1. Lots of definitions to learn Isotopes Isotopes The Mole The Mole Avogadro number Avogadro number Relative Atomic/Molecular Mass Relative Atomic/Molecular Mass Molecular Formula Molecular Formula Empirical Formula Empirical Formula ALL MET AT GCSE

2. Isotopes Atoms (of the same element) that have the same number of protons, but a different number of neutrons. SAME ATOMIC NUMBER, DIFFERENT MASS NUMBER E.g. Carbon-12 and Carbon-14 6,6,6 and 6,6,8

3. The Mole The AMOUNT OF A SUBSTANCE in grams that has the SAME NUMBER OF PARTICLES as there are ATOMS in EXACTLY 12 grams of CARBON-12. Avogadro's number The number of atoms in exactly 12g of Carbon-12 = 6.023 X 10 23

4. Relative atomic/molecular Mass The AVERAGE mass of ONE MOLE of atoms RELATIVE to the mass of ONE MOLE of Carbon- 12 (Which has a mass of 12g) The AVERAGE mass of ONE MOLE of atoms RELATIVE to the mass of ONE MOLE of Carbon- 12 (Which has a mass of 12g) The AVERAGE mass of ONE MOLE of molecules RELATIVE to the mass of ONE MOLE of Carbon- 12 (Which has a mass of 12g) The AVERAGE mass of ONE MOLE of molecules RELATIVE to the mass of ONE MOLE of Carbon- 12 (Which has a mass of 12g) IONIC COMPOUNDS - Can they have RMM?

5. Relative Formula mass The AVERAGE mass of ONE MOLE of the formula RELATIVE to the mass of ONE MOLE of Carbon-12 (Which has a mass of 12g) The AVERAGE mass of ONE MOLE of the formula RELATIVE to the mass of ONE MOLE of Carbon-12 (Which has a mass of 12g) Average mass of one mole of formula/ (1/12(mass of 1 mole C- 12)) Average mass of one mole of formula/ (1/12(mass of 1 mole C- 12))

6. Picture of Machine B A C D E F To computer

8. Determining RAM using Mass spectrometry In a vacuum At A the sample is injected At B the sample is heated to convert it to a gas At C the atoms are ionized by electron bombardment The heated cathode releases electrons which collide with the atoms When the energy of the electrons equals the 1 st IE of the atom X + e -  X + + e - + e - takes place At D the ions are accelerated in an electric field At E the ions path is deflected in a magnetic field At F the ions are identified

9. Data Obtained The percentage composition of each isotope in the sample of the element RAM = ((((%composition x mass)) / 100

10. 90 10 565758 IRON DATA AS GRAPH 90% Fe = 56 8% Fe = 57 2% Fe = 58 RAM Fe =( (90*56)+(8*57)+(2*58))/100 RAM Fe = 5612/100 = 56.12 (NO UNITS)

11. Data obtained The isotopic abundances The isotopic abundances 50% of Bromine = 79 Br 50% of Bromine = 79 Br 50% of Bromine = 81 Br 50% of Bromine = 81 Br 75% of Chlorine = 35 Cl 75% of Chlorine = 35 Cl 25% of Chlorine = 37 Cl 25% of Chlorine = 37 Cl 92.58% of Lithium = 7 Li 92.58% of Lithium = 7 Li 7.40% of Lithium = 6 Li 7.40% of Lithium = 6 Li 0.02% of Lithium = 8 Li 0.02% of Lithium = 8 Li RAM 79.5 35.5 6.93

12. Calculation From RAM From RAM H 2 O H 2 O NH 3 NH 3 C 2 H 4 C 2 H 4 C 6 H 6 C 6 H 6 C 3 H 7 OH C 3 H 7 OH 1+1+ 16 = 18 14+ 1+1+1 =17 12+12+1+1+1+1=28 12+12+12+12+12+12+1+1+1+1+1+1=78 12+12+12+1+1+1+1+1+1+1+1+16=60

