Chapter 10, Problem 10.123QE

Interpretation Introduction

Interpretation:

Value of x in ${\text{SF}}_x$ has to be identified and also the Lewis structure for the compound has to be drawn indicating the bond angles and hybrid orbitals of sulfur.

Concept Introduction:

Molecular formula of a compound can be found if the empirical formula and molar mass of the compound is known.  The molar mass of compound is divided by the molar mass of the empirical formula in order to obtain the factor which is multiplied with the coefficients of empirical formula in order to obtain the molecular formula.

Lewis structure is used for predicting the shape of molecules.  From the steric number obtained in a Lewis structure, the molecular geometry can be predicted.  VSEPR model can predict the shape of molecules considering their Lewis structure.  Certain rules has to be followed in for the VSEPR model.

• The molecule will have a shape where there is minimal electrostatic repulsion between the valence‑shell electron pairs.
• The forces of repulsion between two lone pairs of electrons will be higher than the repulsion between lone pair and bond pair of electrons.  This in turn will be higher than the bond pair‑bond pair of electrons.

The hybridized orbitals and the steric number can be related as shown below;

Steric number	Hybridized orbital
2	$sp$
3	$s{p}^2$
4	$s{p}^3$
5	$s{p}^{3}d$
6	$s{p}^3{d}^2$

Explanation of Solution

Molecular formula:

The percentage composition of the compound was $\text{54}\text{.53}\text{\%}$ of carbon, $\text{9}\text{.15}\text{\%}$ of hydrogen, and $\text{36}\text{.32}\text{\%}$ of oxygen.  Therefore, in $\text{100}\text{g}$ of the compound, there are $\text{54}\text{.53}\text{g}$ of carbon, $\text{9}\text{.15}\text{g}$ of hydrogen and $\text{36}\text{.32}\text{g}$ of oxygen.  Moles of the components in the compound can be calculated as shown below;

Moles of carbon:

    $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{54.53\text{g}}{12.01\text{g}\cdot {\text{mol}}^{-1}}\\ =4.54\text{mol}\end{array}$

Moles of hydrogen:

    $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{91.5\text{g}}{1.008\text{g}\cdot {\text{mol}}^{-1}}\\ =9.08\text{mol}\end{array}$

Moles of oxygen:

    $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{36.32\text{g}}{16.00\text{g}\cdot {\text{mol}}^{-1}}\\ =2.27\text{mol}\end{array}$

Empirical formula can be obtained by dividing the moles of each element with the least mole.  This is done as follows;

    $\begin{array}{c}\text{Carbon}:\frac{4.54\text{mol}}{2.27\text{mol}}=2.00\\ \text{Hydrogen}:\frac{9.08\text{mol}}{2.27\text{mol}}=4.00\\ \text{Oxygen}:\frac{2.27\text{mol}}{2.27\text{mol}}=1.00\end{array}$

Therefore, the ratio of the element can be given as $\text{C}:\text{H}:\text{O}=2:4:1$.  Thus the empirical formula will be ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\text{O}$.

Molar mass of compound is given as $44\text{g}\cdot {\text{mol}}^{-1}$.

Molar mass of the empirical formula is calculated as follows;

$\begin{array}{c}\text{Molar}\text{mass}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\text{O}=2\times 12.01\text{g}\cdot {\text{mol}}^{-1}+4\times 1.008\text{g}\cdot {\text{mol}}^{-1}+1\times 16.00\text{g}\cdot {\text{mol}}^{-1}\\ =24.02\text{g}\cdot {\text{mol}}^{-1}+4.032\text{g}\cdot {\text{mol}}^{-1}+16.00\text{g}\cdot {\text{mol}}^{-1}\\ =44.052\text{g}\cdot {\text{mol}}^{-1}\end{array}$

Molar mass of the compound is divided by the molar mass of empirical formula in order to obtain the factor as shown below;

    $\begin{array}{c}\frac{\text{Molar}\text{mass}\text{of}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{empirical}\text{formula}}=\frac{44\text{g}\cdot {\text{mol}}^{-1}}{44.052\text{g}\cdot {\text{mol}}^{-1}}\\ =1.00\end{array}$

The coefficient of empirical formula is multiplied by the factor $1.00$ in order to obtain the molecular formula as shown below;

    $\begin{array}{c}\text{Molecular}\text{formula}\text{of}\text{compound}=1.00\times ({\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\text{O})\\ ={\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\text{O}\end{array}$

Therefore, the molecular formula of compound is ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\text{O}$.

