### 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**

- Identify the various elements in the compound
- Find the mass of each element in the compound
- Calculate the number of moles of each element in the compound
- 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)
- Multiply each number by an integer so that the results are all whole numbers
- 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) 10.750 9.254 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 Ratio****2****3**- 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 compound 23.3 30.7 46.0 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 Ratio****1****1****3**- 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**

- Find the empirical formula of the compound
- Calculate the empirical formula mass
- Calculate the ratio n
- 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: $\text{C}_{2}\text{H}_{2}\text{O}_{4}$