Question

An element, \(X\) has the following isotopic composition; \({ }^{200} \mathrm{X}: 90 \%\) \({ }^{199} \mathrm{X}: 8.0 \%\) \({ }^{202} \mathrm{X}: 2.0 \%\) The weighted average atomic mass of the naturally occurring element \(\mathrm{X}\) is closest to a. \(200 \mathrm{amu}\) b. 201 amu c. 202 amu d. \(199 \mathrm{amu}\)

Short answer

The weighted average atomic mass of element \( X \) is closest to 200 amu.

Step-by-step solution

Understand the Problem

The goal is to calculate the weighted average atomic mass of an element \( X \) given the isotopic composition of its isotopes. We will consider the natural abundance of each isotope to determine the average atomic mass.

Identify Key Information

We are given the relative abundance and mass number of each isotope of the element \( X \): \({ }^{200} X \) at 90%, \({ }^{199} X \) at 8%, and \({ }^{202} X \) at 2%.

Write the Formula for Weighted Average Atomic Mass

The formula for calculating the weighted average atomic mass is \( \text{Average Atomic Mass} = \sum (\text{fraction of isotope} \times \text{mass of isotope}) \). This means we need to multiply each isotope's mass by its fractional abundance and sum the results.

Convert Percentages to Fractions

Convert the percentages given into fractions of 1 by dividing by 100: - \({ }^{200} X \) : 0.90- \({ }^{199} X \) : 0.08 - \({ }^{202} X \) : 0.02

Calculate Contributions of Each Isotope

Calculate the contribution to the average atomic mass from each isotope:- Contribution of \({ }^{200} X \): \( 0.90 \times 200 = 180 \)- Contribution of \({ }^{199} X \): \( 0.08 \times 199 = 15.92 \)- Contribution of \({ }^{202} X \): \( 0.02 \times 202 = 4.04 \)

Sum to Find the Weighted Average Atomic Mass

Add the contributions from each isotope to get the weighted average:\( 180 + 15.92 + 4.04 = 199.96 \) amu.

Determine the Closest Integer Value

Round the calculated atomic mass of 199.96 amu to the nearest whole number. This number is 200 amu.

Key concepts

Isotopic Composition

Isotopic composition refers to the different types of isotopes that exist for a given element. Isotopes are variants of a particular chemical element that have the same number of protons but different numbers of neutrons. This results in different mass numbers for each isotope.

For example in element \(X\), we have three isotopes:

• \(^{200} \mathrm{X}\)
• \(^{199} \mathrm{X}\)
• \(^{202} \mathrm{X}\)

Each of these isotopes contributes differently to the average atomic mass of the element based on its abundance. The isotopic composition is important because it affects the weighted average atomic mass of an element, which is what you often see on the periodic table.

Fractional Abundance

Fractional abundance represents the proportion of each isotope relative to the total abundance of all isotopes of an element. While the isotopic composition tells us which isotopes are present, the fractional abundance indicates how much of each isotope is there compared to the whole.

Fractional abundance is key in calculating weighted average atomic mass. To find it, you convert the percentage of each isotope into a fraction. For example, if an isotope is present at 90%, its fractional abundance is \(0.90\).

• \(^{200} \mathrm{X}\) has a fractional abundance of \(0.90\)
• \(^{199} \mathrm{X}\) has a fractional abundance of \(0.08\)
• \(^{202} \mathrm{X}\) has a fractional abundance of \(0.02\)

These fractions are used in the formula to calculate the weighted average atomic mass, multiplying each by its respective isotope's mass number.

Mass Number

The mass number of an isotope is the sum of protons and neutrons in its nucleus. For example, isotopes \(^{200} X\), \(^{199} X\), and \(^{202} X\) have mass numbers 200, 199, and 202, respectively.

The mass number is crucial in determining the weighted average atomic mass. It represents the isotope's mass when atoms are grouped together. When calculating the average atomic mass, the mass number of each isotope is multiplied by its fractional abundance. This sum gives us an accurate depiction of the element's mass as per its natural isotopic composition, like how 199.96 amu rounds to 200 amu, illustrating the weighted presence of the isotopes in the element \(X\).