Atomic and Molecular Masses

Understanding atomic and molecular masses is fundamental in chemistry as it helps quantify substances and relate microscopic particles to macroscopic amounts. This section covers the definitions, calculations, and significance of atomic and molecular masses.

1. Atomic Mass

The atomic mass of an element is the mass of a single atom of that element, usually expressed in atomic mass units (amu) or unified atomic mass units (u). One atomic mass unit is defined as exactly \(\frac{1}{12}\) of the mass of a carbon-12 atom.

Since atoms are extremely small, their masses are very tiny and inconvenient to express in grams. Instead, the atomic mass unit provides a convenient scale:

• 1 amu = \(1.6605 \times 10^{-24}\) grams

For example, the atomic mass of a hydrogen atom is approximately 1.0078 amu, and that of oxygen is about 15.999 amu.

Isotopes and Atomic Mass

Elements often exist as mixtures of isotopes — atoms with the same number of protons but different numbers of neutrons. For example, chlorine has two main isotopes: chlorine-35 and chlorine-37.

The atomic mass of an element listed in the periodic table is a weighted average of the masses of its naturally occurring isotopes, weighted by their relative abundances. This is called the relative atomic mass (\(A_r\)) or atomic weight.

For chlorine:\[A_r = (0.7577 \times 34.9689) + (0.2423 \times 36.9659) \approx 35.45\]

2. Molecular Mass

The molecular mass of a molecule is the sum of the atomic masses of all the atoms in the molecule. It is expressed in atomic mass units (amu).

For example, the molecular mass of water (\(H_2O\)) is calculated as:\[Molecular mass = 2 \times 1.0078 + 15.999 = 18.0156 amu\]

The relative molecular mass (\(M_r\)) is the ratio of the molecular mass of a molecule to \(\frac{1}{12}\)th the mass of carbon-12 atom. Numerically, it is the same as the molecular mass expressed in amu.

Empirical and Molecular Formulas

The empirical formula gives the simplest whole-number ratio of atoms in a compound, while the molecular formula gives the actual number of atoms of each element in a molecule.

For example, glucose has an empirical formula \(CH_2O\) and a molecular formula \(C_6H_{12}O_6\). The molecular mass helps determine the molecular formula when the empirical formula and molar mass are known.

3. Mole Concept and Molar Mass

The mole is a bridge between the atomic scale and the macroscopic scale. One mole contains exactly \(6.022 \times 10^{23}\) particles (Avogadro's number).

The molar mass of a substance is the mass of one mole of that substance, expressed in grams per mole (g/mol). It is numerically equal to the relative atomic or molecular mass.

For example, the molar mass of water is:\[Molar mass = 18.0156 g/mol\]

4. Mass Spectrometry

Mass spectrometry is an analytical technique used to determine the atomic or molecular masses of substances and to identify isotopic abundances.

A sample is ionized, and the ions are separated based on their mass-to-charge ratio. The resulting spectrum shows peaks corresponding to different isotopes or molecular fragments, allowing precise calculation of atomic and molecular masses.

Figure: Atomic Mass Scale showing relative masses of common atoms

Figure: Illustration of the mole concept relating atoms to grams

Worked Examples

1. Difficulty: Easy
   Find the relative atomic mass of an element X that has two isotopes: X-10 (mass = 10 amu, abundance = 20%) and X-11 (mass = 11 amu, abundance = 80%).
   
   Solution:
   \[ A_r = (0.20 \times 10) + (0.80 \times 11) = 2 + 8.8 = 10.8 \] So, the relative atomic mass of element X is 10.8 amu.


2. Difficulty: Medium
   Calculate the molecular mass of carbon dioxide (\(CO_2\)). Atomic masses: C = 12.01 amu, O = 16.00 amu.
   
   Solution:
   \[ M_r = 12.01 + 2 \times 16.00 = 12.01 + 32.00 = 44.01 amu \]


3. Difficulty: Medium
   Given the empirical formula \(CH_2\) and molar mass 56 g/mol, find the molecular formula.
   
   Solution:
   Calculate empirical formula mass:
   \[ 12.01 + 2 \times 1.008 = 14.026 g/mol \]
   Find the multiplier:
   \[ n = \frac{56}{14.026} \approx 4 \]
   Molecular formula = \(C_4H_8\).


4. Difficulty: Hard
   Calculate the mass of 3 moles of water molecules.
   
   Solution:
   Molar mass of water = 18.0156 g/mol
   Mass = number of moles \(\times\) molar mass
   \[ = 3 \times 18.0156 = 54.047 g \]


5. Difficulty: Hard
   A sample contains 0.5 moles of oxygen gas (\(O_2\)). Calculate the number of oxygen atoms in the sample.
   
   Solution:
   Number of molecules in 0.5 moles:
   \[ 0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23} \]
   Each molecule has 2 atoms, so total atoms:
   \[ 2 \times 3.011 \times 10^{23} = 6.022 \times 10^{23} atoms \]

Formula Bank

• Atomic mass unit (amu):
  \(1 amu = \frac{1}{12} \times\) mass of carbon-12 atom
• Relative atomic mass:
  \(A_r = \sum (\text{isotopic mass} \times \text{fractional abundance})\)
• Molecular mass:
  \(M_r = \sum A_r \text{ of all atoms in molecule}\)
• Molar mass:
  \(M = \text{mass of 1 mole of substance} = A_r \text{ or } M_r \text{ in g/mol}\)
• Number of particles:
  \(N = n \times N_A\), where \(n\) = moles, \(N_A = 6.022 \times 10^{23}\)
• Mass from moles:
  \(m = n \times M\)
• Empirical formula multiplier:
  \(n = \frac{\text{molar mass}}{\text{empirical formula mass}}\)