Specification Point 1.32:
Know what is meant by the terms Empirical Formula and Molecular Formula
Empirical Formula & Molecular Formula
Empirical Formula: Gives the simplest whole number ratio of atoms of each element in the compound
- Calculated from knowledge of the ratio of masses of each element in the compound
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 |
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 ’
Author: Jamie
Jamie got a First class degree in Chemistry from Oxford University before going on to teach chemistry full time as a professional tutor. He’s put together these handy revision notes to match the Edexcel IGCSE Chemistry specification so you can learn exactly what you need to know for your exams.