# 3) Working out empirical formula

If you know the % components within a compound and want the formula, then divide all the **percentages** by the **mass number** of each element and you will get a series of numbers proportional to the number of atoms.

**Example 1**

**What is the empirical formula of the compound that contains 11.2% hydrogen and 88.8% oxygen by mass?**

Mass number of hydrogen = 1

Mass number of oxygen = 16

**Remember divide % by mass number**

**Hydrogen Oxygen**

__11.2%1__ : 88.8%

16

There is a ratio of:

**Hydrogen: Oxygen**

11.2: 5.55

Now all we have to do is work out a whole number ratio. To do this, divide the large number by the small number:

__11.25.55__= 2

Therefore 11.2 (hydrogen): 5.55 (oxygen) is the same as 2:1.

The formula for the compound is H_{2}O.

*Below is an alternative way of tackling the problem. However, we believe our method is much easier.*

**What is the empirical formula of the compound that contains 11.2% hydrogen and 88.8% oxygen by mass?**

**1. Write what you know**

%H = 11.2% 1 mole of hydrogen = 1g

%O = 88.8% 1 mole of oxygen = 16g

For every 100g of the compound there will be 11.2g of hydrogen and 88.8g of oxygen.

**2.** **Calculate the number of moles of hydrogen and oxygen within 100g of the compound.**

So the number of moles of each element within the compound is calculated as:

__11.21__= 11.2 moles of hydrogen

__88.816__= 5.6 moles of oxygen

**3. Use molar ratio to work out empirical formula of the compound**

The molar ratio of hydrogen to oxygen is therefore 11.2 : 5.6 = 2:1 so there are two hydrogen atoms to every oxygen atoms in the compound. Meaning the formula of the compound is H_{2}O.

**Example 2**

**What is the empirical formula of the compound that contains 40% sulphur and 60% oxygen by mass?**

Mass number of sulphur = 32

Mass number of oxygen = 16

**Remember divide % by mass number**

**Sulphur Oxygen**

__40%32__ : 60%

16

The ratio is **Sulphur: Oxygen**

1.25: 3.75

Now all we have to do is work out a whole number ratio: divide the large number by the small number:

__3.751.25__ = 3

Therefore 1.25 (sulphur): 3.75 (oxygen) is the same as 1:3

So the empirical formula is SO_{3}. Easy isn’t it!

**Example 3**

**Elemental analysis showed 72.36% iron and 27.64% oxygen by mass. What is the empirical formula?**

Mass number of iron = 56

Mass number of oxygen = 16

**Remember divide % by mass number**

** Iron Oxygen**

__72.36%56__ : 27.64

16

The ratio is:

**Iron: Oxygen**

1.292: 1.727

Now all we have to do is work out a whole number ratio: divide the large number by the small number:

__1.7271.292__ = 1.33

Therefore,

**Iron: Oxygen**

1: 1.33

Try multiplying by 2 to get a whole number; if that doesn’t work try multiplying by 3 and if that doesn’t work, try multiplying by 4 etc.

In this case we can multiply by 3 to get

**Iron: Oxygen**

3: 4

So the empirical formula of the compound is Fe_{3}O_{4}. Easy isn’t it!

### Example 4

**Elemental analysis showed 41.09% K, 33.68% S and 25.22% O by mass, what is the formula?**

Mass number of potassium = 39

Mass number of sulphur = 32

Mass number of oxygen = 16

**Remember divide % by mass number**

**Potassium Sulphur Oxygen**

__41.09%39__ : 33.68%

32 : 25.22%

16

1.053 : 1.053 : 1.576

1 : 1 : 1.5

2 : 2 : 3

Therefore the formula is K_{2}S_{2}O_{3}

### Example 5

**A compound contains 92.26% carbon and 7.74% hydrogen by mass and has a molar mass of 78.0 g/mol. What is the molecular formula?**

** **

Mass number of carbon = 12

Mass number of hydrogen = 1

**Remember divide % by mass number**

**Carbon : Hydrogen **

__92.26%12__ : 7.74%

1

7.7 : 7.7

1 : 1

C : H

But, the weight of 1 mole of CH is as follows:

C = there are 12 protons and neutrons in carbon, each weighing 1

6 x 10^{23} g

H = There is 1 proton in hydrogen, weighing 1

6 x 10^{23} g

The total weight of CH is therefore (12+1) x 1

6 x 10^{23}g

The weight of 1 mole of CH is 6 x 10^{23 }x 13 x 1

6 x 10^{23}g

The weight of 1 mole of CH is 13, but because the molar mass = 78.0 g/mole = 78

13= 6, the formula must be (CH)_{6} or C_{6}H_{6.}