Chapter 3, Problem 3.26P

(a)

Interpretation Introduction

Interpretation: Empirical formula and empirical formula mass for ${\text{C}}_4{\text{H}}_8$ should be determined.

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.

Sum of the product of molar masses and number of corresponding atoms results in empirical formula mass.

Answer to Problem 3.26P

Empirical formula and empirical formula mass for ${\text{C}}_4{\text{H}}_8$ is ${\text{C}}_2{\text{H}}_4$ and $14.027\text{ g/mol}$ respectively.

Explanation of Solution

  ${\text{C}}_4{\text{H}}_8$ is the molecular formula that is obtained by multiplication of integral factor to an empirical formula. Since the multiplication of factor 4 by ${\text{CH}}_2$ gives ${\text{C}}_4{\text{H}}_8$ thus the empirical formula is ${\text{C}}_2{\text{H}}_4$ .

The formula to calculate empirical formula mass is as follows:

  ${\text{Empirical formula of CH}}_2=\left(1\right)\left(\text{M of C}\right)+\left(2\right)\left(\text{M of H}\right)$

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

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

Substitute the values in the above equation.

  $\begin{array}{c}{\text{Empirical formula of CH}}_2=\left(1\right)\left(\text{M of C}\right)+\left(2\right)\left(\text{M of H}\right)\\ \text{=}\left(1\right)\left(12.011\text{ g/mol}\right)+\left(2\right)\left(\text{1}\text{.008 g/mol}\right)\\ =14.027\text{ g/mol}\end{array}$

(b)

Interpretation Introduction

Interpretation: Empirical formula and empirical formula mass for ${\text{C}}_3{\text{H}}_{\text{6}}{\text{O}}_3$ should be determined.

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.

Sum of the product of molar masses and number of corresponding atoms results in empirical formula mass.

Answer to Problem 3.26P

Empirical formula and empirical formula mass for ${\text{C}}_3{\text{H}}_{\text{6}}{\text{O}}_3$ is ${\text{CH}}_3\text{O}$ and $30.0264\text{ g/mol}$ respectively.

Explanation of Solution

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}$ is the molecular formula that is obtained by multiplication of integral factor to an empirical formula. Since the multiplication of factor 3 by ${\text{CH}}_2\text{O}$ gives ${\text{C}}_3{\text{H}}_{\text{6}}{\text{O}}_3$ thus the empirical formula is ${\text{CH}}_3\text{O}$ .

The formula to calculate empirical formula mass is as follows:

  ${\text{Empirical formula of CH}}_2\text{O}=\text{M of C}+\left(2\right)\left(\text{M of H}\right)+\left(\text{M of O}\right)$

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

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

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

Substitute the values in above equation.

  $\begin{array}{c}{\text{Empirical formula of CH}}_2\text{O}=\text{M of C}+\left(2\right)\left(\text{M of H}\right)+\left(\text{M of O}\right)\\ =\left[\begin{array}{l}\left(12.011\text{ g/mol}\right)+\left(2\right)\left(\text{1}\text{.008 g/mol}\right)\\ +\left(\text{15}\text{.9994 g/mol}\right)\end{array}\right]\\ =30.0264\text{ g/mol}\end{array}$

(c)

Interpretation Introduction

Interpretation: Empirical formula and empirical formula mass for ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ should be determined.

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.

Sum of the product of molar masses and number of corresponding atoms results in empirical formula mass.

Answer to Problem 3.26P

Empirical formula and empirical formula mass is ${\text{P}}_2{\text{O}}_5$ and $141.9446\text{ g/mol}$ respectively.

Explanation of Solution

  ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ is the molecular formula that is obtained by multiplication of integral factor to an empirical formula. Since the multiplication of factor 2 to ${\text{P}}_2{\text{O}}_5$ gives ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ thus the empirical formula is ${\text{P}}_2{\text{O}}_5$ .

The formula to calculate empirical formula mass is as follows:

  ${\text{Empirical formula of P}}_2{\text{O}}_5=\left(2\right)\left(\text{M of P}\right)+\left(5\right)\left(\text{M of O}\right)$

  $\text{M}$ of $\text{P}$ is $\text{30}\text{.9738 g/mol}$ .

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

Substitute the values in above equation.

