Worksheet 3 Mass measurements—gravimetric techniques
1
a
Materials: heating mat, Bunsen burner, tripod, gauze mat, crucible and lid (or heat-resistant
test tube) and tongs, electronic balance, desiccator, heat resistant gloves.
Risk assessment:



Wear a laboratory coat and goggles.
Solid CuSO4 is harmful if swallowed; irritant to eyes and skin.
Take care with hot crucible and tripod. Use heat-resistant gloves or tongs to
handle the hot equipment.
Procedure:



b
Weigh the crucible. Record its mass.
Add the sample of hydrated copper(II) sulfate to the crucible.
Weigh and record the mass. The mass of the sample is obtained by
subtraction.
 Heat the sample until the blue colour disappears and it turns white.
 Weigh and record the mass of the cooled, heated sample. The mass of the
residue is obtained by subtraction.
Mass of water removed = 3.40 g
n(H2O) = 0.180 mol
Mass of anhydrous copper(II) sulfate = 5.41 g
n(CuSO4) = 0.0339 mol
Ratio of anhydrous solid to water of crystallisation = 0.0339 : 0.180 ≈ 1 : 5
Number of water molecules in formula = 5
Empirical formula = CuSO4.5H2O
2
a
b
1.
Accurately weigh a sample of sausage roll.
2.
Blend the sample in a small amount of de-ionised water and filter, collecting the
filtrate.
3.
To precipitate the chloride ions, add excess silver nitrate.
4.
Filter the precipitate.
5.
To ensure all chloride ions have precipitated, add several drops of silver nitrate solution
to the filtrate. Look for milkiness in the clear solution (above the precipitate). If not all
the chloride ions were precipitated the calculated result would be lower than it should
be.
6.
To remove impurities in the precipitate, wash the residue with de-ionised water. This is
important so that the calculated result will not be higher than it should be.
7.
Dry the precipitate by heating to constant mass.
8.
Weigh the precipitate.
NaCl(aq) + AgNO3(aq) → NaNO3(aq) + AgCl(s)
Mass of precipitate = m(AgCl) = 0.636 g
n(AgCl) = 0.636/143.3 = 0.004 44 mol
n(NaCl) = 0.004 44 mol
m(NaCl) = 0.004 44  58.5 = 0.260 g
% salt =
0.260
 100 = 3.07%
8.45
Worksheet 4 Acids and bases—a concept map
1
2
An amphiprotic substance can act as either an acid or as a base in different reactions.
pH is used to measure the hydrogen ion concentration of a solution; it is defined as pH =
−log10[H3O+].
Worksheet 5 The marvellous mole—key relationships
1
a
Amount, n =
m
, where m is measured in grams.
M
b
Volume, V =
n
, where V is measured in litres.
c
c
For gases, pV = nRT where p is measured in kPa, and V is measured in litres.
d
When temperature is used in calculations, it must be in Kelvin.
e
25°C = 298 K
f
For pressure, 760 mmHg = 1 atm = 101.325 kPa.
g
n=
h
Molar volume of a gas at STP = 22.4 L.
i
Molar volume of a gas, VM =
N
NA
V
, when V and VM are measured at the same temperature and
n
pressure.
2
Multipart question:
For the reaction between 20.00 mL of 0.100 M HCl and 0.200 g of solid Na2CO3:
a
write a balanced equation for the reaction
b
determine the reactant that is in excess
c
calculate the number of millilitres of CO2 produced at a temperature of 28°C and 765
mmHg pressure.
