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Analytical chemistry
Qualitative Analysis
 What substances are present in the product?
    Testing for the presence of pesticide and herbicide residues in milk
    Actual oils used in a margarine
    Ethanol in petrol
    Arsenic in food
Quantitative Analysis
 How much of a particular substance is present in a product?
    Determination of the actual concentration of pesticide and
     herbicide residues in milk
    Percentage of canola oil used in margarine
 Analysis uses modern instrumentation or the old fashioned reactions
Analysis




             Instruments                                      Chemistry




                                spectroscop
                                     y              Gravimetric           Volumetric
chromatography




                 HPL       AA                      UV             Acid base      Redox
TLC    GLC                             IR
                  C
Balance equations
 Na2CO3(aq) + HCl(aq)
     aq) + HNO3(aq)
 NaOH(aq) + HCN(aq)
 HCOOH(l) + NaOH(aq)
Ionic Equations
Full Equation
   NaOH(aq) + HCN(aq)          NaCN(aq) + H2O(l)
Ionic Equation
  Na+(aq) + OH-(aq) + H+(aq) + CN-(aq)   Na+(aq) + CN-(aq) + H2O(l)
Partial Ionic equation
   Just shows the substances that undergo change
   OH-(aq) + H+(aq)     H2O(l)
Units
 Convert
    125oC to K
    23mL to L
    2.4atm to Pa
    740mmHg to Pa
Relative atomic mass
 Mass of an atom or compound relative to 12C having a mass of 12
    MR of Mg =
    MR of MgO =
    MR of Al2O3 =
Examples
 Calculate the volume of
    3.5 mole of helium at STP
    12g of hydrogen at SLC
    12g of hydrogen at 110oC and 200000Pa
Percentage Composition
 % mass = MR of element x number present x 100
                     MR of compound
 Examples
    % S in SO2
    % Al in Al2O3
Empirical Formulae
 A calculation frequently asked for is the determination of an empirical
  formula given the percentage composition of the atoms or just a mass
  ratio
Example
 Determine the empirical formula for cholesterol given that the
  percentage composition of cholesterol is 83.938% carbon, 11.917%
  hydrogen and 4.145% oxygen.




 The empirical formula is therefore C27H46O
Example
 When 0.864g of nitrogen burns, it forms 2.839g of oxide. Find the
  empirical formula of the oxide.
 If the molar mass of the oxide is 92g, determine its molecular formula
  as well.
Example
 When a 0.995g sample of an organic molecule containing
  carbon, hydrogen and oxygen is burnt in air, the only products are
  1.468g of carbon dioxide and 0.602g of water.
 What is the empirical formula?
Example
 10.848g hydrated copper(II) sulfate is dried until there is no further
  change in mass. After drying, the anhydrous salt has a mass of
  6.935g. Determine the degree of hydration of the salt.
The value of MR
 The value of molar mass can sometimes be provided in a round about
  way.
 A 51g sample of a compound occupied the same volume as 16g of
  oxygen at the same temperature and pressure.
    n(O2) = 16 / 32 = 0.5 mol
    Therefore the number of mole of the unknown compound must
     also be 0.5 mol.
    If 0.5 mol has a mass of 51, then the molar mass must be 102g
Mole
 1 mole is the amount of substance that contains the same number of
  particles as there are in 12g of 12C
 The number of particles in 1 mole = 6.02 x 1023.
    This is Avagadro's number
 Examples
    1 mole of aluminium = 26.9g
    1 mole of copper =
    1 mole of Al2O3 =
Mole calculations
      m
  n = MR
 Example
    Calculate the number of mole in
      200g of aluminium

