1.27 carry out mole calculations using volumes and molar concentrations.

Moles / volume = concentration

1.26 calculate percentage yield

Percentage yield is: (actual yield / theoretical yield) x 100

Theoretical yield is what you expect to get, if a reaction doesn't finish you may end up with a lower yield than the expected theoretical yeild.

To work out theoretical yield it is important to understand that an equation gives you a ratio of moles, e.g.
Fe2O3 > 2Fe
tells us that every one mole of iron oxide makes two moles of iron. In this equation the weight of Fe will be your yield.

If we are told the weight of the Fe2O3 is 100g we can easily work out the theoretical yield:
Work out the moles of Fe2O3 by doing the weight (g) divided by the atomic mass: 100/160= 0.625Mol
We know that for every one mole of Fe2O3 there are two of Fe so we do: 0.625 x 2= 1.25 Mol
Now we have moles of Fe we can work out weight by Mol x Ar: 1.25 x 56= 70g

If you in fact got 62g of Fe you'd do (62/80)x100= 77.5%

1.25 calculate reacting masses using experimental data and chemical equations

This means knowing that the mass of A and the mass of B will equal the mass of AB.
Bearing in mind that the mass of one mole of an element is its atomic mass.
If we have 2 moles of A(24) and react it with 5 moles of B(25)
then we have (2x24)+(5x25) and can calculate that A2B5 will weigh 173g

1.24 calculate empirical and molecular formulae from experimental data

Find the masses of the elements in the compound (by weight the compound then weighing it with elements taken out). Make these masses into percentages, divide each percentage by the ar of that element and you will have the number of atoms of it in a molecule.

e.g
Carbon= % mass 80, ar 12
C= 80/12= 6.7
Hydrogen= % mass 20, ar 1
H= 20/1= 20
C6.7H20

To find the empirical formula divide the molecular formula by the smallest number.

e.g
C6.7H20 /6.7
=CH3

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

Weigh you compound, remove one element of the compound through a reaction (break up a metal oxide or salts), then weigh again.

The first weight is AB the second weight is A so doing AB-A= B.

Now that you have the weight of A and B you can work out the formulae by doing the weight divided by the Ar.

e.g

A= 22g with an ar of 11

A= 22/11= 2

B=72g with an ar of 18

B=72/18= 4

A2B4 and to simplify to empirical formula AB2

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

(S) solid
(L) liquid
(G) gas
(Aq) aqueous/ solid dissolved in liquid

In balanced equations these symbols go after a element or compound to show what state it is in.

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

Word equations have just the names of the reactants and products involved:
Hydrogen + Oxygen > Water
Balanced equations are the symbols of the products and reactants including the numbers of each, there must be an equal number of each element on both sides of the equation, if there are not you can alter this by putting the right number infront of a symbol:
2H + O > H2O