Calculate Moles Of Isopropanol A Step By Step Chemistry Guide

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When dealing with chemical substances, it's crucial to understand the concept of moles. Moles provide a standardized way to quantify the amount of a substance, allowing for accurate calculations in chemical reactions and analyses. In this article, we will tackle a common chemistry problem, calculating the number of moles in a given sample of isopropanol ($C_3H_7OH$). We will break down the problem step by step, ensuring a clear understanding of the concepts and calculations involved. Whether you're a student learning chemistry or just refreshing your knowledge, this guide will provide a solid foundation for understanding molar calculations.

Problem Statement

We are given a sample of isopropanol ($C_3H_7OH$) with a mass of 78.6 g. The molar mass of isopropanol is 60.1 g/mol. Our task is to determine how many moles of isopropanol are contained in this sample. This is a classic stoichiometry problem that highlights the relationship between mass, molar mass, and the number of moles. Understanding how to solve this type of problem is fundamental to grasping chemical quantities and reactions. Let’s delve into the solution and understand each step involved in calculating the moles of isopropanol.

Understanding Moles, Mass, and Molar Mass

Before diving into the calculations, let's clarify some key concepts. A mole is the SI unit for the amount of substance. It is defined as the amount of substance that contains as many elementary entities (atoms, molecules, ions, etc.) as there are atoms in 12 grams of carbon-12. This number is known as Avogadro's number, approximately $6.022 \times 10^{23}$. The mass of a substance is a measure of how much matter it contains, typically measured in grams (g) or kilograms (kg). The molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). The molar mass is numerically equivalent to the atomic or molecular weight of the substance in atomic mass units (amu). For example, the molar mass of isopropanol ($C_3H_7OH$) is given as 60.1 g/mol, which means that one mole of isopropanol weighs 60.1 grams. Grasping these definitions is essential for performing stoichiometric calculations accurately. With these concepts in mind, we can proceed to the step-by-step solution of our problem.

Step-by-Step Solution

The relationship between mass, moles, and molar mass is described by the following formula:

Moles=MassMolar Mass\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}

This formula is the cornerstone of many stoichiometry problems and allows us to convert between mass and moles, which is crucial for understanding chemical reactions and quantities. In our case, we have the mass of the isopropanol sample (78.6 g) and the molar mass of isopropanol (60.1 g/mol). We need to calculate the number of moles. Let's apply the formula:

  1. Identify the given values:

    • Mass of isopropanol = 78.6 g
    • Molar mass of isopropanol = 60.1 g/mol

  2. Apply the formula:

    Moles of isopropanol=Mass of isopropanolMolar mass of isopropanol\text{Moles of isopropanol} = \frac{\text{Mass of isopropanol}}{\text{Molar mass of isopropanol}}

  3. Plug in the values:

    Moles of isopropanol=78.6 g60.1 g/mol\text{Moles of isopropanol} = \frac{78.6 \text{ g}}{60.1 \text{ g/mol}}

  4. Calculate the result:

    Moles of isopropanol≈1.3078 mol\text{Moles of isopropanol} \approx 1.3078 \text{ mol}

Thus, the number of moles of isopropanol in the sample is approximately 1.3078 moles. This step-by-step approach ensures clarity and accuracy in solving stoichiometry problems. Now, let's consider the significant figures to provide the final answer in the correct format.

Considering Significant Figures

In scientific calculations, it is crucial to report the answer with the correct number of significant figures. Significant figures are the digits in a number that carry meaning contributing to its precision. When performing calculations, the final answer should be rounded to the same number of significant figures as the least precise measurement used in the calculation. In our problem, the mass of isopropanol (78.6 g) has three significant figures, and the molar mass (60.1 g/mol) also has three significant figures. Therefore, the final answer should be rounded to three significant figures. Our calculated value is approximately 1.3078 mol. Rounding this to three significant figures gives us 1.31 mol. Therefore, the sample contains approximately 1.31 moles of isopropanol. Understanding and applying the rules of significant figures ensures that our results are presented accurately and reflect the precision of the measurements. Let's finalize our answer and summarize the solution.

Final Answer

After performing the calculation and considering significant figures, we find that there are approximately 1.31 moles of isopropanol in the 78.6 g sample. This result answers the problem statement and demonstrates the application of the mole concept in practical calculations. To summarize, we used the formula:

Moles=MassMolar Mass\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}

to find the number of moles of isopropanol. We substituted the given mass (78.6 g) and molar mass (60.1 g/mol) into the formula, performed the division, and rounded the result to three significant figures. This final answer provides a clear and concise solution to the problem. Understanding such calculations is vital in chemistry, as it allows us to quantify and predict the amounts of substances involved in chemical reactions. With a solid grasp of these concepts, you can confidently tackle similar problems in chemistry and related fields.

Conclusion

In conclusion, this article has provided a comprehensive guide on how to calculate the number of moles in a sample of isopropanol. By understanding the concepts of moles, mass, and molar mass, and following a step-by-step approach, we were able to determine that a 78.6 g sample of isopropanol contains approximately 1.31 moles. We emphasized the importance of using the correct formula:

Moles=MassMolar Mass\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}

and highlighted the significance of considering significant figures in the final answer. This problem-solving approach can be applied to a variety of similar stoichiometry problems, making it a valuable skill for students and professionals in chemistry and related fields. Remember, practice is key to mastering these concepts. By working through different examples and problems, you will strengthen your understanding of chemical quantities and calculations. With this knowledge, you can confidently approach more complex chemical problems and analyses. Whether you are studying for an exam or applying these concepts in a research setting, a solid foundation in molar calculations is essential for success.

Practice Problems

To further solidify your understanding of calculating moles, here are a few practice problems:

  1. A sample of ethanol ($C_2H_5OH$) has a mass of 46.07 g. The molar mass of ethanol is 46.07 g/mol. How many moles of ethanol are in the sample?

  2. A chemist has a 100.0 g sample of sodium chloride (NaCl). The molar mass of NaCl is 58.44 g/mol. Calculate the number of moles of NaCl in the sample.

  3. If you have 2.5 moles of glucose ($C_6H_{12}O_6$), and the molar mass of glucose is 180.16 g/mol, what is the mass of the glucose sample?

Working through these problems will help you practice applying the formula and concepts discussed in this article. Check your answers and review the steps if needed. The more you practice, the more confident you will become in solving stoichiometry problems. These practice problems are designed to reinforce your understanding and build your problem-solving skills in chemistry. By tackling a variety of scenarios, you will develop a deeper appreciation for the relationships between mass, moles, and molar mass. Remember, consistent practice is the key to mastering any scientific concept, and stoichiometry is no exception. So, grab your calculator and start solving!