Lecture Notes: Section 3.2 - Percent Composition and Chemical Formulas

Key Topics

Definition: Percentage by mass of each element in a compound.
Example Calculation:
- Given: 10g sample with 2.5g hydrogen and 7.5g carbon.
- Percent Hydrogen: ( \frac{2.5}{10} \times 100 = 25% )
- Percent Carbon: ( \frac{7.5}{10} \times 100 = 75% )
- Sum: 25% + 75% = 100% (confirms composition)
Formula-Based Calculation:
- Use periodic table to find atomic masses.
- Calculate formula mass.
- Example: Ammonia (NH₃)
  - Nitrogen mass: Atomic mass of Nitrogen
  - Hydrogen mass: 3 times atomic mass of Hydrogen
  - Total formula mass: 17.03 amu
  - Percent Nitrogen: ( \frac{\text{Nitrogen mass}}{\text{Total mass}} \times 100%
  - Percent Hydrogen: ( \frac{3 \times \text{Hydrogen mass}}{\text{Total mass}} \times 100%
- Rounding errors may occur.

Empirical Formula:
- Simplest whole number ratio of atoms.
- Example: C₆H₁₂ (\rightarrow) CH₂
Molecular Formula:
- Actual number of atoms in a compound.
- Derived from empirical formula using molar mass.

Process:
1. Start with the mass of each element.
2. Calculate moles: Divide by molar mass.
3. Determine ratio: Divide each by the smallest number of moles.
4. Convert to whole numbers.
Example:
- Given: 1.71g Carbon, 0.383g Hydrogen
- Calculate moles and ratios.
- May need additional steps for fractional subscripts.

From Empirical Formula:
1. Calculate empirical formula mass.
2. Use known molecular mass.
3. Divide molecular mass by empirical formula mass to find multiplication factor (n).
4. Multiply each subscript in empirical formula by n.
Example:
- Empirical formula: CH₂O
- Empirical mass: 30 amu
- Molecular mass: 180 amu
- Calculation: ( n = \frac{180}{30} = 6 )
- Molecular formula: C₆H₁₂O₆