom found in a molecule.

The steps to determine empirical formula are stated as follows:

Divide mass of element by its molar mass to convert mass to moles as follows:

  $\text{Amount}\left(\text{mol}\right)=\left(\text{Given mass}\left(\text{g}\right)\right)\left(\frac{1\text{ mol}}{\text{No}\text{. of grams}}\right)$

Number of moles thus calculated is written as subscript of element’s symbol and this results in a preliminary empirical formula.

In order to convert the moles to the whole number subscripts, each subscript is divided by the smallest subscript. Finally, empirical formula can be simply written from the molar ratio of each element specified at the superscript of each symbol present in the compound.

The formula to calculate the whole number multiple is as follows:

  $\text{Whole}-\text{number multiple}=\frac{\text{Molar mass of compound}}{\text{Empirical formula mass}}$

The product of this whole number with the subscript of each element results in the molecular formula.

Answer to Problem 3.28P

Molecular formula of compound with empirical formula as $\text{CH}$ is ${\text{C}}_6{\text{H}}_6$ .

Explanation of Solution

The expression to calculate empirical formula mass of $\text{CH}$ is as follows:

  $\text{Empirical formula mass of CH}=\left(1\right)\left(\text{M of C}\right)+\left(1\right)\left(\text{M of H}\right)$

  $\text{M}$ of $\text{C}$ is $12.011\text{ g/mol}$ .

  $\text{M}$ of $\text{H}$ is $\text{1}\text{.008 g/mol}$ .

Substitute the value in above formula to get empirical formula mass of $\text{CH}$ .

  $\begin{array}{c}\text{Empirical formula mass of CH}=\left(1\right)\left(\text{M of C}\right)+\left(1\right)\left(\text{M of H}\right)\\ =\left(1\right)\left(12.011\text{ g/mol}\right)+\left(1\right)\left(1.008\text{ g/mol}\right)\\ =13.019\text{ g/mol}\end{array}$

The formula to calculate whole number multiple is as follows:

  $\text{Whole-number multiple}=\frac{\text{Molar mass}}{\text{Empirical formula mass}}$

Molar mass is $\text{78}\text{.11 g/mol}$ .

Empirical formula mass is $13.019\text{ g/mol}$ .

Substitute the values in above equation to compute whole number multiple.

  $\begin{array}{c}\text{Whole}-\text{number multiple}=\frac{\text{Molar mass}}{\text{Empirical formula mass}}\\ =\frac{\text{78}\text{.11 g/mol}}{13.019\text{ g/mol}}\\ \approx 6\end{array}$

Multiply the subscripts in $\text{CH}$ by 6 to obtain molecular formula.

  $\text{Molecular formula}\to \left({\text{C}}_{\left(6\right)\left(1\right)}{\text{H}}_{\left(6\right)\left(1\right)}\right)\to {\text{C}}_6{\text{H}}_6$

Thus molecular formula is ${\text{C}}_6{\text{H}}_6$ .

(b)

Interpretation Introduction

Interpretation:Molecular formula of compound that has empirical formula as ${\text{C}}_{\text{3}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}$ should be written.

Concept introduction:An empirical formula depicts the ratio of simplified whole-number of atoms contained in a molecule. The molecular formula depicts exact number of each atom found in a molecule.

The steps to determine empirical formula are stated as follows:

Divide mass of element by its molar mass to convert mass to moles as follows:

  $\text{Amount}\left(\text{mol}\right)=\left(\text{Given mass}\left(\text{g}\right)\right)\left(\frac{1\text{ mol}}{\text{No}\text{. of grams}}\right)$

Number of moles thus calculated is written as subscript of element’s symbol and this results in a preliminary empirical formula.

In order to convert the moles to the whole number subscripts, each subscript is divided by the smallest subscript. Finally, empirical formula can be simply written from the molar ratio of each element specified at the superscript of each symbol present in the compound.

The formula to calculate the whole number multiple is as follows:

  $\text{Whole}-\text{number multiple}=\frac{\text{Molar mass of compound}}{\text{Empirical formula mass}}$

The product of this whole number with the subscript of each element results in the molecular formula.

Answer to Problem 3.28P

Molecular formula is ${\text{C}}_{\text{3}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}$ .

