Science Unit CC9 Revision Notes
CC9a - Masses & Empirical Formulae:
Empirical Formula = The simplest whole-number ratio of atoms or ions of each element in a compound.
Molecular Formula = The actual number of atoms of each element in 1 molecule of a compound.
❖ Sometimes the empirical formula can be the same as the molecular formula. [E.g. H2O]
❖ A compound’s ‘formula’ is its molecular formula if it is a simple molecule or its empirical formula if it
is has a giant lattice structure.
Relative Formula Mass [Mr]:
❖ The Relative Formula Mass [Mr] of a substance is the sum of the Relative Atomic Masses [RAM / Ar]
of all the atoms or ions in its formula.
Relative Atomic Masses [RAM / Ar] = The mean mass of an atom relative to the mass of an atom of carbon-
12, which is assigned a mass of 12. The RAM of an element is the mean relative mass of the isotopes in the
element. The RAM of each atom is written above its atomic symbol in the Periodic Table.
Isotopes = Atoms of an element with different numbers of neutrons.
❖ Calculating the Relative Formula Mass:
➢ If, for example, we wanted to calculate the Ar of carbon dioxide, then we’d multiply each
element in the compound by its abundance:
■ Ar [C] + [2 x Ar [O]] = 12 + [2 x 16] => The Mr of CO2 = 44
Finding an Empirical Formula:
❖ The empirical formula of a compound can be calculated from the masses of the elements used to make
it.
❖ Calculating the Empirical Formula:
➢ E.g. To calculate the empirical formula of the compound formed when 10.0g of calcium reacts
with 17.8g of chlorine:
1. Find the Ar of each element:
Ca = 40 & Cl = 35.5
2. Divide the mass of each element by its Ar :
10/40 = 0.25 & 17.8/ 35.5 = 0.5
3. Divide the answers by the smallest number to find the simplest ratio:
0.25/0.25 = 1 & 0.5/0.25 = 2
4. Empirical Formula = CaCl2
AP™
, Finding the Molecular Formula from an Empirical Formula:
❖ The molecular formula is determined from its Empirical Formula and its Mr.
❖ Calculating the Molecular Formula:
➢ E.g. Determine the molecular formula for glucose, which has an empirical formula of CH2O
and an Mr of 180:
1. Find the Empirical Formula Mass:
Ar[C] + [2 x Ar[H]] + Ar[O] = 12 + [2 x 1] + 16 = 30
2. Divide the Mr by the Empirical Formula Mass:
180/30 = 6
3. The Molecular Formula is 6x the Empirical Formula, so the
Molecular Formula = C6H12O6.
Finding the Empirical Formula from the Reactants & Products in a Reaction:
❖ Metal oxides can be made by heating metals in a
limited oxygen supply, using the apparatus shown on
the right.
❖ If the reactant and product are ‘weighed’, then the
the empirical formula for the metal oxide can be
calculated.
❖ Determining the empirical formula using an
experiment:
➢ E.g. Find the empirical formula of magnesium oxide by means of an experiment:
1. Find the mass of the empty crucible.
2. Add the magnesium strip.
3. Find the masses of the magnesium and the crucible combined.
4. Place the crucible on top of the pipeclay triangle on the tripod with the lid on.
5. Heat the crucible.
6. Lift the lid at regular intervals to allow air [containing oxygen] to enter.
7. Ensure that the lid is not lifted enough for the magnesium oxide to escape.
8. Stop heating after all the magnesium has reacted [when the magnesium no
longer lights up upon the lifting of the crucible lid].
9. Allow the apparatus to cool.
10. Find the mass of the crucible [containing the magnesium oxide].
11. Subtract the result of step 3 from this mass to give the mass of oxygen.
CC9b - Conservation of Mass:
Solute = A substance that dissolves in a liquid to make a solution.
Solvent = A liquid in which a substance dissolves to make a solution.
Solution = Formed when a substance has dissolved in a liquid.
Law of Conservation of Mass = The idea that mass is never lost or gained during a chemical reaction or a
physical change.
