Empirical formulae
The empirical formula of a compound is the simplest whole number ratio of atoms of each element in the compound. It is determined using data from experiments and therefore empirical.
For example, the molecular formula of glucose is C6H12O6 but the empirical formula is CH2O. This is because we can divide each number in C6H12O6 by 6 to make a simpler whole number ratio.
On the other hand, a compound which has the empirical formula of CH2 could have a molecular formula of C2H4, C3H6, C4H8 or even C13H26.
You can use information about reacting masses to calculate the formula of a compound. Here is an example:
3.2 g of sulfur reacts with oxygen to produce 6.4 g of sulfur oxide. What is the formula of the oxide?
For your calculation you’ll need to use the fact that the Ar (relative atom mass) of sulfur is 32 and the Ar of oxygen is 16.
| Step | Action | Result | |
|---|---|---|---|
| 1 | Write the element symbols | S | O |
| 2 | Write the masses | 3.2 g | 6.4 g – 3.2 g = 3.2 g |
| 3 | Write the Ar values | 32 | 16 |
| 4 | Divide masses by Ar | 3.2 ÷ 32 = 0.1 | 3.2 ÷ 16 = 0.2 |
| 5 | Divide by the smallest number | 0.1 ÷ 0.1 = 1 | 0.2 ÷ 0.1 = 2 |
| 6 | Write the formula | SO2 |
The action at Step 5 usually gives you the simplest whole number ratio straightaway. Sometimes it does not, so you might get 1 and 1.5. In this example, you would multiply both numbers by 2, giving 2 and 3 (instead of rounding 1.5 up to 2).
Converting the empirical formula to a molecular formula
From the empirical formula, you can work out the molecular formula if you know the relative formula mass (Mr) of the compound.
Add up the atomic masses of the atoms in the empirical formula.
For example, the empirical formula of a hydrocarbon is CH2 and its Mr is 42.
- the mass of the atoms in the empirical formula is 14
- 42 ÷ 14 = 3
- so you need to multiply the numbers in the empirical formula by 3
The molecular formula of the hydrocarbon is therefore C3H6.
Empirical formulae experiments
Finding the empirical formula of a metal oxide
The empirical formula of magnesium oxide can be calculated using the following experiment, which finds the mass of the magnesium and oxygen atoms in a sample of the compound.
- Weigh a crucible (with its lid).
- Put a sample of clean magnesium ribbon into the crucible and weigh it with the lid. Calculate the mass of magnesium by subtracting the mass of the empty crucible.
- Strongly heat the crucible over a Bunsen burner for several minutes.
- Carefully lift the lid from time to time to allow sufficient air into the crucible for the magnesium to fully oxidise without letting any magnesium oxide escape.
- Continue heating until the mass of the crucible reaches a constant (maximum) mass, indicating that the reaction is complete.
- Measure the mass of the crucible and contents again. Calculate the mass of the magnesium oxide by subtracting the mass of the empty crucible.
- To work out the empirical formula, you need the mass of the magnesium (from step 2) and the mass of the oxygen atoms as well. To find the mass of the oxygen atoms, subtract the mass of magnesium used from the mass of the magnesium oxide (from step 6).
- Now divide each of the two masses by the relative atomic masses of the elements and simplify the ratio.
Finding the molecular formula – crystallisation
Some ionic compounds have a fixed number of water molecules trapped inside their crystal lattice. The formula of a hydrated ionic compound can be calculated by dehydrating it:
- Record the mass of an evaporating basin.
- Add a known mass of hydrated copper sulfate.
- Heat over a Bunsen burner, gently stirring, until the blue copper sulfate has turned completely white and the mass has reached a constant (minimum) mass, indicating that all the water has been lost.
- Record the mass of the evaporating basin and contents and then use this to calculate the mass of the white anhydrous copper sulfate remaining, and also the mass of the water lost.
