Calculating The Mass Of 3.01 X 10^23 Molecules Of C6H12O6 A Step-by-Step Guide
In the realm of chemistry, understanding the relationship between the number of molecules and their mass is a fundamental concept. This involves using Avogadro's number and molar mass to convert between these quantities. In this comprehensive article, we will delve into the process of calculating the mass of a given number of molecules, specifically 3.01 x 10^23 molecules of glucose (C6H12O6). We will explore the underlying principles, step-by-step calculations, and practical implications of this essential concept in chemistry. Understanding these calculations is crucial for various applications, from laboratory experiments to industrial processes. This detailed explanation will provide you with a solid foundation for tackling similar problems and deepening your understanding of stoichiometry.
Understanding the Concepts
Before diving into the calculations, it's crucial to understand the core concepts that underpin this process. These include Avogadro's number, molar mass, and the relationship between moles, molecules, and mass. Grasping these concepts will make the calculation process more intuitive and less mechanical.
Avogadro's Number
Avogadro's number, approximately 6.022 x 10^23, is a cornerstone of chemistry. It represents the number of entities (atoms, molecules, ions, etc.) in one mole of a substance. A mole is the SI unit for the amount of a substance. To put it simply, just as a dozen represents 12 items, a mole represents 6.022 x 10^23 items. This number provides a bridge between the microscopic world of atoms and molecules and the macroscopic world of grams and kilograms that we can measure in a lab.
Molar Mass
Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is numerically equivalent to the atomic or molecular weight of the substance in atomic mass units (amu). For example, the molar mass of water (H2O) is approximately 18.015 g/mol, which is the sum of the atomic masses of two hydrogen atoms and one oxygen atom. The molar mass serves as a conversion factor between mass and moles, allowing us to convert grams of a substance into moles and vice versa. For the given molecule, glucose (C6H12O6), the molar mass is provided as 180.18 g/mol.
The Mole Concept and Its Importance
The mole concept is central to stoichiometric calculations in chemistry. It allows us to quantify the amount of a substance and relate it to the number of particles (molecules, atoms, ions) present. By knowing the number of moles, we can calculate the mass of a substance using its molar mass, or we can determine the number of particles using Avogadro's number. This concept is essential for performing chemical reactions, preparing solutions, and analyzing chemical compounds. Understanding the relationship between moles, mass, and the number of particles is the key to solving many quantitative problems in chemistry.
Problem Statement and Given Information
The problem at hand is to determine the mass of 3.01 x 10^23 molecules of glucose (C6H12O6). We are given two crucial pieces of information:
- The molar mass of glucose (C6H12O6) is 180.18 g/mol.
- Avogadro's number: 1 mol C6H12O6 = 6.02 x 10^23 molecules.
These two pieces of information are the foundation for our calculation. The molar mass provides the conversion factor between moles and grams, while Avogadro's number provides the conversion factor between moles and the number of molecules. With these values, we can set up a dimensional analysis to solve for the mass.
Step-by-Step Calculation
Now, let's break down the calculation process into a step-by-step approach. This will make the process clearer and easier to follow. The steps involve converting molecules to moles and then moles to grams.
Step 1: Convert Molecules to Moles
The first step is to convert the given number of molecules of glucose to moles. We use Avogadro's number as a conversion factor. The conversion is set up as follows:
Moles of C6H12O6 = (Number of molecules) / (Avogadro's number)
Plugging in the values:
Moles of C6H12O6 = (3.01 x 10^23 molecules) / (6.02 x 10^23 molecules/mol)
Calculation
Performing the division:
Moles of C6H12O6 = 0.5 mol
So, 3.01 x 10^23 molecules of glucose is equivalent to 0.5 moles. This conversion is crucial because we now have the amount of glucose in moles, which can be directly related to its mass using the molar mass.
Step 2: Convert Moles to Grams
The next step is to convert moles of glucose to grams using the molar mass of glucose (180.18 g/mol). The conversion is set up as follows:
Mass of C6H12O6 = (Moles of C6H12O6) x (Molar mass of C6H12O6)
Plugging in the values:
Mass of C6H12O6 = (0.5 mol) x (180.18 g/mol)
Calculation
Performing the multiplication:
Mass of C6H12O6 = 90.09 g
Therefore, 0.5 moles of glucose has a mass of 90.09 grams. This is the final step in determining the mass of the given number of molecules.
