Question

An element has two naturally occurring isotopes. Isotope 1 has a mass of \(120.9038 \mathrm{amu}\) and a relative abundance of \(57.4 \%\), and isotope 2 has a mass of \(122.9042\) amu and a relative abundance of \(42.6 \%\). Find the atomic mass of this element and, referring to the periodic table, identify it.

Short answer

The element is antimony (Sb) with an atomic mass of approximately 121.76 amu.

Step-by-step solution

Convert the Percent Abundance to Decimal Form

Convert the percent abundance of each isotope to decimal form by dividing by 100. For isotope 1 with a percent abundance of 57.4%, the decimal form is 0.574. For isotope 2 with a percent abundance of 42.6%, the decimal form is 0.426.

Calculate the Weighted Average Atomic Mass

Multiply the mass of each isotope by its corresponding abundance in decimal form and add the results together to get the average atomic mass of the element. For isotope 1: \(120.9038 \mathrm{amu} \times 0.574\), and for isotope 2: \(122.9042 \mathrm{amu} \times 0.426\).

Perform the Calculation

Add the products from Step 2 to obtain the average atomic mass: \(120.9038 \mathrm{amu} \times 0.574 + 122.9042 \mathrm{amu} \times 0.426\).

Identify the Element

Compare the calculated atomic mass with the atomic masses listed in the periodic table to identify the element.

Key concepts

Isotopic Abundance

Isotopic abundance refers to the relative amount of each different isotope of an element that exists in a natural sample. Isotopes are atoms of the same element that have the same number of protons but a different number of neutrons, resulting in different mass numbers.

For example, in the given exercise, the element in question has two naturally occurring isotopes. Understanding isotopic abundance involves converting the percentage into a decimal to be used in calculations. The decimal form represents the fraction of each isotope present when compared to the total amount of the element. In the exercise, one isotope's abundance is given as 57.4%, which, when converted to decimal form, becomes 0.574. This implies that out of every 100 atoms of the element, approximately 57 are of isotope 1.

Weighted Average Atomic Mass

The weighted average atomic mass is a calculation that takes into account the mass of each isotope of an element as well as its relative abundance to determine the average mass of an atom of the element as it naturally occurs.

It's akin to calculating the average grade in a class where some assignments weigh more than others. In the context of our exercise, each isotope's mass must be multiplied by its relative abundance (in decimal form). These products are then summed to achieve the weighted average. For instance, the weighted average atomic mass for the exercise's element can be calculated by multiplying isotope 1's mass (\(120.9038 \text{amu}\)) by its abundance (\(0.574\)), and doing the same product for isotope 2 (\(122.9042 \text{amu}\) and an abundance of 0.426), and adding these products together. This gives us a more accurate representation of the element's atomic mass as it typically exists, which is crucial for calculations in chemistry.

Periodic Table

The periodic table is a tabular display of the chemical elements, organized by their atomic number, electron configuration, and recurring chemical properties. Elements are listed in order of increasing atomic number in rows, so that those with similar properties fall under the same column, or group.

As the final step in our exercise, after calculating the weighted average atomic mass, we refer to the periodic table. Each element's atomic weight listed in the periodic table reflects its weighted average atomic mass, accounting for the natural isotopic distribution. By comparing our calculated value with the values on the periodic table, we can identify the element in question. The periodic table is an invaluable tool for chemists, simplifying the identification and classification of elements, understanding their properties, and predicting their reactions.