1. This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Structure of matter seminar: moles and molarity University of Lincoln presentation

2. This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License The mole Definition: 1 mole of any substance contains 6.022x10 23 atoms and/or molecules Amadeo Carlo Avogadro (1776-1856) 6.022 x 10 23

3. This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License For Elements: The relative atomic mass, A r = mass of 1 mole of atoms in grams (g)

4. This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License For Molecules: The relative molecular mass, M r = sum of all A r in the molecule = mass of 1 mole of molecules (g)

5. This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Molar Concentrations, M A molar solution (1M) is a solution containing 1 mole of substance (solute) in every litre of solvent

6. This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Molarity The molarity of a solution is the concentration of the solution expressed as: the number of moles per litre – M or mol L -1 or mol dm -3 (all of these are the same)

7. This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Molarity How to calculate the molarity: 1.Calculate how many moles there are in solution 2.Work out how many moles there are per ml 3.X 1000 to give the number of moles per litre

8. This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Example 1 22 g of CaCO 3 was dissolved in water and made up to give a total volume of 200 cm 3. C Calculate the concentration of the solution in mol dm -3.

9. This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Example 1 1. No mols of CaCO 3 in solution = 22g/100 = 0.22 mols 2. No mols/ml = 0.22/200 = 0.0011 (= 1.1 x 10 -3 ) 3. 1.1 x 10 -3 x 1000 = 1.1M

10. This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Example 2 If we take 15ml of our 1.1M CaCO 3 solution and make it up to 250ml with H 2 O, what is the concentration of the new dilution?

11. This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Example 2 1.N o. mols in 15 ml of 1.1M solution = = 1.1 x 15 = 0.0165 mols 1000 2. This is put into 250 ml. N o. mols per ml = 0.0165/250 = 6.6 x 10 -5 3.6.6 x 10 -5 x 1000 = 0.066M

12. This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Alternatively You can use the following formula: V 1 C 1 = V 2 C 2 15 ml x 1.1 M = 250 ml x C 2 C2 = 15 ml x 1.1 M = 0.066 M 250 ml

13. This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Acknowledgements JISC HEA Centre for Educational Research and Development School of natural and applied sciences School of Journalism SirenFM http://tango.freedesktop.org