13. Formula determination Empirical formula The simplest whole number ratio of elements present in the compound Molecular formula The actual whole number ratio of elements present in the compound BOTH ARE DETERMINED BY EXPERIMENTAL DATA

14. FROM ELEMENTAL ANALYSIS A compound contains A compound contains a. 75% C and 25% H b. 50% O, 37.5% C and 12.5% H c. 85.7% C, 14.3% H d. 71.6% C, 23.9% O, 4.5%H e. 39.3% Na, 60.7% Cl

15. To calculate formula 1 Make a column representing each element in compound Make a column representing each element in compound C H C H Write down the % comp under each element Write down the % comp under each element C H C H 7525 7525 Write down the RAM under each element Write down the RAM under each element C H C H 7525 7525 121 121 Divide the %comp by the RAM (to compensate for the different RAM) Divide the %comp by the RAM (to compensate for the different RAM) C H C H 7525 7525 121 121 6.2525 6.2525

16. To calculate formula 1 Divide the %comp by the RAM (to compensate for the different RAM) Divide the %comp by the RAM (to compensate for the different RAM) C H C H 7525 7525 121 121 6.2525 6.2525 Ratio by dividing by the smallest of the figures (eg by 6.25 above) Ratio by dividing by the smallest of the figures (eg by 6.25 above) C H C H 7525 7525 121 121 6.2525 6.2525 1(6.25/6.25)4(25/6.25) 1(6.25/6.25)4(25/6.25) Empirical formulae = This ratio eg CH 4 Empirical formulae = This ratio eg CH 4

17. To calculate formula if more than 2 elements in compound Make a column representing each element in compound Make a column representing each element in compound C HO C HO Write down the % comp under each element Write down the % comp under each element C HO C HO 37.512.550 37.512.550 Write down the RAM under each element Write down the RAM under each element C HO C HO 37.512.550 37.512.550 12116 12116 Divide the %comp by the RAM (to compensate for the different RAM) Divide the %comp by the RAM (to compensate for the different RAM) C HO C HO 37.512.550 37.512.550 12116 12116 3.12512.53.125 3.12512.53.125

18. To calculate formula 1 Divide the %comp by the RAM (to compensate for the different RAM) Divide the %comp by the RAM (to compensate for the different RAM) C HO C HO 37.512.550 37.512.550 12116 12116 3.12512.53.125 3.12512.53.125 Ratio by dividing by the smallest of the figures (eg by 3.125 above) Ratio by dividing by the smallest of the figures (eg by 3.125 above) C HO C HO 37.512.550 37.512.550 12116 12116 3.12512.53.125 3.12512.53.125 1(3.125/3.125)4(12.5/3.125)1(3.125/3.125) 1(3.125/3.125)4(12.5/3.125)1(3.125/3.125) Empirical formulae = CH 4 O Empirical formulae = CH 4 O

19. Answers to other problems Answers to other problems 85.7% C, 14.3% H Empirical formula = CH 2 71.6% C, 23.9% O, 4.5%H Empirical formula = C 4 H 3 O 39.3% Na, 60.7% Cl Empirical formula = NaCl NB Ratio must be whole number so double if ratio is half Eg 1.5 : 1 becomes 3 : 2

20. Molecular formula from Empirical formula Work out the Mass of the empirical formula Work out the Mass of the empirical formula Look up the RMM Look up the RMM Divide RMM / Mass of empirical formula Divide RMM / Mass of empirical formula Multiply each element ratio by the number formed above Multiply each element ratio by the number formed above Eg Empirical formula = CH 2 Eg Empirical formula = CH 2 Empirical mass = 14 (12+2) Empirical mass = 14 (12+2) If RMM = 28 If RMM = 28 RMM/Empirical mass = 28/14 =2 RMM/Empirical mass = 28/14 =2 Molecular formula = C 2 H 4 Molecular formula = C 2 H 4

