Glucose And NaOH Reaction Yield Calculation A Step-by-Step Solution

by Admin 68 views

In this comprehensive analysis, we will delve into the intricate chemical reaction involving 360g of glucose reacting with sodium hydroxide (NaOH). Our primary objective is to meticulously dissect the process, understand the formation of intermediate products, and, most importantly, calculate the yield percentage of the initial reaction. This calculation is crucial for assessing the efficiency of the chemical transformation and is a fundamental concept in stoichiometry and chemical kinetics. The reaction's complexity stems from the formation of a mixture of salts – specifically, a 228g mixture comprised of both a middle salt and an acidic salt, present in equimolar quantities. Understanding the molar relationships and the stoichiometry of the reactions is paramount to accurately determining the yield. Furthermore, we are given the crucial information that the second reaction proceeds with a 100% yield, which simplifies our calculations by allowing us to focus solely on the initial reaction's efficiency. This article will not only provide a step-by-step solution to the problem but also elucidate the underlying principles and concepts, making it a valuable resource for students and professionals in the field of chemistry.

Problem Statement

We are given 360 grams of glucose (C6H12O6) which reacts with sodium hydroxide (NaOH). The reaction yields 228 grams of a mixture containing a middle salt and an acidic salt in equal molar quantities. Our goal is to calculate the percent yield of the first reaction, given that the second reaction has a 100% yield. This problem encompasses fundamental concepts of stoichiometry, chemical reactions, and yield calculations. It also highlights the practical application of these concepts in determining reaction efficiencies in chemical processes. The equal molar quantities of the middle and acidic salts provide a crucial stoichiometric constraint that guides the solution. Understanding the reactions between glucose and NaOH, and the subsequent formation of salts, is essential to dissecting this problem effectively. Moreover, the knowledge that the second reaction goes to completion simplifies the problem, allowing us to focus solely on the yield of the initial step. This problem serves as a valuable exercise in applying stoichiometric principles to real-world chemical scenarios.

Step-by-Step Solution

  1. Calculating Moles of Glucose:

    The first step involves determining the number of moles of glucose present in the initial sample. This is a fundamental conversion in stoichiometry, allowing us to relate mass to molar quantities. To do this, we divide the given mass of glucose by its molar mass. The molar mass of glucose (C6H12O6) can be calculated by summing the atomic masses of its constituent elements: 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms. This calculation provides the molar mass in grams per mole (g/mol), which is a crucial conversion factor in stoichiometry. Understanding how to calculate molar mass from the chemical formula is a foundational skill in chemistry. Once we have the molar mass, we can use it to convert the given mass of glucose into moles. This conversion is necessary because chemical reactions proceed according to molar ratios, not mass ratios. Therefore, expressing the amount of glucose in moles allows us to relate it directly to the amounts of reactants and products involved in the reaction. This initial step sets the stage for all subsequent calculations and is essential for accurately determining the yield of the reaction. By correctly calculating the moles of glucose, we establish a solid foundation for the rest of our analysis.

    • Molar mass of glucose (C6H12O6) = (6 * 12) + (12 * 1) + (6 * 16) = 72 + 12 + 96 = 180 g/mol
    • Moles of glucose = Mass / Molar mass = 360 g / 180 g/mol = 2 moles

  2. Reactions with NaOH:

    Glucose reacts with sodium hydroxide (NaOH) in a series of neutralization reactions. To fully understand the reaction, we must first consider the nature of glucose and its interaction with a strong base like NaOH. Glucose, a monosaccharide, possesses multiple hydroxyl (-OH) groups, each of which can potentially react with NaOH. These reactions involve the deprotonation of the hydroxyl groups by the hydroxide ions (OH-) from NaOH, leading to the formation of water and a glucose salt. The number of NaOH molecules that react with one molecule of glucose depends on the extent of deprotonation. In this particular scenario, we are told that the reaction produces both a middle salt and an acidic salt. This implies that two distinct reactions are occurring simultaneously, each resulting in a different degree of deprotonation of the glucose molecule. Writing out the balanced chemical equations for these reactions is crucial for understanding the stoichiometry and the molar relationships between reactants and products. The balanced equations not only show the reactants and products but also provide the stoichiometric coefficients, which are essential for calculating the mole ratios involved in the reaction. Accurately representing these reactions is key to correctly solving the problem and determining the yield of the first reaction.

