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Chapter 3: Problem 72

Fermentation is a complex chemical process of wine making in which glucose is converted into ethanol and carbon dioxide: $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+2 \mathrm{CO}_{2}$$ Starting with \(500.4 \mathrm{~g}\) of glucose, what is the maximum amount of ethanol in grams and in liters that can be obtained by this process? (Density of ethanol \(=0.789 \mathrm{~g} / \mathrm{mL} .)\)

Short Answer

Expert verified
The maximum amount of ethanol that can be obtained from 500.4 g of glucose is 256.3 g of ethanol or 0.325 L of ethanol.

Step by step solution

01

Calculate the Molar Mass of Glucose and Ethanol

The molar mass of glucose (C6H12O6) and ethanol (C2H5OH) are calculated as follow: The molar mass of glucose is \(180.16 \mathrm{~g/mol}\) and the molar mass of ethanol is \(46.07 \mathrm{~g/mol}\)
02

Calculate the Amount of Glucose in Moles

The amount of glucose is given in grams and is converted into moles by dividing the given mass by the molar mass of glucose. This gives: \(\frac{500.4 \mathrm{~g}}{180.16 \mathrm{~g/mol}} = 2.78 \mathrm{~mol}\)
03

Calculate the Maximum Amount of Ethanol in Moles

From the stoichiometry of the reaction, 1 mol of glucose produces 2 mol of ethanol. Therefore, the maximum amount of ethanol produced is \(2.78 \mathrm{~mol} * 2 = 5.56 \mathrm{~mol}\)
04

Calculate the Maximum Amount of Ethanol in Grams

The maximum amount of ethanol in grams is given by the product of the number of moles and the molar mass of ethanol. This gives: \(5.56 \mathrm{~mol} * 46.07 \mathrm{~g/mol} = 256.3 \mathrm{~g}\)
05

Calculate the Maximum Amount of Ethanol in Liters

The maximum amount of ethanol in liters is determined by dividing the mass of ethanol by the density. This gives: \(\frac{256.3 \mathrm{~g}}{0.789 \mathrm{~g/mL}} = 325 \mathrm{~mL}\) or \(0.325 \mathrm{~L}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Mass
The concept of molar mass is a fundamental part of chemical stoichiometry and is vital when solving problems about chemical reactions, such as the fermentation process used in wine-making. Molar mass of a compound is the sum of the masses of all the atoms that constitute a molecule of the compound. It is typically expressed in units of grams per mole (g/mol). For example, in the given problem,
  • glucose, which has a chemical formula of \(\text{C}_6\text{H}_{12}\text{O}_6\), has a molar mass of \(180.16\, \text{g/mol}\).
  • ethanol, represented by \(\text{C}_2\text{H}_5\text{OH}\), has a molar mass of \(46.07\, \text{g/mol}\).
Calculating the molar mass consists of adding up the atomic masses of each element multiplied by the number of times the element appears in the molecule. Understanding how to calculate the molar mass is crucial as it allows us to convert between grams and moles, facilitating accurate calculations of reactants and products in a chemical equation.
Fermentation Process
Fermentation is a biological process that converts sugar such as glucose into simpler compounds. It is an anaerobic process, meaning it occurs in the absence of oxygen. In the context of wine-making, the fermentation process involves the conversion of glucose into ethanol and carbon dioxide. This is a vital part of the wine production cycle. The chemical equation for this fermentation process is:\[\text{C}_6\text{H}_{12}\text{O}_6 \rightarrow 2 \text{C}_2\text{H}_5\text{OH} + 2 \text{CO}_2\]Key points to note:
  • For every mole of glucose utilized in the reaction, 2 moles of ethanol and 2 moles of carbon dioxide are produced.
  • This stoichiometric relationship allows us to predict the yields of ethanol and carbon dioxide from a known amount of glucose.
Understanding the fermentation process not only helps in determining the product quantities but also enhances clarity on how energy is conserved during these reactions, as energy stored in glucose is partly transferred to ethanol.
Ethanol Production
The production of ethanol through the fermentation process is a practical application of chemical stoichiometry. From the reaction's stoichiometry, we know that 1 mole of glucose will yield 2 moles of ethanol. Hence, the amount of ethanol that can be produced in a reaction directly depends on the number of moles of the glucose reagent.Here鈥檚 how you can calculate the amount of ethanol produced:
  • Calculate the moles of glucose using its mass and molar mass
  • Use the stoichiometric ratio from the reaction equation to find moles of ethanol
  • Convert moles of ethanol to grams using its molar mass
  • Finally, convert grams to liters using the density of ethanol
For instance, in the original problem, starting with \(500.4\, \text{g}\) of glucose results in a maximum ethanol production of \(256.3\, \text{g}\), which translates to \(0.325\, \text{L}\) when divided by the density of ethanol (\(0.789 \text{g/mL}\)). This step-by-step conversion is crucial for anyone involved in industries like brewing and biofuel production, where precise ethanol yield is desired.

