Practice Isotope Calculations 1 Answer Key

Practice Isotope Calculations: A Comprehensive Guide with Answer Key

Isotope calculations can seem daunting at first, but with a systematic approach and understanding of the underlying principles, they become manageable and even enjoyable. This comprehensive guide provides a step-by-step approach to solving various isotope calculation problems, complete with worked examples and an answer key. We will cover everything from basic calculations involving isotopic abundance to more complex problems involving radioactive decay.

Understanding Isotopes

Before diving into calculations, let's solidify our understanding of isotopes. Isotopes are atoms of the same element that have the same number of protons but a different number of neutrons. This difference in neutron number results in variations in atomic mass. Each isotope is identified by its mass number (A), which is the sum of protons and neutrons. For example, Carbon-12 (¹²C) has 6 protons and 6 neutrons (A = 12), while Carbon-14 (¹⁴C) has 6 protons and 8 neutrons (A = 14).

Key Terminology:

• Atomic Number (Z): The number of protons in an atom's nucleus. This defines the element.
• Mass Number (A): The total number of protons and neutrons in an atom's nucleus.
• Isotopic Abundance: The percentage of a particular isotope relative to the total abundance of all isotopes of that element in a given sample.
• Average Atomic Mass: The weighted average of the masses of all isotopes of an element, taking into account their relative abundances.

Basic Isotope Calculations: Finding Average Atomic Mass

The most common type of isotope calculation involves determining the average atomic mass of an element given the isotopic abundances and their respective masses. This is a weighted average calculation.

Formula:

Average Atomic Mass = Σ (Isotopic Mass × Isotopic Abundance)

Example 1:

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

Solution:

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

Example 2:

Boron has two isotopes, ¹⁰B and ¹¹B. The average atomic mass of Boron is 10.81 amu. The atomic mass of ¹⁰B is 10.01 amu and the atomic mass of ¹¹B is 11.01 amu. Calculate the percent abundance of each isotope.

Solution: Let x be the abundance of ¹⁰B (as a decimal). Then (1-x) is the abundance of ¹¹B.

10.81 amu = (10.01 amu * x) + (11.01 amu * (1-x))

Solving for x:

x = 0.19 or 19% (abundance of ¹⁰B) 1-x = 0.81 or 81% (abundance of ¹¹B)

Isotope Calculations Involving Radioactive Decay

Radioactive decay is the process by which unstable isotopes (radioisotopes) transform into more stable isotopes by emitting radiation. Calculations involving radioactive decay often utilize the concept of half-life.

Half-life (t½): The time it takes for half of a given quantity of a radioactive isotope to decay.

Formula:

Remaining Amount = Initial Amount × (1/2)^(t/t½)

Where:

• t is the elapsed time
• t½ is the half-life

Example 3:

A sample of ¹⁴C initially contains 100 grams. The half-life of ¹⁴C is 5730 years. How much ¹⁴C remains after 11460 years?

Solution:

Remaining Amount = 100 g × (1/2)^(11460 years / 5730 years) = 25 g

More Complex Isotope Calculations: Determining Isotopic Abundance from Average Atomic Mass and Known Isotopes

These calculations involve working backward from the average atomic mass to determine the isotopic abundances. This often requires setting up and solving simultaneous equations.

Example 4:

An element X has two isotopes, ⁶³X and ⁶⁵X. The average atomic mass of X is 63.55 amu. What are the percent abundances of ⁶³X and ⁶⁵X?

Solution:

Let x be the abundance of ⁶³X (as a decimal), and (1-x) be the abundance of ⁶⁵X.

63.55 amu = (63 amu * x) + (65 amu * (1-x))

Solving for x:

x = 0.725 or 72.5% (abundance of ⁶³X) 1-x = 0.275 or 27.5% (abundance of ⁶⁵X)

Advanced Isotope Calculations: Mass Spectrometry and Isotopic Ratios

Mass spectrometry is a powerful technique used to determine the isotopic composition of a sample. The data obtained often involves isotopic ratios, which are the relative abundances of different isotopes.

Example 5:

A mass spectrometry analysis of a sample of iron shows the following isotopic ratios: ⁵⁴Fe:⁵⁶Fe = 0.0584 : 1. Knowing that the natural abundance of ⁵⁶Fe is approximately 91.75%, calculate the percent abundance of ⁵⁴Fe.

Solution:

The ratio 0.0584 : 1 indicates that for every 1 atom of ⁵⁶Fe, there are 0.0584 atoms of ⁵⁴Fe. This translates to approximately 5.84% ⁵⁴Fe relative to the total iron atoms. However, this is only a fraction of the total isotopic abundance. Since the relative abundance of ⁵⁶Fe is given as ~91.75%, we need to correct for this.

The precise calculation involves understanding that the ratio provides the relative proportion within the analyzed portion of the sample. Additional information, or assumptions, are necessary to ascertain the absolute abundances.

Note: This example highlights the importance of understanding the context and limitations of mass spectrometry data. The accurate calculation would require either additional isotopic ratios or knowing the total isotopic composition of the sample.

Practice Problems with Answer Key

Here are some practice problems to test your understanding. Try to solve them before checking the answer key below.

Problem 1: Magnesium has three naturally occurring isotopes: ²⁴Mg (78.99%), ²⁵Mg (10.00%), and ²⁶Mg (11.01%). Their atomic masses are 23.99 amu, 24.99 amu, and 25.98 amu, respectively. Calculate the average atomic mass of magnesium.

Problem 2: Lithium has two isotopes, ⁶Li and ⁷Li. The average atomic mass of lithium is 6.94 amu. The atomic mass of ⁶Li is 6.015 amu, and the atomic mass of ⁷Li is 7.016 amu. Calculate the percent abundance of each isotope.

Problem 3: A sample of ¹⁴C initially contains 500 grams. Its half-life is 5730 years. How much ¹⁴C remains after 17190 years?

Problem 4: An element Y has two isotopes, ⁸⁵Y and ⁸⁷Y. The average atomic mass of Y is 85.47 amu. What are the percent abundances of ⁸⁵Y and ⁸⁷Y?

Problem 5: A mass spectrometry analysis of a sample of copper shows the following isotopic ratio: ⁶³Cu : ⁶⁵Cu = 0.7 : 1. Assuming that these are the only two isotopes present, what is the percent abundance of each isotope?

Answer Key

Problem 1: Average Atomic Mass of Magnesium ≈ 24.31 amu

Problem 2: Abundance of ⁶Li ≈ 7.5%, Abundance of ⁷Li ≈ 92.5%

Problem 3: Amount of ¹⁴C remaining ≈ 62.5 g

Problem 4: Abundance of ⁸⁵Y ≈ 72.5%, Abundance of ⁸⁷Y ≈ 27.5%

Problem 5: Abundance of ⁶³Cu ≈ 41.2%, Abundance of ⁶⁵Cu ≈ 58.8%

Conclusion

Mastering isotope calculations requires a solid understanding of the concepts and a systematic approach to problem-solving. Practice is key to building proficiency. By working through the examples and practice problems, you should gain confidence in tackling a wide range of isotope calculations. Remember to always double-check your work and ensure you understand the underlying principles. This detailed guide provides a strong foundation for further exploration in nuclear chemistry and related fields. Further research into advanced isotopic techniques and applications will enhance your understanding and skills even further.