Question

The fermentation of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) produces ethyl alcohol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and \(\mathrm{CO}_{2}\) $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(g) $$ (a) How many moles of \(\mathrm{CO}_{2}\) are produced when \(0.400\) mol of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) reactsin this fashion? (b) How many grams of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) are needed to form \(7.50 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} ?\) (c) How many grams of \(\mathrm{CO}_{2}\) form when \(7.50 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) are produced?

Short answer

(a) 0.800 moles of CO₂ are produced when 0.400 moles of glucose react. (b) 14.7 grams of glucose are needed to form 7.50 grams of ethyl alcohol. (c) 7.17 grams of CO₂ are formed when 7.50 grams of ethyl alcohol are produced.

Step-by-step solution

(a) Find moles of \(\mathrm{CO}_{2}\) produced

To find the moles of \(\mathrm{CO}_{2}\) produced when 0.400 moles of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) reacts, we can use the mole ratio of glucose to CO₂ from the balanced chemical equation: \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(g)\) This means that for every mole of glucose, two moles of CO₂ are produced. So, moles of CO₂ = 0.400 moles of glucose x 2 moles of CO₂/1 mole of glucose = 0.800 moles of CO₂.

(b) Find grams of glucose needed to form 7.50 g of ethyl alcohol

For this step, we will first find the moles of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) and then the moles of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) using the mole ratio. Finally, convert the moles of glucose to grams. Molar mass of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} = 2(12.01) + 6(1.01) + 1(16.00) = 46.07 \mathrm{~g/mol}\) Moles of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} = \frac{7.50 \mathrm{~g}}{46.07 \mathrm{g/mol}} \approx 0.163 \mathrm{~mol}\) From the balanced chemical equation, we know that 2 moles of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) are produced from 1 mole of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\). So, moles of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} = 0.163 \mathrm{mol} \times \frac{1 \mathrm{mol \space glucose}}{2 \mathrm{mol \space ethyl \space alcohol}} \approx 0.0815 \mathrm{~mol}\) Molar mass of glucose (\(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)) = 6(12.01) + 12(1.01) + 6(16.00) = 180.16 \mathrm{~g/mol}$ Grams of glucose = (0.0815 mol) x (180.16 g/mol) = 14.7 g

(c) Find grams of CO₂ formed when 7.50 g of ethyl alcohol are produced

We have already calculated the moles of ethyl alcohol formed. We will use this and the mole ratio of ethyl alcohol to CO₂ to find moles of CO₂ produced and then convert it to grams. Moles of CO₂ = 0.163 mol x (2 mol CO₂/2 mol ethyl alcohol) = 0.163 mol Molar mass of CO₂ = 1(12.01) + 2(16.00) = 44.01 g/mol Grams of CO₂ = (0.163 mol) x (44.01 g/mol) ≈ 7.17 g So, the answers are: (a) 0.800 moles of CO₂ are produced (b) 14.7 grams of glucose are needed (c) 7.17 grams of CO₂ are formed

Key concepts

Understanding Mole Calculations

Mole calculations are a vital part of stoichiometry, allowing chemists to quantify substances involved in chemical reactions. The mole is a fundamental unit in chemistry that represents a specific number of particles—namely, Avogadro's number, which is approximately \(6.022 \times 10^{23}\). This concept helps bridge the gap between the atomic scale and the macroscopic quantities we can measure.

When performing mole calculations, it's crucial to use balanced chemical equations. These equations indicate the proportions of reactants and products involved in the reaction. In the example of glucose fermentation, the equation \[\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} + 2 \mathrm{CO}_{2}\]shows that each mole of glucose yields two moles of carbon dioxide.

Mole calculations involve finding the number of moles for reactants or products using given quantities and their molar masses. For instance, if 0.400 moles of glucose react, using the mole ratio gives us 0.800 moles of \(\mathrm{CO}_{2}\). This is essential for problems involving conversion between mass and moles, ensuring accurate predictions of chemical behavior.

The Role of Chemical Reactions

Chemical reactions are processes in which substances react to form new products. They are expressed through chemical equations that illustrate the conversion, showcasing the reactants and the products. A balanced chemical equation is necessary to maintain the law of conservation of mass, which states that mass is neither created nor destroyed in a chemical reaction.

Let's take a closer look at the fermentation of glucose, where glucose \((\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6})\) undergoes a chemical change to form ethyl alcohol \((\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH})\) and carbon dioxide \((\mathrm{CO}_{2})\). The balanced equation,

\[\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} + 2 \mathrm{CO}_{2}\]ensures that the number of atoms for each element is the same on both sides of the equation. This balance is crucial for correctly calculating the amounts of reactants needed or the quantities of products formed.

Each component of the reaction has defined roles, with glucose serving as the reactant and its conversion into ethyl alcohol and carbon dioxide highlighting the transformative nature of chemical reactions.

Glucose Fermentation Process

Glucose fermentation is a biological process that converts glucose into smaller organic compounds, producing ethyl alcohol and carbon dioxide as byproducts. It's a naturally occurring reaction facilitated by microorganisms like yeast, which act as catalysts. This process is not only important in metabolic pathways but also in industries such as brewing and biofuel production.

During fermentation, glucose molecules \((\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6})\) are broken down. The chemical equation for this process is:

\[\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(aq) \rightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(aq) + 2 \mathrm{CO}_{2}(g)\]In this reaction, one glucose molecule yields two molecules each of ethanol and carbon dioxide. By understanding this equation, we can determine various quantities, such as the amount of glucose required to produce a specific mass of ethyl alcohol or the amount of \(\mathrm{CO}_{2}\) generated.

The stoichiometric relationships between reactants and products allow for efficient scaling in industrial applications, making glucose fermentation a cornerstone in numerous biochemical processes. The predictive power of such equations helps optimize both laboratory experiments and larger-scale production systems.