Valences of each elemental group
Group | Valency |
I | 1 |
II | 2 |
III | 3 |
IV | +/- 4 |
V | -3 |
VI | -2 |
VII | -1 |
VIII | 0 |
Valences of common ions
Name | Formula | Valency |
Iron (II) | Fe2+ | +2 |
Copper (II) | Cu2+ | +2 |
Ammonium | NH4+ | +1 |
Nitrate | NO3– | -1 |
Hydroxide | OH– | -1 |
Carbonate | CO32- | -2 |
Sulphate | SO42- | -2 |
Silicate | SiO32- | -2 |
Phosphate | PO43- | -3 |
Word equations
- Balancing equations: A chemical equation is balanced when there are equal number of atoms and charges on both sides of the equation
- State symbols:
- (s) = solid
- (l) = liquid
- (g) = gas
- (aq) = aqueous solution
- Names of compounds
- Compound ending with -ide only contain two different elements
- Compound ending with -ate contain oxygen
Masses
- Relative atomic mass (Ar): mass of one atom of an element relative to one twelfth of the mass of one atom of Carbon-12
- Relative molecular mass (Mr): sum of relative atomic masses of all the atoms in one molecule of the compound
The Mole Concept
- A mole of a substance is the amount that contains the same number of units as the number of carbon atoms in 12 grams of carbon-12
- A mole is the Ar or Mr expressed in grams e.g. 1 mole of Carbon-12 is equal to 12 grams.
- It is equal to 6.02 × 1023 particles, this number is called Avogadro’s constant.
Number of Moles
Number of Moles = mass/ molar mass
Moles in Gases
Volume = No. of Moles×24dm3 at r.t.p (room temperature and pressure)
Concentration
concentration = (no. of moles) / volume
- Moles per dm3
- 1mol/dm3 = 1M
- Grams per dm3 or g/dm3
Molecular Formulae
- It shows the actual number of atoms in one molecule of a substance.
Empirical Formulae
- This is the simplest ratio of the atoms in a compound
- For example:
- Molecular formula of ethane= C2H6
- Empirical formula of ethane = CH3
- To find out the empirical formula you:
- Make the percent ratio into the simplest whole number ratio (NOTE: if given %s, use them as grams)
- Divide the coefficients of each element symbol by the lowest coefficient
Percentages
- Percentage purity = (mass of product / mass of compound) x 100
- Percentage yield = actual mass obtained / calculated mass