Percentage Composition and Empirical and Molecular Formulas

Percentage Composition is a fundamental concept in chemistry that expresses the relative amount of each element present in a compound as a percentage of the total mass. It helps chemists understand the makeup of substances and is essential for determining formulas of compounds.

1. Percentage Composition

The percentage composition of an element in a compound is calculated by dividing the total mass of that element in the compound by the molar mass of the compound, then multiplying by 100.

Mathematically,

\[ \text{Percentage of element} = \frac{\text{Mass of element in 1 mole of compound}}{\text{Molar mass of compound}} \times 100 \]

Example: Calculate the percentage composition of carbon in methane (\( \mathrm{CH_4} \)).

The molar mass of methane is \(12 + 4 \times 1 = 16 , \mathrm{g/mol}\). The mass of carbon is 12 g.

So, percentage of carbon = \( \frac{12}{16} \times 100 = 75% \).

This means 75% of methane's mass is due to carbon.

2. Empirical Formula

The empirical formula of a compound is the simplest whole-number ratio of atoms of each element present in the compound. It does not necessarily represent the actual number of atoms in a molecule but gives the simplest ratio.

Steps to determine empirical formula from percentage composition:

1. Assume a 100 g sample of the compound. This converts percentages directly to grams.
2. Convert the mass of each element to moles by dividing by its atomic mass.
3. Divide all mole values by the smallest number of moles calculated.
4. If necessary, multiply all ratios by a whole number to get whole-number subscripts.

Example: A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Find its empirical formula.

Assuming 100 g sample:

• Carbon: 40 g, moles = \( \frac{40}{12} = 3.33 \)
• Hydrogen: 6.7 g, moles = \( \frac{6.7}{1} = 6.7 \)
• Oxygen: 53.3 g, moles = \( \frac{53.3}{16} = 3.33 \)

Divide by smallest moles (3.33):

• C: \( \frac{3.33}{3.33} = 1 \)
• H: \( \frac{6.7}{3.33} = 2 \)
• O: \( \frac{3.33}{3.33} = 1 \)

Empirical formula = \( \mathrm{CH_2O} \).

3. Molecular Formula

The molecular formula shows the actual number of atoms of each element in a molecule of the compound. It is always a whole-number multiple of the empirical formula.

To find the molecular formula, you need the molar mass of the compound and the molar mass of the empirical formula.

Formula:

\[ \text{Molecular formula} = (\text{Empirical formula})_n \quad \text{where} \quad n = \frac{\text{Molar mass of compound}}{\text{Molar mass of empirical formula}} \]

Example: A compound has an empirical formula \( \mathrm{CH_2O} \) and a molar mass of 180 g/mol. Find its molecular formula.

Molar mass of empirical formula = \(12 + 2 \times 1 + 16 = 30 , \mathrm{g/mol}\).

Calculate \( n = \frac{180}{30} = 6 \).

Molecular formula = \( (\mathrm{CH_2O})_6 = \mathrm{C_6H_{12}O_6} \).

4. Applications

• Determining chemical formulas: Percentage composition data from elemental analysis helps deduce empirical and molecular formulas.
• Pharmaceuticals: Knowing exact molecular formulas is critical for drug formulation.
• Material science: Composition analysis guides the synthesis of new materials.

Worked Examples

Formula Bank

• Percentage Composition:
  \[ \text{Percentage of element} = \frac{\text{Mass of element in 1 mole of compound}}{\text{Molar mass of compound}} \times 100 \]
• Moles from mass:
  \[ \text{Moles} = \frac{\text{Mass of element}}{\text{Atomic mass of element}} \]
• Empirical formula ratio:
  \[ \text{Divide moles of each element by smallest mole value} \]
• Molecular formula:
  \[ \text{Molecular formula} = (\text{Empirical formula})_n, \quad n = \frac{\text{Molar mass of compound}}{\text{Molar mass of empirical formula}} \]