Chapter 9, Problem 52A

Interpretation Introduction

Interpretation: The balanced chemical equation for burning of glucose in air to produce water and carbon dioxide should be determined and the theoretical yield of carbon dioxide needs to be calculated on burning 1.00 g of glucose.

Concept introduction:The reactions for those the number of atoms of each element in the reactant and in the product, side are equal, such reactions are said to be a balanced chemical equation.

The ratio of mass of substance to its molar mass is said to be number of moles of that substance.

Answer to Problem 52A

The balanced chemical equation for burning of glucose in air to produce water and carbon dioxide is:

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}(s)+{\text{ 6O}}_2(g)\to {\text{ 6CO}}_2(g){\text{ + 6H}}_2\text{O}(g)$

The mass of carbon dioxide produced from burning of 1.0 g glucose is 1.48 g.

Explanation of Solution

An exothermic reaction that is where energy is produced in the reaction by reacting compound with oxygen which results in the formation of water and carbon dioxide is said to be combustion reaction.

The molecular formula for glucose is ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ which on burning in air, ${\text{O}}_2$ produces carbon dioxide, ${\text{CO}}_2$ and water, ${\text{H}}_{\text{2}}\text{O}$ . This reaction can be written in terms of chemical equation as:

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}(s)+{\text{ O}}_2(g)\to {\text{ CO}}_2(g){\text{ + H}}_2\text{O}(g)$

The above equation is not balanced as the number of C, H and O on the reactant side is 6, 12 and 8 respectively whereas the number of C, H and O on the product side is 1, 2 and 3 respectively. So, in order to balance the reaction, the coefficient 6 is used before ${\text{O}}_2$ on the reactant side andcoefficient 6 is used for both ${\text{CO}}_2$ and ${\text{H}}_2\text{O}$ on the product side. Thus, the balanced reaction is:

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}(s)+{\text{ 6O}}_2(g)\to {\text{ 6CO}}_2(g){\text{ + 6H}}_2\text{O}(g)$

The molar mass of glucose is 180.156 g/mol. The number of moles of glucose is calculated as shown:

  $\begin{array}{l}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ \text{Number of moles}=\frac{\text{1}\text{.0 g}}{\text{180}\text{.156 g/mol}}\\ \text{Number of moles}=0.0056\text{ mol}\end{array}$

From the balanced chemical reaction, the mole ratio of glucose and carbon dioxide is 1:6 so, the number of moles of carbon dioxide for 0.0056 mol of glucose is:

  $\text{0}{\text{.0056 mol of C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{×}\frac{{\text{6 mol of CO}}_{\text{2}}}{{\text{1 mol of C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}}\text{ = 0}{\text{.0336 mol of CO}}_{\text{2}}$

Now, the theoretical yield of carbon dioxideproduced from burning of 1.0 g glucose is calculated as:

The molar mass of carbon dioxideis 44.01 g/mol. So,

  $\begin{array}{l}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ \text{0}\text{.0336 mol}=\frac{\text{Mass}}{\text{44}\text{.01 g/mol}}\\ \text{Mass = 0}\text{.0336 mol}\times \text{44}\text{.01 g/mol}\\ \text{Mass = }1.48\text{ g}\end{array}$

Hence, the mass of carbon dioxideproduced from burning of 1.0 g glucose is 1.48 g.