Chapter 5, Problem 92P

Answer the following questions about diethyl ether $({\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{O})$ , the first widely used general anesthetic. Diethyl ether can be prepared from ethanol according to the following unbalanced equation.

   $\begin{array}{l}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O(l) }\to {\text{ C}}_{\text{4}}{\text{H}}_{\text{10}}{\text{O(l) + H}}_{\text{2}}\text{O(l)}\\ \text{ethanol diethyl ether}\\ \end{array}$

1. What is the molar mass of diethyl ether?
2. Balance the given equation.
3. How many moles of diethyl ether are formed from 2 mol of ethanol?
4. How many moles of water are formed from 10 mol ethanol?
5. How many grams of diethyl ether are formed from 0.55 mol of ethanol?
6. How many grams of diethyl ether are formed from 4.60 g of ethanol?
7. What is the theoretical yield of diethyl ethr in grams from 2.30 g of ethanol?
8. If 1.80 g of diethyl ether are formed in the reaction in part (g), what is the percent yield of diethyl ether?

Interpretation Introduction

(a)

Interpretation:

The molar mass of diethyl ether should be predicted.

Concept Introduction:

Molar mass of a substance is sum of atomic masses of all the individual atoms present in it. Mole is the amount of the substance that contains the same number of particles or atoms or molecules.

Answer to Problem 92P

The molar mass of diethyl ether is 74.03 g/mol.

Explanation of Solution

The reaction is given as shown below:

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\left(l\right)\to {\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{O}\left(l\right)+{\text{H}}_2\text{O}\left(l\right)$

The molecular formula of diethyl ether is given as ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{O}$ . The molar mass of carbon, hydrogen and oxygen is 12.01 g/mol, 1.00 g/mol and 15.99 g/mol respectively.

The molar mass of diethyl ether is calculated as follows:

  $\begin{array}{c}\text{Molar}\text{mass}\text{of}\text{diethyl}\text{ether}=\left[4\left(\text{mass}\text{of}\text{C}\right)+10\left(\text{mass}\text{of}\text{H}\right)+\left(\text{mass}\text{of}\text{O}\right)\right]\\ =\left[4\left(\text{12}\text{.01}\text{g}/\text{mol}\right)+10\left(\text{1}\text{.00}\text{g}/\text{mol}\right)+\left(\text{15}\text{.99}\text{g}/\text{mol}\right)\right]\\ =48.04+10.00+\text{15}\text{.99}\\ =74.03\text{g}/\text{mol}\end{array}$

Therefore, the molar mass of diethyl ether is $74.03\text{g}/\text{mol}$.

Interpretation Introduction

(b)

Interpretation:

The ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\left(l\right)\to {\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{O}\left(l\right)+{\text{H}}_2\text{O}\left(l\right)$ reaction should be balanced.

Concept Introduction:

A balance equation has the equal number of atoms on the left-hand side as well as on the right-hand side.

Answer to Problem 92P

The balanced equation is ${\text{2C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\left(l\right)\to {\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{O}\left(l\right)+{\text{H}}_2\text{O}\left(l\right)$.

Explanation of Solution

The reaction is given as shown below:

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\left(l\right)\to {\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{O}\left(l\right)+{\text{H}}_2\text{O}\left(l\right)$

In the given reaction, the number of carbon atoms, hydrogen atoms and oxygen atoms are not equal on both sides. To balance the reaction, coefficient 2 is placed before ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}$.

Therefore, the balanced equation is ${\text{2C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\left(l\right)\to {\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{O}\left(l\right)+{\text{H}}_2\text{O}\left(l\right)$.

Interpretation Introduction

(c)

Interpretation:

The moles of diethyl ether formed from 2 mol of ethanol should be predicted.

Concept Introduction:

Molar mass of a substance is sum of atomic masses of all the individual atoms present in it. Mole is the amount of the substance that contains the same number of particles or atoms or molecules.

Answer to Problem 92P

The moles of diethyl ether formed from 2 mol of ethanol are 1.

Explanation of Solution

The balanced reaction is given as shown below:

  ${\text{2C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\left(l\right)\to {\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{O}\left(l\right)+{\text{H}}_2\text{O}\left(l\right)$

In the given reaction, 2 moles of diethyl ether are formed by 1 mole of ethanol. Therefore, moles of diethyl ether formed from 2 mol of ethanol are 1.

Interpretation Introduction

(d)

Interpretation:

The moles of water needed to react with 10 mol of ethanol should be predicted.

Concept Introduction:

Molar mass of a substance is sum of atomic masses of all the individual atoms present in it. Mole is the amount of the substance that contains the same number of particles or atoms or molecules.

Answer to Problem 92P

The moles of water needed to react with 10 mol of ethanol are 5.

Explanation of Solution

The balanced reaction is given as shown below:

  ${\text{2C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\left(l\right)\to {\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{O}\left(l\right)+{\text{H}}_2\text{O}\left(l\right)$

In the given reaction, 2 moles of ethanol give 1 mole of water. Therefore, 10 moles of ethanol give 5 moles of water.

Therefore, the moles of water needed to react with 10 mol of ethanol are 5.

Interpretation Introduction

(e)

Interpretation:

The grams of diethyl ether formed from 0.55 mol of ethanol should be predicted.

Concept Introduction:

Molar mass of a substance is sum of atomic masses of all the individual atoms present in it. Mole is the amount of the substance that contains the same number of particles or atoms or molecules.

