Understanding Understanding Chemical Chemical ReactionsReactions

Lesson: Calculations in Lesson: Calculations in Chemistry 2Chemistry 2

Page 175 Q 4Page 175 Q 4a) One mole of ZnO has a RAM of

65 + 16 = 81g0.2 moles will have a mass of0.2 X 81 = 16.2 g

Page 175 Q 4Page 175 Q 4b) 1 mole of H2S has a mass of

(2x1) + 32 = 34g2.5 moles will have a mass of2.5 X 34 = 85 g

Page 175 Q 4Page 175 Q 4c) 1 mole of CuSO4 has a mass of

63.5 + 32 + (16 x 4) = 259.5g0.45 moles will have a mass of0.45 X 259.5 = 116.8 g

Calculating Calculating formulaeformulae

Page 178 of Chemistry Page 178 of Chemistry texttext

Working out the formula of Working out the formula of magnesium oxidemagnesium 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 Working out the formula of magnesium oxidemagnesium 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 Working out the formula of magnesium oxidemagnesium 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 Working out the formula of magnesium oxidemagnesium 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 Working out the formula of magnesium oxidemagnesium oxide

Step 4 – Work out the ratio of moles

Mg : O 0.01 : 0.01

1 : 1

Therefore the formula of magnesium oxide is MgO

One for you!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!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!One for you!Step 2 – work out the ratio of the number of

moles to the lowest whole numbersN : H0.1 : 0.31 : 3

Therefore there is 3 times as many H atoms as N atoms.

Its formula must be NH3

Reacting iron with copper Reacting iron with copper sulphatesulphate

• A more reactive metal will displace a less reactive metal from its solution:

• Iron + copper sulphate iron sulphate + copper

• Fe(s) + CuSO4(aq) FeSO4(aq) + Cu(s)

• How much copper will 0.5g of iron produce?

Reacting iron with copper Reacting iron with copper sulphatesulphate

• 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 Reacting iron with copper sulphatesulphate

• 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