# Empirical & Molecular Formula

Show/Hide Sub-topics (Stoichiometry | O Level)

### Empirical Formula

Empirical formula of a compound shows the simplest whole number ratio of the different types of atoms in a compound.

Steps to determine empirical formula of a compound

1. Identify the various elements in the compound
2. Find the mass of each element in the compound
3. Calculate the number of moles of each element in the compound
4. Divide each value of the number of moles by the smallest number. (E.g. if a compound has 0.02 moles of X and 0.06 moles of Y, we will have $\frac{0.02}{0.02} = 1$ and $\frac{0.06}{0.02} = 3$ after this step)
5. Multiply each number by an integer so that the results are all whole numbers
6. The whole numbers are then the subscripts of the elements in the formula

Examples:

A sample of iron sulphide contains 10.750 g of iron and 9.254 g of sulphur. What is the empirical formula of the compound?

• Iron (Fe)Sulphur (S)
Mass (g)10.7509.254
No. of moles$\frac{10.750}{56} = 0.192$$\frac{9.254}{32} = 0.289 Divide by smallest no.\frac{0.192}{0.192} = 1$$\frac{0.289}{0.192} = 1.51$
Multiply to get whole number$1 \times 2 = 2$$1.51 \times 2 \approx 3 Simplest Ratio23 • Empirical formula of the compound is \text{Fe}_{2}\text{S}_{3} Consider a compound that contains 23.3 \% magnesium, 30.7 \% sulphur and 46.0 \% oxygen by mass. Find the empirical formula • Magnesium (Mg)Sulphur (S)Oxygen (O) Mass per 100g of compound23.330.746.0 No. of moles\frac{23.3}{24} = 0.97$$\frac{30.7}{32} = 0.96$$\frac{46}{16} = 2.88 Divide by smallest no.\frac{0.97}{0.96} \approx 1$$\frac{0.96}{0.96} = 1$$\frac{2.88}{0.96} = 3$
Simplest Ratio113
• Empirical formula of compound is $\text{MgSO}_{3}$

### Molecular Formula

The molecular formula shows the actual number of the component elements in a compound. It may be same as or a multiple of the empirical formula. (Remember that empirical formula is the simplest)

• $\text{Molecular formula} = n \times \text{Empirical formula}$
• n is calculated from: $n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}}$
• E.g. The empirical formula of $\text{C}_{6}\text{H}_{12}\text{O}_{6}$ is $\text{C}\text{H}_{2}\text{O}$

Steps to determine molecular formula of a compound

1. Find the empirical formula of the compound
2. Calculate the empirical formula mass
3. Calculate the ratio n
4. Multiply the subscripts in empirical formula by n

Example:

Consider a compound with an empirical formula of $\text{CHO}_{2}$ and a molecular mass of 90.1. Find the molecular formula of the compound.

• Empirical formula mass: $12 + 1 + 2(16) = 45$
• n: $n = \frac{90.1}{45} = 2$
• Molecular formula: $\text{C}_{2}\text{H}_{2}\text{O}_{4}$