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Some Basic Concepts of Chemistry - Atomic and Molecular Masses

Grade 11CBSEChemistry

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

🔑Concepts

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Atomic Mass: Defined as the mass of an atom relative to the mass of a carbon-12 atom. One atomic mass unit (1 amu1 \text{ amu} or 1 u1 \text{ u}) is defined as exactly 1/12th1/12^{\text{th}} the mass of one 12C^{12}C atom.

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Average Atomic Mass: Most elements exist as a mixture of isotopes. The average atomic mass is calculated by taking the sum of the products of the isotopic masses and their relative fractional abundances.

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Molecular Mass: It is the sum of atomic masses of the elements present in a molecule. It is obtained by multiplying the atomic mass of each element by the number of its atoms and adding them together, e.g., for H2OH_2O, mass is 2×1.008 u+16.00 u=18.016 u2 \times 1.008 \text{ u} + 16.00 \text{ u} = 18.016 \text{ u}.

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Formula Mass: Used for ionic compounds like NaClNaCl, which do not exist as discrete molecules but as a 3D crystal lattice. It is the sum of atomic masses of all atoms in the formula unit.

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Molar Mass: The mass of one mole of a substance in grams. It is numerically equal to atomic/molecular/formula mass but expressed in g mol−1\text{g mol}^{-1}.

📐Formulae

1 amu=112×mass of one 12C atom≈1.66056×10−24 g1 \text{ amu} = \frac{1}{12} \times \text{mass of one } ^{12}C \text{ atom} \approx 1.66056 \times 10^{-24} \text{ g}

Average Atomic Mass=∑i=1n(Abundancei×Isotopic Massi)\text{Average Atomic Mass} = \sum_{i=1}^{n} (\text{Abundance}_i \times \text{Isotopic Mass}_i)

Molecular Mass=∑(Number of atoms of element×Atomic mass of element)\text{Molecular Mass} = \sum (\text{Number of atoms of element} \times \text{Atomic mass of element})

Number of moles (n)=Mass of substance (m)Molar mass (M)\text{Number of moles (n)} = \frac{\text{Mass of substance (m)}}{\text{Molar mass (M)}}

💡Examples

Problem 1:

Calculate the average atomic mass of Chlorine using the following data: 35Cl^{35}Cl (Isotopic mass: 34.9689 u34.9689 \text{ u}, Abundance: 75.77%75.77\%) and 37Cl^{37}Cl (Isotopic mass: 36.9659 u36.9659 \text{ u}, Abundance: 24.23%24.23\%).

Solution:

Average Atomic Mass=(34.9689×75.77)+(36.9659×24.23)100=35.45 u\text{Average Atomic Mass} = \frac{(34.9689 \times 75.77) + (36.9659 \times 24.23)}{100} = 35.45 \text{ u}

Explanation:

The average atomic mass is the weighted average of the isotopes based on their natural occurrence.

Problem 2:

Calculate the molecular mass of Glucose (C6H12O6C_6H_{12}O_6).

Solution:

Molecular mass of C6H12O6=6(12.011)+12(1.008)+6(16.00)=72.066+12.096+96.00=180.162 u\text{Molecular mass of } C_6H_{12}O_6 = 6(12.011) + 12(1.008) + 6(16.00) = 72.066 + 12.096 + 96.00 = 180.162 \text{ u}

Explanation:

The molecular mass is the sum of the atomic masses of 6 Carbon atoms, 12 Hydrogen atoms, and 6 Oxygen atoms.

Atomic and Molecular Masses - Revision Notes & Key Formulas | CBSE Class 11 Chemistry