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Structure of matter seminar: moles and molarity

Structure of matter seminar: moles and molarity. University of Lincoln presentation. 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

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Structure of matter seminar: moles and molarity

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

  2. The mole Definition: 1 mole of any substance contains 6.022x1023atoms and/or molecules Amadeo Carlo Avogadro (1776-1856) 6.022 x 1023 This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License

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

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

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

  6. 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) This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License

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

  8. Example 1 22 g of CaCO3 was dissolved in water and made up to give a total volume of 200 cm3. C Calculate the concentration of the solution in mol dm-3. This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License

  9. Example 1 1. No mols of CaCO3 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 This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License

  10. Example 2 If we take 15ml of our 1.1M CaCO3 solution and make it up to 250ml with H2O, what is the concentration of the new dilution? This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License

  11. Example 2 • No. mols in 15 ml of 1.1M solution = = 1.1 x 15 = 0.0165 mols 1000 • This is put into 250 ml. No. mols per ml = 0.0165/250 = 6.6 x 10-5 • 6.6 x 10-5 x 1000 = 0.066M This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License

  12. Alternatively You can use the following formula: V1 C1 = V2 C2 15 ml x 1.1 M = 250 ml x C2 C2 = 15 ml x 1.1 M = 0.066 M 250 ml This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License

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

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