21. Working out formulae from experimental data A Formula to remember No of Moles =Mass present/Mass of 1 mole (RAM)

22. Results in Air Mass of Iron turnings before heating = 1.9g Mass of iron turnings after heating = 2.38g Mass of oxygen combined with iron = Moles of Iron = Moles of Oxygen = FORMULA = ? ? ? ? 0.48g 0.034 (=1.9 /56) 0.03 ( = 0.48/16) FeO

23. Results in oxygen Mass of Iron turnings before heating = 1.9g Mass of iron turnings after heating = 2.73g Mass of oxygen combined with iron = Moles of Iron = Moles of Oxygen = FORMULA = ? ? ? ? 0.83 g 0.034 (= 1.9 / 56) 0.051 (=0.83/16) Fe 2 O 3

24. Results in oxygen Mass of Mg turnings before heating = 2.1g Mass of Mg turnings after heating = 3.5g Mass of oxygen combined with Mg = Moles of Mg = Moles of Oxygen = FORMULA = ? ? ? ? 1.4g 0.0875 (=2.1/24) 0.0875 (=1.4/16) MgO

25. Results in chlorine Mass of Al turnings before heating = 3.7g Mass of Al turnings after heating = 18.3g Mass of oxygen combined with Al = Moles of Al = Moles of Chlorine = FORMULA = ? ? ? ? 14.6 g 0.137 (= 3.7/27) 0.411 (= 0.411/35.5) AlCl 3

26. MOLE CALCULATIONS Things you must get right 1)Correct formulae 2)Balanced Equations 3)The correct mole equation 4)The numbers on your calculator

27. Ionic compounds Write down the charges on the ions Write down the charges on the ions The charges become the subscripted number for the other ion. The charges become the subscripted number for the other ion. This results in the charges balancing out. This results in the charges balancing out. E.g. Sodium Na + Oxide O -2 You need 2 Na + to one O -2 Formula = Na 2 O

28. Covalent Molecules LEARN THE FOLLOWING LEARN THE FOLLOWING MethaneHydrochloric acid Carbon dioxideSulphuric acid WaterAmmonia Sulphur dioxideNitric acid Sulphur trioxideAll elements

29. Complex ions Ammonium = Sulphate = Nitrate = Carbonate = Hydroxide = Phosphate = Ethanoate = Iron (II) = Iron (III) = NH 4 + SO 4 2- NO 3 - CO 3 2- OH - PO 4 3- CH 3 COO - Fe 2+ Fe 3+

30. Sodium Carbonate = Sodium Carbonate = Sodium Oxide = Sodium Oxide = Lithium Sulphate = Lithium Sulphate = Calcium Nitrate = Calcium Nitrate = Aluminium Sulphate = Aluminium Sulphate = Potassium Phosphate = Potassium Phosphate = Ammonium Sulphate = Ammonium Sulphate = Sodium Ethanoate = Sodium Ethanoate = Iron (II) Hydroxide = Iron (II) Hydroxide = Iron (III) Chloride = Iron (III) Chloride = Na 2 CO 3 Na 2 O Li 2 SO 4 Ca(NO 3 ) 2 Al 2 (SO 4 ) 3 K 3 PO 4 (NH 4 ) 2 SO 4 CH 3 COONa Fe(OH) 2 FeCl 3

31. Putting formulae together to get an equation A balanced equation will have Same number of atoms of each element on both sides Same number of atoms of each element on both sides Eg Mg + H 2 O  Mg(OH) 2 + H 2 isn’t balanced as there are 2 O on one side and 1 O on the other Mg + 2H 2 O  Mg(OH) 2 + H 2 Same charge on both sides Same charge on both sides Zn +2 + Na  Na +1 + Zn isn’t balanced as there are +2 charge on one side and +1 charge on the other Zn +2 + 2Na  2Na +1 + Zn

