Question

A fictitious element has two naturally occurring isotopes and has an atomic weight of \(29.5 \mathrm{amu}\). a. If the natural abundance of isotope 1 is \(33.7 \%\), what is the natual abundance of isotope 2 ? b. If the mass of isotope 2 is \(30.0 \mathrm{amu}\), what is the mass of isotope 1 ?

Short answer

The natural abundance of Isotope 2 is 66.3%. The mass of Isotope 1 is calculated using the equation: mass of isotope 1 = (atomic weight - (mass of isotope 2 * abundance of isotope 2)) / abundance of isotope 1.

Step-by-step solution

Determine the Abundance of Isotope 2

Since there are only two isotopes and the natural abundance must sum up to 100%, subtract the abundance of isotope 1 from 100%. So, the natural abundance of isotope 2 is 100% - 33.7%.

Set up the Equation for Isotope Masses

Use the formula for atomic weight, which is the sum of the products of the mass and abundance of each isotope. For two isotopes: atomic weight = (mass of isotope 1 * abundance of isotope 1) + (mass of isotope 2 * abundance of isotope 2).

Substitute Known Values

Replace the known values: atomic weight (29.5 amu), mass of isotope 2 (30.0 amu), and abundance of isotope 1 (33.7%). Convert the percentage of abundance to decimal form to use in calculations.

Calculate the Mass of Isotope 1

Rearrange the equation to solve for the mass of isotope 1 by subtracting the product of mass of isotope 2 and its abundance from the atomic weight and then dividing by the abundance of isotope 1.

Calculate the Abundance and Mass

Calculate the abundance of isotope 2 from Step 1 and then the mass of isotope 1 using the equation derived in Step 4.

Key concepts

Atomic Weight

Understanding the concept of atomic weight is essential for students delving into chemistry. Simply put, the atomic weight, or atomic mass, of an element represents the average mass of all the isotopes that compose that element, taking into account their relative natural abundance. Each isotope has its own distinct mass, and these masses differ slightly due to the variance in the number of neutrons in the nucleus.

When calculating the atomic weight, you're looking at a weighted average, where each isotope's mass is multiplied by its natural abundance (as a decimal), and the results are summed up. It's important to note that the atomic weight of an element listed on the periodic table isn't just the mass of a single atom but rather this calculated average value. This distinct concept is vital to grasp because it explains why atomic weights are not whole numbers—reflecting the mixture of isotopes found in nature.

Natural Abundance of Isotopes

Isotopes are atoms of the same element that differ in the number of neutrons. The natural abundance of isotopes refers to the percentage of each isotope present in a naturally occurring sample of an element. It plays a key role in determining an element's atomic weight because isotopes do not exist in equal quantities. Some isotopes are more prevalent, while others may be rather rare.

Natural abundance is expressed as a percentage and should always add up to 100% because it represents the complete makeup of the element's isotopes in nature. Knowing the natural abundance allows scientists and students to solve complex chemistry problems, as seen in the original exercise. Understanding and using this concept is crucial for accurate isotope abundance calculation, and correctly converting percentage abundance into decimal form for computations is a fundamental skill in this domain.

Isotope Mass Calculation

In the realm of chemistry, the mass of isotopes isn't something we can guess; instead, it needs to be calculated carefully, especially when not all information is readily available. For a two-isotope system, the calculation becomes an exercise in understanding linear equations. After determining the natural abundances, we set up an equation where the atomic weight equals the sum of each isotope's mass times its relative abundance.

In our exercise, with the mass of one isotope and the atomic weight provided, we can calculate the missing mass of the other isotope. This is achieved by reorganizing our original equation to isolate the unknown variable. This step logically follows the conversion of abundance percentages into decimals, which is a crucial transitional step in the calculation process. A firm understanding of the linear relationship between mass and abundance is fundamental to solving such problems.

Chemistry Problem Solving

When approaching chemistry problem solving, a step-by-step methodology, as shown in our textbook solution, is invaluable. Clear and logical progression of steps prevents confusion and errors in calculations. Firstly identifying what is known and unknown paves the way forward. In our case, continuity in using a consistent unit (such as atomic mass units, amu) and correctly translating percentage abundances into decimals form the foundation of accurate calculation.

Applying these concepts with attention to detail and a systematic approach leads to reliable solutions in the realm of chemistry. Breaking down complex problems into smaller, manageable parts and understanding the relationship between atomic weight, isotope masses, and natural abundance can turn a challenging concept into an achievable exercise. It's imperative to practice these steps to enhance problem-solving skills that are essential for any aspiring chemist.