Explanation of Solution

The expression to calculate the empirical formula mass of ${\text{C}}_{\text{3}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}$ is as follows:

  $\text{Empirical formula mass}=\left(3\right)\left(\text{M of C}\right)+\left(6\right)\left(\text{M of H}\right)+\left(2\right)\left(\text{M of O}\right)$

  $\text{M}$ of $\text{C}$ is $\text{12}\text{.011 g/mol}$ .

  $\text{M}$ of $\text{H}$ is $\text{1}\text{.008 g/mol}$ .

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

Substitute the value in above formula to get empirical formula mass of ${\text{C}}_{\text{3}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}$ .

  $\begin{array}{c}\text{Empirical formula mass}=\left(3\right)\left(\text{M of C}\right)+\left(6\right)\left(\text{M of H}\right)+\left(2\right)\left(\text{M of O}\right)\\ =\left[\begin{array}{c}\left(3\right)\left(\text{12}\text{.011 g/mol}\right)+\left(6\right)\left(1.008\text{ g/mol}\right)\\ +\left(2\right)\left(15.9994\text{ g/mol}\right)\end{array}\right]\\ =74.0798\text{ g/mol}\end{array}$

The formula to calculate the whole number multiple is as follows:

  $\text{Whole-number multiple}=\frac{\text{Molar mass}}{\text{Empirical formula mass}}$

Molar mass is $\text{74}\text{.08 g/mol}$ .

Empirical formula mass is $74.0798\text{ g/mol}$ .

Substitute the values in the above equation to compute whole number multiple.

  $\begin{array}{c}\text{Whole}-\text{number multiple}=\frac{\text{Molar mass}}{\text{Empirical formula mass}}\\ =\frac{\text{74}\text{.08 g/mol}}{74.0798\text{ g/mol}}\\ =1\end{array}$

Multiply the subscripts in ${\text{C}}_{\text{3}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}$ by 1to obtain molecular formula as ${\text{C}}_{\text{3}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}$ .

(c)

Interpretation Introduction

Interpretation:Molecular formula of compound that has empirical formula as $\text{HgCl}$ should be written.

Concept introduction:An empirical formula depicts the ratio of simplified whole-number of atoms contained in a molecule. The molecular formula depicts exact number of each atom found in a molecule.

The steps to determine empirical formula are stated as follows:

Divide mass of element by its molar mass to convert mass to moles as follows:

  $\text{Amount}\left(\text{mol}\right)=\left(\text{Given mass}\left(\text{g}\right)\right)\left(\frac{1\text{ mol}}{\text{No}\text{. of grams}}\right)$

Number of moles thus calculated is written as subscript of element’s symbol and this results in a preliminary empirical formula.

In order to convert the moles to the whole number subscripts, each subscript is divided by the smallest subscript. Finally, empirical formula can be simply written from the molar ratio of each element specified at the superscript of each symbol present in the compound.

The formula to calculate the whole number multiple is as follows:

  $\text{Whole}-\text{number multiple}=\frac{\text{Molar mass of compound}}{\text{Empirical formula mass}}$

The product of this whole number with the subscript of each element results in the molecular formula.

Answer to Problem 3.28P

Molecular formula of compound with empirical formula as $\text{HgCl}$ is ${\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}$ .

Explanation of Solution

The expression to calculate the empirical formula mass of $\text{HgCl}$ is as follows:

  $\text{Empirical formula mass}=\left(1\right)\left(\text{M of Hg}\right)+\left(1\right)\left(\text{M of Cl}\right)$

  $\text{M}$ of $\text{Hg}$ is $\text{200}\text{.59 g/mol}$ .

  $\text{M}$ of $\text{Cl}$ is $\text{35}\text{.4527 g/mol}$ .

Substitute the value in above formula to get empirical formula mass of $\text{HgCl}$ .

  $\begin{array}{c}\text{Empirical formula mass}=\left(1\right)\left(\text{M of Hg}\right)+\left(1\right)\left(\text{M of Cl}\right)\\ =\left(1\right)\left(\text{200}\text{.59 g/mol}\right)+\left(1\right)\left(\text{35}\text{.4527 g/mol}\right)\\ =236.0427\text{ g/mol}\end{array}$

The formula to calculate the whole number multiple is as follows:

  $\text{Whole-number multiple}=\frac{\text{Molar mass}}{\text{Empirical formula mass}}$

Molar mass is $\text{472}\text{.1 g/mol}$ .

Empirical formula mass is $236.0427\text{ g/mol}$ .

Substitute the values in above equation to compute whole number multiple.

