1.27 carry out mole calculations using volumes and molar concentrations.

CALCULATING MOLES:

- To calculate moles using volumes and molar concentrations, you simply do:

- Moles = Concentration x Volume.

An example from (http://www.docbrown.info/page04/4_73calcs11msc.htm)
What mass of sodium hydroxide (NaOH) is needed to make up 500 cm³ (0.500 dm³) of a 0.500 mol dm^-3 (0.5M) solution? [A_r's: Na = 23, O = 16, H = 1]
- 1 mole of NaOH = 23 + 16 + 1 = 40g
- molarity = moles / volume, so mol needed = molarity x volume in dm³
- 500 cm³ = 500/1000 = 0.50 dm³
- mol NaOH needed = 0.500 x 0.500 = 0.250 mol NaOH
- therefore mass = mol x formula mass
- = 0.25 x 40 = 10g NaOH required 

1.26 calculate percentage yield

CALCULATING PERCENTAGE YIELD:

- To calculate percentage yield you do:
- Percentage yield = Actual yield (grams) / Theoretical Yield (grams) x 100
- Always between 0% and 100%, never above.

1.25 calculate reacting masses using experimental data and chemical equations

CALCULATING REACTING MASSES:

- 1. Write out the balanced equation.
- 2. For the bits you need, given in the question, write their relative formula mass and multiply them by the balancing numbers (numbers in front of the formulae).
- 3. The next bit is confusing, if the ratio was 1X : 2Y, you would divide 1X by 1 and 2Y by 1 to get how much 1 gram of X to reacts to make ... grams of Y.

To see a better explanation with examples goto: http://www.docbrown.info/page04/4_73calcs14other5.htm

1.24 calculate empirical and molecular formulae from experimental data

EMPIRICAL FORMULA:

- It is the smallest whole number ratio of atoms in a compound.

- 1. Write all the elements in compound.
- 2. Write their masses / percentages from the question.
- 3. Divide them by relative atomic mass.
- 4. Multiply or divide each number to get a whole number.
- 5. Obtain ratio in simplest form and write into formulae.

MOLECULAR FORMULAE:

- It is the exact value of atoms of each element in a single molecule.
- They are whole number multiples of empirical formulae.

- 1. Find the mass of empirical formulae.
- 2. That gives you how many empirical units are in the molecule.
- 3. Therefore you do the given empirical formulae x empirical units.

1.23 understand how the formulae of simple compounds can be obtained experimentally, including metal oxides, water and salts containing water of crystallisation

FORMULAE:

- 1. Find the mass of the multiple compounds.
- 2. Calculate the moles of water lost. That is mass / Mr.
- 3. Calculate the moles of anhydrous salt made. That is mass / Mr.
- 4. Work out ratio of anhydrous and water. Must be a whole number, e.g. if you get 0.02 moles of X and 0.01 moles of Y the ratio would be 2X : 1Y (since 0.02 / 0.01)

Another example: (thank you - http://www.docbrown.info/page04/4_73calcs14other4.htm)

Calculate the % of water in hydrated magnesium sulphate MgSO₄.7H₂O salt crystals

- Relative atomic masses: Mg = 24, S = 32, O = 16 and H = 1
- relative formula mass = 24 + 32 + (4 x 16) + [7 x (1 + 1 + 16)] = 246
- 7 x 18 = 126 is the mass of water
- so % water = 126 x 100 / 246 = 51.2%

1.22 use the state symbols (s), (l), (g) and (aq) in chemical equations to represent solids, liquids, gases and aqueous solutions respectively

STATE SYMBOLS:

- They tell you the physical state of what the reactants and products are.
- (s) is solid.
- (l) is liquid.
- (g) is gas.
- (aq) is aqueous - which means that it has been dissolved in water.

1.21 write word equations and balanced chemical equations to represent the reactions studied in this specification

BALANCED EQUATIONS:

- A chemical equation can be used to show reactants to products in the form of word or symbol equations.

- Symbol equations must be balanced; there must be equal number of atoms on each side. To do this you add numbers in front of the formulae.
E.g:
copper + oxygen --> copper(II) oxide
Cu + O2 --> CuO
- However this isn't balanced so you add numbers.
- There are 2 oxygens so you need to make CuO x2 ; it becomes 2CuO. This still isn't balanced as we have Cu + O2 --> 2CuO. This now tells us that there are 2 coppers so we simply do Cu x2 to make 2Cu.
- The final correct equation is 2Cu + O2 --> 2CuO