Problem 86 - Chemistry: Principles and Reactions

Question

Most wine is prepared by the fermentation of the glucose in grape juice by yeast: $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(g) $$ How many grams of glucose should there be in grape juice to produce \(725 \mathrm{~mL}\) of wine that is \(11.0 \%\) ethyl alcohol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\left(d=0.789 \mathrm{~g} / \mathrm{cm}^{3}\right)\), by volume?

Step-by-Step Solution

Answer

Answer: 123.09 grams

Step 1: Calculate the volume of ethanol in wine

We are given that the wine is 11.0% ethanol by volume. So, we'll calculate the volume of ethanol in the wine. Ethanol_volume = Total_wine_volume × Ethanol_percentage_by_volume Ethanol_volume = 725 mL × 0.11 = 79.75 mL

Step 2: Calculate the mass of ethanol in wine

We are given the density of ethanol. We can use it to find the mass of ethanol in wine. Ethanol_mass = Ethanol_volume × Density Ethanol_mass = 79.75 mL × 0.789 g/mL = 62.97 g

Step 3: Calculate the moles of ethanol in wine

Now, we'll calculate the moles of ethanol in wine using its molar mass (46.07 g/mol). Moles_ethanol = Ethanol_mass / Molar_mass_ethanol Moles_ethanol = 62.97 g / 46.07 g/mol = 1.367 mol

Step 4: Find the moles of glucose required using stoichiometry

We will use the balanced chemical equation (C6H12O6 → 2C2H5OH + 2CO2) to figure out the moles of glucose required to produce the given moles of ethanol. Mole_ratio_glucose_to_ethanol = 1:2 (from balanced chemical equation) Moles_glucose = Moles_ethanol / 2 Moles_glucose = 1.367 mol / 2 = 0.6835 mol

Step 5: Calculate the grams of glucose required

Finally, we will convert the moles of glucose to grams using the molar mass of glucose (180.16 g/mol). Grams_glucose = Moles_glucose × Molar_mass_glucose Grams_glucose = 0.6835 mol × 180.16 g/mol = 123.09 g So, there should be 123.09 grams of glucose in the grape juice to produce 725 mL of wine that is 11.0% ethyl alcohol by volume.

Key Concepts

Chemical Equation Balancing

Understanding how chemical equations are balanced is essential for nearly all stoichiometric calculations in chemistry. For example, in fermentation, the biochemical reaction converts glucose to ethyl alcohol and carbon dioxide. The balanced equation for this process is: \[ \mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6(aq) \longrightarrow 2\mathrm{C}_2\mathrm{H}_5\mathrm{OH}(aq) + 2\mathrm{CO}_2(g) \]Here, one glucose molecule (\( \mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 \)) is converted into two molecules each of ethanol (\( \mathrm{C}_2\mathrm{H}_5\mathrm{OH} \)) and carbon dioxide (\( \mathrm{CO}_2 \)). This balance is crucial because it represents the law of conservation of mass, indicating that the number of atoms for each element is conserved through the reaction. Balancing equations allows us to directly relate reactants and products through their mole ratios, which are used to calculate how much of one substance is needed to react with a certain amount of another substance.

Molar Mass Calculation

The molar mass of a substance is the weight of one mole (6.022 \( \times \) 10^23 particles) of that substance. For instance, glucose has a molar mass of 180.16 g/mol, and ethanol has a molar mass of 46.07 g/mol. These values are found by summing up the atomic weights of each atom in the molecule, which can be gathered from the periodic table.For ethanol (\( \mathrm{C}_2\mathrm{H}_5\mathrm{OH} \)), the molar mass calculation would look like this:\[ \text{Molar mass of ethanol} = (2 \times \text{Carbon}) + (6 \times \text{Hydrogen}) + (1 \times \text{Oxygen}) \]\[ \text{Carbon: } 12.01 \text{ g/mol, Hydrogen: } 1.01 \text{ g/mol, Oxygen: } 16.00 \text{ g/mol } \]\[ \text{Molar mass of ethanol} = (2 \times 12.01 \text{ g/mol}) + (6 \times 1.01 \text{ g/mol}) + (16.00 \text{ g/mol}) = 46.07 \text{ g/mol} \]Knowing how to calculate molar mass is necessary to convert between grams and moles, a step that is often required in stoichiometric calculations.

Stoichiometric Calculations

Stoichiometric calculations are used to predict the amounts of products and reactants in a chemical reaction based on the mole ratio from the balanced chemical equation. In our fermentation example, the balanced equation informs us that one mole of glucose will produce two moles of ethanol.The steps in the example you're studying show a process by which the amount of glucose needed to produce a certain volume of ethanol is determined. First, we calculate the volume and mass of ethanol produced, then we use the molar mass to find the number of moles of ethanol. With the mole ratio from the balanced reaction, we arrive at the number of moles of glucose required. Finally, multiplying the moles of glucose by its molar mass gives us the mass in grams.To improve understanding of Stoichiometric calculations, one should remember to:

• Confirm the chemical equation is balanced.
• Understand the mole concept and how to use it for relating different substances in a reaction.
• Know how to convert between grams and moles using the molar mass.
• Recognize and use the mole ratios provided by the balanced equation.

These concepts apply universally in stoichiometry, making it a foundation for problem-solving in chemistry.