Empirical and Molecular Formulas, Percent Composition, and Moles in General Chemistry

Empirical and Molecular Formulas

Definitions and Calculations

The empirical formula of a compound represents the simplest whole-number ratio of the elements present, while the molecular formula shows the actual number of atoms of each element in a molecule. Determining these formulas is fundamental in chemical analysis and synthesis.

• Empirical Formula: The simplest integer ratio of elements in a compound.

• Molecular Formula: The actual number of atoms of each element in a molecule; it is a multiple of the empirical formula.

• Example: For glucose, the molecular formula is C6H12O6, and the empirical formula is CH2O.

Steps to Determine Empirical Formula:

1. Obtain the mass (or percent composition) of each element in the compound.

2. Convert masses to moles using the molar mass of each element.

3. Divide each mole value by the smallest number of moles calculated.

4. If necessary, multiply all ratios by an integer to obtain whole numbers.

Formula for Empirical Formula Calculation:

• Convert mass to moles: $\text{moles} = \frac{\text{mass}}{\text{molar mass}}$

• Find the simplest ratio: $\text{ratio} = \frac{\text{moles of each element}}{\text{smallest number of moles}}$

Formula for Molecular Formula:

• $\text{Molecular Formula} = n \times \text{Empirical Formula}$

• $n = \frac{\text{Molar Mass}}{\text{Empirical Formula Mass}}$

Percent Composition

Calculating Percent by Mass

Percent composition expresses the mass percentage of each element in a compound. This is useful for determining empirical formulas and for chemical analysis.

• Percent by Mass: The percentage of the total mass of a compound that is due to a particular element.

• Formula: $\%\,\text{element} = \frac{\text{mass of element in 1 mole of compound}}{\text{molar mass of compound}} \times 100$

• Example: For CuSO4:

  • Mass of Cu = 64 g

  • Molar mass of CuSO4 = 156 g/mol

  • $\%\,\text{Cu} = \frac{64}{156} \times 100 = 41.0\%$

Sample Calculation for Oxygen in CuSO4:

• Mass of O = 64 g

• Molar mass of CuSO4 = 156 g/mol

• $\%\,\text{O} = \frac{64}{156} \times 100 = 41.0\%$

Sample Calculation for Sulfur in CuSO4:

• Mass of S = 32 g

• Molar mass of CuSO4 = 156 g/mol

• $\%\,\text{S} = \frac{32}{156} \times 100 = 20.5\%$

Moles and Avogadro's Number

Fundamental Concepts

The mole is a standard scientific unit for measuring large quantities of very small entities such as atoms, molecules, or other specified particles. Avogadro's number defines the number of particles in one mole.

• Mole (mol): The amount of substance containing as many entities as there are atoms in 12 g of carbon-12.

• Avogadro's Number ($N_A$): $6.02 \times 10^{23}$ particles/mol.

• Molar Mass (M): The mass of one mole of a substance, usually expressed in g/mol.

• Relationship: $\text{Number of particles} = \text{moles} \times N_A$

• Formula: $n = \frac{\text{given mass}}{\text{molar mass}}$

Summary Table: Key Formulas and Relationships

Concept	Formula	Example
Percent Composition	$\%\,\text{element} = \frac{\text{mass of element}}{\text{molar mass of compound}} \times 100$	Cu in CuSO4: $\frac{64}{156} \times 100 = 41.0\%$
Empirical Formula	Convert mass to moles, divide by smallest, get ratio	CH2O from C6H12O6
Molecular Formula	$n = \frac{\text{Molar Mass}}{\text{Empirical Formula Mass}}$	Glucose: $\frac{180}{30} = 6$; C6H12O6
Moles	$n = \frac{\text{mass}}{\text{molar mass}}$	$\frac{32\,g}{16\,g/mol} = 2\,mol$
Particles	$\text{Number of particles} = n \times N_A$	$2\,mol \times 6.02 \times 10^{23}$

Additional info: Some calculations and ratios were inferred from context and standard chemistry procedures for empirical and molecular formula determination, as well as percent composition calculations.