Worked solution:
a
2HCl(aq) + Na2CO3(s) → 2NaCl(aq) + CO2(g) + H2O(l)
b
To determine excess: n(HCl)/coefficient =
0.02000  0.100
0.00200
=
= 0.001 00
2
2
n(Na2CO3)/coefficient = (
0.200
)/1 = 0.001 89
106
Na2CO3 is in excess
c
n(CO2)/n(HCl) =
1
2
n(CO2) =
1
 0.0200  0.100 = 0.001 00 mol
2
V(CO2) =
nRT
0.001 00  8.31  301
=
= 0.0245 L = 24.5 mL
p
(765 760 )  101 .325
Worksheet 6 Electrons on the move—redox equations
1
a
Reduction half-equation:
MnO4–(aq) + 8H+(aq) + 5e– → Mn2+(aq) + 4H2O(l)
Oxidation half-equation:
Fe2+(aq) → Fe3+(aq) + e–
Overall ionic equation:
5Fe3+(aq)
MnO4–(aq) + 8H+(aq) + 5Fe2+(aq) → Mn2+(aq) + 4H2O(l) +
In the above reaction, MnO4–(aq) is the oxidant and is reduced by Fe2+(aq).
b
Overall ionic equation:
2S2O42–(aq) + H2O(l) → S2O32–(aq) + 2HSO3–(aq)
Reduction half-equation:
S2O42–(aq) + 2H+(aq) + 2e– → S2O32–(aq) + H2O(l)
Oxidation half-equation:
S2O42–(aq) + 2H2O(l) → 2HSO3–(aq) + 2H+(aq) + 2e–
In S2O42–, let x = ON, oxidation number, of S:
So 2x + 4(–2) = –2
2x – 8 = –2
2x = +6
x = +3
Whereas in S2O32–:
2x + 3(–2) = –2
x = +2
Therefore, S has been reduced (ON has decreased) when forming S2O32–.
c
Overall equation:
2HCHO(aq) + O2(g) → 2HCOOH(aq)
Reduction half-equation:
O2(g)+ 4H+(aq) + 4e– → 2H2O(l)
Oxidation half-equation:
HCHO(aq) + H2O(l) → HCOOH(aq) + 2H+(aq) + 2e–
In HCHO, the C has ON = 0 and it is oxidised to HCOOH where the ON of C is +2. The O
in O2 has an ON = 0, so it has been reduced in H2O(l) where the ON of O is –2.
d
Reduction half-equation: Cr2O72–(aq) + 14H+(aq) + 6e– → 2Cr3+(aq) + 7H2O(l)
Oxidation half-equation: C2H6O(aq) + H2O(l) → C2H4O2(aq) + 4H+(aq) + 4e–
Overall ionic equation:
+ 3C2H4O2(aq)
2Cr2O72–(aq) + 16H+(aq) + 3C2H6O(aq) → 4Cr3+(aq) + 11H2O(l)
i
In Cr2O72– the ON of Cr = +6; in Cr3+ the ON of Cr = +3; in K2CrO4 the ON of Cr = +6
ii
In C2H6O the ON of C = –4 and in C2H4O2 the ON of C = 0
2
3
a
ON of Pb in PbO and in PbCl2 = +2, no change, nor does ON of O, H or Cl change, so it is
not a redox reaction.
b
ON of Pb in PbO2 = +4 and in PbCl2 = +2, so it is a redox reaction.
c
ON of Fe in Fe2O3 = +3 and in Fe = 0, so it is a redox reaction. (The presence of an element
indicates that a redox reaction has occurred.)
d
ON of Fe in FeO = +2 and in FeSiO3 = +2; since ON of O and Si do not change either, it is
not a redox reaction.
Worksheet 7 An Australian invention—atomic absorption
spectroscopy
1
The absorption of light energy of specific wavelengths causes electrons to jump from lower to
higher energy levels.
2
The number of protons in the nucleus of the atoms of each element is different. The attraction
between the nucleus and the electrons in their energy levels therefore will be different for each
element. The energy absorbed in order to move to a higher energy level will therefore be unique
for every element.
3
It is used for qualitative and quantitative analysis to detect the presence and concentration of 68
metals in blood, urine, air, soil, food and drinks.
4
a
A graph of the results.
b
If the absorbance is 3.5, the mass of cadmium in 10 mL sample 8.5 ppm.
c
8.5 ppm = 8.5 g g–1 = concentration in the 10 mL
So there are 85 g of Cd in 10 mL that contained 1.20 g of soil
So concentration of Cd in the soil = 85/1.2 g g–1 = 71 g g–1
Worksheet 8 Particles on the move—chromatography
1
a
Chromatography can be used for both qualitative and quantitative analysis. In thin layer
chromatography (TLC) the mixture of components is separated by placing a concentrated
spot of the mixture on the origin and placing the TLC plate in the solvent.
