1. Maths and Chemistry for Biologists

2. Chemistry 2 Solutions, Moles and Concentrations This section of the course covers – why some molecules dissolve in water to form solutions how we measure the amounts of atoms and molecules - the mole The relative molar mass of molecules how the concentrations of solutions are described and how solutions of a given concentration are made

3. The bonds in some molecules are polarised C – HN – HO – HCl – H increasing electronegativity of the heavy atom As electronegativity increases the pair of electrons in the bond is displaced towards the heavy atom and hence the bond is polarised Electrons spend a larger proportion of their time near the electronegative atom. Shown by e.g.  - O – H  + The  indicates a fraction of a charge

4. Water is a bent, polar molecule The oxygen in a water molecule has two lone pairs of electrons which take up space. The two bonds and two lone pairs point to the corners of a tetrahedron – hence molecule is bent. The bonds are polar so the molecule can be drawn as In water the value of  is about 0.3

5. Consequences of the structure of water (1) It forms hydrogen bonds – electrostatic interaction between  + on H and  - on O of adjacent molecules O – H  +  - O – H hydrogen bond H-bonds are weak (about 1/20 th strength of covalent O – H bond) and long (about 0.176 nm compared with O – H bond) but are enormously important for the properties of water. H-bonds also occur in compounds containing N – H

6. Consequences of the structure of water (2) Water dissolves ionic compounds such as NaCl because it interacts electrostatically with the ions Water dissolves polar molecules such as ethanol because it interacts electrostatically with the partial charges

7. Consequences of the structure of water (3) In liquid water molecules stick together to give a partially ordered structure

8. The need to “weigh” atoms and molecules Consider the reaction Na + F  NaF One meaning of this is that 1 atom of sodium reacts with 1 atom of fluorine to give 1 molecule of NaF Can’t weigh atoms so how can we know how much sodium to take to react exactly with a given amount of fluorine?

9. Weighing atoms continued Na has a mass number of 23 (it contains 11 protons and 12 neutrons) so its relative atomic mass is 23 F has a mass number of 19 (9 protons and 10 neutrons) so its relative atomic mass is 19 So 23 g of sodium will react exactly with 19 g of fluorine to give 42 g of sodium fluoride

10. The mole 23 g of Na and 19 g of F contain the same numbers of atoms 23 g of Na is a mole of sodium 19 g of F is a mole of fluorine Definition:- a mole is the amount of a substance that contains the same number of particles as there are atom in 12 g of the isotope 12 C This is the Avogadro constant equal to 6.022 x 10 23

11. The mole continued Na + F  NaF Also means that 1 mol of Na reacts with 1 mol of F to give 1 mol of NaF (mol is the abbreviation of mole) What the mol gives us the weights of atoms that will exactly react together to give molecules

12. One for you to do Relative atomic mass of fluorine = 19 What mass of F is 0.1 mol? How many mol of F is 38 g?

13. Answers 1 mol = 19 g  0.1 mol = 1.9 g 19 g = 1 mol  38 g = 2 mol

14. Relative molar mass (M r ) of compounds Calculate these by adding the relative atomic masses (A r ) of the constituent atoms H 2 O: 1 x 2 + 16 = 18 Urea (NH 2 CONH 2 ): (2 x 14) + (4 x 1) + 12 + 16 = 60 Hence 18 g of water is 1 mol 1.8 g of water is 0.1 mol 18  g of water is 18  mol 60 g of urea is 1 mol

15. Molecular weight You will see this term used sometimes. It is numerically equal to the relative molar mass but it needs units. The correct unit is the Dalton (Da) which is 1/12 th of the mass of an atom of 12 C Better to stick to relative molar mass

16. One for you to do C + O 2  CO 2 (means that 1 mol of carbon reacts with 1 mol of oxygen molecules to give 1 mol of carbon dioxide) What weight of O 2 is required to react completely with 1.2 g of C?

17. Answer 1.2 g of C is 0.1 mol. Hence we require 0.1 mol of O 2 M r of 0 2 is 2 x 16 = 32 Hence we need 3.2 g of O 2 Product is 4.4 g (0.1 mol) of CO 2

18. Concentrations of solutions Expressed in terms of MOLARITY – the number of moles per litre A 1 MOLAR solution contains 1 MOLE in 1 LITRE or a 1 M solution contains 1 mol/L NB the abbreviations for mole (mol) and molar (M) – NEVER confuse them. One is an amount of substance (mol) the other is a concentration (M)

19. Concentrations of solutions (contd) A 1 MOLAR solution contains 1 MOLE in 1 LITRE or 1 mol in 1 L gives a 1 M solution 1 mol/L solution is 1M 0.1 mol/L solution is 0.1 M 0.1 mol/100 ml solution is 1 M 1 mol/100ml solution is 10 M 1 mmol/ml solution is 1M

20. Examples of how to make solutions How much NaOH is needed to make 500 ml of a 0.2 M solution? M r of NaOH is 23 + 16 + 1 =40. 1L of a 0.2 M solution contains 0.2 mol. Hence 500 ml contains 0.1 mol. 0.1 mol of NaOH = 4 g How much KCl is there in 100 ml of a 2 M soln? M r of KCl is 39 + 35.5 = 74.5. 1L of 2M soln contains 2 mol hence 100 mol contains 0.2 mol. 0.2 mol of KCl is 0.2 x 74.5 = 14.9 g

21. More examples What is the conc of a solution of NaH 2 PO 4 containing 2g/L? M r is 23 + 2 + 31 + 64 = 120. Hence 2 g is 2/120 of a mol or 0.0167 mol. This is contained in 1 L so conc = 0.0167 M (or 1.67 x 10 -2 M or 16.7 mM) How many mol is 1 mg of a substance with M r = 250? 1 mg = 10 -3 g. Hence 10 -3 /250 = 4 x 10 -6 mol or 4  mol

22. One for you to do What is the concentration of a solution of NaOH containing 20 g in 500 ml? (M r = 40) How much NaOH is required to make 10 ml of a 1M solution?

23. Answer Solution has 20 g in 500 ml and hence 40 g in 1 L. 40 g is 1 mol and hence conc = 1 M For 1 M need 40 g / L 10 ml = 0.01 L. Hence need 0.01 x 40 g = 0.4 g (or 400 mg)

24. Dilutions If you dilute a solution (add more solvent) the amount of solute remains the same but the concentration decreases Take 250 ml of a 0.1 M solution and dilute to 1L New conc = 0.1 x = 0.025 M Amount of solute remains as 0.025 mol

25. Example How would you make 50 ml of a 10 mM solution by dilution of a 1M stock solution? New solution has conc 1/100 that of stock. So need 50/100 = 0.5 ml of stock solution plus 49.5 ml water or 1 L of 10mM solution contains 10 mmol of solute so 50 ml (1/20 L) contains 0.5 mmol. Stock has 1 mol/L or 1 mmol/ml. So need 0.5 ml of stock plus 49.5 ml water as above.