Chapter 11, Problem 11.92P

Interpretation Introduction

Interpretation:

Empirical formula of compound when $30.450\text{mg}$ of sample on combustion gives $54.246\text{mg}{\text{CO}}_2$ 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 11.92P

Empirical formula of compound when $30.450\text{mg}$ of sample on combustion gives $54.246\text{mg}{\text{CO}}_2$ is ${\text{H}}_{12}{\text{O}}_2{\text{C}}_6\text{S}$.

Explanation of Solution

The compound contains carbon, hydrogen oxygen and sulfur. Therefore, mass of carbon in carbon dioxide is equal to mass of carbon in the compound. Also, mass of hydrogen in water is equal to mass of hydrogen in compound.

The formula to calculate mass of $\text{C}$ is as follows:

  $\text{Mass of C}=\left({\text{Mass of CO}}_{\text{2}}\right)\left(\frac{\text{molar mass of C}}{{\text{molar mass of CO}}_{\text{2}}}\right)$        (1)

Substitute $54.246\text{ mg}$ for mass of ${\text{CO}}_{\text{2}}$, $\text{44}\text{.01 g/mol}$ for molar mass of ${\text{CO}}_{\text{2}}$ and $\text{12}\text{.0107 g/mol}$ for molar mass of $\text{C}$ in equation (1).

  $\begin{array}{c}\text{Mass of C}=\left(54.246\text{ mg}\right)\left(\frac{\text{1 g}}{\text{1000 mg}}\right)\left(\frac{\text{12}\text{.0107 g/mol}}{\text{44}\text{.01 g/mol}}\right)\\ =\text{0}\text{.01480 g}\end{array}$

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

  $\text{Mass of H}=\left({\text{Mass of H}}_{\text{2}}\text{O}\right)\left(\frac{\text{2}\left(\text{molar mass of H}\right)}{{\text{molar mass of H}}_{\text{2}}\text{O}}\right)$        (2)

Substitute $22.206\text{ mg}$ for mass of ${\text{H}}_2\text{O}$, $\text{18}\text{.0152 g/mol}$ for molar mass of ${\text{H}}_2\text{O}$ and $\text{1}\text{.00784 g/mol}$ for molar mass of $\text{H}$ in equation (2).

  $\begin{array}{c}\text{Mass of H}=\left(22.206\text{ mg}\right)\left(\frac{\text{1 g}}{\text{1000 mg}}\right)\left(\frac{\text{2}\left(\text{1}\text{.00784 g/mol }\right)}{\text{18}\text{.0152 g/mol}}\right)\\ =0.00248\text{ g}\end{array}$

The mass of sulfur in sulfur dioxide is equal to mass of sulfur in compound.

The formula to calculate mass of $\text{S}$ is as follows:

  $\text{Mass of S}=\left({\text{Mass of SO}}_2\right)\left(\frac{\text{molar mass of S}}{{\text{molar mass of SO}}_2}\right)$        (3)

Substitute $10.255\text{ mg}$ for mass of ${\text{SO}}_2$, $\text{66}\text{.066 g/mol}$ for molar mass of ${\text{SO}}_2$ and $\text{32}\text{.065 g/mol}$ for molar mass of $\text{S}$ in equation (3).

  $\begin{array}{c}\text{Mass of S}=\left(10.255\text{ mg}\right)\left(\frac{\text{1 g}}{\text{1000 mg}}\right)\left(\frac{\text{32}\text{.065 g/mol}}{\text{66}\text{.066 g/mol}}\right)\\ =0.0049\text{77 g}\end{array}$

The formula to calculate mass $\%$ of $\text{C}$ in sample is as follows:

  $\text{Mass \% of C}=\left(\frac{\text{Mass of C}}{\text{Mass of sample}}\right)\left(100\right)$        (4)

Substitute $\text{0}\text{.01480 g}$ for mass of $\text{C}$ and $30.450\text{ mg}$ for mass of sample in equation (4).

  $\begin{array}{c}\text{Mass \% of C}=\left(\frac{\text{0}\text{.01480 g}}{30.450\text{ mg }}\right)\left(\frac{\text{1000 mg}}{\text{1 g}}\right)\left(100\%\right)\\ =48.604\%\end{array}$

The formula to calculate mass $\%$ of $\text{H}$ in sample is as follows:

  $\text{Mass \% of H}=\left(\frac{\text{Mass of H}}{\text{Mass of sample}}\right)\left(100\right)$        (5)

Substitute $0.00248\text{ g}$ for mass of $\text{H}$ and $30.450\text{ mg}$ for mass of sample in equation (5).

  $\begin{array}{c}\text{Mass \% of H}=\left(\frac{0.00248\text{ g}}{30.450\text{ mg }}\right)\left(\frac{\text{1000 mg}}{\text{1 g}}\right)\left(100\%\right)\\ =8.144\%\end{array}$

The formula to calculate mass $\%$ of $\text{S}$ in sample is as follows:

  $\text{Mass \% of S}=\left(\frac{\text{Mass of S}}{\text{Mass of sample}}\right)\left(100\right)$        (6)

Substitute $0.0049\text{77 g}$ for mass of $\text{S}$ and $23.725\text{ mg}$ for mass of sample in equation (6).

