Fundamentals of Chemistry: Relative and Average Atomic Mass

Relative Atomic Masses of Selected Elements

• The relative atomic mass ($\text{a.m.u.}$) of an element is a measure of the mass of an atom of that element expressed relative to the mass of an atom of carbon-12 ($^{12}C$).

• Table 1.6 provides the relative atomic masses for specific elements as follows:

  • Hydrogen: $1.0079\text{ a.m.u.}$

  • Oxygen: $15.8994\text{ a.m.u.}$

  • Sodium: $22.9897\text{ a.m.u.}$

  • Magnesium: $24.3050\text{ a.m.u.}$

  • Silver: $107.8082\text{ a.m.u.}$

Concept and Definition of Average Atomic Mass

• Observations on Atomic Mass Values: As observed in the specific data for elements like Hydrogen and Oxygen, atomic masses are rarely found to be exact whole numbers.

• The Role of Isotopes: The non-integer nature of atomic masses is due to the fact that most elements are composed of two or more naturally occurring isotopes.

• Weighted Averaging: The relative atomic mass provided for an element represents the weighted average of the atomic masses of all its naturally occurring isotopes. This calculation takes into account the specific abundance (percentage of occurrence in nature) of each individual isotope.

• Definition: Average atomic mass is defined as the weighted average of the atomic masses of the naturally occurring isotopes of an element.

Mathematical Formula for Average Atomic Mass

• The average atomic mass of an element can be calculated by summing the contributions of each isotope based on its mass and its relative percentage in nature.

• General Formula: $\text{Average atomic mass} = \frac{\text{Atomic mass of } 1^{st} \text{ Isotope} \times \text{\% abundance}}{100} + \frac{\text{Atomic mass of } 2^{nd} \text{ Isotope} \times \text{\% abundance}}{100}$

Case Study: Average Atomic Mass of Chlorine

• Chlorine acts as a primary example for calculating average atomic mass because it exists as two distinct isotopes in nature.

• Isotope 1: Chlorine-35 ($Cl-35$)

  • Atomic Mass: $35\text{ a.m.u.}$

  • Percentage occurrence in nature: $75\%$

• Isotope 2: Chlorine-37 ($Cl-37$)

  • Atomic Mass: $37\text{ a.m.u.}$

  • Percentage occurrence in nature: $25\%$

• Calculation Procedure:

  • Contribution from $Cl-35$: $\frac{35 \times 75}{100}$

  • Contribution from $Cl-37$: $\frac{37 \times 25}{100}$

• Detailed Mathematical Step: $\text{Average atomic mass of Chlorine} = \frac{35 \times 75}{100} + \frac{37 \times 25}{100}$

• Final Result: $\text{Average atomic mass of Chlorine} = 35.5\text{ a.m.u.}$

Basis of Comparison

• The relative atomic mass of each isotope and the subsequent average atomic mass are determined relative to the mass of an atom of the standard reference: Carbon-12 ($^{12}C$).

Average weight, often referred to in scientific terms as average atomic mass, is the weighted mean of the weights of different forms of an element, accounting for how often each form appears in nature. This means that the average weight considers both the mass of the isotopes and their relative abundances. For example, if an element has isotopes, the average weight is determined by calculating how much each isotope contributes to the total average based on its occurrence in nature.

Average atomic mass is the average weight of all the different forms (isotopes) of an element, based on how common each form is in nature.

• Isotopes are types of the same element that have different numbers of neutrons in their nuclei.

• To find the average atomic mass, you can use this simple formula:
  $\text{Average atomic mass} = \frac{\text{Weight of 1st Isotope} \times \text{Abundance}}{100} + \frac{\text{Weight of 2nd Isotope} \times \text{Abundance}}{100}$

• This average is usually not a whole number because it combines the weights of different isotopes.

• For example, Chlorine has two forms: Chlorine-35 (more common) and Chlorine-37 (less common).

• The average atomic mass helps us understand how heavy an atom of the element is compared to a standard atom called carbon-12.