      0.34g of Al2O3
Significant figures
 Your answer must only have the same number of significant figures
  as the least acurate data given
 The zeros before and immediately after a decimal point are not
  counted as significant
Masses of solids
 Take special care with questions that give mass in kilograms or
  milligrams. Moles are calculated using grams so you must convert kg
  and mg to grams before you start.
Stoichiometry
 Calculate the number of mole of each substance for the following
  reactions
    2FeCl3    2FeCl2 + Cl2
    3 mole    _____ ___
    ______    0.45mole ____
    4Al + 3O2        2Al2O3
    10mole ___      ______
    _____ 0.45mole ______
Excess
 20g of magnesium is added to 200mL of 2M hydrochloric acid.
  Which chemical is in excess?
 Mg(s) + 2HCl(aq)       MgCl2(aq) + H2(g)
 The amount of product will be determined by the reactant which is
  scarce, as it all reacts.
 The volume of H2(g) must be calculated from the limiting reactant.
Gravimetric Analysis
 Is a method of analysis that involves accurate measurement of masses
  in a precipitation reaction
 An ion in solution is caused to precipitate out of solution. The
  precipitate is filtered, then dried. Its mass is related to the
  concentration of the original ion.
 Gravimetric analysis relies on stoichiometry
 Gravimetric analysis will work if either of the ions present can be
  precipitated.
Simplified solubility table
 All nitrates, acetates (ethanoates), group 1 and ammonium
  compounds are soluble
 All chlorides, bromides and iodides are soluble except Ag+
  and Pb2+ compounds
 All sulfates are soluble except BaSO4 and PbSO4
 All carbonate compounds are of low solubility except group
  1 carbonates and (NH4)2CO3
 All hydroxide compounds are of low solubility except
  group 1 hydroxides and NH4OH, Sr(OH)2 and Ba(OH)2
Gravimetric Analysis
 A chemist determined that the salt (NaCl) content of food by
  precipitating chloride ions as silver chloride. A 8.45g sample of food
  yielded 0.636g of precipitate.
An example of the steps involved in
       gravimetric analysis
 The equation needs to be written down first
 Find the molar mass of the precipitate and salt
 Find the number of moles of the precipitate and equate this to the
  number moles of the salt
 Find the mass in grams of the salt from n = m/MR
 Determine the percentage of salt
What can go wrong in gravimetric
           analysis?
 Although precipitates used are of low solubility, a very
  small amount will remain in solution
 Other insoluble compounds may also be precipitated
 The precipitate must be washed to prevent any other
  chemicals such as solute particles from crystallising out
  during the drying process. The washing must be
  limited, however, so as not to re-dissolve any precipitate
 The balance has an uncertainty that may be tiny for very
  precise instruments or quite large for simpler models. For
  this reason the mass of precipitate should be reasonably
  large.
Test yourself
 The formation of an insoluble salt in solution is known as
  __________________. Insoluble salts can be removed from solution
  by ________________.
 When two soluble salts, sodium chloride and silver nitrate, are reacted
  together they form a white precipitate of silver chloride. The overall
  equation is: NaCl(aq) + AgNO3(aq)    AgCl(s) + NaNO3(aq) The
  precipitate is ______________ The ionic equation is: Ag+(aq) + Cl-(aq)
     AgCl(s) The spectator ions are: _______________ +
  ________________
Test yourself
 An analytical procedure in which the masses of solids are measured in
  order to determine the concentration of a particular substance in a
  mixture is called _________________
 Steps involved in gravimetric analysis are: (i) add an _____________
  of a reactant to form a precipitate, (ii) ______________ the
  precipitate and wash it, (iii) dry the precipitate to ____________
  mass.
Chicken soup worksheet
Example
 A 5.40g sample of potato chips is crushed and mixed in water to
  dissolve the potassium chloride that is used instead of sodium
  chloride. To analyse the KCl, excess silver nitrate solution is added.
  The precipitate is filtered, dried and weighed. Its mass is 0.192g.
  Calculate the % mass of potassium chloride in the chips (assume no
  NaCl is used.)
Tips
 Set out each question carefully
 Write the equation out, with each piece of data placed under the
  chemical it refers to
 Include the units
 Where to start will often be evident from what you have written