    • Glucose + 2NaOH → Middle salt + 2H2O
    • Glucose + 1NaOH → Acidic salt + H2O

  3. Defining Variables:

    To solve this problem efficiently, we introduce variables to represent the unknowns. This is a common and powerful strategy in problem-solving, especially in quantitative fields like chemistry and physics. By assigning variables, we can translate the word problem into a mathematical form, making it easier to manipulate and solve. In this case, we define 'x' as the number of moles of the middle salt formed and, since the mixture contains equal molar quantities of the middle and acidic salts, 'x' also represents the number of moles of the acidic salt formed. This definition is crucial because it directly incorporates the information about the equimolar ratio of the two salts, which is a key piece of information provided in the problem statement. Defining variables clearly and accurately is essential for setting up the correct equations. A well-defined variable acts as a placeholder for the unknown quantity, allowing us to track and ultimately determine its value. This step is a crucial bridge between the chemical problem and its mathematical solution.

    • Let 'x' be the moles of middle salt and also the moles of acidic salt.

  4. Setting up Equations:

    The next critical step involves translating the chemical information into mathematical equations. This is where our understanding of stoichiometry and the molar relationships in the reactions comes into play. The equations we set up are based on two key pieces of information: the total mass of the salt mixture and the total moles of glucose consumed in the reactions. The 228 grams of the salt mixture provides a mass-based constraint, which we can express as an equation using the molar masses of the middle and acidic salts. To do this, we need to know the chemical formulas of these salts, which we can deduce from the reactions between glucose and NaOH. The fact that we started with 2 moles of glucose provides a molar constraint, allowing us to relate the moles of glucose consumed in each reaction to the moles of salts formed. This molar constraint is essential because the overall reaction involves two competing pathways, leading to different products (the middle and acidic salts). By setting up these equations correctly, we create a mathematical model of the chemical system, which we can then solve to determine the unknowns. This step requires a solid grasp of stoichiometry and the ability to translate chemical information into a mathematical framework.

    • Moles of glucose reacted: x (for middle salt) + x (for acidic salt) = 2 moles (total glucose)
    • Mass of salt mixture: (Molar mass of middle salt * x) + (Molar mass of acidic salt * x) = 228 g

  5. Determining Molar Masses:

    To solve the equations we've set up, we need the molar masses of the middle and acidic salts. This requires understanding the chemical formulas of these salts, which are derived from the reactions between glucose and NaOH. The middle salt is formed when two hydroxyl groups of glucose react with NaOH, while the acidic salt is formed when only one hydroxyl group reacts. Determining the chemical formulas and calculating molar masses are fundamental skills in chemistry. The chemical formula of a salt indicates the number and type of atoms present in the molecule, which is essential for calculating the molar mass. The molar mass, in turn, is the mass of one mole of the substance and is expressed in grams per mole (g/mol). Accurately calculating the molar masses of the salts is crucial because these values are used in the mass balance equation, which relates the moles of salts formed to the total mass of the salt mixture. Without these accurate molar masses, we cannot correctly solve for the unknowns and determine the yield of the reaction. This step highlights the importance of understanding chemical formulas and their relationship to molar masses in quantitative chemical calculations.

    • Molar mass of middle salt (C6H10O6Na2) = (6 * 12) + (10 * 1) + (6 * 16) + (2 * 23) = 72 + 10 + 96 + 46 = 224 g/mol
    • Molar mass of acidic salt (C6H11O6Na) = (6 * 12) + (11 * 1) + (6 * 16) + (1 * 23) = 72 + 11 + 96 + 23 = 202 g/mol

  6. Solving the Equations:

    Now we substitute the molar masses into our equations and solve for 'x'. This step involves algebraic manipulation and is a crucial part of the problem-solving process. We have two equations and one unknown ('x'), so the system is solvable. The first equation is derived from the mass balance of the salt mixture, and the second equation reflects the total moles of glucose consumed. The process of solving the equations may involve substitution, elimination, or other algebraic techniques. The goal is to isolate the variable 'x' and determine its numerical value. This value represents the number of moles of both the middle and acidic salts formed in the reaction. The accurate solution of these equations is critical because it directly impacts the subsequent yield calculation. Any error in the algebraic manipulation will propagate through the rest of the problem, leading to an incorrect answer. Therefore, careful and systematic execution of this step is essential for obtaining the correct result.

    • Moles of glucose reacted: x + x = 2
    • 2x = Total moles of glucose reacted
    • Total moles of glucose can also be calculated from the moles of each salt produced:
    • x + x = 2

  7. Mass balance equation:

    • (224 g/mol * x) + (202 g/mol * x) = 228 g
    • 426x = 228
    • x = 228 / 426 ≈ 0.535 moles

  8. Calculating Theoretical Yield:

    The theoretical yield represents the maximum amount of product that can be formed in a reaction, assuming that all of the limiting reactant is converted into product. This is a crucial concept in chemistry because it provides a benchmark against which we can compare the actual yield of a reaction. To calculate the theoretical yield, we need to consider the stoichiometry of the reaction, specifically the mole ratio between the limiting reactant (in this case, glucose) and the desired product. Since we are interested in the yield of the first reaction (glucose reacting with NaOH to form the salts), we need to relate the moles of glucose to the moles of salts that could be formed if the reaction went to completion. We've already calculated the moles of glucose that reacted (based on the amount of salts formed), so we can use this information to determine the theoretical yield. This calculation provides us with an ideal scenario, representing the perfect conversion of reactants to products. By comparing the actual yield (the amount of product obtained in the experiment) to the theoretical yield, we can assess the efficiency of the reaction and identify factors that may have limited the yield.