AP™
CC9a - Masses & Empirical Formulae:
Empirical Formula = The simplest whole-number ratio of atoms or ions of each element in a compound.
Molecular Formula = The actual number of atoms of each element in 1 molecule of a compound.
❖ Sometimes the empirical formula can be the same as the molecular formula. [E.g. H2O]
❖ A compound’s ‘formula’ is its molecular formula if it is a simple molecule or its empirical formula if it
is has a giant lattice structure.
Relative Formula Mass [Mr]:
❖ The Relative Formula Mass [Mr] of a substance is the sum of the Relative Atomic Masses [RAM / Ar]
of all the atoms or ions in its formula.
Relative Atomic Masses [RAM / Ar] = The mean mass of an atom relative to the mass of an atom of carbon-
12, which is assigned a mass of 12. The RAM of an element is the mean relative mass of the isotopes in the
element. The RAM of each atom is written above its atomic symbol in the Periodic Table.
Isotopes = Atoms of an element with different numbers of neutrons.
❖ Calculating the Relative Formula Mass:
➢ If, for example, we wanted to calculate the Ar of carbon dioxide, then we’d multiply each
element in the compound by its abundance:
■ Ar [C] + [2 x Ar [O]] = 12 + [2 x 16] => The Mr of CO2 = 44
Finding an Empirical Formula:
❖ The empirical formula of a compound can be calculated from the masses of the elements used to make
it.
❖ Calculating the Empirical Formula:
➢ E.g. To calculate the empirical formula of the compound formed when 10.0g of calcium reacts
with 17.8g of chlorine:
1. Find the Ar of each element:
Ca = 40 & Cl = 35.5
2. Divide the mass of each element by its Ar :
10/40 = 0.25 & 17.8/ 35.5 = 0.5
3. Divide the answers by the smallest number to find the simplest ratio:
0.25/0.25 = 1 & 0.5/0.25 = 2
4. Empirical Formula = CaCl2
AP™
, Finding the Molecular Formula from an Empirical Formula:
❖ The molecular formula is determined from its Empirical Formula and its Mr.
❖ Calculating the Molecular Formula:
➢ E.g. Determine the molecular formula for glucose, which has an empirical formula of CH2O
and an Mr of 180:
1. Find the Empirical Formula Mass:
Ar[C] + [2 x Ar[H]] + Ar[O] = 12 + [2 x 1] + 16 = 30
2. Divide the Mr by the Empirical Formula Mass:
180/30 = 6
3. The Molecular Formula is 6x the Empirical Formula, so the
Molecular Formula = C6H12O6.
Finding the Empirical Formula from the Reactants & Products in a Reaction:
❖ Metal oxides can be made by heating metals in a
limited oxygen supply, using the apparatus shown on
the right.
❖ If the reactant and product are ‘weighed’, then the
the empirical formula for the metal oxide can be
calculated.
❖ Determining the empirical formula using an
experiment:
➢ E.g. Find the empirical formula of magnesium oxide by means of an experiment:
1. Find the mass of the empty crucible.
2. Add the magnesium strip.
3. Find the masses of the magnesium and the crucible combined.
4. Place the crucible on top of the pipeclay triangle on the tripod with the lid on.
5. Heat the crucible.
6. Lift the lid at regular intervals to allow air [containing oxygen] to enter.
7. Ensure that the lid is not lifted enough for the magnesium oxide to escape.
8. Stop heating after all the magnesium has reacted [when the magnesium no
longer lights up upon the lifting of the crucible lid].
9. Allow the apparatus to cool.
10. Find the mass of the crucible [containing the magnesium oxide].
11. Subtract the result of step 3 from this mass to give the mass of oxygen.
CC9b - Conservation of Mass:
Solute = A substance that dissolves in a liquid to make a solution.
Solvent = A liquid in which a substance dissolves to make a solution.
Solution = Formed when a substance has dissolved in a liquid.
Law of Conservation of Mass = The idea that mass is never lost or gained during a chemical reaction or a
physical change.
AP™