Example
 A 50.0mL solution containing iron(III) nitrate, Fe(NO3)3, has excess
  sodium hydroxide solution, NaOH, added to it to precipitate the iron
  as Fe(OH)3. After heating, the hydroxide decomposes to iron oxide,
  Fe2O3. The mass of precipitate obtained was 0.533g. Calculate the
  iron concentration in the original solution.
Fertiliser (Exam question)
 A soluble fertiliser contains phosphorus in the form of phosphate ions
  (PO43-) content by gravimetric analysis, 5.97g of the fertiliser powder
  was completely dissolved in water to make a volume of 250.0mL. A
  20.00mL volume of this solution was pipetted into a conical flask and
  the PO43- ions in the solution were precipitated as MgNH4PO4. The
  precipitate was filtered, washed with water and then converted by
  heating into Mg2P2O7. The mass of Mg2P2O7 was 0.0352g.
Questions
 Calculate the amount, in mole, of Mg2P2O7.
 Calculate the amount, in mole, of phosphorus in the 20.00mL volume
  of solution
 Calculate the amount, in mole, of phosphorus in 5.97g of fertiliser
 Calculate the percentage of phosphate ions by mass in the fertiliser.
  Ensure you express your answer to an appropriate number
  of significant figures.
Continued
 Several actions which could occur during this analytical
  procedure are listed below. For each action indicate the likely
  effect on the calculated percentage of phosphate ions in the
  fertiliser.
    The MgNH4PO4 precipitate was not washed with water
    The conical flask had been previously washed with water but
     not dried
    A 25.00mL pipette was unknowingly used instead of a
     20.00mL pipette
    The mass of the fertiliser was recorded incorrectly. The
     recorded mass was 0.2g higher than the actual mass
Method again
 Step 1: A known mass of the sample is dissolved in a
 suitable solvent
   Polar substances are dissolved in
   Non polar substances are dissolved in
   Alloys are dissolved in
 Step 2: the substance to be analysed is precipitated by
 the addition of an appropriate chemical species. The
 chosen solution is one that exclusively precipitates the
 ion of interest.
  If the precipitating solution co-precipitates other ions, the
     weighed mass will be                  and therefore, the
     calculated percentage by mass will be
 Step 3: the precipitate is collected by filtration and
  thoroughly washed to remove substances that would
  otherwise contribute to the mass of the precipitate.
  Washing is usually performed using deionised water.
  If the washing step is omitted, the weighed mass will be
               and the calculated percentage by mass of
  precipitate will be
 Step 4: the precipitate is then dried in an oven at
  110oC.
  If the drying step is incomplete, water will contribute
  to the mass; the weighed mass will be
  and the calculated percentage by mass of the
  precipitate will be
 Step 5: the sample is cooled in a dessicator and
  weighed. The dessicator removes moisture from the
  atmosphere and minimised the amount of moisture
  absorbed by the sample during cooling
 Step 6: steps 4 and 5 are repeated until there is no
  significant change in mass. This ensures that all the
  water has been evaporated.
The precipitate formed must
  exhibit the following properties:
 The precipitate must have a low solubility so that it does
  not go back into solution
  If the precipitate has a relatively high solubility, the
  weighed mass will be                   and the calculated
  percentage by mass of precipitate will be
 The precipitate must have a molar mass that does not
  vary, so that stoichiometric calculations can be accurately
  preformed
  If a sample absorbs substances from the atmosphere, the
  weighed mass will be                      and the calculated
  percentage by mass of precipitate will be
  If a sample gives off substances to the atmosphere, the
  weighed mass will be                  and the claculated
  percentage by mass of precipitate will be
 The precipitate must be stable when heated and
  dessicated
 The precipitate must be pure
 The precipitate must be easy to recover by filtration
 The molar mass should be high so that weighing
  errors are minimised
Effects of incorrect techniques on
             calculations
        ERROR         Mass of collected   Mass of sample       %
                        precipitate        component       composition
Co-precipitation of
ions
High solubility of
precipitate
Loss of precipitate
during filtration
Incomplete drying