The stationary phase is the solid material. Separation occurs as a result of the differing
abilities of the components to adsorb to the stationary phase and desorb in the mobile phase.
The more strongly components adsorb to the stationary phase, the slower the rate of
movement of the component over the stationary phase and the smaller the Rf value will be.
Before the solvent reaches the top of the stationary phase, the process is stopped and the
solvent front is marked to allow calculation of Rf values.
b
In high performance liquid chromatography (HPLC) and gas chromatography (GC) the time
taken for a component to pass through the instrument is measured by the Rt value.
Components that adsorb well to the stationary phase have large Rt values.
HPLC is used to separate organic compounds such as proteins. Because these compounds
have high molecular masses, it is often too difficult to vaporise them as required in GC. In
order to increase the degree of separation in the HPLC, the components are passed through a
column packed with fine particles under pressure.
In GC the degree of separation is increased by using a very long, coiled column. The
stationary phase is a porous finely divided, solid coated with a high boiling point liquid
hydrocarbon called the liquid stationary phase. The gas often used as the mobile phase is
nitrogen and it is known as the carrier gas.
2
a
It is proportional to the amount of each substance present.
b
The Rt values were compared to standards of the same compounds whose chromatograms
were run under the same conditions.
c
Both qualitative and quantitative analysis
d
On the basis of peak heights, the ratio of palmitate to stearate is 76 : 54 = 1.4. This is within
the limits for linseed oil.
Worksheet 9 A chemical mystery—salicylic acid
Identifying salicylic acid
1
a
b
c
Initial preparation of the TLC plate:
 Mark origin line on plate, ensuring that it is at a height greater than the depth
of the solvent.
 Concentrate a spot of the sample solution on the origin to one side of the
plate.
 Concentrate a spot of the standard solution on the origin on the other side of
the plate.
Producing the chromatogram:
 Stand the plate in the solvent vertically.
 Allow the mobile phase to move up the plate until it almost reaches the top.
 Remove the plate, mark the solvent front and dry quickly.
Examination of the TLC plate:
 Measure the distance the sample has moved up the plate.
 Measure the distance the standard has moved up the plate.
 If they are identical, it is likely that the sample is the same compound as the
standard - salicylic acid. (You could calculate the Rf of each, but it is not
necessary here because the distance each sample moved can be compared
directly on the plate.)
d
TLC chromatogram
Both spots moved the same distance from the origin under the same conditions. Therefore,
this compound can be assumed to be salicylic acid.
2
a
Molecular ion is at mass = 138.
b
Likely molecular mass = 138 g mol–1.
c
The molecular formula of salicylic acid is C7H6O3. The calculated molecular mass is 138
[(7×12) + (6×1) + (3×16) = 138]. The white powder is likely to be salicylic acid because the
calculated molecular mass equals the mass of the molecular ion in the mass spectrum.
Determining the structure
3
The band at about 1700 cm–1 is possibly produced by C=O group in a carboxylic acid and one
around 3300 cm–1 is due to the O–H group.
It therefore possibly contains an alcohol and a carboxylic acid group, probably attached to a
benzene ring.
4
The seven peaks in the 13C NMR spectrum indicate that there are seven non-equivalent carbon
atoms in salicylic acid.
