Chapter 13, Problem 13.128P

(a)

Interpretation Introduction

Interpretation:

The molar mass of the compound is to be calculated.

Concept introduction:

The freezing point is the temperature at which both the solid and liquid phases coexist in equilibrium. It is the temperature at which the vapor pressure of the substance in the liquid state becomes equal to the vapor pressure in a solid state.

The formula to calculate the change in freezing point is as follows:

  $\Delta T_{\text{f}}=i{k}_{\text{f}}m$        (1)

Here,

$\Delta T_{\text{f}}$ is the change in freezing point.

$i$ is van’t Hoff factor.

$k_{\text{f}}$ is the freezing point depression constant.

$m$ is the molality of the solution.

Answer to Problem 13.128P

$\text{89}\text{.9}\text{g/mol}$ is the molar mass of the compound.

Explanation of Solution

The formula to calculate the change in freezing point is as follows:

  $\Delta T_{\text{f}}=T_{\text{f}}{}_{\left(\text{solvent}\right)}-T_{\text{f}}{}_{\left(\text{solution}\right)}$        (2)

Substitute $0°\text{C}$ for $T_{\text{f}}{}_{\left(\text{solvent}\right)}$ and $-0.201°\text{C}$ for $T_{\text{f}}{}_{\left(\text{solution}\right)}$ in equation (2).

  $\begin{array}{c}\Delta T_{\text{f}}=0°\text{C}-\left(-0.201°\text{C}\right)\\ =0.201°\text{C}\end{array}$

The solute is a nonvolatile non-electrolyte so its van’t Hoff factor is 1.

Rearrange equation (1) to calculate the molarity of the solution as follows:

  $m=\frac{\Delta T_{\text{f}}}{i{k}_{\text{f}}}$        (3)

Substitute 1 for $i$, $0.201°\text{C}$ for $\Delta T_{\text{f}}$, $\text{1}\text{.86}°\text{C/}m$ for $k_{\text{b}}$ in equation (3).

  $\begin{array}{c}m=\frac{\text{0}\text{.201}°\text{C}}{\left(1\right)\left(\text{1}\text{.86}°\text{C/}m\right)}\\ =0.1080645m\end{array}$

The density of the solution is calculated as follows:

  $\text{Density of solution}\left(\rho \right)=\frac{\text{Mass }\left(M\right)\text{of solution}}{\text{Volume }\left(V\right)\text{of solution}}$        (4)

Rearrange equation (4) to calculate the mass of the solution as follows:

  $\text{Mass of solution}=\left(\text{Density of solution}\right)\left(\text{Volume of solution}\right)$        (5)

Substitute $25\text{mL}$ for the volume of solution and $1\text{g/mL}$ for the density of the solution in equation (5) to calculate the mass of water.

  $\begin{array}{c}\text{Mass of water}=\left(25\text{mL}\right)\left(\frac{1\text{g}}{1\text{mL}}\right)\left(\frac{1\text{kg}}{{10}^3\text{g}}\right)\\ =0.0250\text{kg}\end{array}$

The formula to calculate the molality of the solution is as follows:

  $\text{Molality}=\frac{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}}{\text{mass}\left(\text{kg}\right)\text{of}\text{}\text{solvent}}$        (6)

Rearrange equation (6) to calculate the moles of solute as follows:

  $\text{Amount}\text{of}\text{solute}=\left(\text{Molality}\right)\left(\text{mass}\text{of}\text{}\text{solvent}\right)$        (7)

Substitute $0.1080645m$ for the molality and $0.0250\text{kg}$ for the mass of solvent in equation (7) to calculate the amount of solute.

  $\begin{array}{c}\text{Amount of solute}=\left(0.1080645m\right)\left(0.0250\text{kg}\right)\\ =0.0027016\text{mol}\end{array}$

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

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

Rearrange equation (8) to calculate the molar mass as follows:

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

Substitute $0.243\text{g}$ for the given mass and $0.0027016\text{mol}$ for the number of moles in equation (9).

  $\begin{array}{c}\text{Molar mass}=\frac{0.243\text{g}}{0.0027016\text{mol}}\\ =89.946698\text{}\text{g/mol}\\ \approx \text{89}\text{.9}\text{g/mol}\text{.}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The empirical and the molecular formula of the compound are to be determined.

Concept introduction:

An empirical formula gives the simplest whole number ratio of atoms of each element present in a molecule. The molecular formula tells the exact number of atoms of each element present in a molecule.

Answer to Problem 13.128P

The empirical and molecular formula of the compound is ${\text{C}}_2{\text{H}}_5\text{O}$ and ${\text{C}}_4{\text{H}}_{10}{\text{O}}_2$.

Explanation of Solution

Consider the mass of the compound to be $100\text{g}$. The mass of carbon is $53.31\text{g}$, the mass of hydrogen is $11.18\text{g}$.

The formula to calculate the mass of the compound is as follows:

  $\text{Mass of compound}=\text{Mass of C}+\text{Mass of H}+\text{Mass of O}$        (10)

Rearrange equation (10) to calculate the mass of oxygen as follows:

  $\text{Mass of O}=\text{Mass of compound}-\left(\text{Mass of C}+\text{Mass of H}\right)$        (11)

Substitute $100\text{g}$ for the mass of the compound, $53.31\text{g}$ for the mass of carbon and $11.18\text{g}$ for the mass of hydrogen in equation (11).

