Empirical formulae

The empirical formula of a substance is the simplest whole number ratio of the atoms of each element it contains.

Examples of empirical formula

The molecular formula of ethane is C2H6. It shows the actual number of atoms of each element in a molecule of ethane. This formula does not show the simplest whole number ratio because each number can be divided by two. This gives the empirical formula of ethane: CH3.

The molecular formula and empirical formula of some substances are the same. For example, both types of formula for carbon dioxide are CO2.

The formulae given for ionic compounds and giant covalent structures are all empirical formulae. This is because the actual numbers of ions or atoms they contain are huge and variable.

Converting empirical formula to molecular formula

The molecular formula for a substance can be worked out using:

Example

The empirical formula for a compound is CH2 and its relative formula mass is 42.0. Deduce its molecular formula. (Relative atomic masses: C = 12.0, H = 1.0)

Mr of CH2 = 12.0 + (2 × 1.0) = 14.0

Factor to apply = 42.0/14.0 = 3

Multiply the numbers in the empirical formula by the factor 3:

Molecular formula = C3H6

Question

The empirical formula for a compound is C2H5 and its relative formula mass is 58.0. Deduce its molecular formula. (Relative atomic masses: C = 12.0, H = 1.0)

Mr of C2H5 = (2 ×12.0) + (5 × 1.0) = 29.0

Factor to apply = 58.0/29.0 = 2

Multiply the numbers in the empirical formula by the factor 2:

Molecular formula = C4H10

Calculating an empirical formula

Information about reacting masses is used to calculate empirical formulae. This is obtained from experiments.

Example

A hydrocarbon is found to contain 4.8 g of carbon and 1.0 g of hydrogen. Calculate its empirical formula. (Ar of C = 12, Ar of H = 1)

StepActionResult(s)
1Write the element symbols C, H
2Write the masses4.8 g , 1.0 g
3Write the Ar values12.0 , 1.0
4Divide masses by Ar4.8 ÷ 12.0 = 0.4, 1.0 ÷ 1.0 = 1
5Divide by the smallest number0.4 ÷ 0.4 = 1, 1.0 ÷ 0.4 = 2.5
6Multiply (if needed) to get whole numbers1 × 2 = 2, 2.5 × 2 = 5
7Write the formulaC2H5

The action at step 5 usually gives the simplest whole number ratio straight away. Sometimes it does not, so both numbers may need to be multiplied to get a whole number (step 6). For example, by 2 if you have .5, by 3 if you have .33, or by 4 if you have .25 in a number.

Question

3.21 g of sulfur reacts completely with oxygen to produce 6.41 g of an oxide of sulfur. Calculate the empirical formula of the oxide of sulfur. (Relative atomic masses: S = 32.1, O = 16.0)

StepActionResult(s)
1Write the element symbolsS, O
2Write the masses3.21 g, 6.41 g - 3.21 g = 3.20 g
3Write the Ar values32.1, 16.0
4Divide masses by Ar3.21 ÷ 32.1 = 0.1, 3.20 ÷ 16.0 = 0.2
5Divide by the smallest number0.1 ÷ 0.1 = 1, 0.2 ÷ 0.1 = 2
6Multiply (if needed)(not needed - already have whole numbers)
7Write the formulaSO2
Question

In an experiment, 1.27 g of hot copper reacts with iodine vapour to form 3.81 g of copper iodide. Calculate the empirical formula of copper iodide. (Relative atomic masses: Cu = 63.5, I = 126.9)

StepActionResult(s)
1Write the element symbolsCu, I
2Write the masses1.27 g, 3.81 g - 1.27 g = 2.54 g
3Write the Ar values63.5, 127
4Divide masses by Ar1.27 ÷ 63.5 = 0.02, 2.54 ÷ 126.9 = 0.02
5Divide by the smallest number0.02 ÷ 0.02 = 1, 0.02 ÷ 0.02 = 1
6Multiply (if needed)(not needed - already have whole numbers)
7Write the formulaCuI