Incomplete washing
Question 1
 A solution containing 10.0g of sodium chloride is
 mixed with a solution of silver nitrate. What mass of
 precipitate will be formed?
Question 2a
 A 1.595 g sample of silver alloy is dissolved in 50.00 ml
  (an excess) of nitric acid. A 10.00 ml sample was then
  treated with excess sodium chloride solution to
  produce a precipitate of silver chloride. The
  precipitate was filtered, dried and weighed.
 If the mass of silver chloride precipitated is 0.25g, find
  the percentage of silver in the alloy.
Question 2b,c
 State the assumptions that were necessary to
 determine the percentage of silver in the alloy.




 What are the principal sources of experimental error
 in a gravimetric analysis?
Question 2d, e
 What conditions could be arranged to precipitate as
 much of the substance to be analysed as possible?




 Why was excess sodium chloride added to the
 reaction mixture?
Question 2 f
 Consider the following reaction:
  BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + NaCl(aq)
 Discuss why you would choose filtration techniques in
  preference to evaporation when collecting the
  precipitate of this reaction.
Question 3 a
 A sample of contaminated hydrated copper sulphate
  (CuSO4.5H2O) was tested for purity. A 15.0 g sample
  was dissolved in water and filtered to remove
  insoluble impurities. The sulphate ions were
  precipitated by the addition of excess barium chloride
  and the resulting precipitate was collected, dries and
  weighed. If the final mass is 4.95 g:
 Find the percentage purity of the sample.
Question 3 b, c
 If the precipitate was not completely dry when
 weighed, what effect would this have on the
 calculated percentage purity of the sample?




 What would be the effect on the calculated percentage
 if barium nitrate was used instead of barium chloride?
Question 3 d, e
 What would be the effect on the calculated percentage
 if silver ions were present in the sample?




 Find the mass of contaminated hydrated copper
 sulphate that would be required to produce 100.00 ml
 of a 0.250 M copper sulphate solution.
Question 4
 A 0.500 g sample of sodium sulphate and a 0.500 g of
 aluminium sulphate were dissolved in a volume of
 water, and excess barium chloride added to
 precipitate barium sulphate. What was the mass of
 barium sulphate produced?
Question 5
 0.6238 g of copper(II) sulphate crystals with formula
 CuSO4.xH2O was dissolved in water, and the black
 copper (II) oxide was precipitated by treatment with
 boiling NaOH solution. The precipitate was collected
 by filtration, washed, dried and weighed. If the
 precipitate weighs 0.1988 g, calculate the value of x in
 the formula CuSO4.xH2O.
Question 6
 In order to determine the molecular formula of a
  compound known to contain only carbon, hydrogen
  and oxygen, the following experiments were carried
  out.
1. A 0.60 g sample of the compound was burnt in
   excess oxygen. When the gases evolved were passed
   through anhydrous CaCl2, its mass increased by
   0.36g due to the absorption of H2O. The remaining
   gas(es) when bubbled through a NaOH
   solution, increased its mass by 0.88 g.
Question 6 a, b
2. A 1.21 g sample of the compound was vaporised. The
  vapour occupied 0.403L at 150oC and 1.17 x 105 Pa
 Determine the mass of the gaseous products



 Determine the mass of carbon in the sample
Question 6 c, d
 Determine the mass of hydrogen in the sample




 Determine the mass of oxygen in the sample
Question 6 e, f
 Determine the empirical formula of CxHyOz




 Determine the relative molecular mass of CxHyOz
Question 6 g
 Determine the molecular formula of CxHyOz

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Analytical instruments week 1