Final Answer
Combining the two steps, we find that the mass of 3.01 x 10^23 molecules of C6H12O6 is 90.09 grams. This result provides a quantitative answer to the problem, illustrating the connection between the number of molecules and their mass.
Importance of Dimensional Analysis
Throughout the calculation, we used dimensional analysis to ensure that the units cancelled out correctly, leading to the final answer in the desired unit (grams). Dimensional analysis is a powerful problem-solving technique that involves treating units as algebraic quantities that can be multiplied, divided, and cancelled. This method helps in setting up the problem correctly and verifying the final result.
For instance, in the first step, we divided the number of molecules by Avogadro's number (molecules/mol), which resulted in the unit moles. In the second step, we multiplied moles by the molar mass (g/mol), which resulted in the unit grams. By keeping track of the units, we can avoid errors and ensure the accuracy of our calculations. This approach is fundamental in chemistry and other scientific disciplines.
Practical Applications and Implications
The ability to calculate the mass of a given number of molecules has numerous practical applications in chemistry and related fields. Here are a few examples:
Laboratory Experiments
In laboratory settings, chemists often need to prepare solutions of specific concentrations. This requires calculating the mass of a solute needed to dissolve in a given volume of solvent. For instance, if a chemist needs to prepare a 1 M solution of glucose, they must first calculate the mass of glucose needed using its molar mass. The ability to convert between the number of molecules and mass is essential for accurate solution preparation.
Stoichiometry
Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. Knowing the mass of reactants allows chemists to predict the amount of products formed in a reaction. For example, in a reaction involving glucose, knowing the mass of glucose allows one to calculate the mass of carbon dioxide and water produced during its combustion or metabolism. This is crucial in various applications, including industrial chemical synthesis and biological studies.
Materials Science
In materials science, understanding the composition and mass of different compounds is essential for designing and synthesizing new materials. Calculating the mass of a specific number of molecules helps in determining the exact composition of a material, which is crucial for its properties and applications. For example, in polymer chemistry, knowing the molar mass and the number of molecules is important for controlling the molecular weight and properties of the polymer.
Pharmaceutical Industry
In the pharmaceutical industry, accurate calculations of the mass of drug molecules are critical for drug formulation and dosage. The efficacy and safety of a drug depend on administering the correct amount, which requires precise calculations involving molar mass and Avogadro's number. For instance, determining the mass of the active pharmaceutical ingredient (API) in a tablet or capsule is essential for ensuring the correct dosage.
Common Mistakes to Avoid
When performing these types of calculations, it is easy to make mistakes if one is not careful. Here are some common mistakes to avoid:
Using the Wrong Units
One of the most common mistakes is using the wrong units or mixing up units. For example, using grams instead of moles or vice versa can lead to significant errors. Always ensure that the units are consistent throughout the calculation and that they cancel out correctly in dimensional analysis.
Incorrectly Applying Avogadro's Number
Avogadro's number is a specific conversion factor between moles and the number of particles. Misunderstanding or misapplying this number can lead to errors in calculations. Remember that 6.022 x 10^23 particles (molecules, atoms, etc.) are present in one mole of a substance. Make sure to use it correctly in the conversion process.
Rounding Errors
Rounding numbers prematurely during the calculation can lead to inaccuracies in the final answer. It is best to carry out the calculations with as many significant figures as possible and round off the final answer to the appropriate number of significant figures.
Misunderstanding Molar Mass
The molar mass is a critical conversion factor between mass and moles. Confusing molar mass with molecular weight or using the wrong molar mass for a substance will result in incorrect calculations. Always use the correct molar mass for the substance involved in the calculation.
Conclusion
In summary, calculating the mass of 3.01 x 10^23 molecules of glucose (C6H12O6) involves converting molecules to moles using Avogadro's number and then converting moles to grams using the molar mass. The calculation yields a mass of 90.09 grams. This process highlights the importance of understanding fundamental concepts such as the mole, molar mass, and Avogadro's number in chemistry. The ability to perform these calculations is essential for various applications in the laboratory, industry, and research. By mastering these concepts and avoiding common mistakes, one can confidently tackle a wide range of quantitative problems in chemistry.
This article has provided a comprehensive guide to calculating the mass of a given number of molecules, with specific emphasis on glucose. By understanding the principles, following the step-by-step calculations, and recognizing the practical applications, you can enhance your understanding of stoichiometry and its significance in the world of chemistry.