  $\begin{array}{c}\text{Mass \% of S}=\left(\frac{0.0049\text{77 g}}{23.725\text{ mg }}\right)\left(\frac{\text{1000 mg}}{\text{1 g}}\right)\left(100\%\right)\\ =20.977\%\end{array}$

The formula to calculate mass $\%$ of $\text{O}$ is as follows:

    $\text{Mass }\%\text{ of O}=\left[\left(\text{100 }\%\right)-\left(\begin{array}{l}\text{mass }\%\text{ of H}+\text{mass }\%\text{ of C}+\\ \text{mass }\%\text{ of S}\end{array}\right)\right]$        (7)

Substitute, $48.604\%$ for mass $\%$ of $\text{C}$, $8.144\%$ for mass $\%$ of $\text{H}$ and $20.977\%$ for mass $\%$ of in $\text{S}$ equation (7).

  $\begin{array}{c}\text{Mass }\%\text{ of O}=\left[\left(\text{100 }\%\right)-\left(8.144\%+48.604\%+20.977\%\right)\right]\\ =22.275\text{ \%}\end{array}$

Consider the sample to be $100\text{ g}$. Therefore, mass of $\text{H}$ is $8.144\text{ g}$, mass of $\text{S}$ is $20.977\text{ g}$, mass of $\text{C}$ is $48.604\text{ g}$ and mass of $\text{O}$ is $22.275\text{ g}$.

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

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

Substitute $8.144\text{ g}$ for mass of $\text{H}$ and $\text{1}\text{.00784 g/mol}$ for molar mass of $\text{H}$ in equation (8).

  $\begin{array}{c}\text{number of moles of H}=\frac{8.144\text{ g}}{\text{1}\text{.00784 g/mol}}\\ =\text{8}\text{.0806 mol}\end{array}$

The formula to calculate number of moles of carbon is as follows:

  $\text{Number of moles of C}=\frac{\text{Given mass of C}}{\text{molar mass of C}}$        (9)

Substitute $48.604\text{ g}$ for mass of $\text{C}$ and $\text{12}\text{.0107 g/mol}$ for molar mass of $\text{C}$ in equation (9).

  $\begin{array}{c}\text{number of moles of C}=\frac{48.604\text{ g}}{\text{12}\text{.0107 g/mol}}\\ =\text{4}\text{.0467 mol}\end{array}$

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

  $\text{Number of moles of O}=\frac{\text{Given mass of O}}{\text{molar mass of O}}$        (10)

Substitute $22.275\text{ g}$ for mass of $\text{O}$ and $\text{15}\text{.999 g/mol}$ for molar mass of $\text{O}$ in equation (10).

  $\begin{array}{c}\text{number of moles of O}=\frac{22.275\text{ g}}{\text{15}\text{.999 g/mol}}\\ =\text{1}\text{.39227 mol}\end{array}$

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

  $\text{Number of moles of S}=\frac{\text{Given mass of S}}{\text{molar mass of S}}$        (11)

Substitute $20.977\text{ g}$ for mass of $\text{S}$ and $\text{32}\text{.065 g/mol}$ for molar mass of $\text{S}$ in equation (11).

  $\begin{array}{c}\text{number of moles of S}=\frac{20.977\text{ g}}{\text{32}\text{.065 g/mol}}\\ =\text{0}\text{.6542 mol}\end{array}$

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

  $\text{Preliminary formula}={\text{H}}_{\text{8}\text{.080}6}{\text{O}}_{\text{1}\text{.39227}}{\text{C}}_{\text{4}\text{.0467}}{\text{S}}_{\text{0}\text{.6542}}$

Each of subscript of $\text{H}$, $\text{O}$, $\text{S}$ and $\text{C}$ is divided by smallest value to determine empirical formula of compound. The smallest value is 0.6542. Therefore, empirical formula of given compound is as follows:

  $\begin{array}{c}\text{Empirical formula}={\text{H}}_{{}_{\frac{\text{8}\text{.0806}}{\text{0}\text{.6542}}}}{\text{O}}_{{}_{\frac{\text{1}\text{.39227}}{\text{0}\text{.6542}}}}{\text{C}}_{{}_{\frac{\text{4}\text{.0467}}{\text{0}\text{.6542}}}}{\text{S}}_{{}_{\frac{\text{0}\text{.6542}}{\text{0}\text{.6542}}}}\\ ={\text{H}}_{12.35}{\text{O}}_{2.12}{\text{C}}_{6.18}\text{S}\\ \approx {\text{H}}_{12}{\text{O}}_2{\text{C}}_6\text{S}\end{array}$