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Atoms, Elements and Compounds - Isotopes

Grade 11IGCSEChemistry

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 have the same number of protons but a different number of neutrons.

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Isotopes of an element have the same atomic number (ZZ) but different nucleon/mass numbers (AA).

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Chemical properties of isotopes are identical because they have the same number of electrons and the same electronic configuration (e.g.e.g., same number of outer-shell electrons).

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Physical properties of isotopes, such as density, melting point, and boiling point, differ slightly because these properties depend on the mass of the atom.

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The Relative Atomic Mass (ArA_r) is the weighted average mass of the isotopes of an element on a scale where an atom of carbon-12 (12C^{12}C) has a mass of exactly 1212 units.

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Nuclide notation is written as ZAX^{A}_{Z}X, where AA is the nucleon number and ZZ is the atomic number.

📐Formulae

Nucleon Number (A)=Protons (Z)+Neutrons (n)\text{Nucleon Number (A)} = \text{Protons (Z)} + \text{Neutrons (n)}

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

Ar=(m1×a1)+(m2×a2)+⋯+(mn×an)100A_r = \frac{(m_1 \times a_1) + (m_2 \times a_2) + \dots + (m_n \times a_n)}{100}

💡Examples

Problem 1:

Chlorine exists as two isotopes: 35Cl^{35}Cl with an abundance of 75%75\% and 37Cl^{37}Cl with an abundance of 25%25\%. Calculate the relative atomic mass (ArA_r) of chlorine.

Solution:

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

Explanation:

The relative atomic mass is the weighted average of the isotopic masses based on their natural abundance. Multiplying each mass by its percentage and dividing by 100100 gives the ArA_r found on the Periodic Table.

Problem 2:

An isotope of Copper is represented as 2963Cu^{63}_{29}Cu. State the number of protons, neutrons, and electrons in a neutral atom of this isotope.

Solution:

Protons = 2929, Electrons = 2929, Neutrons = 63−29=3463 - 29 = 34.

Explanation:

The lower number (Z=29Z=29) represents the number of protons. In a neutral atom, electrons equal protons. The upper number (A=63A=63) is the sum of protons and neutrons, so neutrons are calculated as A−ZA - Z.

Problem 3:

Boron has an ArA_r of 10.810.8. It consists of two isotopes, 10B^{10}B and 11B^{11}B. Calculate the percentage abundance of each isotope.

Solution:

Let the abundance of 10B^{10}B be x%x\% and 11B^{11}B be (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. So, 10B=20%^{10}B = 20\% and 11B=80%^{11}B = 80\%.

Explanation:

By setting up an algebraic equation where the total abundance sums to 100%100\%, we can solve for the individual isotopic contributions using the provided relative atomic mass.

Isotopes - Revision Notes & Key Formulas | IGCSE Grade 11 Chemistry