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Most popular questions from this chapter

Certain race cars use methanol (CH \(_{3} \mathrm{OH}\), also called wood alcohol) as a fuel. The combustion of methanol occurs according to the following equation:$$2 \mathrm{CH}_{3} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)$$In a particular reaction, 9.8 moles of \(\mathrm{CH}_{3} \mathrm{OH}\) are reacted with an excess of \(\mathrm{O}_{2}\). Calculate the number of moles of \(\mathrm{H}_{2} \mathrm{O}\) formed.

Titanium(IV) oxide \(\left(\mathrm{TiO}_{2}\right)\) is a white substance produced by the action of sulfuric acid on the mineral ilmenite \(\left(\mathrm{FeTiO}_{3}\right)\) $$\mathrm{FeTiO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow\mathrm{TiO}_{2}+\mathrm{FeSO}_{4}+\mathrm{H}_{2} \mathrm{O}$$Its opaque and nontoxic properties make it suitable as a pigment in plastics and paints. In one process, \(8.00 \times 10^{3} \mathrm{~kg}\) of \(\mathrm{FeTiO}_{3}\) yielded \(3.67 \times 10^{3} \mathrm{~kg}\) of \(\mathrm{TiO}_{2} .\) What is the percent yield of the reaction?

In a natural product synthesis, a chemist prepares a complex biological molecule entirely from nonbiological starting materials. The target molecules are often known to have some promise as therapeutic agents, and the organic reactions that are developed along the way benefit all chemists. The overall synthesis, however, requires many steps, so it is important to have the best possible percent yields at each step. What is the overall percent yield for such a synthesis that has 24 steps with an 80 percent yield at each step?

Ethylene reacts with hydrogen chloride to form ethyl chloride: $$\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{HCl}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}(g) $$Calculate the mass of ethyl chloride formed if \(4.66 \mathrm{~g}\) of ethylene reacts with an 89.4 percent yield.

Compounds containing ruthenium(II) and bipyridine, \(\mathrm{C}_{10} \mathrm{H}_{8} \mathrm{~N}_{2},\) have received considerable interest because of their role in systems that convert solar energy to electricity. The compound \(\left[\mathrm{Ru}\left(\mathrm{C}_{10} \mathrm{H}_{8} \mathrm{~N}_{2}\right)_{3}\right]$$\mathrm{Cl}_{2}\) is synthesized by reacting \(\mathrm{RuCl}_{3} \cdot 3 \mathrm{H}_{2} \mathrm{O}(s)\) with three molar equivalents of \(\mathrm{C}_{10} \mathrm{H}_{8} \mathrm{~N}_{2}(s),\) along with an excess of triethylamine, \(\mathrm{N}\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3}(l),\) to convert ruthenium(III) to ruthenium(II). The density of triethylamine is \(0.73 \mathrm{~g} / \mathrm{mL},\) and typically eight molar equivalents are used in the synthesis. (a) Assuming that you start with \(6.5 \mathrm{~g}\) of \(\mathrm{RuCl}_{3} \cdot 3 \mathrm{H}_{2} \mathrm{O},\) how many grams of \(\mathrm{C}_{10} \mathrm{H}_{8} \mathrm{~N}_{2}\) and what volume of \(\mathrm{N}\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3}\) should be used in the reaction? (b) Given that the yield of this reaction is 91 percent, how many grams of \(\left[\mathrm{Ru}\left(\mathrm{C}_{10} \mathrm{H}_{8} \mathrm{~N}_{2}\right)_{3}\right] \mathrm{Cl}_{2}\) will be obtained?
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