Lewis structure of first compound:

The compound is analyzed as ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\text{O}$.  This can be an aldehyde and an alcohol.  Lewis structure for aldehyde is drawn as given below;

The total number of valence electrons in ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\text{O}$ is calculated as shown below;

    $\begin{array}{c}\text{Total}\text{number}\text{of}\text{valence}\text{electrons}=(2\times 4)+(4\times 1)+(1\times 6)\\ =8+4+6\\ =18\end{array}$

Skeletal structure of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\text{O}$ is drawn as given below considering aldehyde;

A total of twelve electrons are involved in the skeletal structure.  Six electrons are placed over the oxygen atom as lone pair of electrons.  Thus the Lewis structure for ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\text{O}$ is drawn as given below;

Bond angles and hybridization:

The first carbon atom does not have any lone pair of electrons while it is bonded to four other atoms.  Therefore, the steric number is calculated as shown below;

    $\begin{array}{c}\text{Steric}\text{number}=0+4\\ =4\end{array}$

As the steric number is four, the arrangement is tetrahedral and the bond angle will be $\text{109}\text{.5}\text{°}$.  As the steric number is four, the hybridization of first carbon atom is $s{p}^3$.

The second carbon atom does not have any lone pair of electrons while it is bonded to three other atoms.  Therefore, the steric number is calculated as shown below;

    $\begin{array}{c}\text{Steric}\text{number}=0+3\\ =3\end{array}$

As the steric number is three, the arrangement is trigonal planar and the bond angle will be $\text{120}\text{°}$.  As the steric number is three, the hybridization of second carbon atom is $s{p}^2$.

Lewis structure of second compound:

The compound is analyzed as ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\text{O}$.  This can be an aldehyde and an alcohol.  Lewis structure for aldehyde is drawn as given below;

The total number of valence electrons in ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\text{O}$ is calculated as shown below;

    $\begin{array}{c}\text{Total}\text{number}\text{of}\text{valence}\text{electrons}=(2\times 4)+(4\times 1)+(1\times 6)\\ =8+4+6\\ =18\end{array}$

Skeletal structure of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\text{O}$ is drawn as given below considering alcohol;

A total of twelve electrons are involved in the skeletal structure.  Four electrons are placed over the oxygen atom as lone pair of electrons.  Thus the Lewis structure for ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\text{O}$ is drawn as given below;

Bond angles and hybridization:

The first carbon atom does not have any lone pair of electrons while it is bonded to three other atoms.  Therefore, the steric number is calculated as shown below;

    $\begin{array}{c}\text{Steric}\text{number}=0+3\\ =3\end{array}$

As the steric number is three, the arrangement is trigonal planar and the bond angle will be $\text{120}\text{°}$.  As the steric number is three, the hybridization of second carbon atom is $s{p}^2$.

The second carbon atom does not have any lone pair of electrons while it is bonded to three other atoms.  Therefore, the steric number is calculated as shown below;

    $\begin{array}{c}\text{Steric}\text{number}=0+3\\ =3\end{array}$

As the steric number is three, the arrangement is trigonal planar and the bond angle will be $\text{120}\text{°}$.  As the steric number is three, the hybridization of second carbon atom is $s{p}^2$.

The oxygen atom does have two lone pair of electrons while it is bonded to two other atoms.  Therefore, the steric number is calculated as shown below;

    $\begin{array}{c}\text{Steric}\text{number}=2+2\\ =4\end{array}$

As the steric number is four, the arrangement is tetrahedral and the bond angle will be $\text{109}\text{.5}\text{°}$.  As the steric number is four, the hybridization of oxygen atom is $s{p}^3$.