Problem 85 - Chemistry: An Atoms-Focused Approach

Question

Chemistry of Fermentation Yeast converts glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) in aqueous solution into ethanol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right.\) \(d=0.789 \mathrm{g} / \mathrm{mL})\) in a process called fermentation. Carbon dioxide is also produced. a. Write a balanced chemical equation for the fermentation reaction. b. If 100.0 grams of glucose yields \(50.0 \mathrm{mL}\) of ethanol, what is the percent yield for the reaction?

Step-by-Step Solution

Answer

Answer: The balanced chemical equation for the fermentation of glucose by yeast is C6H12O6 -> 2 C2H5OH + 2 CO2. The percent yield of ethanol is approximately 77.13%.

Step 1: (Step 1: Write the balanced chemical equation for the fermentation reaction.)

To write a balanced chemical equation for the fermentation reaction, we need to know the reactants and products involved and their respective coefficients. Reactants: Glucose (C6H12O6) Products: Ethanol (C2H5OH) and Carbon Dioxide (CO2) Now let's balance the equation: C6H12O6 → 2 C2H5OH + 2 CO2 Hence, the balanced chemical equation for the fermentation reaction is: C6H12O6 -> 2 C2H5OH + 2 CO2

Step 2: (Step 2: Calculate the theoretical yield)

To calculate the theoretical yield, we need the molecular weights of glucose (C6H12O6) and ethanol (C2H5OH). We can then use stoichiometry to find the amount of ethanol produced by 100 g of glucose. Molecular weight of glucose: C: 6 × 12.01 g/mol = 72.06 g/mol H: 12 × 1.01 g/mol = 12.12 g/mol O: 6 × 16.00 g/mol = 96.00 g/mol Total = 180.18 g/mol Molecular weight of ethanol: C: 2 × 12.01 g/mol = 24.02 g/mol H: 6 × 1.01 g/mol = 6.06 g/mol O: 1 × 16.00 g/mol = 16.00 g/mol Total = 46.08 g/mol Now, convert 100 g of glucose to moles: 100 g × (1 mol / 180.18 g) = 0.555 moles Using the balanced equation, 1 mole of glucose produces 2 moles of ethanol. Therefore, 0.555 moles of glucose will produce: 0.555 moles × 2 = 1.11 moles of ethanol. Now, convert moles of ethanol to grams: 1.11 moles × 46.08 g/mol = 51.13 g The theoretical yield of ethanol from 100 g of glucose is 51.13 g.

Step 3: (Step 3: Calculate the percent yield)

We are given that 100 g of glucose yields 50.0 mL of ethanol. We will first convert the given volume of ethanol into grams using the given density (0.789 g/mL). Mass of ethanol = Volume × Density Mass of ethanol = 50.0 mL × 0.789 g/mL = 39.45 g Now, we will calculate the percent yield using the theoretical yield and the actual yield: Percent Yield = (Actual Yield / Theoretical Yield) × 100 Percent Yield = (39.45 g / 51.13 g) × 100 = 77.13 % The percent yield for the reaction is approximately 77.13%.

Key Concepts

Balanced Chemical Equation

A balanced chemical equation shows the relationship between reactants and products in a chemical reaction. It reflects the conservation of mass, meaning the number of atoms per element in reactants must equal the number in the products. In the fermentation reaction, glucose (\(\text{C}_6\text{H}_{12}\text{O}_6\)) is converted into ethanol (\(\text{C}_2\text{H}_5\text{OH}\)) and carbon dioxide (\(\text{CO}_2\)).
The unbalanced form of this reaction might look like: - \[\text{C}_6\text{H}_{12}\text{O}_6 \rightarrow \text{C}_2\text{H}_5\text{OH} + \text{CO}_2\] By balancing, we adjust coefficients to ensure equality of each element on both sides. Therefore, the balanced equation becomes:

• \[\text{C}_6\text{H}_{12}\text{O}_6 \rightarrow 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2\]

This shows that one glucose molecule produces two molecules of ethanol and two of carbon dioxide. Having a balanced equation is crucial for proceeding further with yield calculations.

Theoretical Yield Calculation

Theoretical yield is the maximum amount of product expected from a reaction. It is calculated using the balanced equation and the reactant amounts. By using molecular weights and stoichiometry, you'll determine the potential output, assuming perfect conditions.
**Step-by-Step Process**

• First, calculate the molar mass of glucose by summing the atomic masses of its constituents: - Carbon: 6 × 12.01 g/mol = 72.06 g/mol- Hydrogen: 12 × 1.01 g/mol = 12.12 g/mol- Oxygen: 6 × 16.00 g/mol = 96.00 g/mol- Total for glucose = 180.18 g/mol
• For ethanol, sum similarly: - Carbon: 2 × 12.01 g/mol = 24.02 g/mol- Hydrogen: 6 × 1.01 g/mol = 6.06 g/mol- Oxygen: 1 × 16.00 g/mol = 16.00 g/mol- Total for ethanol = 46.08 g/mol
• Convert 100 g of glucose to moles: \[100 \text{ g} \times \left( \frac{1 \text{ mol}}{180.18 \text{ g}} \right) = 0.555 \text{ moles} \]
• Using the balanced equation (1 mole glucose = 2 moles ethanol), these 0.555 moles yield \(0.555 \times 2 = 1.11\) moles of ethanol.
  
• Finally, convert to grams: \[1.11 \text{ moles} \times 46.08 \text{ g/mol} = 51.13 \text{ g} \]

This 51.13 g is the theoretical yield of ethanol.

Percent Yield Calculation

Percent yield assesses the efficiency of a chemical reaction, comparing the actual yield to the theoretical yield. It indicates how much product was actually obtained from the reaction relative to the maximum possible amount. This is a valuable measure in experiments.
**Steps to Calculate Percent Yield:**

• Determine the actual yield of ethanol. We're given 50.0 mL of ethanol, which is then converted to mass using the density provided (0.789 g/mL): \[\text{Mass} = 50.0 \text{ mL} \times 0.789 \text{ g/mL} = 39.45 \text{ g}\]
• Compare this to the theoretical yield determined earlier: 51.13 g.
  
• Apply the percent yield formula: \[\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\]
- Substituting the values: \[\text{Percent Yield} = \left( \frac{39.45 \text{ g}}{51.13 \text{ g}} \right) \times 100 \approx 77.13\%\]

This helps in identifying losses or inefficiencies during the process and improving future reactions.