32. BALANCED EQUATIONS Cu +2 + K  Cu + K + Cu +2 + K  Cu + K + CH 4 + O 2  CO 2 + H 2 O CH 4 + O 2  CO 2 + H 2 O Ammonia + sulphuric acid  ammonium sulphate Ammonia + sulphuric acid  ammonium sulphate Calcium hydrogen carbonate + Nitric acid  Calcium nitrate + carbon dioxide + water Calcium hydrogen carbonate + Nitric acid  Calcium nitrate + carbon dioxide + water Calcium ions + Aluminium  Aluminium ions and calcium Calcium ions + Aluminium  Aluminium ions and calcium Iron(II) chloride + Chlorine  Iron (III) chloride Iron(II) chloride + Chlorine  Iron (III) chloride Copper (II) sulphate + Sodium hydroxide  Copper (II) hydroxide + sodium chloride Copper (II) sulphate + Sodium hydroxide  Copper (II) hydroxide + sodium chloride Fe 2 O 3 + Al  Al 2 O 3 + Fe Fe 2 O 3 + Al  Al 2 O 3 + Fe ANSWERS Cu 2+ + 2K  Cu + 2K + CH 4 + 2O 2  CO 2 + 2H 2 O 2NH 3 + H 2 SO 4  (NH 4 ) 2 SO 4 Ca(HCO 3 ) 2 + 2HNO 3  Ca(NO 3 ) 2 + CO 2 + H 2 O 3Ca 2+ + 2Al  2Al +3 + 3Ca FeCl 2 + ½ Cl 2  FeCl 3 CuSO 4 + 2NaOH  Na 2 SO 4 + Cu(OH) 2 Fe 2 O 3 + 2Al  Al 2 O 3 + 2Fe

33. WHAT DOES A BALANCED EQUATION TELL US? The reaction taking place The reaction taking place The state the reactants / products are in The state the reactants / products are in The Mole ratio of reactants / Products The Mole ratio of reactants / Products The equation should not include any spectator ions unless the full equation is specifically requested The equation should not include any spectator ions unless the full equation is specifically requested.

34. Let’s write some ionic equations HCl + NaOH  NaCl + H 2 O HCl + NaOH  NaCl + H 2 O 2 KI + Cl 2  2 KCl + I 2 2 KI + Cl 2  2 KCl + I 2 Mg + CuSO 4  MgSO 4 + Cu Mg + CuSO 4  MgSO 4 + Cu Mg + 2 HCl  MgCl 2 + H 2 Mg + 2 HCl  MgCl 2 + H 2 Na 2 CO 3 + H 2 SO 4  Na 2 SO 4 + H 2 O + CO 2 Na 2 CO 3 + H 2 SO 4  Na 2 SO 4 + H 2 O + CO 2 H 2 SO 4 + Ca(OH) 2  CaSO 4 + 2H 2 O H 2 SO 4 + Ca(OH) 2  CaSO 4 + 2H 2 O Ca + 2 HNO 3  Ca(NO 3 ) 2 + H 2 Ca + 2 HNO 3  Ca(NO 3 ) 2 + H 2 H + + OH -  H 2 O 2I - + Cl 2  2Cl - + I 2 Mg + Cu 2+  Mg 2+ + Cu Mg + 2 H +  Mg 2+ + H 2 CO 3 -2 + 2H +  CO 2 + H 2 O H + + OH -  H 2 O Ca + 2 H +  Ca 2+ + H 2

35. Mole Calculations THREE EQUATIONS TO LEARN AND USE Moles = mass / rfm Moles = mass / rfm Moles = volume/24dm 3 Moles = volume/24dm 3 Moles = (volume*concentration)/1000 Moles = (volume*concentration)/1000 In any question use a marker pen to highlight the key information given in the questions so you can work out which equation to use

36. Solving mole equations 1. Use the appropriate mole equation to calculate the number of moles of one of the substances in the equation 2. Use the balanced equation to work out the mole ratio 3. Using the mole ratio calculate the number of moles of the other substance. 4. Rearrange the appropriate mole equation to answer the question. SEEMS COMPLICATED BUT IT ISN’T WITH PRACTICE!!!!!

