Practice Isotope Calculations #2 Answer Key

Practice Isotope Calculations #2: Answer Key and Detailed Explanations

This comprehensive guide provides detailed solutions and explanations for a second set of practice problems focused on isotope calculations. Understanding isotope calculations is crucial in various scientific fields, including chemistry, geology, and nuclear physics. This guide aims to solidify your understanding of these calculations and enhance your problem-solving skills. We'll cover various scenarios and approaches, ensuring a thorough grasp of this essential scientific concept.

Understanding Isotopes and Isotope Calculations

Before diving into the answer key, let's refresh our understanding of isotopes. Isotopes are atoms of the same element that have the same number of protons but differ in the number of neutrons. This difference in neutron number results in different mass numbers (the sum of protons and neutrons). Isotope calculations often involve determining the abundance of different isotopes or calculating the average atomic mass based on isotopic abundances.

Key Concepts and Formulae

• Average Atomic Mass: This is the weighted average of the masses of all naturally occurring isotopes of an element. It's calculated considering the abundance of each isotope. The formula is:

  Average Atomic Mass = (fractional abundance of isotope 1 × mass of isotope 1) + (fractional abundance of isotope 2 × mass of isotope 2) + ...

• Fractional Abundance: This represents the percentage abundance of an isotope divided by 100. It's always expressed as a decimal between 0 and 1.

• Mass Number: The sum of protons and neutrons in an atom's nucleus.

Practice Isotope Calculations #2: Answer Key

Let's proceed with the answer key and explanations for the practice problems. Assume each problem involves a naturally occurring element with two isotopes unless otherwise stated.

Problem 1: Element X has two isotopes: ¹²X (mass = 12.000 amu) and ¹³X (mass = 13.003 amu). The average atomic mass of element X is 12.011 amu. Calculate the percent abundance of each isotope.

Answer 1:

Let's denote the fractional abundance of ¹²X as 'a' and the fractional abundance of ¹³X as 'b'. We know that a + b = 1 (since they represent the total abundance). We can set up the following equation based on the average atomic mass formula:

12.011 amu = (a × 12.000 amu) + (b × 13.003 amu)

Since a + b = 1, we can express b as (1 - a). Substituting this into the equation:

12.011 amu = (a × 12.000 amu) + ((1 - a) × 13.003 amu)

Solving for 'a':

12.011 = 12a + 13.003 - 13.003a 12.011 - 13.003 = -a -0.992 = -a a = 0.992

Therefore, the fractional abundance of ¹²X is 0.992. Since a + b = 1, the fractional abundance of ¹³X (b) is 1 - 0.992 = 0.008.

To express these as percentages:

¹²X abundance: 0.992 × 100% = 99.2% ¹³X abundance: 0.008 × 100% = 0.8%

Problem 2: Element Y has an average atomic mass of 63.55 amu. It has two isotopes: ⁶³Y (mass = 62.93 amu) and ⁶⁵Y (mass = 64.93 amu). Determine the percent abundance of each isotope.

Answer 2:

Again, let 'a' represent the fractional abundance of ⁶³Y and 'b' represent the fractional abundance of ⁶⁵Y. We have: a + b = 1 and:

63.55 amu = (a × 62.93 amu) + (b × 64.93 amu)

Substitute b = (1 - a):

63.55 = 62.93a + 64.93(1 - a) 63.55 = 62.93a + 64.93 - 64.93a 63.55 - 64.93 = -2a -1.38 = -2a a = 0.69

Therefore, the fractional abundance of ⁶³Y is 0.69. The fractional abundance of ⁶⁵Y is 1 - 0.69 = 0.31.

Percent abundances:

⁶³Y abundance: 0.69 × 100% = 69% ⁶⁵Y abundance: 0.31 × 100% = 31%

Problem 3: A sample of chlorine contains 75.77% ³⁵Cl (mass = 34.97 amu) and 24.23% ³⁷Cl (mass = 36.97 amu). Calculate the average atomic mass of chlorine.

Answer 3:

We can directly apply the average atomic mass formula:

Average Atomic Mass = (0.7577 × 34.97 amu) + (0.2423 × 36.97 amu) Average Atomic Mass = 26.50 amu + 8.96 amu Average Atomic Mass = 35.46 amu

Problem 4: Element Z has three isotopes: ⁹⁰Z (mass = 89.90 amu, abundance = 51.45%), ⁹¹Z (mass = 90.91 amu, abundance = 11.22%), and ⁹²Z (mass = 91.91 amu). Calculate the percent abundance of ⁹²Z and the average atomic mass of element Z.

Answer 4:

Let 'c' be the fractional abundance of ⁹²Z. The sum of fractional abundances must equal 1:

0.5145 + 0.1122 + c = 1 c = 1 - 0.5145 - 0.1122 c = 0.3733

Percent abundance of ⁹²Z: 0.3733 × 100% = 37.33%

Now, calculate the average atomic mass:

Average Atomic Mass = (0.5145 × 89.90 amu) + (0.1122 × 90.91 amu) + (0.3733 × 91.91 amu) Average Atomic Mass = 46.27 amu + 10.19 amu + 34.30 amu Average Atomic Mass = 90.76 amu

Problem 5 (Advanced): A sample of an unknown element contains two isotopes. Isotope A has a mass of 10.01 amu, and isotope B has a mass of 11.01 amu. If the average atomic mass of the element is 10.81 amu, and the abundance of isotope A is 20%, what is the abundance of isotope B?

Answer 5:

Let's denote the fractional abundance of isotope A as 0.20 (20%). Since there are only two isotopes, the fractional abundance of isotope B is (1 - 0.20) = 0.80 (80%). We can verify this using the average atomic mass formula:

10.81 amu = (0.20 × 10.01 amu) + (0.80 × 11.01 amu) 10.81 amu = 2.002 amu + 8.808 amu 10.81 amu ≈ 10.81 amu (Slight discrepancy due to rounding)

Therefore, the abundance of isotope B is 80%.

Further Practice and Advanced Concepts

This answer key provides a solid foundation for understanding isotope calculations. To further enhance your skills, consider exploring more complex scenarios:

Advanced Isotope Calculation Scenarios:

• Isotopes with more than two isotopes: Practice problems involving elements with three or more isotopes will challenge your understanding of weighted averages.
• Calculating isotopic abundances from mass spectrometry data: Mass spectrometry data provides peak intensities that are directly proportional to isotopic abundances. Learning to interpret such data is a valuable skill.
• Applications in radiometric dating: Understand how isotope ratios are used to determine the age of rocks, fossils, and other materials.
• Isotope effects in chemical reactions: Explore how the different masses of isotopes can subtly affect reaction rates and equilibrium constants.

By practicing a diverse range of problems and exploring these advanced concepts, you'll develop a comprehensive understanding of isotope calculations and their applications in various scientific disciplines. Remember, consistent practice and a clear understanding of the underlying principles are key to mastering this important topic.