  • 2. Qualitative Analysis  What substances are present in the product?  Testing for the presence of pesticide and herbicide residues in milk  Actual oils used in a margarine  Ethanol in petrol  Arsenic in food
  • 3. Quantitative Analysis  How much of a particular substance is present in a product?  Determination of the actual concentration of pesticide and herbicide residues in milk  Percentage of canola oil used in margarine  Analysis uses modern instrumentation or the old fashioned reactions
  • 4. Analysis Instruments Chemistry spectroscop y Gravimetric Volumetric chromatography HPL AA UV Acid base Redox TLC GLC IR C
  • 5. Balance equations  Na2CO3(aq) + HCl(aq)  aq) + HNO3(aq)  NaOH(aq) + HCN(aq)  HCOOH(l) + NaOH(aq)
  • 6. Ionic Equations Full Equation NaOH(aq) + HCN(aq) NaCN(aq) + H2O(l) Ionic Equation Na+(aq) + OH-(aq) + H+(aq) + CN-(aq) Na+(aq) + CN-(aq) + H2O(l) Partial Ionic equation Just shows the substances that undergo change OH-(aq) + H+(aq) H2O(l)
  • 7. Units  Convert  125oC to K  23mL to L  2.4atm to Pa  740mmHg to Pa
  • 8. Relative atomic mass  Mass of an atom or compound relative to 12C having a mass of 12  MR of Mg =  MR of MgO =  MR of Al2O3 =
  • 9. Examples  Calculate the volume of  3.5 mole of helium at STP  12g of hydrogen at SLC  12g of hydrogen at 110oC and 200000Pa
  • 10. Percentage Composition  % mass = MR of element x number present x 100 MR of compound  Examples  % S in SO2  % Al in Al2O3
  • 11. Empirical Formulae  A calculation frequently asked for is the determination of an empirical formula given the percentage composition of the atoms or just a mass ratio
  • 12. Example  Determine the empirical formula for cholesterol given that the percentage composition of cholesterol is 83.938% carbon, 11.917% hydrogen and 4.145% oxygen.  The empirical formula is therefore C27H46O
  • 13. Example  When 0.864g of nitrogen burns, it forms 2.839g of oxide. Find the empirical formula of the oxide.  If the molar mass of the oxide is 92g, determine its molecular formula as well.
  • 14. Example  When a 0.995g sample of an organic molecule containing carbon, hydrogen and oxygen is burnt in air, the only products are 1.468g of carbon dioxide and 0.602g of water.  What is the empirical formula?
  • 15. Example  10.848g hydrated copper(II) sulfate is dried until there is no further change in mass. After drying, the anhydrous salt has a mass of 6.935g. Determine the degree of hydration of the salt.
  • 16. The value of MR  The value of molar mass can sometimes be provided in a round about way.  A 51g sample of a compound occupied the same volume as 16g of oxygen at the same temperature and pressure.  n(O2) = 16 / 32 = 0.5 mol  Therefore the number of mole of the unknown compound must also be 0.5 mol.  If 0.5 mol has a mass of 51, then the molar mass must be 102g
  • 17. Mole  1 mole is the amount of substance that contains the same number of particles as there are in 12g of 12C  The number of particles in 1 mole = 6.02 x 1023.  This is Avagadro's number  Examples  1 mole of aluminium = 26.9g  1 mole of copper =  1 mole of Al2O3 =
  • 18. Mole calculations m n = MR  Example  Calculate the number of mole in  200g of aluminium  0.34g of Al2O3
  • 19. Significant figures  Your answer must only have the same number of significant figures as the least acurate data given  The zeros before and immediately after a decimal point are not counted as significant
  • 20. Masses of solids  Take special care with questions that give mass in kilograms or milligrams. Moles are calculated using grams so you must convert kg and mg to grams before you start.
  • 21. Stoichiometry  Calculate the number of mole of each substance for the following reactions  2FeCl3 2FeCl2 + Cl2  3 mole _____ ___  ______ 0.45mole ____  4Al + 3O2 2Al2O3  10mole ___ ______  _____ 0.45mole ______
  • 22. Excess  20g of magnesium is added to 200mL of 2M hydrochloric acid. Which chemical is in excess?  Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)  The amount of product will be determined by the reactant which is scarce, as it all reacts.  The volume of H2(g) must be calculated from the limiting reactant.
  • 23. Gravimetric Analysis  Is a method of analysis that involves accurate measurement of masses in a precipitation reaction  An ion in solution is caused to precipitate out of solution. The precipitate is filtered, then dried. Its mass is related to the concentration of the original ion.  Gravimetric analysis relies on stoichiometry  Gravimetric analysis will work if either of the ions present can be precipitated.
  • 24. Simplified solubility table  All nitrates, acetates (ethanoates), group 1 and ammonium compounds are soluble  All chlorides, bromides and iodides are soluble except Ag+ and Pb2+ compounds  All sulfates are soluble except BaSO4 and PbSO4  All carbonate compounds are of low solubility except group 1 carbonates and (NH4)2CO3  All hydroxide compounds are of low solubility except group 1 hydroxides and NH4OH, Sr(OH)2 and Ba(OH)2
  • 25. Gravimetric Analysis  A chemist determined that the salt (NaCl) content of food by precipitating chloride ions as silver chloride. A 8.45g sample of food yielded 0.636g of precipitate.
