1. IC5.2.3 Mole calculations © Oxford University Press Mole calculations

2. IC5.2.3 Mole calculations © Oxford University Press It is possible to calculate the masses of substances that are reacting using  moles  the balanced equation  the relative formula masses or relative atomic masses of the substances.

3. IC5.2.3 Mole calculations © Oxford University Press For example: hydrogen reacts with oxygen to form water vapour. 2H 2 + O 2  2H 2 O What mass of water can be made from 6 g of hydrogen gas?

4. IC5.2.3 Mole calculations © Oxford University Press  6 g of H 2 = 6 ÷ 2 = 3 mol  Looking at the equation, 2 mol of H 2 would make 2 mol of H 2 O,  so 3 mol of H 2 would make 3 mol of H 2 O.  M r of H 2 O = (2 × 1) + 16 = 18 so its molar mass is 18 g/mol.  mass = number of moles × molar mass  mass of H 2 O = 3 mol × 18 g/mol = 54 g 2H 2 + O 2  2H 2 O A r of H = 1, O = 16 M r of H 2 = (2 × 1) = 2

5. IC5.2.3 Mole calculations © Oxford University Press Now try these questions yourself. Nitrogen reacts with hydrogen to form ammonia. N 2 + 3H 2  2NH 3 A r of H = 1, N = 14 1.Work out the molar masses of nitrogen, hydrogen, and ammonia.  Molar mass of nitrogen is 14 × 2 = 28 g/mol  Molar mass of hydrogen is 1 × 2 = 2 g/mol  Molar mass of ammonia is 14 + (1 × 3) = 17 g/mol

6. IC5.2.3 Mole calculations © Oxford University Press N 2 + 3H 2  2NH 3 N = 28 g/mol, H = 2 g/mol, NH 3 = 17 g/mol 2.How many moles are there in 2.8 g of nitrogen gas?  Number of moles = mass ÷ molar mass = 2.8 g ÷ 28 g/mol = 0.1 mol

7. IC5.2.3 Mole calculations © Oxford University Press N 2 + 3H 2  2NH 3 3.How many moles of ammonia would 2.8 g of nitrogen gas form?  1 mol of nitrogen gas forms 2 mol of ammonia, so 0.1 mol of nitrogen gas would form 0.2 mol of ammonia.

8. IC5.2.3 Mole calculations © Oxford University Press N 2 + 3H 2  2NH 3 N = 28 g/mol, H = 2 g/mol, NH 3 = 17 g/mol 4. For 0.1 mol of nitrogen gas, work out the mass of ammonia formed.  Mass = molar mass × number of moles = 17 g/mol × 0.2 mol = 3.4 g