37. Moles = mass / rfm What mass of Calcium oxide is formed when 25g of Calcium Carbonate is decomposed? CaCO 3  CaO + CO 2 Using the equation moles = mass/rfm calculate the moles of CaCO 3 Moles of CaCO 3 = 25/100 (from 40+12+16+16+16) = 0.25 Now use the balanced equation to work out the mole ratio 1 CaCO 3  1 CaO + CO 2 ( Hence mole ratio is 1:1) Moles of CaO = Moles of CaCO 3 = 0.25 Using the Using the equation moles = mass/rfm calculate the mass of CaO 0.25 = Mass/Rfm = mass / 56 (from 40+16) Mass = 0.25 x 56 = 14 g

38. Moles = mass / rfm USE WHEN ASKED ABOUT THE MASS OF REACTANT REQUIRED. What mass of Potassium is required to form 0.94g of Potassium oxide? 2K + ½ O 2  K 2 O Use the equation Moles = mass/rfm calculate the moles of K 2 O present Rfm K 2 O = 39+39+16 = 94 Moles K 2 O = 0.94 /94 = 0.01 Use the equation to work out the mole ratio 2 K + ½ O 2  1 K 2 O Mole ratio = 1:2 hence moles of K = 2xmoles of K 2 O Use the equation Moles = mass/rfm calculate the mass of K present 0.02 = mass / 39 Mass K = 0.02 x 39 = 0.78

39. Moles = volume/24dm 3 USE WHEN ASKED ABOUT THE VOLUME OF A GAS BEING FORMED. USE WHEN ASKED ABOUT THE VOLUME OF A GAS BEING FORMED. You will almost certainly have to use it along side another mole equation. You will almost certainly have to use it along side another mole equation. What volume of carbon dioxide is formed when 25g of Calcium Carbonate is decomposed? CaCO 3  CaO + CO 2 Using the equation moles = mass/rfm calculate the moles of CaCO 3 Moles of CaCO 3 = 25/100 (from 40+12+16+16+16) = 0.25 Now use the balanced equation to work out the mole ratio 1 CaCO 3  CaO + 1 CO2 ( Hence mole ratio is 1:1) Moles of CO 2 = Moles of CaCO 3 = 0.25 Using the Using the equation moles = Volume of gas / 24dm 3 calculate the volume of CO 2 0.25 = Volume of gas / 24dm 3 Volume = 0.25 x 24= 6dm 3

40. Moles = volume/24dm 3 USE WHEN ASKED ABOUT THE VOLUME OF A GAS required. USE WHEN ASKED ABOUT THE VOLUME OF A GAS required. You will almost certainly have to use it along side another mole equation. You will almost certainly have to use it along side another mole equation. What volume of oxygen is required to form 18.8g of Potassium oxide? 2K + ½ O 2  K 2 O Moles of K 2 O = 18.8 / 94 = 0.2 Moles of O 2 = 0.1 (mole ratio = ½ O 2 : 1 K 2 O Volume of O 2 = 2.4dm 3

41. Moles = volume/24dm 3 You need to make sure that the units of the gas has been converted to dm 3 You need to make sure that the units of the gas has been converted to dm 3 1dm 3 = 1000cm 3 so 1cm 3 = 0.001dm 3 1dm 3 = 1000cm 3 so 1cm 3 = 0.001dm 3 How many moles of Cl 2 are present in 24cm 3 ? Volume = 24cm 3 = 0.024dm 3 Moles = volume /24 = 0.024/24 = 0.001 moles