Answer to Problem 92P

The grams of diethyl ether formed from 0.55 mol of ethanol are $20.36\text{g}$.

Explanation of Solution

The balanced reaction is given as shown below:

  ${\text{2C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\left(l\right)\to {\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{O}\left(l\right)+{\text{H}}_2\text{O}\left(l\right)$

In the given reaction, 2 moles of ethanol give 1 mole of diethyl ether. Therefore, 0.55 mole of ethanol give 0.55/2 = 0.275 mol of diethyl ether.

The grams of diethyl ether are calculated as follows:

  $\begin{array}{c}\text{grams}\text{of}\text{diethyl}\text{ether}=\text{moles}\times \text{molar}\text{mass}\\ =0.275\text{mol}\times 74.03\text{g}/\text{mol}\\ =20.36\text{g}\end{array}$

Therefore, the grams of diethyl ether formed from 0.55 mol of ethanol are $20.36\text{g}$.

Interpretation Introduction

(f)

Interpretation:

The grams of diethyl ether formed from 4.60 g of ethanol should be predicted.

Concept Introduction:

Molar mass of a substance is sum of atomic masses of all the individual atoms present in it. Mole is the amount of the substance that contains the same number of particles or atoms or molecules.

Answer to Problem 92P

The grams of diethyl ether formed from 4.60 g of ethanol are 3.69 g.

Explanation of Solution

The molar mass and given mass of ethanol is 46.07 g/mol and 4.60 g respectively. The number of moles is calculated as follows:

  $n=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$

Substitute given mass and molar mass in the formula.

  $\begin{array}{c}n=\frac{4.60\text{g}}{46.07\text{g}/\text{mol}}\\ =0.0998\text{mol}\end{array}$

Therefore, the number of moles of ethanol is 0.0998 mol.

The balanced reaction is given as shown below:

  ${\text{2C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\left(l\right)\to {\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{O}\left(l\right)+{\text{H}}_2\text{O}\left(l\right)$

In the given reaction, 2 moles of ethanol give 1 mole of diethyl ether. Therefore, 0.0998 mole of ethanol give 0.0998/2 = 0.0499 moles of diethyl ether.

The grams of ethanol are calculated as follows:

  $\begin{array}{c}\text{grams}\text{of}\text{diethyl}\text{ether}=\text{moles}\times \text{molar}\text{mass}\\ =0.0499\text{mol}\times 74.03\text{g}/\text{mol}\\ =3.69\text{g}\end{array}$

Therefore, the grams of diethyl ether formed from 4.60 g of ethanol are 3.69 g.

Interpretation Introduction

(g)

Interpretation:

The theoretical yield of diethyl ether in grams formed from 2.30 g of ethanol should be predicted.

Concept Introduction:

Molar mass of a substance is sum of atomic masses of all the individual atoms present in it. Mole is the amount of the substance that contains the same number of particles or atoms or molecules.

Answer to Problem 92P

The theoretical yield of diethyl ether in grams formed from 2.30 g of ethanol is 1.81 g.

Explanation of Solution

The molar mass and given mass of ethanol is 46.07 g/mol and 2.30 g respectively. The number of moles is calculated as follows:

  $n=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$

Substitute given mass and molar mass in the formula.

  $\begin{array}{c}n=\frac{2.30\text{g}}{46.07\text{g}/\text{mol}}\\ =0.049\text{mol}\end{array}$

Therefore, the number of moles of ethanol is 0.049 mol.

The balanced reaction is given as shown below:

  ${\text{2C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\left(l\right)\to {\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{O}\left(l\right)+{\text{H}}_2\text{O}\left(l\right)$

In the given reaction, 2 moles of ethanol give 1 mole of diethyl ether. Therefore, 0.049 mole of ethanol give 0.049/2 = 0.0245 moles of diethyl ether.

The grams of ethanol are calculated as follows:

  $\begin{array}{c}\text{grams}\text{of}\text{diethyl}\text{ether}=\text{moles}\times \text{molar}\text{mass}\\ =0.0245\text{mol}\times 74.03\text{g}/\text{mol}\\ =1.81\text{g}\end{array}$

Therefore, the theoretical yield of diethyl ether in grams formed from 2.30 g of ethanol is 1.81 g.

Interpretation Introduction

(h)

Interpretation:

The percent yield of ethanol if 1.80 g of diethyl ether formed in part (g) should be predicted.

Concept Introduction:

The amount that is predicted by the calculation of stoichiometry of the reaction is known as theoretical yield. The amount that is produced by a product in a reaction is known as the actual yield. The ratio of actual yield to the theoretical yield is called percentage yield of a reaction.

Answer to Problem 92P

The percent yield of ethanol if 1.80 g of diethyl ether formed in part (g) is 100.5%.

Explanation of Solution

The theoretical yield and actual yield of diethyl ether is 1.81 g and 1.80 g respectively.

The percentage yield for the reaction is calculated as follows:

  $\%\text{yield}=\frac{\text{Actual}\text{yield}}{\text{Theoretical}\text{yield}}\times 100$

Substitute observed and theoretical yield in the above formula.

  $\begin{array}{c}\%\text{yield}=\frac{1.81\text{g}}{1.80\text{g}}\times 100\\ =100.5\%\end{array}$

Therefore, the percent yield of ethanol if 1.80 g of diethyl ether formed in part (g) is 100.5%.