  $\begin{array}{c}{\text{Empirical formula of P}}_2{\text{O}}_5=\left(2\right)\left(\text{M of P}\right)+\left(5\right)\left(\text{M of O}\right)\\ =\left(2\right)\left(\text{30}\text{.9738 g/mol}\right)+\left(5\right)\left(15.9994\text{ g/mol}\right)\\ =141.9446\text{ g/mol}\end{array}$

(d)

Interpretation Introduction

Interpretation: Empirical formula and empirical formula mass for ${\text{Ga}}_2{\left({\text{SO}}_{\text{4}}\right)}_3$ should be determined.

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.

Sum of the product of molar masses and number of corresponding atoms results in empirical formula mass.

Answer to Problem 3.26P

Empirical formula and empirical formula mass is ${\text{Ga}}_2{\left({\text{SO}}_{\text{4}}\right)}_3$ and $427.656\text{ g/mol}$ respectively.

Explanation of Solution

  ${\text{Ga}}_2{\left({\text{SO}}_{\text{4}}\right)}_3$ is the molecular formula that is obtained by multiplication of integral factor to an empirical formula. Since the multiplication of only factor 1 to the coprime subscripts 2 and 3 in ${\text{Ga}}_2{\left({\text{SO}}_{\text{4}}\right)}_3$ gives ${\text{Ga}}_2{\left({\text{SO}}_{\text{4}}\right)}_3$ thus the empirical formula is same as ${\text{Ga}}_2{\left({\text{SO}}_{\text{4}}\right)}_3$ .

The formula to calculate empirical formula mass is as follows:

  ${\text{Empirical formula of Ga}}_2{\left({\text{SO}}_{\text{4}}\right)}_3=\left(2\right)\left({\text{M of Ga}}^{3+}\right)+\left(3\right)\left({\text{M of SO}}_4^{2-}\right)$

  $\text{M}$ of ${\text{Ga}}^{3+}$ is $\text{69}\text{.723 g/mol}$ .

  $\text{M}$ of ${\text{SO}}_4^{2-}$ is $\text{96}\text{.07 g/mol}$ .

Substitute the values in above equation.

  $\begin{array}{c}{\text{Empirical formula of Ga}}_2{\left({\text{SO}}_{\text{4}}\right)}_3=\left(2\right)\left({\text{M of Ga}}^{3+}\right)+\left(3\right)\left({\text{M of SO}}_4^{2-}\right)\\ =\left(2\right)\left(\text{69}\text{.723 g/mol}\right)+\left(3\right)\left(\text{96}\text{.07 g/mol}\right)\\ =427.656\text{ g/mol}\end{array}$

(e)

Interpretation Introduction

Interpretation: Empirical formula and empirical formula mass for ${\text{Al}}_{\text{2}}{\text{Br}}_{\text{6}}$ should be determined.

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.

Sum of the product of molar masses and number of corresponding atoms results in empirical formula mass.

Answer to Problem 3.26P

Empirical formula and empirical formula mass is ${\text{AlBr}}_3$ and $266.68\text{ g/mol}$ respectively.

Explanation of Solution

  ${\text{Al}}_{\text{2}}{\text{Br}}_{\text{6}}$ is the molecular formula that is obtained by multiplication of integral factor to an empirical formula. Since the multiplication of only factor 2 to ${\text{AlBr}}_3$ gives ${\text{Al}}_{\text{2}}{\text{Br}}_{\text{6}}$ thus the empirical formula is ${\text{AlBr}}_3$ .

The formula to calculate empirical formula mass is as follows:

  ${\text{Empirical formula of AlBr}}_3=\left(1\right)\left({\text{M of Al}}^{3+}\right)+\left(3\right)\left({\text{M of Br}}^{-}\right)$

  $\text{M}$ of ${\text{Al}}^{3+}$ is $\text{26}\text{.98 g/mol}$ .

  $\text{M}$ of ${\text{Br}}^{-}$ is $\text{79}\text{.9 g/mol}$ .

Substitute the values in above equation.

  $\begin{array}{c}{\text{Empirical formula of AlBr}}_3=\left(1\right)\left({\text{M of Al}}^{3+}\right)+\left(3\right)\left({\text{M of Br}}^{-}\right)\\ =\left(1\right)\left(\text{26}\text{.98 g/mol}\right)+\left(3\right)\left(\text{79}\text{.9 g/mol}\right)\\ =266.68\text{ g/mol}\end{array}$