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?

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.

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 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.