1. p.13 Q1 Write balanced equations including state symbols for the following: (a)zinc + oxygen → zinc oxide 2 Zn (s) + O 2 (g) → 2 ZnO (s) (b)potassium + water → potassium hydroxide + hydrogen 2 K (s) + 2 H 2 O (l) → 2 KOH (aq) + H 2 (g) (c)calcium carbonate → calcium oxide + carbon dioxide CaCO 3 (s) → CaO (s) + CO 2 (g) Chemical Equations

2. Ionic Equations These show what happens which ions actually react. –spectator ions do not change and are omitted. They just maintain electrical neutrality. Start with the balanced full formula equation. Covalent substances are written normally in the ionic equation. Ionic substances which are in solution or molten have their ions written out separately. Cancel out any ions appearing on both sides of the equation unchanged

3. Ionic Equations 2 K (s) + 2 H 2 O (l) → 2 KOH (aq) + H 2 (g) 2 K (s) + 2 H 2 O (l) → 2 K + (aq) + 2 OH - (aq) + H 2 (g) Fe 2 (SO 4 ) 3 (aq) + 3 Zn (s) → 3 ZnSO 4 (aq) + 2 Fe (s) 2 Fe 3+ (aq) + 3 SO 4 2- (aq) + 3 Zn (s) → 3 Zn 2+ (aq) + 3 SO 4 2- (aq) + 2 Fe (s) 2 Fe 3+ (aq) + 3 Zn (s) → 3 Zn 2+ (aq) + 2 Fe (s) Fe 2 (SO 4 ) 3 (aq) + 6 KOH (aq) → 2 Fe(OH) 3 (s) + 3 K 2 SO 4 (aq) 2 Fe 3+ (aq) + 3 SO 4 2- (aq) + 6 K + (aq) + 6 OH - (aq) → 2 Fe(OH) 3 (s) + 6 K + (aq) + 3 SO 4 2- (aq) Fe 3+ (aq) + 3 OH - (aq) → Fe(OH) 3 (s) 3

4. Ionic Equations for precipitation reactions Start by writing the formula of the precipitate with its (s) state symbol on the product (right hand) side. Write the 2 reactant ions needed to form this product on the left side of the equations with their (aq) state symbols.

5. p. 23 Q. 2 Al 2 O 3 (s) + 6 HI (g) → 2 AlI 3 (aq) + 3 H 2 O (l) What masses of aluminium oxide and hydrogen iodide would be required to produce 102 g aluminium iodide? R.F.M. of AlI 3 = 27 + (3 × 126.9) = 407.7 Moles of AlI 3 = 102 / 407.7 = 0.250 Moles Al 2 O 3 = 0.250 / 2 = 0.125 R.F.M. of Al 2 O 3 = (2 × 27) + (3 × 16) =102 Mass of Al 2 O 3 = 0.125 × 102= 12.8 g Moles of HI = 0.250 × 3 = 0.750 R.F.M of HI = 127.9 Mass of HI = 0.750 × 127.9 = 95.9 g

6. p. 23 Q.3 The substance ATP is important for transporting energy in living cells. a sample of 1.6270 g of ATP was analysed and found to contain 0.3853 g of carbon, 0.05178 g of hydrogen, 0.2247 g nitrogen and 0.2981 g of phosphorus. The rest was oxygen. The relative molecular mass of ATP is 507. What are its empirical and molecular formulae? Oxygen mass= 1.6270 - 0.3853 – 0.05178 – 0.2247 – 0.2981 = 0.6671 g C : H : N : P : O = (0.3853/12):(0.05178/1):(0.2247/14):(0.2981/31):(0.6671/16) = 0.03211 : 0.05178 : 0.01605 : 0.009616 : 0.04169 0.009616 0.009616 0.009616 0.009616 0.009616 = 3.339 : 5.385 : 1.669 : 1.000 : 4.335 =10 : 16 : 5 : 3 : 13 empirical formula = C 10 H 16 N 5 P 3 O 13 empirical formula mass = 507 molecular formula mass = 1 × empirical formula mass so molecular formula = C 10 H 16 N 5 P 3 O 13

7. p.27 Q. 1 A stock solution contains 10 mol dm -3 of hydrochloric acid. What volume of this solution should be dissolved in water to make 6 dm 3 of acid of concentration 0.1 mol dm -3 ? Moles of HCl = 6 dm 3 × 0.1 mol dm -3 = 0.6 Volume = 0.6 mole / 10 mol dm -3 = 0.06 dm 3 = 60 cm 3.