  $\begin{array}{c}\text{Whole}-\text{number multiple}=\frac{\text{Molar mass}}{\text{Empirical formula mass}}\\ =\frac{\text{472}\text{.1 g/mol}}{236.0427\text{ g/mol}}\\ =2\end{array}$

Multiply the subscripts in $\text{HgCl}$ by 2 to obtain molecular formula.

  $\text{Molecular formula}\to \left({\text{Hg}}_{\left(\text{2}\right)\left(\text{1}\right)}{\text{Cl}}_{\left(\text{2}\right)\left(1\right)}\right)\to {\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}$

Thus molecular formula is ${\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}$ .

(c)

Interpretation Introduction

Interpretation:Molecular formula of compound with empirical formula as ${\text{C}}_{\text{7}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}$ should be written.

Concept introduction:An empirical formula depicts the ratio of simplified whole-number of atoms contained in a molecule. The molecular formula depicts exact number of each atom found in a molecule.

The steps to determine empirical formula are stated as follows:

Divide mass of element by its molar mass to convert mass to moles as follows:

  $\text{Amount}\left(\text{mol}\right)=\left(\text{Given mass}\left(\text{g}\right)\right)\left(\frac{1\text{ mol}}{\text{No}\text{. of grams}}\right)$

Number of moles thus calculated is written as subscript of element’s symbol and this results in a preliminary empirical formula.

In order to convert the moles to the whole number subscripts, each subscript is divided by the smallest subscript. Finally, empirical formula can be simply written from the molar ratio of each element specified at the superscript of each symbol present in the compound.

The formula to calculate the whole number multiple is as follows:

  $\text{Whole}-\text{number multiple}=\frac{\text{Molar mass of compound}}{\text{Empirical formula mass}}$

The product of this whole number with the subscript of each element results in the molecular formula.

Answer to Problem 3.28P

Molecular formula of compound with empirical formula as ${\text{C}}_{\text{7}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}$ is ${\text{C}}_{14}{\text{H}}_8{\text{O}}_4$ .

Explanation of Solution

The expression to calculate empirical formula mass of ${\text{C}}_{\text{7}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}$ is as follows:

  $\text{Empirical formula mass}=\left(7\right)\left(\text{M of C}\right)+\left(4\right)\left(\text{M of H}\right)+\left(2\right)\left(\text{M of O}\right)$

  $\text{M}$ of $\text{C}$ is $\text{12}\text{.011 g/mol}$ .

  $\text{M}$ of $\text{H}$ is $\text{1}\text{.008 g/mol}$ .

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

Substitute the value in the above formula to get empirical formula mass of ${\text{C}}_{\text{7}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}$ .

  $\begin{array}{c}\text{Empirical formula mass}=\left(7\right)\left(\text{M of C}\right)+\left(4\right)\left(\text{M of H}\right)+\left(2\right)\left(\text{M of O}\right)\\ =\left[\begin{array}{l}\left(7\right)\left(\text{12}\text{.011 g/mol}\right)+\left(4\right)\left(\text{1}\text{.008 g/mol}\right)\\ +\left(2\right)\left(\text{15}\text{.9994 g/mol}\right)\end{array}\right]\\ =120.1078\text{ g/mol}\end{array}$

The formula to calculate the whole number multiple is as follows:

  $\text{Whole-number multiple}=\frac{\text{Molar mass}}{\text{Empirical formula mass}}$

Molar mass is $\text{240}\text{.20 g/mol}$ .

Empirical formula mass is $120.1078\text{ g/mol}$ .

Substitute the valuesin the above equation to compute whole number multiple.

  $\begin{array}{c}\text{Whole}-\text{number multiple}=\frac{\text{Molar mass}}{\text{Empirical formula mass}}\\ =\frac{\text{240}\text{.20 g/mol}}{120.1078\text{ g/mol}}\\ =1.99\\ \approx 2\end{array}$

Multiply the subscripts in ${\text{C}}_{\text{7}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}$ by 2 to obtain molecular formula.

  $\text{Molecular formula}\to \left({\text{C}}_{\left(7\right)\left(2\right)}{\text{H}}_{\left(4\right)\left(2\right)}{\text{O}}_{\left(2\right)\left(2\right)}\right)\to {\text{C}}_{14}{\text{H}}_8{\text{O}}_4$

Thus molecular formula is ${\text{C}}_{14}{\text{H}}_8{\text{O}}_4$ .