Chapter 5, Problem 5.100AP

(a)

Interpretation Introduction

Interpretation:

The molar mass of diethyl ether has to be given.

Answer to Problem 5.100AP

The molar mass of diethyl ether is $\text{74}\text{.1 g/mol}\text{.}$

Explanation of Solution

The formula weight of diethyl ether is calculated as,

  $\begin{array}{l}\text{4}\text{C}\text{atoms}\text{×}\text{12}\text{.01}\text{amu}\text{=}\text{48}\text{.04}\text{amu}\\ \text{10}\text{H}\text{atoms×1}\text{.008}\text{amu}\text{=}\text{10}\text{.08}\text{amu}\\ \text{1}\text{O}\text{atoms×16}\text{.00}\text{amu}\text{=}\underset{\_}{\text{16}\text{.00}\text{amu}}\\ \text{Formula weight}\text{of}{\text{C}}_{\text{4}}{\text{H}}_{10}\text{O=74}\text{.12}\text{amu}\end{array}$

Since, the formula weight of diethyl ether is $\text{74}\text{.1}\text{amu,}$ the molar mass of diethyl ether is $\text{74}\text{.1 g/mol}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The equation of ethanol to give diethyl ether and water has to be balanced.

Answer to Problem 5.100AP

The balanced equation of ethanol to give diethyl ether and carbon dioxide is,

  $\text{2}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\left(\text{s}\right)\stackrel{}{\to }{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{O}\left(\text{l}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Explanation of Solution

The given equation is,

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\left(\text{s}\right)\stackrel{}{\to }{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{O}\left(\text{l}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right)$

The atoms in ethanol have to be balanced.

Place the coefficient $\text{2}$ to balance carbons in the reactant side (before ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}$).

Place the coefficient $\text{2}$ to balance hydrogens in the reactant side (before ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}$).

Place the coefficient $\text{2}$ to balance oxygens in the reactant side (before ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}$).

The balanced equation of ethanol to give diethyl ether and carbon dioxide is,

  $\text{2}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\left(\text{s}\right)\stackrel{}{\to }{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{O}\left(\text{l}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right)$

(c)

Interpretation Introduction

Interpretation:

The number of moles of diethyl ether formed from $\text{2}$ mole of ethanol has to be given.

Answer to Problem 5.100AP

The number of moles of diethyl ether formed from $\text{2}$ mole of ethanol is $\text{1 mole}\text{.}$

Explanation of Solution

Given,

Moles of ethanol are $\text{2}$ moles.

The balanced equation of ethanol with water to give diethyl ether and carbon dioxide is,

  $\text{2}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\left(\text{s}\right)\stackrel{}{\to }{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{O}\left(\text{l}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right)$

The number of moles of diethyl ether formed from $\text{2}$ mole of ethanol is calculated as,

  $\text{2}\cancel{\text{mol}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}}\text{×}\frac{\text{1}\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{O}}{\text{2}\cancel{\text{mol}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}}}\text{=1}\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{O}$

The number of moles of diethyl ether formed from $\text{2}$ mole of ethanol is $\text{1 mole}\text{.}$

(d)

Interpretation Introduction

Interpretation:

The number of grams of diethyl ether that is formed from $\text{4}\text{.60 g}$ of ethanol has to be given.

Answer to Problem 5.100AP

The number of grams of diethyl ether that is formed from $\text{4}\text{.60 g}$ of ethanol is $\text{3}\text{.70}\text{g}\text{.}$

Explanation of Solution

Given,

Grams of ethanol are $\text{4}\text{.60 g}\text{.}$

The balanced chemical equation is,

  $\text{2}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\left(\text{s}\right)\stackrel{}{\to }{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{O}\left(\text{l}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Grams can be converted into moles using the conversion factor:

  $\frac{\text{1}\text{mol}}{\text{Molar}\text{mass}}$

The number of moles present in $\text{4}\text{.60 g}$ of ethanol is calculated as,

  $\text{4}\text{.60}\cancel{\text{g}}{\text{C}}_2{\text{H}}_6\text{O×}\frac{\text{1}\text{mol}}{\text{46}\cancel{\text{g}}}\text{=0}\text{.1}\text{mol}$

The number of moles present in $\text{4}\text{.60 g}$ of ethanol is $\text{0}\text{.1}\text{mol}\text{.}$

The number of moles of diethyl ether that is formed from $\text{4}\text{.60 g}$ of ethanol is calculated as,

  $\text{0}\text{.1}\cancel{\text{mol}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}}\text{×}\frac{\text{1}\text{mol}{\text{C}}_4{\text{H}}_{10}\text{O}}{\text{2}\cancel{\text{mol}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}}}\text{=0}\text{.05}\text{mol}{\text{C}}_4{\text{H}}_{10}\text{O}$

The number of moles of diethyl ether that is formed from $\text{4}\text{.60 g}$ of ethanol is $\text{0}\text{.05}\text{moles}\text{.}$

Moles can be converted into grams using the conversion factor:

  $\frac{\text{Molar}\text{mass}}{\text{1}\text{mol}}$

The number of grams of diethyl ether that is formed from $\text{4}\text{.60 g}$ of ethanol is calculated as,

  $\text{0}\text{.05}\cancel{\text{mol}}\text{×}\frac{74.1\text{g}{\text{C}}_4{\text{H}}_{10}\text{O}}{\text{1}\cancel{\text{mol}}}\text{=3}\text{.70}\text{g}$

The number of grams of diethyl ether that is formed from $\text{4}\text{.60 g}$ of ethanol is $\text{3}\text{.70}\text{g}\text{.}$