- Divide the mass of the white copper sulfate by the Mr (relative formula mass) of CuSO4 (159.5) and divide the mass of water by the Mr of water (18). Simplify the ratio of waters to copper sulfates to find the whole number represented here by ‘x’: CuSO4.xH2O
Example:
A compound that contains 10 g of Hydrogen and 80 g of Oxygen has an Empirical Formula of H2O. This can be shown by the following calculations:
Amount of Hydrogen Atoms = Mass in grams ÷ Ar of Hydrogen = (10 ÷ 1) = 10 moles
Amount of Oxygen Atoms = Mass in grams ÷ Ar of Oxygen = (80 ÷ 16) = 5 moles
The Ratio of Moles of Hydrogen Atoms to Moles of Oxygen Atoms:
Hydrogen Oxygen
Moles 10 : 5
Ratio 2 : 1
Since equal numbers of Moles of Atoms contain the same number of atoms, the Ratio of Hydrogen Atoms to Oxygen Atoms is 2: 1
Hence the Empirical Formula is H2O
Molecular Formula: Gives the exact numbers of atoms of each element present in the formula of the compound
- Divide the relative formula mass of the molecular formula by the relative formula mass of the empirical formula
- Multiply this to each number of elements
Relationship between Empirical and Molecular Formula:
| Name of compound | Empirical formula | Molecular formula |
|---|---|---|
| Methane | CH4 | CH4 |
| Ethane | CH3 | C2H6 |
| Ethene | CH2 | C2H4 |
| Benzene | CH | C6H6 |
REVISION
Example:
The Empirical Formula of X is C4H10S1 and the Relative Formula Mass of X is 180
What is the Molecular Formula of X?
Relative Formula Mass: Carbon : 12 Hydrogen : 1 Sulfur : 32
Step 1 – Calculate Relative Formula Mass of Empirical Formula
( C x 4 ) + ( H x 10 ) + ( S x 1) = ( 12 x 4 ) + ( 1 x 10 ) + ( 32 x 1) = 90
Step 2 – Divide Relative Formula Mass of X by Relative Formula Mass of Empirical
Formula
180 / 90 = 2
Step 3 – Multiply Each Number of Elements by 2
( C 4 x 2 ) + ( H 10 x 2 ) + ( S 1 x 2 ) = ( C 8 ) + ( H 20 ) + ( S 2 )
Molecular Formula of X = C8H20S2
Specification Point 1.33:
Calculate Empirical and Molecular Formulae from Experimental Data
- Find number of moles by dividing mass by relative formula mass
- Find ratio of moles
- Gives empirical formula
- To find molecular formula divide relative formula mass given by relative formula mass of empirical formula.
Metal Oxides
The apparatus needed to find the formulae of a Metal Oxide
Method:
- Measure mass of crucible with lid
- Add sample of metal into crucible and measure mass with lid (calculate the mass of metal by subtracting the mass of empty crucible)
- Strong heat the crucible over a Bunsen burner for several minutes
- Lift the lid frequently to allow sufficient air into the crucible for the metal to fully oxidise without letting magnesium oxide escape
- Continue heating until the mass of crucible remains constant (maximum mass), indicating that the reaction is complete
- Measure the mass of crucible and contents (calculate the mass of metal oxide by subtracting the mass of empty crucible)
Working out Empirical Formula / Formulae:
Mass of Metal: Subtract mass of crucible from metal and mass of empty crucible
Mass of Oxygen: Subtract mass of metal used from the mass of magnesium oxide
STEP 1 – Divide Each of the two masses by the relative atomic masses of elements
STEP 2 – Simplify the ratio
Metal Oxygen
Mass x y
Mole x / Mr y / Mr
= a = b
Ratio a : b
STEP 3 – Represent the Ratio into the ‘ Metal O ‘ E.g, MgO
Water and Salts containing Water of Crystallisation
The apparatus needed to find the formulae of crystals
Method:
- Measure mass of evaporating dish
- Add a known mass of hydrated salt
- Heat over a Bunsen burner, gently stirring, until the blue salt turns completely white, indicating that all the water has been lost
- Record the mass of the evaporating dish and contents
Working out Empirical Formula / Formulae:
Mass of White Anhydrous Salt: Measure Mass of White Anhydrous Salt Remaining
Mass of Water: Subtract Mass of White Anhydrous Salt Remaining from the Mass of Known Hydrated Salt
STEP 1 – Divide Each of the Two Masses by the Relative Atomic Masses of Elements
STEP 2 – Simplify the Ratio of Water to Anhydrous Salt
Anhydrous Salt Water
Mass a b
Mole a / Mr b / Mr
= y = x
Ratio 1 : x
STEP 3 – Represent the Ratio into ‘ Salt.xH2O ’