42. Moles = (volume*concentration)/1000 USE WHEN GIVEN INFORMATION REGARDING THE CONCENTRATION OF A SOLUTION USE WHEN GIVEN INFORMATION REGARDING THE CONCENTRATION OF A SOLUTION Key questions What is concentration? Whatt units are concentration measured in? Concentration is defined as the number of moles (or the mass) of a substance dissolved in 1dm 3 (1000cm 3 ) of water. A concentration of NaCl of 0.5Molsdm -3 means 0.5 moles of NaCl dissolved in 1dm 3 H 2 O The units are Moles / dm 3 = moldm -3 = Molar = M OR g/dm 3 or gdm -3

43. Calculating concentrations A student added 5 g of NaOH into 250mls of water. What was the concentration of NaOH in A student added 5 g of NaOH into 250mls of water. What was the concentration of NaOH in A) g/dm 3 A) g/dm 3 B) mols/dm 3 Another student added 0.0074g of Ca(OH) 2 into 50cm 3 of water. What was the concentration of Another student added 0.0074g of Ca(OH) 2 into 50cm 3 of water. What was the concentration of A) Ca(OH) 2 in g/dm 3 A) Ca(OH) 2 in g/dm 3 B) Ca(OH) 2 in mols/dm 3 C) OH - ions in mols/dm 3 It is easy to determine the conc in g / dm 3 using simple ratios 5 grams in 250cm 3 so in 1000 cm 3 there must be 20g Hence conc = 20g/dm 3 Use the equation Moles = mass/rfm to calculate the number of NaOH dissolved in the water Rfm NaOH = 23+16+1 = 40 Hence 5g of NaOH = 5 / 40 = 0.125 moles Volume of solution = 250 Moles = 0.125 Use the equation Moles = Volume x conc /1000 to calculate the concentration of the NaOH 0.125 = 250xconc/1000 hence 0.125x1000/250 = conc Conc = 0.5Moldm -3 It is easy to determine the conc in g / dm 3 using simple ratios 0.0074 grams in 50cm 3 so in 1000 cm 3 there must be 0.148g Hence conc = 0.148g/dm 3 To convert from g/dm 3 to molsdm -3 use the equation moles = mass/rfm 0.148gdm 3 = 0.148/(40+16+16+1+1)molsdm 3 = 0.148/74molsdm -3 = 0.002moldm -3 ASIDE To convert from mols/dm 3 to g/dm 3 you use the equation mass = moles x rfm Hence 0.1Molddm -3 Ca(OH) 2 = 0.1x74gdm 3 = 7.4gdm 3 In 1 mole of Ca(OH) 2 there are 2 moles of OH - hence the conc of the OH - ions is twice that of the Ca(OH) 2 = 0.004M

