1. 7 The Mole

2. Number of moles = mass in grams relative mass How many moles are present in 0.35g of ammonium hydroxide? RMM = 14 + 4 + 16 + 1 = 35  1 mole = 35g  No of moles = 0.35 = 0.01 35

3. Find the mass of a)1 mole of lead (II) nitrate b)4.30 moles of methane c)0.24 moles of hydrated sodium carbonate Na 2 CO 3.10H 2 O

4. a)lead(II) nitrate is Pb(NO 3 ) 2  1 mole = 207 + (2x14) + (6x16) = 331g b) Methane is CH 4  1 mole = 12 + (1x4) = 16g  4.30 moles = 16 x 4.30 = 68.8g c) 1 mole of Na 2 CO 3.10H 2 O = (2x23) + 12 + (3x16) + (10x18) = 286g  0.24 moles = 0.24x286 = 68.6g

5. Avogadros’s no = 6.022 x 10 23 It is the number of atoms in 12g of 12 C The RAM of an element contains Avogadros number of atoms i.e if we weigh out 12g of carbon-12 we will have 6.022 x 10 23 atoms of carbon-12 and 12g of carbon-12 = 1 mole

6. Similarly the RMM of a substance contains Avogadro’s number of molecules  A mole of hydrogen molecules (H 2 ) = 2g and contains 6.022 x 10 23 molecules of hydrogen and 2 x 6.022 x 10 23 atoms of hydrogen The abbreviation for moles is mol

7. A mole of substance is the amount of that substance that contains the same number of stated elementary units as there are atoms in 12g of 12 C Stated elementary units can mean atoms, molecules ions, electrons

8. For example 16g (1mole) of oxygen atoms O 32g (1mole) of oxygen molecules O 2 18g (1mole) of water molecules H 2 O 24g of magnesium ions Mg 2+

9. Which of the following contains the greatest number of the stated particles? a)Molecules of hydrogen in 1g of hydrogen gas b)Atoms of helium in 1g of helium gas c)Atoms of beryllium in 1g of beryllium

10. 1g of hydrogen gas = 1 mol = 0.5mol 2  1g hydrogen gas contains 0.5x 6.022x10 23 molecules = 3.011x 10 23 molecules 1g of helium gas contains 1 mol = 0.25 mol 4  1g helium gas contains 0.25x 6.022x1023atoms = 1.51x10 23 atoms 1g beryllium contains 1 mol = 0.11mol 9  1g beryllium contains 0.11x 6.022x10 23 atoms = 6.69x 10 22 atoms  1g hydrogen gas contains the greatest number of the stated particles

11. How many atoms are there in 1 mol of carbon dioxide? 1 molecule of CO 2 contains 3 atoms  1 mol contains 6.022x10 23 x 3 atoms =1.81x 10 24 atoms

12. How many hydrogen ions will 0.5 mols of sulphuric acid release on dissociation? H 2 SO 4  2H + + SO 4 2-  1mol of sulphuric acid releases 2 mols of hydrogen ions  0.5 mols releases 0.5x2mols hydrogen ions = 1mol hydrogen ions

13. Concentrations of solutions are measured in moles per L (or dm 3 ) 1000ml = 1 L 1000cm 3 = 1 dm 3 How many moles are there in 20ml of a solution of concentration 0.5mol/L? 1000ml contains 0.5mol  1ml contains 0.5 mol 1000 And 20ml contains 0.5 x 20 mol = 0.01mol 1000

14. What mass of solute must be used to prepare 750ml of 0.100M aqueous sodium carbonate from solid sodium carbonate? RMM Na 2 CO 3 = 23x2 + 12 + (16x3) = 106  106g = 1mol  1000ml of 0.100M solution would need 106 x 0.100 g = 10.6g  750ml needs 10.6 x 750 g = 7.95g 1000