5
Molecular mass: 138
Molecular formula: C7H6O3
Functional groups present: C=O in a carboxylic acid group and O–H on a benzene ring
Number of carbon atoms in different environments: 7
6
Structure of salicylic acid
Worksheet 10 Booze buses—forensic testing
1
Oxidation: CH3CH2OH(aq) + H2O(l) → CH3COOH(aq) + 4H+(aq) + 4e−
Reduction: O2(g) + 4H+(aq) + 4e− → 2H2O(l)
2
0.05 mg per 100 mL = 0.000 05 g per 100 mL = 0.0005 g per 1000 mL (1L) =
0.0005
mol per L
46
= 0.000 011 mol L–1
3
4
a
Because it is different from the absorption bands for water vapour and therefore
distinguishable. Furthermore, it is a strong absorption, so small amounts are detectable.
b
They both contain an –OH group that will stretch and bend in a similar way when absorbing
energy.
c
i
C–H bond
ii
O–H bond
iii
C–O bond
a
Oven: maintains a constant temperature throughout the analysis so that the mobile phase and
the samples are in the gas phase.
Column: a long loop that is packed with a porous solid that is coated with the liquid
stationary phase.
Injection port: where the sample is injected before it is vaporised.
b
Standard solutions of different concentrations of ethanol are prepared. Each standard is run
through the GLC to obtain chromatograms. The area under the peak is calculated for each
standard solution. A graph of peak area versus concentration is plotted.
c
The peak due to ethanol can be identified by its Rt value. The peak area of the sample is
measured and its concentration is found from the calibration graph. From the concentration,
the BAC can be calculated.
Worksheet 11 Decision-making—choosing an analytical technique
Aim
Technique
Reason
1 Determine the
concentration of red
colour in cordial.
UV-visible
Coloured organic compounds absorb light in the UV–visible
region
2 Determine the
concentration of mercury
in oysters.
AAS
Metals readily absorb light of a wavelength that is unique to
that particular metal.
3 Identify the low-mass
carboxylic acids in wine.
GC
Carboxylic acids of low molar mass are easily vaporised in
the oven of the chromatograph and would produce different
Rt values.
4 Determine the molecular
mass of a new
insecticide.
mass
spectroscopy
The parent ion peak in the mass spectrum would give the
relative molecular mass.
5 Identify the functional
groups in a plant extract.
infrared
spectroscopy
Different functional groups absorb different energies in the
IR spectrum.
6 Determine the
concentration of a protein
marker in blood.
HPLC
Proteins are too large to vaporise in a GC but are readily
separated under pressure in a HPLC column.
7 Determine the
concentration of ethanol
in methylated spirits.
GC
Methanol and ethanol are easily vaporised and readily pass
through the coil of a GC, producing a chromatogram. The
area under the peak due to ethanol can be used to
determine the concentration of ethanol.
8 Identify pigments in
forgeries and counterfeit
banknotes without
destroying the sample.
synchrotron IR
The sample size of a particular pigment would be very
small and the note may be needed again for further testing.
Pigments contain different functional groups that may be
identified by IR spectroscopy.
Worksheet 12 High energy spectroscopy—synchrotron infrared
1
When molecules absorb energy in the IR region of the spectrum, the molecule moves to higher
vibrational energy levels. The energy absorbed depends on the strength of the bond and the mass
of the atoms attached in the bond. A single bond absorbs lower energy radiation than a triple
bond and bonds involving heavier atoms absorb lower energy radiation than those between light
atoms. The radiation causes molecules to stretch (change distance between atoms) and bend
(change angle between bonds).
2
A very tiny sample is needed, providing an excellent IR spectrum and the tissue is not destroyed
and can be reanalysed.
3
The strength of the bonds and the size of the atoms involved in the bond.
Worksheet 13 Atoms in a spin—nuclear magnetic resonance
1
Nuclei with an odd number of protons or with an odd number of neutrons that are in a magnetic
field can absorb radio waves of particular energies. The nuclei move from the lower-energy state
to higher-energy state are said to be in resonance.
2
Compound
Groups in which
protons are found
No. of different
types of protons
No. of different
types of carbon
atoms
Propane
CH3 and CH2
2
2
Ethanoic acid
CH3 and OH
2
2
3
1 E, 2 C, 3 G, 4 H, 5 B, 6 A
4
Chemical
Corresponding 1H spectrum
Corresponding 13C spectrum
CH3CH2OH
D
J
CH3COOCH2CH3
C
F
HCOOH
A
H
CH3CH2CH2Cl
E
G
CH3CH2CH2CH2CH2OH
B
I