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Lots of definitions to learn

Lots of definitions to learn. Isotopes The Mole Avogadro number Relative Atomic/Molecular Mass Molecular Formula Empirical Formula ALL MET AT GCSE. Isotopes. Atoms (of the same element) that have the same number of protons, but a different number of neutrons.

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Lots of definitions to learn

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  1. Lots of definitions to learn • Isotopes • The Mole • Avogadro number • Relative Atomic/Molecular Mass • Molecular 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 1023

  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 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) • Average mass of one mole of formula/ (1/12(mass of 1 mole C-12))

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

  7. 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 1st 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

  8. Data Obtained • The percentage composition of each isotope in the sample of the element RAM = (S (%composition x mass)) / 100

  9. 90 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) 10 56 57 58

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

  11. RMM Calculation 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 From RAM (add up each individual RAM) • H2O • NH3 • C2H4 • C6H6 • C3H7OH

  12. 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

  13. FROM ELEMENTAL ANALYSIS • A compound contains • 75% C and 25% H • 50% O, 37.5% C and 12.5% H • 85.7% C, 14.3% H • 71.6% C, 23.9% O, 4.5%H • 39.3% Na, 60.7% Cl

  14. To calculate formula 1 • Make a column representing each element in compound • C H • Write down the % comp under each element • C H • 75 25 • Write down the RAM under each element • C H • 75 25 • 12 1 • Divide the %comp by the RAM (to compensate for the different RAM) • C H • 75 25 • 12 1 • 6.25 25

  15. To calculate formula 1 • Divide the %comp by the RAM (to compensate for the different RAM) • C H • 75 25 • 12 1 • 6.25 25 • Ratio by dividing by the smallest of the figures (eg by 6.25 above) • C H • 75 25 • 12 1 • 6.25 25 • 1(6.25/6.25) 4(25/6.25) • Empirical formulae = This ratio eg CH4

  16. To calculate formula if more than 2 elements in compound • Make a column representing each element in compound • C H O • Write down the % comp under each element • C H O • 37.5 12.5 50 • Write down the RAM under each element • C H O • 37.5 12.5 50 • 12 1 16 • Divide the %comp by the RAM (to compensate for the different RAM) • C H O • 37.5 12.5 50 • 12 1 16 • 3.125 12.5 3.125

  17. To calculate formula 1 • Divide the %comp by the RAM (to compensate for the different RAM) • C H O • 37.5 12.5 50 • 12 1 16 • 3.125 12.5 3.125 • Ratio by dividing by the smallest of the figures (eg by 3.125 above) • C H O • 37.5 12.5 50 • 12 1 16 • 3.125 12.5 3.125 • 1(3.125/3.125) 4(12.5/3.125) 1(3.125/3.125) • Empirical formulae = CH4O

  18. Answers to other problems 85.7% C, 14.3% H Empirical formula = CH2 71.6% C, 23.9% O, 4.5%H Empirical formula = C4H3O 39.3% Na, 60.7% Cl Empirical formula = NaCl NB Ratio must be whole number so double if ratio is half Eg 82.76 % C and 17.24% H 6.90 and 17.24 1 C to 2.5 H Hence Empirical formula = C2H5

  19. Molecular formula from Empirical formula • Work out the Mass of the empirical formula • Look up the RMM • Divide RMM / Mass of empirical formula • Multiply each element ratio by the number formed above • Eg Empirical formula = CH2 • Empirical mass = 14 (12+2) • If RMM = 28 • RMM/Empirical mass = 28/14 =2 • Molecular formula = C2H4

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

  21. 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 ?

  22. 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) ? Fe2O3 ?

  23. 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 ?

  24. Results in chlorine for Al • 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) ? AlCl3 ?

  25. Empirical Formula to Molecular Formula • Mass of Al turnings before heating = 3.7g • Mass of Al turnings after heating = 18.3g • Mass of chlorine combined with Al = • Moles of Al = • Moles of Chlorine = • FORMULA = • If RFM = 267 Molecular formula of Aluminium Chloride MUST be Al2Cl6 • As Empirical mass (mass of AlCl3) =133.5 (27+ 3*35.5) • Hence RFM/EM = 2 so there are twice the number of Al and Cl in real compound as there are in the Empirical formula ? 14.6 g ? 0.137 (= 3.7/27) ? 0.411 (= 0.411/35.5) AlCl3 ?

