Properties and Calculations of Molecules

Molecular Definitions and Classification

• Definition: A molecule consists of a group of atoms connected by covalent bonds.

• Molecular Element: Composed of only one type of atom (e.g., $O_2$, $F_2$).

• Molecular Compound: Composed of two or more different atoms (e.g., $H_2O$, $HCl$).

Molar Mass and Moles of Molecules

• Molar Mass ($M$): The sum of the molar masses of all atoms within a molecule, expressed in $g/mol$.

  • Example: $M_{H_2O} = 2(1) + 16 = 18\,g/mol$

  • Example: $M_{SO_2} = 32 + 2(16) = 64\,g/mol$

  • Example: $M_{C_6H_{12}O_6} = 6(12) + 12(1) + 6(16) = 180\,g/mol$

• Amount of Substance ($n$): Calculated using the formula $n = \frac{m}{M}$, where $m$ is the mass in grams.

  • Example: For a $4\,g$ sample of $H_2O$, $n = \frac{4}{18} = 0.22\,mol$.

Particle Calculations and Avogadro's Number

• Number of Molecules ($N$): Calculated using the relation $n = \frac{N}{N_A}$.

• Avogadro’s Number ($N_A$): $6.022 \times 10^{23}\,molecule/mol$.

  • Example: In a $2\,mol$ sample of $H_2O$, $N = 2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{22}$ molecules.

• Atomic Counting: The number of specific atoms is deduced by the molecular ratio.

  • For $1.2044 \times 10^{22}$ water molecules:

  • $H$ atoms: $2 \times 1.2044 \times 10^{22} = 2.008 \times 10^{22}$ atoms.

  • $O$ atoms: $1 \times 1.2044 \times 10^{22} = 1.2044 \times 10^{22}$ atoms.

• Combined Relation: $\frac{m}{M} = \frac{N}{N_A}$. This allows finding the number of molecules directly from mass.

  • Example: In $3\,g$ of $SO_2$, $\frac{3}{64} = \frac{N}{6.022 \times 10^{23}}$, resulting in $N = 2.82 \times 10^{22}$ molecules.

Empirical and Molecular Formulas

• Empirical Formula: Represents the simplest positive integer ratio of atoms in a compound (e.g., $CH_2O$ for glucose).

  • Calculation Steps:

  1. Assume a $100\,g$ sample to convert mass percentages to grams.

  2. Divide mass by atomic molar mass ($M$) to find moles for each element.

  3. Divide each molar value by the smallest calculated value to find the ratio ($x, y, z$).

  • Example: A compound with $38.67\%\,C$, $16.22\%\,H$, and $45.11\%\,N$ yields an empirical formula of $(CH_5N)_n$.

• Molecular Formula: Represents the actual number of atoms of each element in a molecule (e.g., $C_6H_{12}O_6$).

  • Deduction via $n$: $n = \frac{Molar\,mass\,of\,Molecular}{Molar\,mass\,of\,empirical}$.

  • Using $(CH_5N)_n$ with a molecular molar mass of $62\,g/mol$: $n = \frac{62}{31} = 2$, leading to the formula $C_2H_{10}N_2$.

  • Law of Proportions Method: $\frac{12x}{\%\,C} = \frac{y}{\%\,H} = \frac{16z}{\%\,O} = \frac{Molar\,mass\,of\,compound}{100}$.

  • Example for a compound with $M = 256.26\,g/mol$, $56.25\%\,C$, $6.29\%\,H$, and $37.46\%\,O$: results in $x \approx 12$, $y \approx 16$, $z \approx 6$, providing the molecular formula $C_{12}H_{16}O_6$.