  • 26. An example of the steps involved in gravimetric analysis  The equation needs to be written down first  Find the molar mass of the precipitate and salt  Find the number of moles of the precipitate and equate this to the number moles of the salt  Find the mass in grams of the salt from n = m/MR  Determine the percentage of salt
  • 27. What can go wrong in gravimetric analysis?  Although precipitates used are of low solubility, a very small amount will remain in solution  Other insoluble compounds may also be precipitated  The precipitate must be washed to prevent any other chemicals such as solute particles from crystallising out during the drying process. The washing must be limited, however, so as not to re-dissolve any precipitate  The balance has an uncertainty that may be tiny for very precise instruments or quite large for simpler models. For this reason the mass of precipitate should be reasonably large.
  • 28. Test yourself  The formation of an insoluble salt in solution is known as __________________. Insoluble salts can be removed from solution by ________________.  When two soluble salts, sodium chloride and silver nitrate, are reacted together they form a white precipitate of silver chloride. The overall equation is: NaCl(aq) + AgNO3(aq) AgCl(s) + NaNO3(aq) The precipitate is ______________ The ionic equation is: Ag+(aq) + Cl-(aq) AgCl(s) The spectator ions are: _______________ + ________________
  • 29. Test yourself  An analytical procedure in which the masses of solids are measured in order to determine the concentration of a particular substance in a mixture is called _________________  Steps involved in gravimetric analysis are: (i) add an _____________ of a reactant to form a precipitate, (ii) ______________ the precipitate and wash it, (iii) dry the precipitate to ____________ mass.
  • 31. Example  A 5.40g sample of potato chips is crushed and mixed in water to dissolve the potassium chloride that is used instead of sodium chloride. To analyse the KCl, excess silver nitrate solution is added. The precipitate is filtered, dried and weighed. Its mass is 0.192g. Calculate the % mass of potassium chloride in the chips (assume no NaCl is used.)
  • 32. Tips  Set out each question carefully  Write the equation out, with each piece of data placed under the chemical it refers to  Include the units  Where to start will often be evident from what you have written
  • 33. Example  A 50.0mL solution containing iron(III) nitrate, Fe(NO3)3, has excess sodium hydroxide solution, NaOH, added to it to precipitate the iron as Fe(OH)3. After heating, the hydroxide decomposes to iron oxide, Fe2O3. The mass of precipitate obtained was 0.533g. Calculate the iron concentration in the original solution.
  • 34. Fertiliser (Exam question)  A soluble fertiliser contains phosphorus in the form of phosphate ions (PO43-) content by gravimetric analysis, 5.97g of the fertiliser powder was completely dissolved in water to make a volume of 250.0mL. A 20.00mL volume of this solution was pipetted into a conical flask and the PO43- ions in the solution were precipitated as MgNH4PO4. The precipitate was filtered, washed with water and then converted by heating into Mg2P2O7. The mass of Mg2P2O7 was 0.0352g.
  • 35. Questions  Calculate the amount, in mole, of Mg2P2O7.  Calculate the amount, in mole, of phosphorus in the 20.00mL volume of solution  Calculate the amount, in mole, of phosphorus in 5.97g of fertiliser  Calculate the percentage of phosphate ions by mass in the fertiliser. Ensure you express your answer to an appropriate number of significant figures.
  • 36. Continued  Several actions which could occur during this analytical procedure are listed below. For each action indicate the likely effect on the calculated percentage of phosphate ions in the fertiliser.  The MgNH4PO4 precipitate was not washed with water  The conical flask had been previously washed with water but not dried  A 25.00mL pipette was unknowingly used instead of a 20.00mL pipette  The mass of the fertiliser was recorded incorrectly. The recorded mass was 0.2g higher than the actual mass
  • 37. Method again  Step 1: A known mass of the sample is dissolved in a suitable solvent  Polar substances are dissolved in  Non polar substances are dissolved in  Alloys are dissolved in  Step 2: the substance to be analysed is precipitated by the addition of an appropriate chemical species. The chosen solution is one that exclusively precipitates the ion of interest. If the precipitating solution co-precipitates other ions, the weighed mass will be and therefore, the calculated percentage by mass will be
  • 38.  Step 3: the precipitate is collected by filtration and thoroughly washed to remove substances that would otherwise contribute to the mass of the precipitate. Washing is usually performed using deionised water. If the washing step is omitted, the weighed mass will be and the calculated percentage by mass of precipitate will be  Step 4: the precipitate is then dried in an oven at 110oC. If the drying step is incomplete, water will contribute to the mass; the weighed mass will be and the calculated percentage by mass of the precipitate will be