  26. MOLE CALCULATIONS • Things you must get right • Correct formulae • Balanced Equations • The correct mole equation • The numbers on your calculator

  27. Ionic compounds • Write down the charges on the ions • The charges become the subscripted number for the other ion. • This results in the charges balancing out. E.g. Sodium Na+ Oxide O-2 You need 2 Na+ to one O-2 Formula = Na2O

  28. Complex ions NH4+ SO42- NO3- CO32- OH- PO43- CH3COO- Fe2+ Fe3+ Ammonium = Sulphate = Nitrate = Carbonate = Hydroxide = Phosphate = Ethanoate = Iron (II) = Iron (III) =

  29. Na2CO3 Na2O Li2SO4 Ca(NO3)2 Al2(SO4)3 K3PO4 (NH4)2SO4 CH3COONa Fe(OH)2 FeCl3 • Sodium Carbonate = • Sodium Oxide = • Lithium Sulphate = • Calcium Nitrate = • Aluminium Sulphate = • Potassium Phosphate = • Ammonium Sulphate = • Sodium Ethanoate = • Iron (II) Hydroxide = • Iron (III) Chloride =

  30. Covalent Molecules • LEARN THE FOLLOWING Methane Hydrochloric acid Carbon dioxide Sulphuric acid Water Ammonia Sulphur dioxide Nitric acid Sulphur trioxide All elements

  31. Putting formulae together to get an equation A balanced equation will have • Same number of atoms of each element on both sides Eg Mg + H2O  Mg(OH)2 + H2 isn’t balanced as there are 2 O on one side and 1 O on the other Mg + 2H2O  Mg(OH)2 + H2 • 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 ANSWERS Cu2+ + 2K  Cu + 2K+ CH4 + 2O2  CO2 + 2H2O 2NH3 + H2SO4  (NH4)2SO4 Ca(HCO3)2 + 2HNO3 Ca(NO3)2 + CO2 + H2O 3Ca2+ + 2Al  2Al+3 + 3Ca FeCl2 + ½ Cl2  FeCl3 CuSO4 + 2NaOH  Na2SO4 + Cu(OH)2 Fe2O3 + 2Al Al2O3 + 2Fe • Cu+2 + K  Cu + K+ • CH4 + O2  CO2 + H2O • Ammonia + sulphuric acid  ammonium sulphate • Calcium hydrogen carbonate + Nitric acid  Calcium nitrate + carbon dioxide + water • Calcium ions + Aluminium  Aluminium ions and calcium • Iron(II) chloride + Chlorine  Iron (III) chloride • Copper (II) sulphate + Sodium hydroxide  Copper (II) hydroxide + sodium chloride • Fe2O3 + Al  Al2O3 + Fe

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

  34. Let’s write some ionic equations H+ + OH- H2O 2I- + Cl2  2Cl- + I2 Mg + Cu2+  Mg2+ + Cu Mg + 2 H+  Mg2+ + H2 CO3-2 + 2H+  CO2 + H2O H+ + OH- H2O Ca + 2 H+  Ca2+ + H2 • HCl + NaOH  NaCl + H2O • 2 KI + Cl2  2 KCl + I2 • Mg + CuSO4  MgSO4 + Cu • Mg + 2 HCl  MgCl2 + H2 • Na2CO3 + H2SO4  Na2SO4 + H2O + CO2 • H2SO4 + Ca(OH)2  CaSO4 + 2H2O • Ca + 2 HNO3  Ca(NO3)2 + H2

  35. Mole CalculationsTHREE EQUATIONS TO LEARN AND USE • Moles = mass / rfm • Moles = volume/24dm3 • 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 • Use the appropriate mole equation to calculate the number of moles of one of the substances in the equation • Use the balanced equation to work out the mole ratio • Using the mole ratio calculate the number of moles of the other substance. • 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? CaCO3 CaO + CO2 Using the equation moles = mass/rfm calculate the moles of CaCO3 Moles of CaCO3 = 25/100 (from 40+12+16+16+16) = 0.25 Now use the balanced equation to work out the mole ratio 1 CaCO3 1 CaO + CO2 ( Hence mole ratio is 1:1) Moles of CaO = Moles of CaCO3 = 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 O2 K2O Use the equation Moles = mass/rfm calculate the moles of K2O present Rfm K2O = 39+39+16 = 94 Moles K2O = 0.94 /94 = 0.01 Use the equation to work out the mole ratio 2K + ½ O2 1K2O Mole ratio = 1:2 hence moles of K = 2xmoles of K2O 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/24dm3 • USE WHEN ASKED ABOUT THE VOLUME OF A GAS BEING FORMED. • 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? CaCO3 CaO + CO2 Using the equation moles = mass/rfm calculate the moles of CaCO3 Moles of CaCO3 = 25/100 (from 40+12+16+16+16) = 0.25 Now use the balanced equation to work out the mole ratio 1 CaCO3 CaO + 1 CO2 ( Hence mole ratio is 1:1) Moles of CO2 = Moles of CaCO3 = 0.25 Using the Using the equation moles = Volume of gas / 24dm3 calculate the volume of CO2 0.25 = Volume of gas / 24dm3 Volume = 0.25 x 24= 6dm3