44. A student took 10cm 3 of 1.08M HCl. 13cm 3 of NaOH neutralised the acid. What is the concentration of the acid? A student took 10cm 3 of 1.08M HCl. 13cm 3 of NaOH neutralised the acid. What is the concentration of the acid? A student took 10cm 3 of 0.1M Ca(OH) 2 solution and neutralised it with 0.25M HCl. What volume of HCl did he add? A student took 10cm 3 of 0.1M Ca(OH) 2 solution and neutralised it with 0.25M HCl. What volume of HCl did he add? Another student took 5cm 3 of an unknown concentration of H 2 SO 4 and diluted it to 1000cm 3. 10cm 3 of the diluted solution was exactly neutralised with 18cm 3 of 0.05M NaOH. Another student took 5cm 3 of an unknown concentration of H 2 SO 4 and diluted it to 1000cm 3. 10cm 3 of the diluted solution was exactly neutralised with 18cm 3 of 0.05M NaOH. What was the concentration of the diluted acid? What was the concentration of the diluted acid? What was the concentration of the undiluted acid? What was the concentration of the undiluted acid? Use the equation Moles = Vol x conc to determine the number of moles of HCl present Moles = 10 x1.08/1000 = 0.0108 moles Use the balanced equation to work out the mole ratio 1HCl + 1NaOH  NaCl + H 2 O (Mole ratio = 1:1) Moles NaOH = 0.0108 Rearrange the equation Moles = Vol x conc to determine conc of NaOH Moles = Vol x conc / 1000 so Moles x 1000/ vol = conc Conc = 0.0108x1000/13 = 0.831M Use the equation Moles = Vol x conc to determine the number of moles of Ca(OH) 2 present Moles = 10 x0.1/1000 = 0.001 moles Use the balanced equation to work out the mole ratio 2HCl + 1Ca(OH) 2  CaCl 2 + 2H 2 O (Mole ratio = 2:1) Moles HCl = 0.002 (twice that of the Ca(OH) 2 ) Rearrange the equation Moles = Vol x conc to determine vol of HCl Moles = Vol x conc / 1000 so Moles x 1000/ conc = vol Vol = 0.002 x 1000 / 0.25 = 8cm 3 Use the equation Moles = Vol x conc to determine the number of moles NaOH present Moles = 18 x 0.05 / 1000 = 0.0009 Use the balanced equation to work out the mole ratio 1 H 2 SO 4 + 2 NaOH  Na 2 SO 4 + 2H 2 O (Mole ratio = 1:2) Moles diluted H 2 SO 4 = 0.000045 (half of 0.0009) Rearrange the equation Moles = Vol x conc to determine conc of the diluted H 2 SO 4 Conc = 0.000045*1000/10 = 0.045Mol/dm 3 The student originally diluted the sulphuric acid from 5cm 3 to 1000cm 3 making it 200 times more dilute Hence the original undiluted acid must be 200 times more concentrated than the original Hence conc of undiluted acid = 200x0.045 = 9Mol/dm -3

45. All the equations in same question A student took 2.3g limestone and reacted it with excess 0.5M HCl. He collected 0.5dm 3 of CO 2 A student took 2.3g limestone and reacted it with excess 0.5M HCl. He collected 0.5dm 3 of CO 2 Assuming that the limestone was pure CaCO 3. What is the minimum volume of HCl needed to react with the limestone? Assuming that the limestone was pure CaCO 3. What is the minimum volume of HCl needed to react with the limestone? What is the percentage of CaCO 3 in the limestone given the results obtained by the student. What is the percentage of CaCO 3 in the limestone given the results obtained by the student. A lot of information here highlight relevant and think which mole equations you can use This might seem a complicated question, but is quite easy if the data given is used sensibly

46. You are given a mass of limestone (2.3g) (and hence CaCO 3 if pure) so the moles of CaCO 3 can be determined using moles = mass/rfm RFm CaCO 3 = 100 (40 + 12 + 16 + 16 + 16) Moles of CaCO 3 = 2.3/100 = 0.023 Balanced equation 1 CaCO 3 + 2 HCl  CaCl 2 + CO 2 + H 2 O Mole ratio = 1:2 Moles of HCl = 2 x moles CaCO 2 = 0.046 You are given a conc of HCl so you can work out the volume of HCl required by rearranging Moles = vol x conc / 1000 Vol = 0.046*1000 / 0.5 = 92 cm 3 As Limestone IS NOT pure CaCO 3 this amount of HCL was be an excess (assuming other materials in Limestone do not themselves react with HCl.

47. The student collected 0.5dm 3 of CO 2 in the experiment so the number of moles of CO 2 can be determined Moles = volume of gas/24 Moles = 0.5 / 24 = 0.020833 1 CaCO 3 + 2 HCl  CaCl 2 + 1 CO 2 + H 2 O Mole ratio = 1:1 Moles of CaCO 3 present = Moles CO 2 Moles CaCO 3 = 0.020833 Mass CaCO 3 present in the limestone = 100 x 0.020833 Mass of CaCO 3 present in the limestone = 2.0833 Mass of limestone = 2.3g % CaCO 3 in limestone = 2.0833/2.3 x 100 = 90.6%

57. 1.Limestone (CaCO 3 ) is thermally decomposed to CaO 1.Limestone (CaCO 3 ) is thermally decomposed to CaO a.Write a balanced equation (with state symbols) for this reaction