Calculating Average Atomic Mass Worksheet Answer Key

Calculating Average Atomic Mass: A Comprehensive Worksheet and Answer Key

The concept of average atomic mass is fundamental to chemistry, bridging the gap between the theoretical world of atomic masses and the practical reality of naturally occurring elements. This worksheet will guide you through the calculations, providing a thorough understanding and equipping you with the skills to solve various problems related to average atomic mass. We'll cover the essential concepts, step-by-step calculations, and provide a detailed answer key to ensure you master this important topic.

Understanding Isotopes and Average Atomic Mass

Before diving into calculations, let's establish a firm grasp of the underlying principles. Atoms of the same element can exist as different isotopes. Isotopes are atoms of the same element that have the same number of protons but different numbers of neutrons. This difference in neutron number results in variations in their mass.

For example, consider carbon. The most common isotope is carbon-12 (¹²C), with 6 protons and 6 neutrons. However, another stable isotope exists, carbon-13 (¹³C), with 6 protons and 7 neutrons. Each isotope has its own isotopic mass, which is the mass of a single atom of that specific isotope.

The average atomic mass is a weighted average of the isotopic masses of all the naturally occurring isotopes of an element. It reflects the relative abundance of each isotope in a naturally occurring sample. This average atomic mass is what's typically reported on the periodic table.

The Formula for Calculating Average Atomic Mass

The calculation of average atomic mass relies on a straightforward formula:

Average Atomic Mass = (Isotope 1 Abundance × Isotope 1 Mass) + (Isotope 2 Abundance × Isotope 2 Mass) + ...

Where:

• Isotope Abundance: This is the percentage abundance of each isotope, often expressed as a decimal (e.g., 75% becomes 0.75).
• Isotope Mass: This is the isotopic mass of each isotope, usually expressed in atomic mass units (amu).

Step-by-Step Calculation Examples

Let's work through some examples to solidify your understanding.

Example 1: Calculating the Average Atomic Mass of Boron

Boron has two naturally occurring isotopes: ¹⁰B (19.9% abundance, 10.01 amu) and ¹¹B (80.1% abundance, 11.01 amu). Calculate the average atomic mass of boron.

Step 1: Convert percentages to decimals:

• ¹⁰B abundance: 19.9% = 0.199
• ¹¹B abundance: 80.1% = 0.801

Step 2: Apply the formula:

Average Atomic Mass = (0.199 × 10.01 amu) + (0.801 × 11.01 amu)

Step 3: Perform the calculations:

Average Atomic Mass = 1.99199 amu + 8.81801 amu = 10.81 amu

Therefore, the average atomic mass of boron is approximately 10.81 amu.

Example 2: A More Complex Scenario

Let's consider an element with three isotopes:

• Isotope 1: 60% abundance, 20.00 amu
• Isotope 2: 30% abundance, 22.00 amu
• Isotope 3: 10% abundance, 24.00 amu

Step 1: Convert percentages to decimals:

• Isotope 1: 60% = 0.60
• Isotope 2: 30% = 0.30
• Isotope 3: 10% = 0.10

Step 2: Apply the formula:

Average Atomic Mass = (0.60 × 20.00 amu) + (0.30 × 22.00 amu) + (0.10 × 24.00 amu)

Step 3: Perform the calculations:

Average Atomic Mass = 12.00 amu + 6.60 amu + 2.40 amu = 21.00 amu

The average atomic mass of this element is 21.00 amu.

Worksheet: Practice Problems

Now, let's put your knowledge to the test with a series of practice problems. Remember to show your work for each problem.

Problem 1: Chlorine has two isotopes: ³⁵Cl (75.77% abundance, 34.97 amu) and ³⁷Cl (24.23% abundance, 36.97 amu). Calculate the average atomic mass of chlorine.

Problem 2: Magnesium has three isotopes: ²⁴Mg (78.99%, 23.99 amu), ²⁵Mg (10.00%, 24.99 amu), and ²⁶Mg (11.01%, 25.98 amu). Calculate its average atomic mass.

Problem 3: An element has two isotopes. One isotope has a mass of 62.93 amu and an abundance of 69.17%. The other isotope has a mass of 64.93 amu. Calculate the average atomic mass and determine the identity of the element (hint: use the periodic table).

Problem 4: An element's average atomic mass is 14.01 amu. It has two isotopes: one with a mass of 14.00 amu and an abundance of 99.63%. What is the mass of the second isotope?

Problem 5: A newly discovered element has three isotopes with the following data: Isotope A (mass = 180.95 amu, abundance = 37.4%), Isotope B (mass = 182.96 amu, abundance = 62.6%). Calculate the average atomic mass. (This problem tests your understanding of missing data, making you think about the significance of percentage abundance always adding to 100%).

Answer Key

Problem 1:

Average Atomic Mass = (0.7577 × 34.97 amu) + (0.2423 × 36.97 amu) = 35.45 amu

Problem 2:

Average Atomic Mass = (0.7899 × 23.99 amu) + (0.1000 × 24.99 amu) + (0.1101 × 25.98 amu) = 24.31 amu

Problem 3:

Let x be the abundance of the second isotope. Since the abundances must add up to 100%, we have: 0.6917 + x = 1. Therefore, x = 0.3083.

Average Atomic Mass = (0.6917 × 62.93 amu) + (0.3083 × 64.93 amu) ≈ 63.55 amu. This is the average atomic mass of Copper (Cu).

Problem 4:

Let x be the mass of the second isotope. The abundance of the second isotope is 100% - 99.63% = 0.37%. Converting to decimal: 0.0037.

14.01 amu = (0.9963 × 14.00 amu) + (0.0037 × x)

Solving for x: x ≈ 15.04 amu

Problem 5:

Notice that only two isotopes are given. This problem is designed to highlight the importance of complete data. If we assume these are the only isotopes present, then we can solve as before:

Average Atomic Mass = (0.374 × 180.95 amu) + (0.626 × 182.96 amu) ≈ 182.01 amu

Conclusion

Calculating average atomic mass is a crucial skill in chemistry. Through understanding the principles of isotopes, isotopic abundance, and the application of the weighted average formula, you can confidently tackle problems related to this essential concept. Remember to always double-check your calculations and ensure your understanding of the underlying chemistry. This worksheet and answer key provide a solid foundation for further exploration of atomic structure and related topics. Continue practicing, and you will master this skill!