  40. Moles = volume/24dm3 • USE WHEN ASKED ABOUT THE VOLUME OF A GAS required. • 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 O2 K2O Moles of K2O = 18.8 / 94 = 0.2 Moles of O2 = 0.1 (mole ratio = ½ O2 : 1 K2O Volume of O2 = 2.4dm3

  41. Moles = volume/24dm3 • You need to make sure that the units of the gas has been converted to dm3 • 1dm3 = 1000cm3 so 1cm3 = 0.001dm3 How many moles of Cl2 are present in 24cm3 ? Volume = 24cm3 = 0.024dm3 Moles = volume /24 = 0.024/24 = 0.001 moles

  42. Moles = (volume*concentration)/1000 • 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 1dm3 (1000cm3) of water. A concentration of NaCl of 0.5Molsdm-3 means 0.5 moles of NaCl dissolved in 1dm3 H2O The units are Moles / dm3 = moldm-3 = Molar = M OR g/dm3 or gdm-3

  43. It is easy to determine the conc in g / dm3 using simple ratios 5 grams in 250cm3 so in 1000 cm3 there must be 20g Hence conc = 20g/dm3 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 / dm3 using simple ratios 0.0074 grams in 50cm3 so in 1000 cm3 there must be 0.148g Hence conc = 0.148g/dm3 To convert from g/dm3 to molsdm-3 use the equation moles = mass/rfm 0.148gdm3 = 0.148/(40+16+16+1+1)molsdm3 = 0.148/74molsdm-3 = 0.002moldm-3 ASIDE To convert from mols/dm3 to g/dm3 you use the equation mass = moles x rfm Hence 0.1Molddm-3 Ca(OH)2 = 0.1x74gdm3 = 7.4gdm3 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 Calculating concentrations • A student added 5 g of NaOH into 250mls of water. What was the concentration of NaOH in • A) g/dm3 B) mols/dm3 • Another student added 0.0074g of Ca(OH)2 into 50cm3 of water. What was the concentration of • A) Ca(OH)2 in g/dm3 B) Ca(OH)2 in mols/dm3 C) OH- ions in mols/dm3

  44. 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 1H2SO4 + 2NaOH  Na2SO4 + 2H2O (Mole ratio = 1:2) Moles diluted H2SO4 = 0.000045 (half of 0.0009) Rearrange the equation Moles = Vol x conc to determine conc of the diluted H2SO4 Conc = 0.000045*1000/10 = 0.045Mol/dm3 The student originally diluted the sulphuric acid from 5cm3 to 1000cm3 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 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 CaCl2 + 2H2O (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 = 8cm3 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 + H2O (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 • A student took 10cm3 of 1.08M HCl. 13cm3 of NaOH neutralised the acid. What is the concentration of the acid? • A student took 10cm3 of 0.1M Ca(OH)2 solution and neutralised it with 0.25M HCl. What volume of HCl did he add? • Another student took 5cm3 of an unknown concentration of H2SO4 anddiluted it to 1000cm3. 10cm3 of the diluted solution was exactly neutralised with 18cm3 of 0.05M NaOH. • What was the concentration of the diluted acid? • What was the concentration of the undiluted acid?

  45. All the equations in same question • A student took 2.3g limestone and reacted it with excess 0.5M HCl. He collected 0.5dm3 of CO2 • Assuming that the limestone was pure CaCO3. What is the minimum volume of HCl needed to react with the limestone? • What is the percentage of CaCO3 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 CaCO3 if pure) so the moles of CaCO3 can be determined using moles = mass/rfm RFm CaCO3 = 100 (40 + 12 + 16 + 16 + 16) Moles of CaCO3 = 2.3/100 = 0.023 Balanced equation1 CaCO3 + 2HCl CaCl2 + CO2 + H2O Mole ratio = 1:2 Moles of HCl = 2 x moles CaCO2 = 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 cm3 As Limestone IS NOT pure CaCO3 this amount of HCL was be an excess (assuming other materials in Limestone do not themselves react with HCl.

  47. The student collected 0.5dm3 of CO2 in the experiment so the number of moles of CO2 can be determined Moles = volume of gas/24 Moles = 0.5 / 24 = 0.020833 1 CaCO3 + 2 HCl CaCl2 + 1 CO2 + H2O Mole ratio = 1:1 Moles of CaCO3 present = Moles CO2 Moles CaCO3 = 0.020833 Mass CaCO3 present in the limestone = 100 x 0.020833 Mass of CaCO3 present in the limestone = 2.0833 Mass of limestone = 2.3g % CaCO3 in limestone = 2.0833/2.3 x 100 = 90.6%

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