8. p.27 Q. 2 When solutions of potassium chloride and silver nitrate are mixed together, a precipitate of silver chloride forms. (a) Write a balanced molecular equation for this reaction. KCl (aq) + AgNO 3 (aq) → AgCl (s) + KCl (aq) (b) In an experiment, 24.0 cm 3 of 0.05 M silver nitrate solution exactly reacts with 15.0 cm 3 of potassium chloride solution. Calculate the molar concentration of the potassium chloride. Moles of AgNO 3 = 24.0/1000 dm 3 × 0.05 mol dm -3 = 0.0012 1:1 AgNO 3 :KCl so 0.0012 mol KCl Concentration = 0.0012 mol / (15/1000) dm 3 = 0.0800 mol dm -3 (c) How much KCl would be needed to produce 100 cm 3 of this concentration? Moles = 100/1000 dm 3 × 0.0800 mol dm -3 = 0.008 Mass = 0.008 mol × 74.5 g mol -1 = 0.596 g

9. p.31 Question 1 A chemist mixed 12.0 g of phosphorus with 35.5 g of chlorine gas to synthesise phophorus (III) chloride. The yield was 42.4 g of PCl 3. the equation for the reaction is: 2 P (s) + 3 Cl 2 (g) → 2 PCl 3 (l) Moles of P = 12 / 31 = 0.3871;Moles of Cl 2 = 35.5 / 71 =0.5 Moles of PCl 3 = 42.4 / 137.5 = 0.3084. This amount of PCl 3 required 3/2 × 0.3084 moles Cl 2 which is 0.463 moles, and 0.3084 moles P. Therefore P was used in excess and Cl 2 was limiting. Percentage yield in terms of Cl 2 = (0.463 / 0.5)*100 = 92.6%

10. p.31 Q2 2-bromo-2-methylpropane, C 4 H 9 Br can be used to produce methylpropene, C 4 H 8. 2-bromo-2-methyl propane reacts with sodium ethoxide, NaOC 2 H 5 to produce ethanol C 2 H 5 OH and sodium bromide, along with methylpropene. Work out the atom economy for the reaction. Ratio C 4 H 9 Br : C 2 H 5 OH : C 4 H 8 = 1 : 1 : 1 R.F.M. 137 46 56 Atom economy = (56/[137 + 46]) × 100 = 30.6%

11. Volume of 1.0M KI (aq) (cm 3 ) Moles of KI Volume of 1.0M Pb(NO 3 ) 2 (aq) (cm 3 ) Moles of Pb(NO 3 ) 2 Precipitate depth (mm) 50.0050.50.00055 50.00510.00106 50.0051.50.00158 50.00520.002010 50.0052.50.002512 50.00530.003012 50.0053.50.003512

12. p.29 Q1 0.025 mol Pb(NO3)2 to 0.50 mol KI. 1:2 ratio Pb(NO 3 ) 2 (aq) + 2 KI (aq) → PbI 2 (aq) + 2 KNO 3 (aq)

13. Isotopes An isotope is a sample of an element in which the atoms have the same number of neutrons as well as the same number of neutrons. Isotopes are samples of the same element in which the atoms have different numbers of neutrons whilst having the same number of protons.

14. Relative atomic mass R.A.M. = (weighted average mass of an atom) / ( 1 / 12 mass of an atom of 12 C) Relative isotopic mass, relative formula mass and relative molecular mass are defined on the same scale. –Relative isotopic mass is not necessarily an integer and identical to mass number R.A.M., R.F.M. and R.M.M. have no units –ratios of masses Molar mass has the same numerical value as R.F.M. but has units g mole -1 (‘the mass of 1 mole’)

15. Mass spectrometer

16. Gaseous sample needed (vaporisation by heating or laser dye absorption) Vacuum so ions move freely without colliding Electron gun bombards sample molecule with high energy electrons causing ionisation, e.g. CH 4 (g) + e - → CH 4 + (g) + 2 e - Negative plates accelerate sample ions and select for uniform velocity Variable electromagnet deflects the ions Detector, connected to computer, generates a graph of ion abundance against mass / charge (a function of electromagnetic field strength) Molecular ions can fragment by random breaking of any covalent bond within the molecule. Every species detected in the mass spectrum is a positive ion.

17. Calculating relative atomic mass Bromine (50.5% 79 Br, 49.5% 81 Br) RAM = 80.0 Silver (51.3% 107 Ag, 48.7% 109 Ag RAM = {(51.3 × 107) + (48.7 × 109)} / 100 = 108.0 Chromium ( 4.3% 50 Cr, 83.8% 52 Cr, 9.6% 53 Cr, 2.3% 54 Cr) RAM = {(4.3 × 50) + (83.8 × 52) + (9.6 × 53) + (2.3 × 54)} / 100 = 52.1 Tips: SUS, check answer is plausible.

18. p. 55 Q2 IsotopeRelative isotopic massPercentage abundance % 78 Kr77.920.12 80 Kr79.922.0 82 Kr81.9112 83 Kr82.9112 84 Kr83.9157 86 Kr85.9117 Notice that relative isotopic mass is not identical to mass number (due to nuclear binding energy). R.A.M. = {(0.12 × 77.92) + (2.0 × 79.92) + (12 × 81.91) + (12 × 82.91) + (57 × 83.91) + (17 × 85.91)} / 100 = 83.90 (SUS)

19. Some other uses of mass spectrometry Determining 12 C / 13 C in plant derived material –hence photosynthetic pathway and type of plant Dating rocks 14 C dating (age of materials of biological origin) Isotopic composition of hormones in athletes (is the molecule made ‘recently’ from material of photosynthetic origin and is therefore ‘natural’ or was it made from petroleum derived material by an organic chemist? Analysis of atmospheres on other planets by spacecraft.