Definition

The empirical formula shows the simplest whole-number ratio of atoms of each element in a compound. It is derived from experimental data such as percentage composition or masses of elements.

Steps to Determine Empirical Formula

1. Convert percentages to masses (if given as percentages). Assume a 100 g sample so that percentages become grams directly.

2. Calculate moles of each element by dividing the mass of each element by its relative atomic mass (R.A.M.).

3. Divide by the smallest value obtained in Step 2 to get a simple ratio.

4. Convert to whole numbers by dividing each value by the smallest number obtained. If decimals result, multiply all values by a common factor to get whole numbers.

Worked Example 1: From Mass Data

A compound contains 4.9 g of magnesium and 3.2 g of oxygen. Find its empirical formula.

Solution

Divide by the smallest value (0.20):

Mg : O = 0.20 ÷ 0.20 : 0.20 ÷ 0.20 = 1 : 1

Empirical formula = MgO

Worked Example 2: From Percentage Composition

A compound contains 69.59% barium, 6.09% carbon, and the rest is oxygen. Find the empirical formula.

Solution

Step 1: Assume 100 g of compound.

• Barium = 69.59 g
• Carbon = 6.09 g
• Oxygen = 100 − (69.59 + 6.09) = 24.32 g

Step 2: Calculate moles.

Step 3: Divide by smallest value (0.51).

• Ba: 0.51 ÷ 0.51 = 1
• C: 0.51 ÷ 0.51 = 1
• O: 1.52 ÷ 0.51 ≈ 3

Empirical formula = BaCO₃

Definition

The molecular formula shows the actual number of atoms of each element in one molecule. It is related to the empirical formula by:

$\text{Molecular formula} = n \times \text{Empirical formula}$

where $n$ is a whole number.

To find $n$, use:

$n = \frac{\text{Relative Molecular Mass (R.M.M.)}}{\text{Sum of R.A.M. from empirical formula}}$

Worked Example: From Empirical Formula to Molecular Formula

A compound contains 15.8% carbon and 84.2% sulfur. Its relative molecular mass is 76. Find: (a) The empirical formula (b) The molecular formula

Solution

(a) Empirical formula

Assume 100 g:

• Carbon = 15.8 g
• Sulfur = 84.2 g

Moles:

• C: 15.8 ÷ 12 = 1.32
• S: 84.2 ÷ 32 = 2.63

Divide by smallest (1.32):

• C: 1.32 ÷ 1.32 = 1
• S: 2.63 ÷ 1.32 ≈ 2

Empirical formula = CS₂

(b) Molecular formula

R.M.M. of CS₂ = 12 + (2 × 32) = 12 + 64 = 76

$n = \frac{76}{76} = 1$

Since $n = 1$, the molecular formula equals the empirical formula.

Molecular formula = CS₂

Worked Example 3: With Different n Value

A compound has empirical formula CH and relative molecular mass of 78. Find its molecular formula.

Solution

R.A.M. sum from CH = 12 + 1 = 13

$n = \frac{78}{13} = 6$

Molecular formula = 6 × (CH) = C₆H₆

To find empirical formula from % composition:

1. Assume 100 g → convert % to grams
2. Divide by R.A.M. to get moles
3. Divide by smallest mole value
4. Round to nearest whole numbers

To find molecular formula:

1. Find empirical formula
2. Calculate $n$ = R.M.M. ÷ (sum of R.A.M. in empirical formula)
3. Multiply empirical formula by $n$

In Tanzania, farmers use fertilizers such as urea (CO(NH₂)₂) and ammonium sulfate ((NH₄)₂SO₄) to boost crop yields. Understanding empirical and molecular formulae helps extension officers and farmers calculate the exact proportion of nitrogen, phosphorus, or potassium in fertilizer bags, ensuring accurate application rates that improve harvests while avoiding waste of expensive agricultural inputs.