    • From the reaction stoichiometry, 1 mole of glucose can produce 1 mole of middle salt and 1 mole of acidic salt.
    • Theoretical yield (if all 2 moles of glucose reacted to form middle salt): 2 moles * 224 g/mol = 448g

  9. Calculating Actual Yield Based on Middle Salt:

    The actual yield represents the amount of product that is actually obtained from a chemical reaction. This is a practical measurement, reflecting the real-world outcome of the reaction process. To calculate the actual yield based on the middle salt, we use the moles of middle salt calculated earlier ('x') and multiply it by the molar mass of the middle salt. This calculation gives us the mass of the middle salt that was actually produced in the reaction. The actual yield is often less than the theoretical yield due to various factors, such as incomplete reactions, side reactions, loss of product during purification, and experimental errors. Understanding the difference between the theoretical and actual yields is crucial for assessing the efficiency of a chemical process and identifying areas for improvement. By calculating the actual yield, we gain a quantitative measure of the amount of product we successfully obtained from the reaction, which is essential for determining the percent yield and evaluating the overall success of the chemical transformation.

    • Actual mass of middle salt = 0.535 moles * 224 g/mol = 119.84 g

  10. Calculating Theoretical yield based on Acidic Salt:

    To calculate the theoretical yield, it is crucial to consider the stoichiometry of the reaction accurately. In this scenario, we focus on the production of the acidic salt. The stoichiometry of the reaction dictates that for every mole of glucose that reacts, one mole of the acidic salt is formed. Given this 1:1 molar relationship, we can directly relate the amount of glucose initially present to the maximum amount of acidic salt that could theoretically be produced. This calculation provides a benchmark against which we compare the actual yield to assess the efficiency of the reaction. Understanding and applying stoichiometric principles correctly is essential for determining theoretical yields accurately. By identifying the limiting reactant and using the balanced chemical equation, we can calculate the maximum possible yield of the desired product under ideal conditions. This theoretical value serves as a crucial reference point for evaluating the experimental outcome and understanding the factors that may have influenced the actual yield obtained.

    • Theoretical yield (if all 2 moles of glucose reacted to form Acidic salt): 2 moles * 202 g/mol = 404g

  11. Calculating Actual Yield Based on Acidic Salt:

    The actual yield represents the tangible quantity of product harvested from the reaction, reflecting real-world outcomes subject to experimental factors. To ascertain the actual yield specific to the acidic salt, we leverage the moles of acidic salt computed earlier ('x') and multiply it by the molar mass of the acidic salt. This computation yields the mass of the acidic salt practically synthesized in the reaction, providing a quantitative measure of the reaction's productivity. Often, the actual yield falls short of the theoretical yield due to various influences such as incomplete reactions, concurrent side reactions, product dissipation during purification, and inherent experimental inaccuracies. Grasping the distinction between theoretical and actual yields is paramount for gauging the efficacy of a chemical process and pinpointing avenues for enhancement. By meticulously calculating the actual yield, we secure a tangible metric of the product amassed, vital for ascertaining the percent yield and appraising the overall triumph of the chemical transformation.

    • Actual mass of Acidic salt = 0.535 moles * 202 g/mol = 108.07 g

  12. Calculating % Yield:

    Finally, we calculate the percent yield of the first reaction using the formula: Percent Yield = (Actual Yield / Theoretical Yield) * 100%. This calculation provides a quantitative measure of the efficiency of the reaction, expressing the actual amount of product obtained as a percentage of the maximum amount that could theoretically be obtained. The percent yield is a crucial metric for evaluating the success of a chemical reaction and for comparing the effectiveness of different reaction conditions or procedures. A higher percent yield indicates a more efficient reaction, meaning that a larger proportion of the reactants was converted into the desired product. In this calculation, we use the actual yield of the limiting reactant (glucose) and the theoretical yield calculated based on the stoichiometry of the reaction. The percent yield can be influenced by various factors, including reaction conditions, purity of reactants, and experimental techniques. By accurately calculating the percent yield, we can assess the overall effectiveness of the reaction and identify areas for optimization.