  • 39.  Step 5: the sample is cooled in a dessicator and weighed. The dessicator removes moisture from the atmosphere and minimised the amount of moisture absorbed by the sample during cooling  Step 6: steps 4 and 5 are repeated until there is no significant change in mass. This ensures that all the water has been evaporated.
  • 40. The precipitate formed must exhibit the following properties:  The precipitate must have a low solubility so that it does not go back into solution If the precipitate has a relatively high solubility, the weighed mass will be and the calculated percentage by mass of precipitate will be  The precipitate must have a molar mass that does not vary, so that stoichiometric calculations can be accurately preformed If a sample absorbs substances from the atmosphere, the weighed mass will be and the calculated percentage by mass of precipitate will be If a sample gives off substances to the atmosphere, the weighed mass will be and the claculated percentage by mass of precipitate will be
  • 41.  The precipitate must be stable when heated and dessicated  The precipitate must be pure  The precipitate must be easy to recover by filtration  The molar mass should be high so that weighing errors are minimised
  • 42. Effects of incorrect techniques on calculations ERROR Mass of collected Mass of sample % precipitate component composition Co-precipitation of ions High solubility of precipitate Loss of precipitate during filtration Incomplete drying Incomplete washing
  • 43. Question 1  A solution containing 10.0g of sodium chloride is mixed with a solution of silver nitrate. What mass of precipitate will be formed?
  • 44. Question 2a  A 1.595 g sample of silver alloy is dissolved in 50.00 ml (an excess) of nitric acid. A 10.00 ml sample was then treated with excess sodium chloride solution to produce a precipitate of silver chloride. The precipitate was filtered, dried and weighed.  If the mass of silver chloride precipitated is 0.25g, find the percentage of silver in the alloy.
  • 45. Question 2b,c  State the assumptions that were necessary to determine the percentage of silver in the alloy.  What are the principal sources of experimental error in a gravimetric analysis?
  • 46. Question 2d, e  What conditions could be arranged to precipitate as much of the substance to be analysed as possible?  Why was excess sodium chloride added to the reaction mixture?
  • 47. Question 2 f  Consider the following reaction: BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + NaCl(aq)  Discuss why you would choose filtration techniques in preference to evaporation when collecting the precipitate of this reaction.
  • 48. Question 3 a  A sample of contaminated hydrated copper sulphate (CuSO4.5H2O) was tested for purity. A 15.0 g sample was dissolved in water and filtered to remove insoluble impurities. The sulphate ions were precipitated by the addition of excess barium chloride and the resulting precipitate was collected, dries and weighed. If the final mass is 4.95 g:  Find the percentage purity of the sample.
  • 49. Question 3 b, c  If the precipitate was not completely dry when weighed, what effect would this have on the calculated percentage purity of the sample?  What would be the effect on the calculated percentage if barium nitrate was used instead of barium chloride?
  • 50. Question 3 d, e  What would be the effect on the calculated percentage if silver ions were present in the sample?  Find the mass of contaminated hydrated copper sulphate that would be required to produce 100.00 ml of a 0.250 M copper sulphate solution.
  • 51. Question 4  A 0.500 g sample of sodium sulphate and a 0.500 g of aluminium sulphate were dissolved in a volume of water, and excess barium chloride added to precipitate barium sulphate. What was the mass of barium sulphate produced?
  • 52. Question 5  0.6238 g of copper(II) sulphate crystals with formula CuSO4.xH2O was dissolved in water, and the black copper (II) oxide was precipitated by treatment with boiling NaOH solution. The precipitate was collected by filtration, washed, dried and weighed. If the precipitate weighs 0.1988 g, calculate the value of x in the formula CuSO4.xH2O.
  • 53. Question 6  In order to determine the molecular formula of a compound known to contain only carbon, hydrogen and oxygen, the following experiments were carried out. 1. A 0.60 g sample of the compound was burnt in excess oxygen. When the gases evolved were passed through anhydrous CaCl2, its mass increased by 0.36g due to the absorption of H2O. The remaining gas(es) when bubbled through a NaOH solution, increased its mass by 0.88 g.
  • 54. Question 6 a, b 2. A 1.21 g sample of the compound was vaporised. The vapour occupied 0.403L at 150oC and 1.17 x 105 Pa  Determine the mass of the gaseous products  Determine the mass of carbon in the sample
  • 55. Question 6 c, d  Determine the mass of hydrogen in the sample  Determine the mass of oxygen in the sample
  • 56. Question 6 e, f  Determine the empirical formula of CxHyOz  Determine the relative molecular mass of CxHyOz
  • 57. Question 6 g  Determine the molecular formula of CxHyOz