This document provides an overview of analytical chemistry, including qualitative and quantitative analysis techniques. It discusses various instruments and methods used in analytical chemistry, such as chromatography, spectroscopy, titration, and gravimetric analysis. Key concepts covered include units, molar mass, moles, stoichiometry, empirical formulas, and significant figures. Examples are provided to demonstrate calculations for composition percentages, empirical formulas, hydration states, and concentration determination through gravimetric analysis.

1. Analytical chemistry

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.

30. Chicken soup worksheet

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