Chapter 5, Problem 5.103P

Answer the following questions about the conversion of the sucrose $({\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}})$ in sugarcane to ethanol $({\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O})$ and CO₂ according to the following unbalanced equation. In this way sugarcane is a renewable source of ethanol, which is used as a fuel additive in gasoline.

   $\begin{array}{l}{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}{\text{(S) + H}}_{\text{2}}\text{O(l)}\to {\text{ C}}_{\text{2}}{\text{H}}_{\text{6}}{\text{O(l)+ CO}}_{\text{2}}\text{(g) }\\ \text{sucrose ethanol}\\ \end{array}$

1. What is the molar mass of sucrose?
2. Balance the given equation.
3. How many moles of ethanol are formed from 2 mol of sucrose?
4. How many moles of water are needed to react with 10 mol of sucrose?
5. How many grams of ethanol are formed from 0.550 mol of sucrose?
6. How many grams of ethanol are formed from 34.2 of sucrose?
7. What is the theoretical yeild of ethanol in grams from 17.1 g of sucrose?
8. If 1.25 g of ethanol are formed in the reaction in part (g). what is the percent yield of ethanol?

Interpretation Introduction

(a)

Interpretation:

The molar mass of sucrose 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 5.103P

The molar mass of sucrose is 342.01 g/mol.

Explanation of Solution

The reaction is given as shown below:

  ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{11}\left(\text{s}\right)+{\text{H}}_2\text{O}\left(l\right)\to {\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\left(l\right)+{\text{CO}}_2\left(g\right)$

The molecular formula of sucrose is given as ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{11}$ . 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 sucrose is calculated as follows:

  $\begin{array}{c}\text{Molar}\text{mass}\text{of}\text{sucrose}=\left[\text{12}\left(\text{mass}\text{of}\text{C}\right)+2\text{2}\left(\text{mass}\text{of}\text{H}\right)+\text{11}\left(\text{mass}\text{of}\text{O}\right)\right]\\ =\left[\text{12}\left(\text{12}\text{.01}\text{g}/\text{mol}\right)+2\text{2}\left(\text{1}\text{.00}\text{g}/\text{mol}\right)+\text{11}\left(\text{15}\text{.99}\text{g}/\text{mol}\right)\right]\\ =144.12+22.00+175.89\\ =342.01\text{g}/\text{mol}\end{array}$

Therefore, the molar mass of sucrose is 342.01 g/mol.

Interpretation Introduction

(b)

Interpretation:

The ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{11}\left(\text{s}\right)+{\text{H}}_2\text{O}\left(l\right)\to {\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\left(l\right)+{\text{CO}}_2\left(g\right)$ 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 5.103P

The balanced equation is ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{11}\left(\text{s}\right)+{\text{H}}_2\text{O}\left(l\right)\to 4{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\left(l\right)+4{\text{CO}}_2\left(g\right)$.

Explanation of Solution

The reaction is given as shown below:

  ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{11}\left(\text{s}\right)+{\text{H}}_2\text{O}\left(l\right)\to {\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\left(l\right)+{\text{CO}}_2\left(g\right)$

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 4 is placed before ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}$ and ${\text{CO}}_2$.

Therefore, the balanced equation is ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{11}\left(\text{s}\right)+{\text{H}}_2\text{O}\left(l\right)\to 4{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\left(l\right)+4{\text{CO}}_2\left(g\right)$.

Interpretation Introduction

(c)

Interpretation:

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

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules.

Answer to Problem 5.103P

The moles of ethanol formed from 2 mol of sucrose are 8.

Explanation of Solution

The balanced reaction is given as shown below:

  ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{11}\left(\text{s}\right)+{\text{H}}_2\text{O}\left(l\right)\to 4{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\left(l\right)+4{\text{CO}}_2\left(g\right)$

In the given reaction, 1 mole of sucrose gives 4 moles of ethanol. Therefore, 2 moles of sucrose give 8 moles of ethanol.

Therefore, moles of ethanol formed from 2 mol of sucrose are 8.

Interpretation Introduction

(d)

Interpretation:

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

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules.

Answer to Problem 5.103P

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

Explanation of Solution

The balanced reaction is given as shown below:

  ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{11}\left(\text{s}\right)+{\text{H}}_2\text{O}\left(l\right)\to 4{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\left(l\right)+4{\text{CO}}_2\left(g\right)$

In the given reaction, 1 mole of sucrose reacts with 1 mole of water. Therefore, 10 moles of sucrose reacts with 10 moles of water.

Therefore, moles of water needed to react with 10 mol of sucrose are 10.

