krit.club logo

Atomic Structure - Relative Atomic Mass (Ar)

Grade 9IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Relative Atomic Mass (ArA_r) is defined as the weighted average mass of an atom of an element, taking into account its naturally occurring isotopes, relative to 112\frac{1}{12} of the mass of an atom of Carbon-12.

•

Since ArA_r is a ratio of masses, it is a dimensionless quantity and has no units.

•

Isotopes are atoms of the same element that contain the same number of protons but a different number of neutrons, leading to different mass numbers (AA).

•

The relative abundance of an isotope is the percentage of that specific isotope found in a natural sample of the element.

•

The value of ArA_r on the Periodic Table is rarely a whole number because it is an average of all stable isotopes weighted by their abundance.

•

Carbon-12 is used as the standard reference because it is a stable isotope and was assigned a mass of exactly 1212 units.

📐Formulae

Ar=∑(isotope mass×percentage abundance)100A_r = \frac{\sum (\text{isotope mass} \times \text{percentage abundance})}{100}

Ar=(m1×%1)+(m2×%2)+⋯+(mn×%n)100A_r = \frac{(m_1 \times \%_1) + (m_2 \times \%_2) + \dots + (m_n \times \%_n)}{100}

💡Examples

Problem 1:

Chlorine exists as two isotopes: 35Cl^{35}\text{Cl} with an abundance of 75%75\% and 37Cl^{37}\text{Cl} with an abundance of 25%25\%. Calculate the Relative Atomic Mass (ArA_r) of Chlorine.

Solution:

Ar=(35×75)+(37×25)100A_r = \frac{(35 \times 75) + (37 \times 25)}{100} Ar=2625+925100A_r = \frac{2625 + 925}{100} Ar=3550100A_r = \frac{3550}{100} Ar=35.5A_r = 35.5

Explanation:

To find the weighted average, multiply each isotopic mass by its percentage abundance, sum the products, and divide by 100100.

Problem 2:

An element XX has two isotopes. Isotope 1 has a mass of 1010 and an abundance of 20%20\%. Isotope 2 has a mass of 1111 and an abundance of 80%80\%. Calculate the ArA_r of element XX.

Solution:

Ar=(10×20)+(11×80)100A_r = \frac{(10 \times 20) + (11 \times 80)}{100} Ar=200+880100A_r = \frac{200 + 880}{100} Ar=1080100A_r = \frac{1080}{100} Ar=10.8A_r = 10.8

Explanation:

The result 10.810.8 is closer to 1111 than 1010 because the isotope with mass 1111 has a much higher relative abundance (80%80\%).