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Atoms, Elements and Compounds - Isotopes

Grade 12A LevelChemistry

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

🔑Concepts

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Isotopes are defined as atoms of the same element that possess the same number of protons (atomic number ZZ) but different numbers of neutrons, leading to different mass numbers (AA).

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Isotopes of the same element have identical chemical properties because they have the same number of electrons and the same electronic configuration, which determines chemical reactivity.

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Physical properties of isotopes, such as density, rate of diffusion, and melting/boiling points, differ slightly due to the variation in atomic mass.

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The relative atomic mass (ArA_r) of an element is a weighted average of the masses of its naturally occurring isotopes relative to 112\frac{1}{12} of the mass of a 12C^{12}C atom.

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Notation for isotopes is usually written as ZAX^A_Z X, where AA is the nucleon number (protons + neutrons) and ZZ is the proton number.

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Radioisotopes are unstable isotopes that undergo radioactive decay to reach a more stable state, emitting particles such as α\alpha, β\beta, or γ\gamma radiation.

📐Formulae

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

A=Z+NA = Z + N

Abundance Percentage=number of atoms of a specific isotopetotal number of atoms of all isotopes×100%\text{Abundance Percentage} = \frac{\text{number of atoms of a specific isotope}}{\text{total number of atoms of all isotopes}} \times 100\%

💡Examples

Problem 1:

Naturally occurring Chlorine consists of two isotopes: 35Cl^{35}Cl with an abundance of 75.77%75.77\% and 37Cl^{37}Cl with an abundance of 24.23%24.23\%. Calculate the relative atomic mass (ArA_r) of Chlorine to two decimal places.

Solution:

Ar=(35×75.77)+(37×24.23)100A_r = \frac{(35 \times 75.77) + (37 \times 24.23)}{100} Ar=2651.95+896.51100=35.4846A_r = \frac{2651.95 + 896.51}{100} = 35.4846 Ar≈35.48A_r \approx 35.48

Explanation:

The relative atomic mass is calculated by multiplying each isotopic mass by its percentage abundance, summing the results, and dividing by 100 to find the weighted mean.

Problem 2:

An element XX has two isotopes, 10X^{10}X and 11X^{11}X. If the relative atomic mass (ArA_r) of the element is 10.810.8, determine the percentage abundance of each isotope.

Solution:

Let the abundance of 10X^{10}X be x%x\%. Then the abundance of 11X^{11}X is (100−x)%(100 - x)\%. 10.8=(10×x)+(11×(100−x))10010.8 = \frac{(10 \times x) + (11 \times (100 - x))}{100} 1080=10x+1100−11x1080 = 10x + 1100 - 11x −20=−x⇒x=20-20 = -x \Rightarrow x = 20 Abundance of 10X=20%^{10}X = 20\%, Abundance of 11X=80%^{11}X = 80\%.

Explanation:

By setting up an algebraic equation where the sum of abundances equals 100%100\%, we can solve for the unknown percentage based on the given average relative atomic mass.