15. Percentage Composition From the formula of a compound and the relative atomic masses of the elements in it, the percentage of each element in it can be calculated e.g. Calculate the percentage composition of magnesium oxide MgO RAM Mg = 24 RAM O = 16 RMM MgO = 24 + 16 = 40 %Mg = 24 x 100% = 60% 40 %O = 16 x 100% = 40% 40

16. Find the percentage of water of crystallisation in CUSO 4.5H 2 O RMM = 249.5 5H 2 O = 5x (2 + 16) = 90 %water = 90 x 100% = 36.1% 249.5 Find the percentage oxygen in the CUSO 4.5H 2 O RAM O = 16  total RM O = 16 x 9 = 144  %O = 144 x 100% = 57.7% 249.5

17. What is the percentage compostion by mass of the elements Cu, S and in CuSO 4.5H 2 O? %Cu = 63.5 x 100 = 25.45% 249.5 %S = 32 x 100 = 12.82% 249.5

18. Calculations based on Chemical Formulas Equations tell us what amounts of substances react together. e.g. the equation 2NaHCO 3(s)  Na 2 CO 3 + CO 2(g) + H 2 O (g ) tells us that 2 moles of NaHCO 3 give 1 mole of Na 2 CO 3 2 moles of NaHCO 3 = 2 x 84 g = 168g 1 mole Na 2 CO 3 = 106g  168g of NaHCO 3 will give 106g of Na 2 CO 3

19. The amounts of substances undergoing reaction, as given by the balanced equation, are called the stoichiometric amounts Stoichiometry is the relationship between the amounts of reactants and products in a chemical reaction If one reactant is present in excess of the stoichiometric amount required for the reaction with another of the reactants, then the excess of one reactant will be left unused at the end of the reaction.

20. On heating potassium chlorate decomposes into potassium chloride and oxygen. 2KClO 3  2KCl (s) + 3O 2 2 moles of potassium chlorate give 2 moles of potassium chloride. Find the mass of potassium chloride and oxygen formed when 9.8g of potassium chlorate are heated. Method 1) find the RMM of potassium chlorate 2) find the moles of chlorate heated 3) find the moles of potassium chloride formed 4) find the mass of potassium chloride formed.

21. 1)Moles KClO 3 = 9.8 = 0.08moles 122.5 2) Moles of KCl formed = moles of KClO 3 heated = 0.08 3) Mass of KCl formed = 0.08 x RMM = 0.08 x 74.5 = 5.96g Moles of oxygen formed = 3 x 0.08 = 0.12 2 Mass Oxygen formed = 0.12 x 32 = 3.84g

22. How many moles of iodine can be obtained from 1/6 mole of potassium iodate? KIO 3(aq) + 5KI (aq) + 6H +  3I 2(aq) + 6K + (aq) + 3H 2 O (l) The equation tells us that 1 mole of KIO 3 gives 3 moles of I 2.  1/6 mole of KIO 3 gives 1/6 x 3 moles of I 2 = ½ mole I 2

23. What mass of sodium carbonate can be obtained by heating 100g of sodium hydrogen carbonate? On heating the mixture the following reaction occurs 2NaHCO 3(s)  Na 2 CO 3(s) + CO 2(g) + H 2 O (g). RMM NaHCO 3 = 84  100g = 100 = 1.190mols 84 1.190 mols NaHCO 3 will give 1.190/2 mols Na 2 CO 3 = 0.590mols RMM Na 2 CO 3 = 106  106 x 0.600g = 63.1g

24. What is the maximum mass of ethyl ethanoate that can be obtained from 0.1mol of ethanol according to the following equation? C 2 H 5 OH (l) + CH 3 CO 2 H (l)  CH 3 CO 2 C 2 H 5(l) + H 2 O ethanol ethanoic acid ethyl ethanoate From the equation 1 mol of ethanol gives 1 mole ethyl ethanoate  0.1mol ethanol give 0.1mol ethyl ethanoate RMM ethyl ethanoate = 88  0.1mol = 88 x 0.1g = 8.8g  8.8g ethyl ethanoate can be produced from 0.1mol ethanol