    • The actual yield is derived from the amount of glucose consumed in forming the salt mixture. From our mole balance, we used 2 * 0.535 moles of glucose which is 1.07moles.
    • % Yield = (Actual Moles of Glucose Used / Initial Moles of Glucose) * 100
    • % Yield = (1.07 moles / 2 moles) * 100
    • % Yield ≈ 53.5% (This answer is not within the options provided and there is likely an error in the problem statement or the options provided.)

  13. An alternative approach

Let's consider the yield based on the total mass of product formed versus the theoretical possible based on the mass of Glucose initially used and that all reacted to the middle salt. We will recalculate the theoretical yield.

*   From 2 moles of glucose, if 2 moles of middle salt **C6H10O6Na2** are formed the mass should be 2 mol * 224 g/mol = 448g
*   From 2 moles of glucose, if 2 moles of acidic salt **C6H11O6Na** are formed the mass should be 2 mol * 202 g/mol = 404g

The 228g product formed is a combination of the middle and acidic salts.

Let us re-examine the glucose and their respective salt forms. Glucose reacting with 2 NaOH becomes C6H10O6Na2, also known as the middle salt, and glucose reacting with 1 NaOH becomes C6H11O6Na, also known as the acidic salt. The total number of moles of salts produced = moles of middle salt + moles of acidic salt which is x + x = 2x.

Going back to previous equations.

Mass balance equation:

*   (224 g/mol * x) + (202 g/mol * x) = 228 g
*   426x = 228
*   x = 228 / 426 ≈ 0.535 moles

Moles of glucose converted in total = x moles (for middle salt) + x moles (for acidic salt) = 0.535 + 0.535 = 1.07 moles.

Percent yield is then calculated as:

*   Percentage Yield = (Moles of glucose reacted / Initial moles of glucose) × 100
*   Percentage Yield = (1.07 moles / 2 moles) × 100 ≈ 53.5 %

*This value does not match any of the given options, suggesting there might be a mistake either in the provided options or in the problem statement itself.* However, let's attempt another method by reverse calculating to see if any of the option can be a probable answer.

Reverse Calculation Method

Given options are 80%, 70%, 60%, and 75%. We could work backwards assuming those were the answers and check if the math works up.

Let’s assume the yield is 75%:

  • Moles of glucose reacted at 75% yield = 0.75 * 2 moles = 1.5 moles

If x moles of middle salt C6H10O6Na2 and x moles of acidic salt C6H11O6Na are formed, then:

  • x + x = 1.5 (total moles of glucose reacted)
  • This equation is incorrect as it assumes that two glucose molecules react, one for forming the middle salt and other for acidic salt.

The correct equation:

  • Glucose reacted to form middle salt C6H10O6Na2 can be written as C6H12O6 + 2NaOH -> C6H10O6Na2
  • Moles of glucose reacted = moles of middle salt which is x.
  • Glucose reacted to form acidic salt C6H11O6Na can be written as C6H12O6 + NaOH -> C6H11O6Na
  • Moles of glucose reacted = moles of acidic salt which is x.
  • So the total moles of glucose reacted can be written as x + x = 1.5
  • 2x = 1.5
  • x = 0.75 moles

The mass of the mixture of salts should be:

  • mass of middle salt = 0.75 * 224 = 168g
  • mass of acidic salt = 0.75 * 202 = 151.5g
  • Total mass = 168 + 151.5 = 319.5g

Since 319.5g is not matching 228g provided in the question, we can conclude that 75% is not the correct answer. This process could be repeated for the other options too. Since none of the answers appears to fit, there may be a problem with the question itself.

Conclusion

In summary, we have undertaken a comprehensive analysis of the reaction between 360g of glucose and NaOH, aiming to calculate the yield of the first reaction. Our step-by-step solution involved calculating moles of glucose, outlining the relevant reactions, defining variables, setting up and solving equations, and finally, calculating the percent yield. The calculated yield of approximately 53.5% does not match any of the provided options, suggesting a potential error within the problem statement or the answer choices. While this discrepancy exists, the process outlined in this analysis provides a robust framework for solving similar stoichiometric problems. The application of these principles is crucial in various chemical processes, from laboratory synthesis to industrial production. This exercise highlights the importance of meticulous calculation, attention to stoichiometric relationships, and critical evaluation of results. Even when encountering inconsistencies, the analytical skills honed through such problem-solving are invaluable for chemical understanding and practice. Furthermore, the reverse calculation approach underscores the value of verifying solutions and questioning assumptions when confronted with unexpected outcomes.

Based on the calculations, none of the given options (A) 80, (B) 70, (C) 60, (D) 75 is correct. There might be an error in the problem statement or the given options.

Calculate the percent yield of the reaction between 360g of glucose and NaOH, which produces 228g of a mixture containing equal molar amounts of middle salt and acidic salt. Assume the second reaction has a 100% yield.

Glucose and NaOH Reaction Yield Calculation A Step-by-Step Solution