Empirical formulae

The of a substance is the simplest whole number of the of each it contains.

Examples of empirical formula

The 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 and giant covalent structures are all empirical formulae. This is because the actual numbers of 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:

the empirical formula
the , M_r

Example

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

M_r 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)

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. (A_r of C = 12, A_r of H = 1)

Step	Action	Result(s)
1	Write the element symbols	C, H
2	Write the masses	4.8 g , 1.0 g
3	Write the A_r values	12.0 , 1.0
4	Divide masses by A_r	4.8 ÷ 12.0 = 0.4, 1.0 ÷ 1.0 = 1
5	Divide by the smallest number	0.4 ÷ 0.4 = 1, 1.0 ÷ 0.4 = 2.5
6	Multiply (if needed) to get whole numbers	1 × 2 = 2, 2.5 × 2 = 5
7	Write the formula	C₂H₅

Step	1
Action	Write the element symbols
Result(s)	C, H

Step	2
Action	Write the masses
Result(s)	4.8 g , 1.0 g

Step	3
Action	Write the A_r values
Result(s)	12.0 , 1.0

Step	4
Action	Divide masses by A_r
Result(s)	4.8 ÷ 12.0 = 0.4, 1.0 ÷ 1.0 = 1

Step	5
Action	Divide by the smallest number
Result(s)	0.4 ÷ 0.4 = 1, 1.0 ÷ 0.4 = 2.5

Step	6
Action	Multiply (if needed) to get whole numbers
Result(s)	1 × 2 = 2, 2.5 × 2 = 5

Step	7
Action	Write the formula
Result(s)	C₂H₅

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)

Step	Action	Result(s)
1	Write the element symbols	S, O
2	Write the masses	3.21 g, 6.41 g - 3.21 g = 3.20 g
3	Write the A_r values	32.1, 16.0
4	Divide masses by A_r	3.21 ÷ 32.1 = 0.1, 3.20 ÷ 16.0 = 0.2
5	Divide by the smallest number	0.1 ÷ 0.1 = 1, 0.2 ÷ 0.1 = 2
6	Multiply (if needed)	(not needed - already have whole numbers)
7	Write the formula	SO₂

Step	1
Action	Write the element symbols
Result(s)	S, O

Step	2
Action	Write the masses
Result(s)	3.21 g, 6.41 g - 3.21 g = 3.20 g

Step	3
Action	Write the A_r values
Result(s)	32.1, 16.0

Step	4
Action	Divide masses by A_r
Result(s)	3.21 ÷ 32.1 = 0.1, 3.20 ÷ 16.0 = 0.2

Step	5
Action	Divide by the smallest number
Result(s)	0.1 ÷ 0.1 = 1, 0.2 ÷ 0.1 = 2

Step	6
Action	Multiply (if needed)
Result(s)	(not needed - already have whole numbers)

Step	7
Action	Write the formula
Result(s)	SO₂

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)

Step	Action	Result(s)
1	Write the element symbols	Cu, I
2	Write the masses	1.27 g, 3.81 g - 1.27 g = 2.54 g
3	Write the A_r values	63.5, 127
4	Divide masses by A_r	1.27 ÷ 63.5 = 0.02, 2.54 ÷ 126.9 = 0.02
5	Divide by the smallest number	0.02 ÷ 0.02 = 1, 0.02 ÷ 0.02 = 1
6	Multiply (if needed)	(not needed - already have whole numbers)
7	Write the formula	CuI

Step	1
Action	Write the element symbols
Result(s)	Cu, I

Step	2
Action	Write the masses
Result(s)	1.27 g, 3.81 g - 1.27 g = 2.54 g

Step	3
Action	Write the A_r values
Result(s)	63.5, 127

Step	4
Action	Divide masses by A_r
Result(s)	1.27 ÷ 63.5 = 0.02, 2.54 ÷ 126.9 = 0.02

Step	5
Action	Divide by the smallest number
Result(s)	0.02 ÷ 0.02 = 1, 0.02 ÷ 0.02 = 1

Step	6
Action	Multiply (if needed)
Result(s)	(not needed - already have whole numbers)

Step	7
Action	Write the formula
Result(s)	CuI

Introducing chemical reactions - OCR GatewayEmpirical formulae