1.27 carry out mole calculations using volumes and molar concentrations

For this we need to use a few equations (it is best to show with an example)...

Calculate the volume of carbon dioxide produced when 50g of calcium carbonate is decomposed by heating.


Step 1 - Calculate the amount, in moles, of calcium carbonate reacted

work out the Mr of CaCO... it's 100 

Therefore, we can work out the amount, in moles, of calcium carbonate that was reacted...

Amount, in moles, of CaCO₃ = (50 ÷ 100) = 0.5 mol



Step 2 - Calculate the amount, in moles, of carbon dioxide formed

CaCO₃ ------> CaO + CO₂
 
1 mol of CaCO₃ produces 1 mol of CO₂, therefore, 0.5 mol of CaCO₃ must produce 0.5 mol of CO₂

Step 3 - Calculate the volume of CO₂ formed

Volume of CO₂ (in dm3) = (0.5 x 24) dm³ = 12dm³

So, the volume of carbon dioxide produced when 50g of calcium carbonate is decomposed by heating, is 12dm3

1.26 calculate percentage yield

Percentage yield is calculated with the equation...

Percentage yield = (actual yield (grams) / theoretical yield (grams)) x 100

NOTE: a 100% yield means you got all the product you expected to get; a 0% yield means that no reactants were converted into product

1.25 calculate reacting masses using experimental data and chemical equations

1 - write out the balanced equation (you may be given this in an exam)

2 - for the two bits you want, work out the relative formula mass (Mr) and multiply them by the balancing numbers in the equation

3 - Apply the rule : divide to get one, them multiply to get all (but you have to apply the first to the substance they give information about, and then the other one)

For example: What mass of magnesium oxide is produced when 60g of magnesium is burnt is air?

Step 1: balanced equation...

2Mg + O2 ----->  2MgO

Step 2: work out the relative formula masses of each component and multiply them by the balancing numbers in the equation (sounds confusing, sorry)

2Mg: 2 x 24 = 48                                             2MgO: 2 x (24 x 16) = 80

Step 3: apply the rule 'divide to get one, multiply to get all'...

The two numbers, 48 and 80, tell us that 48g of Mg react to give 80g of MgO. However, the question asks us how much MgO is produced when 60g of Mg is reacted, so we need to work out how much MgO is produced when 60g of Mg is reacted...

48g of Mg........80g of MgO     (divide by the reactant, in this case, divide by 48)

1g of Mg..........1.67g of MgO           ( x by 60, as thats how much Mg we have)

60g of Mg........100g of MgO

Therefore, 60g of Mg reacts with air to produce 100g of MgO

1.24 calculate empirical and moleculer formulae from experimental data

Empirical (and molecular) formula can also be calculated from experimental data. This is how its done...

- List all the elements in the compound
- Underneath, write their experimental masses or percentages
- Divide each mass by the Ar (relative atomic mass) of that particular element
- Turn the numbers into a ratio
- Simplify the ratio


For example: In an experiment, some iron oxide powder is reduced to pure metallic iron. Use the following experimental data to find the empirical formula of the iron oxide used.

Mass of empty container = 32.0g
Mass of container + mass of iron oxide = 96.0g
Mass of container + iron = 76.8g

METHOD - 
During this experiment, oxygen is lost. To find the mass of oxygen lost, minus the 'mass of container + iron' from 'mass of container + mass of iron oxide'...

96 - 76.8 = 19.2g

The mass of iron made is 'mass of container + iron' minus 'mass of container'...

76.8 - 32 = 44.8g

Now, list the elements in iron oxide...                       
    Fe                                O

write their experimental masses                          
   44.8                            19.2

divide by their Ar                                   
   44.8 / 56 = 0.8         19.2 / 16 = 1.2

multiply by 10 to put into ratio          
    8              :               12

cancel down                                            
   2              :                3

This means that the simplest formula is 2 Fe atoms to every 3 O atoms, so the empirical formula is Fe2O3

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

One way of working out formulae in an experiment is... 

- Weigh your compound
- Remove one element of the compound through a reaction (break up a metal oxide or salts)
- Weigh again

The first weight is AB the second weight is A so do 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.

For example...

A= 22g with an Ar of 11

A= 22/11= 2

B=72g with an Ar of 18

B=72/18= 4

This means the formula is, A2B4 which can 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

If a compound/atom/molecule is...

liquid, the state symbol is (l)

solid, the state symbol is (s)

gas, the state symbol is (g)

aqueous, the state symbol is (aq)

How to tell which state symbol to use... if your just given an equation, it may be hard to work out what is solid/liquid/gas/aqueous. Unfortunately there is no rule to learn that will give you the answer, but there are a few ways of working it out...

if a liquid (l) is heated, it will become a gas (g)
if a precipitate is formed, it will be a solid (s)
aqueous is anything dissolved in water, so if the compound contains an OH, it is most likely aqueous.

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

To balance an equation, there has to be the same number of each atom/element on each side of the equation.

For example...
Aluminum + copper (II) oxide ----> Aluminum oxide + copper

      Al       +         CuO           ---->          Al₂O₃        +    Cu

on this side there is...                    on this side there is...

1 Al                                                      2 Al

1 Cu                                                     1Cu

1 O                                                       3 O



You need to make both sides even... there needs to be the same amount of each elements on both sides, so we need to balance it.

2Al + 3CuO ----> Al₂O₃ + 3Cu

on this side there is...                    on this side there is...

2 Al                                                      2 Al

3 Cu                                                     3 Cu

3 O                                                       3 O


Now the equation is balanced :)