Understanding Chemical Reactions Lesson: Calculations in Chemistry 2

Page 175 Q 4 a)One mole of ZnO has a RAM of = 81g 0.2 moles will have a mass of 0.2 X 81 = 16.2 g

Page 175 Q 4 b)1 mole of H 2 S has a mass of (2x1) + 32 = 34g 2.5 moles will have a mass of 2.5 X 34 = 85 g

Page 175 Q 4 c)1 mole of CuSO 4 has a mass of (16 x 4) = 259.5g 0.45 moles will have a mass of 0.45 X = g

Calculating formulae Page 178 of Chemistry text

Working out the formula of magnesium oxide A student heated some Mg as shown. When Mg burns in air it combines with oxygen to make magnesium oxide.

Working out the formula of magnesium oxide Step 1 - Results Mass of crucible + lid + Mg before heating= 25.24g Mass of crucible and lid = 25.00g Therefore, mass of Mg = 0.24g

Working out the formula of magnesium oxide Step 2 - Results Mass of crucible + lid + magnesium oxide after heating= 25.40g Mass of crucible + lid + Mg before heating= 25.24g Therefore, mass of oxygen in magnesium oxide = 0.16g

Working out the formula of magnesium oxide Step 3 – Change the masses into moles Magnesium = 0.24 / 24 = 0.01 mole Oxygen = 0.16 / 16 = 0.01 mole

Working out the formula of magnesium oxide Step 4 – Work out the ratio of moles Mg :O 0.01 : : 1 Therefore the formula of magnesium oxide is MgO

One for you! A compound of nitrogen and hydrogen was broken down into its elements. It was found that 1.4g of nitrogen had combined with 0.3g of hydrogen in the compound. What was the formula of the compound? (R.A.M.s N = 14, H = 1)

One for you! Step 1 – work out the number of moles Moles of N = 1.4 / 14 = 0.1 Moles of H = 0.3 / 1 = 0.3

One for you! Step 2 – work out the ratio of the number of moles to the lowest whole numbers N:HN:H 0.1:0.3 1:31:3 Therefore there is 3 times as many H atoms as N atoms. Its formula must be NH 3

Reacting iron with copper sulphate A more reactive metal will displace a less reactive metal from its solution: Iron + copper sulphate iron sulphate + copper Fe (s) + CuSO 4(aq) FeSO 4(aq) + Cu (s) How much copper will 0.5g of iron produce?

Reacting iron with copper sulphate Fill your beaker with copper sulphate solution. Add precisely 5g of iron to the beaker and stir gently for 4 minutes. Filter the mixture – making sure all the solid is removed from the beaker – rinse beaker with water if needed.

Reacting iron with copper sulphate Rinse the filter paper with propanone to dry the copper. When dry – re-weigh the solid.

1 mole of Fe should produce 1 mole of Cu 5 g of Fe is 5 / 56 = 0.09 moles 0.09 moles of Fe should produce 0.09 moles of Cu 0.09 moles of Cu will have a mass of 0.09 x 64 = 5.7 g