Interpretation Introduction

(e)

Interpretation:

The grams of ethanol formed from 0.550 mol of sucrose should be predicted.

Concept Introduction:

The number of moles is calculated as follows:

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

Answer to Problem 5.103P

The grams of ethanol formed from 0.550 mol of sucrose are 101.354 g.

Explanation of Solution

The balanced reaction is given as shown below:

  ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{11}\left(\text{s}\right)+{\text{H}}_2\text{O}\left(l\right)\to 4{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\left(l\right)+4{\text{CO}}_2\left(g\right)$

In the given reaction, 1 mole of sucrose gives 4 moles of ethanol. Therefore, 0.550 mole of sucrose give 0.550×4 = 2.2 moles of ethanol.

The grams of ethanol are calculated as follows:

  $\begin{array}{c}\text{rams}\text{of}\text{ethanol}=\text{moles}\times \text{molar}\text{mass}\\ =2.2\text{mol}\times 46.07\text{g}/\text{mol}\\ =101.354\text{g}\end{array}$

Therefore, the grams of ethanol formed from 0.550 mol of sucrose are 101.354 g.

Interpretation Introduction

(f)

Interpretation:

The grams of ethanol formed from 34.2 g of sucrose should be predicted.

Concept Introduction:

The number of moles is calculated as follows:

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

Answer to Problem 5.103P

The grams of ethanol formed from 34.2 g of sucrose are 18.24 g.

Explanation of Solution

The molar mass and given mass of sucrose is 342.01 g/mol and 34.2 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.

  $\begin{array}{c}n=\frac{\text{34}\text{.2}\text{g}}{\text{342}\text{.01}\text{g}/\text{mol}}\\ =0.099\text{mol}\end{array}$

Therefore, the number of moles of sucrose is 0.099 mol.

The balanced reaction is given as shown below:

  ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{11}\left(\text{s}\right)+{\text{H}}_2\text{O}\left(l\right)\to 4{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\left(l\right)+4{\text{CO}}_2\left(g\right)$

In the given reaction, 1 mole of sucrose gives 4 moles of ethanol. Therefore, 0.099 mole of sucrose give 0.099×4 = 0.396 moles of ethanol.

The grams of ethanol are calculated as follows:

  $\begin{array}{c}\text{grams}\text{of}\text{ethanol}=\text{moles}\times \text{molar}\text{mass}\\ =0.396\text{mol}\times 46.07\text{g}/\text{mol}\\ =18.24\text{g}\end{array}$

Therefore, the grams of ethanol formed from 34.2 g of sucrose are 18.24 g.

Interpretation Introduction

(g)

Interpretation:

The theoretical yield of ethanol in grams formed from 17.1 g of sucrose should be predicted.

Concept Introduction:

The number of moles is calculated as follows:

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

Answer to Problem 5.103P

The theoretical yield of ethanol in grams formed from 17.1 g of sucrose is 9.02 g.

Explanation of Solution

The molar mass and given mass of sucrose is 342.01 g/mol and 17.1 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.

  $\begin{array}{c}n=\frac{17.1\text{g}}{\text{342}\text{.01}\text{g}/\text{mol}}\\ =0.049\text{mol}\end{array}$

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

The balanced reaction is given as shown below:

  ${\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{11}\left(\text{s}\right)+{\text{H}}_2\text{O}\left(l\right)\to 4{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\left(l\right)+4{\text{CO}}_2\left(g\right)$

In the given reaction, 1 mole of sucrose gives 4 moles of ethanol. Therefore, 0.049 mole of sucrose give 0.049×4 = 0.196 moles of ethanol.

The grams of ethanol are calculated as follows:

  $\begin{array}{c}\text{grams}\text{of}\text{ethanol}=\text{moles}\times \text{molar}\text{mass}\\ =0.196\text{mol}\times 46.07\text{g}/\text{mol}\\ =9.02\text{g}\end{array}$

Therefore, the theoretical yield of ethanol in grams formed from 17.1 g of sucrose is 9.02 g.

Interpretation Introduction

(h)

Interpretation:

The percent yield of ethanol if 1.25 g of ethanol 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 5.103P

The percent yield of ethanol if 1.25 g of ethanol formed in part (g) is 13.86%.

Explanation of Solution

The theoretical yield and actual yield of ethanol is 9.02 g and 1.25 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.

  $\begin{array}{c}\%\text{yield}=\frac{1.25\text{g}}{9.02\text{g}}\times 100\\ =13.86\%\end{array}$

Therefore, the percent yield of ethanol if 1.25 g of ethanol formed in part (g) is 13.86%.