1.3i analyse and evaluate the results obtained from finding a formula or confirming an equation by experiment, eg the reaction of lithium with water and deducing the equation from the amounts in moles of lithium and hydrogen

CALCULATION OF FORMULAE FROM EXPERIMENTAL DATA
Calculate the empirical formula of the compound from the given data. This may be as a percentage composition or as the masses of materials found in an experiment. (This was my summer homework, and i was only required to do questions 5, 6, and 10 in each section)

5. Pb 90.66%, O 9.34%
- assume that 100g of the compound is present
Pb 90.66g, O 9.34g
- convert masses to moles
moles = mass/RAM = 90.66g/207 = 0.438 moles
                                = 9.34g/16 = 0.584 moles
- divide both by the lowest moles
0.438/0.438 = 1
0.584/0.438 = 1.333
Pb3O4

6. H 3.66%, P 37.8%, O 58.5%
H 3.66g, P 37.8g, O 58.5g
- convert masses to moles
moles = mass/RAM = 3.66/1 = 3.66 moles
                                = 37.8/31 = 1.22 moles
                                = 58.5/16 = 3.66 moles
- divide by the lowest moles
3.66/1.22 = 3
1.22/1.22 = 1
3.66/1.22 = 3
H3PO3

10. H 5.88%, O 94.12%
H 5.88g, O 94.12
- convert masses into moles
moles = mass/RFM = 5.88/1 = 5.88
                               = 94.12/16 = 5.88
- divide by the lowest amount of moles
5.88/5.88 = 1
5.88/5.88 = 1
HO (H2O2)

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5. 50cm3 of a solution of citric acid, Mr = 192, containing 19.2dm-3 reacted with 50cm3 of a solution of sodium hydroxide containing 12gdm-3. Citric acid can be represented by the formula HxA, where x represents the number of hydrogen atoms in the molecule. Use the data above to calculate the number of moles of sodium hydroxide that react with one mole of citric acid and hence find the value of x.

6. When 12.475g of hydrated copper (II) sulphate, CuSO4.xH2O, was heated, 7.980g of anhydrous salt was produced. Use the data to find the value of x and write out the equation for the reaction.
- calculate the masses of the copper (II) sulphate, the anhydrous salt, and the water
hydrated copper (II) sulphate: 12.475g
anhydrous copper (II) sulphate: 7.980g
water: 4.495g
- calculate the moles 
moles = mass/RFM = 7.980/(63.5+32.1+64) = 0.05 moles
                               = 4.495/(2+16) = 0.25 moles
- divide by the smallest moles
0.05/0.05 = 1
0.25/0.05 = 5
x = 5
CuSO4.5H2O --> CuSO4 + 5H2O

10. When 13.9g of FeSO4.xH2O is heated, 4g of solid iron (III) oxide is produced, together with the loss of 1.6g of sulphur dioxide and 2g of sulphur trioxide. The rest of the mass loss being due to the water of crystallisation lost. Use the data to write out the full equation for the action of heat.
FeSO4.xH2O: 13.9g
Fe: 4g
SO3: 2g
SO2: 1.6g
H2O: 6.3g
13.9g - 6.3 = 7.6g
- calculate the moles 
moles = mass/RFM = 7.6/(55.8+32.1+64) = 0.05 moles
                               = 6.3/(2+16) = 0.35 moles
- divide by the smallest amount of moles
0.05/0.05 = 1
0.35/0.05 = 7
x = 7
FeSO4.7H2O --> Fe2O3 + SO3 + SO2 + 14H2O