  $\begin{array}{c}\text{Mass of O}=\text{100 g}-\left(\text{53}\text{.31 g}+\text{11}\text{.18 g}\right)\\ =35.51\text{g}\end{array}$

The formula to calculate the moles of the compound is as follows:

  $\text{Moles of compound}=\frac{\text{Given mass}}{\text{Molar mass}}$        (12)

Substitute $53.31\text{g}$ for the given mass and $\text{12}\text{.01 g/mol}$ for the molar mass in equation (12) to calculate the moles of carbon.

  $\begin{array}{c}\text{Moles of C}=\left(\text{53}\text{.31 g}\right)\left(\frac{1\text{mol}}{\text{12}\text{.01 g}}\right)\\ =4.43880\text{mol}\end{array}$

Substitute $11.18\text{g}$ for the given mass and $\text{1}\text{.008 g/mol}$ for the molar mass in equation (12) to calculate the moles of hydrogen.

  $\begin{array}{c}\text{Moles of H}=\left(\text{11}\text{.18 g}\right)\left(\frac{1\text{mol}}{\text{1}\text{.008 g}}\right)\\ =11.09127\text{mol}\end{array}$

Substitute $35.51\text{g}$ for the given mass and $\text{16 g/mol}$ for the molar mass in equation (12) to calculate the moles of oxygen.

  $\begin{array}{c}\text{Moles of O}=\left(\text{35}\text{.51 g}\right)\left(\frac{1\text{mol}}{\text{16 g}}\right)\\ =2.219375\text{mol}\end{array}$

Write the amount of carbon, hydrogen, and oxygen as subscripts of their symbols to obtain a preliminary formula as follows:

  ${\text{C}}_{4.43880}{\text{H}}_{11.09127}{\text{O}}_{2.219375}$

The smallest subscript is $2.219375$. Therefore, divide each subscript by $2.219375$ as follows:

  ${\text{C}}_{\frac{4.43880}{2.219375}}{\text{H}}_{\frac{11.09127}{2.219375}}{\text{O}}_{\frac{2.219375}{2.219375}}\to {\text{C}}_2{\text{H}}_5\text{O}$

The subscripts are in the whole number. Hence, the empirical formula of the compound is ${\text{C}}_2{\text{H}}_5\text{O}$.

The expression to calculate the empirical formula mass of ${\text{C}}_2{\text{H}}_5\text{O}$ is as follows:

  $\begin{array}{c}{\text{Empirical formula mass of C}}_2{\text{H}}_5\text{O}=\left(2\right)\left(\text{M of C}\right)+\left(5\right)\left(\text{M of H}\right)\\ +\left(1\right)\left(\text{M of O}\right)\end{array}$        (13)

Substitute $12.01\text{ g/mol}$ for $\text{M}$ of $\text{C}$, $1.008\text{ g/mol}$ for $\text{M}$ of $\text{H}$, and $16.00\text{ g/mol}$ for $\text{M}$ of $\text{O}$ in equation (13) as follows:

  $\begin{array}{c}{\text{Empirical formula mass of C}}_2{\text{H}}_5\text{O}=\left(2\right)\left(12.01\text{ g/mol}\right)+\left(5\right)\left(1.008\text{ g/mol}\right)\\ +\left(1\right)\left(16.00\text{ g/mol}\right)\\ =45.06\text{ g/mol}\text{.}\end{array}$

The molar mass of the compound is $89.9\text{g/mol}$ that is double its empirical mass. So the molecular formula of the compound is ${\text{C}}_4{\text{H}}_{10}{\text{O}}_2$.

(c)

Interpretation Introduction

Interpretation:

The Lewis structures for the compound that forms hydrogen bonds and one that does not form hydrogen bonds are to be drawn.

Concept introduction:

Lewis structure is basically a simplified representation of the structure of any molecule or atom. Lewis structure shows the bonding between the atoms and the lone pairs of electrons as dot.

The steps to draw the Lewis structure of any molecule are as follows:

1. Write the letter symbol for each element present in the compound and draw the skeletal structure. Generally, the least electronegative element is considered as the central element.

2. Count the total number of valence electrons in the molecule. In case of charged molecules subtract the positive charge from the total number of valence electrons and add the negative charge to the total number of valence electrons.

3. Assign two electrons between two atoms and join them via a single bond. Place the remaining valence electrons as lone pairs such that octet of each element is achieved. Use multiple bonds to complete the octet.

Answer to Problem 13.128P

The Lewis structure of the compound that forms hydrogen bonds is as follows:

The Lewis structure of the compound that does not form hydrogen bonds is as follows:

Explanation of Solution

The compound will form hydrogen bonds with the electronegative element is present at the end positions. But no hydrogen bonding occurs when the electronegative element is placed in between the other elements.

So the Lewis structure of the compound that forms hydrogen bonds is as follows:

The Lewis structure of the compound that does not form hydrogen bonds is as follows: