trans:it science
trans:it science
APPENDIX: PRACTICAL WORK
Chemistry, like other sciences, is a practical subject. To become successful
chemists, students need to master a range of laboratory techniques. University
chemistry courses provide plenty of chances to practice the required skills, but to get
the most out of these opportunities you need to prepare well.
Planning an experiment
Even if you are provided with a fully comprehensive lab script, it is worth spending
the time to carefully read it through before you start any practical work. This provides
you with a chance to read up on any techniques you are unfamiliar with or have not
done for a while, as well as a chance for you to understand the chemistry involved.
Sometimes you may be required to do some pre-planning for an experiment, such as
calculating amounts of reagents needed.
Even if it is not required, it is often useful to draw up a plan for the experiment – this
will help you use your time more effectively when in the laboratory, as well as help
prevent you missing any vital steps! Careful planning allows you to spot where time
can be saved e.g. when a reaction is being heated under reflux, things for the next
step can sometimes be prepared.
Try to make yourself familiar with common laboratory techniques such as methods of
purification (recrystallisation, separation and chromatography) and standard
apparatus setups (e.g. reflux and distillation).
Carrying out the experiment

Record any molar calculations, observations and results in a lab book.

Remember it is better to record too much information than too little.

If unsure about anything, ask, especially if it relates to using something safely.
101
trans:it science

Make sure your glassware is clean before you start. It is much better to spend
time ensuring everything is properly clean at the beginning, than finding out
that your reaction has been contaminated by dirty glassware later on.
Useful book:
Dean, J. et al (2002). Practical Skills in Chemistry. Harlow: Pearson.
A book explaining various practical techniques and methods including the running
and interpretation of spectra, purification techniques, distillation and reflux.
Useful units
To make it easier for scientists to communicate with each other and to simplify
equations, a standard set of units, the SI units, was adopted by the scientific
community around the middle of the 20th century. However, you will still find that nonstandard units are used sometimes for convenience on a day-to-day basis. If you
look at glassware in the laboratory you may find that it is marked in millilitres (ml)
rather than in cm3 (the SI unit for volume is the cubic metre, m3). Thankfully, this is a
very simple conversion, as 1 ml = 1 cm3, but some others are not quite so
straightforward.
Whenever you are writing up a report, or carrying out a calculation you will need to
use SI units, so if you have a volume in cm3, you will need to convert it to m3 first. At
the very least, you should have all your units the same (for example, you shouldn’t
have a mixture of m3, dm3 and cm3 in the same equation).
quantity
SI unit
Mass
kg (kilogram)
Length
m (metre)
Time
s (second)
Temperature
K (kelvin)
Amount of substance
mol (mole)
Volume
m3 (cubic metre)
Energy or work
J (joule)
Force
N (newton)
Density
kg m-3 (kilograms per cubic metre)
Concentration
mol m-3(moles per cubic metre)
102
trans:it science
SI prefixes and multiplication factors
It is not always sensible to use the appropriate SI unit, if the quantity you are dealing
with is very large or very small it is convenient to use a prefix to denote multiples of
the unit. Some of these prefixes may be familiar to you, others less so:
multiplication factor
prefix
symbol
1 000 000 000 000 000 000 000 000 = 10 24
yotta
Y
1 000 000 000 000 000 000 000 = 1021
zetta
Z
1 000 000 000 000 000 000 = 1018
exa
E
1 000 000 000 000 000 = 1015
peta
P
1 000 000 000 000 = 1012
tera
T
1 000 000 000 = 109
giga
G
1 000 000 = 106
mega
M
kilo
k
100 = 102
hecto
h
10 = 101
deca
da
0.1 = 10-1
deci
d
0.01 = 10-2
centi
c
0.001 = 10-3
milli
m
0.000 001 = 10-6
micro

0.000 000 001 = 10-9
nano
n
0.000 000 000 001 = 10-12
pico
p
femto
f
atto
a
0.000 000 000 000 000 000 001 = 10-21
zepto
z
0.000 000 000 000 000 000 000 001 = 10 -24
yocto
y
1 000 = 103
0.000 000 000 000 001 = 10-15
0.000 000 000 000 000 001 = 10-18
103
trans:it science
Some useful conversions and equations
To convert a temperature in °C into K, simply add 273.15.
For example, 22°C in kelvin = 22°C + 273.15 = 295.15 K
Number of moles (mol) 
Density (kg m- 3 ) 
mass of substance (g)
mass of one mole (g mol -1 )
mass (kg)
volume (m3 )
Concentrat ion (mol m- 3 ) 
Percentage yield (%) 
molar amount (mol)
volume (m3 )
molar amount of product
 100
maximum molar yield from limiting reagent
Determining the yield of a reaction…
Example One:
For the preparation of copper(II) sulfate from copper(II) carbonate:
CuCO3 + H2SO4 + 5H2O  CuSO4.5H2O + H2O + CO2(g)
We react 5 g copper carbonate (assume this to be anhydrous CuCO3) with 90 cm3 of
1 mol dm-3 H2SO4, and get 9.2 g of the product, CuSO4.5H2O. What is the reaction
yield?
To start, we can calculate the relative molecular masses and hence the molar
amounts of the reagents:
Relative atomic masses:
Cu
63.546
C
12.011
O
15.9994
Relative molecular mass of CuCO3 = 63.546 + 12.011 + (3  15.9994) = 123.5552
104
trans:it science
So the molar mass of CuCO3 is 123.56 g mol-1
Number of moles of CuCO 3 
5.00 g
 0.04 mol
123.56 g mol -1
As the volume of sulfuric acid is given in cm3, we need to convert to dm3 by dividing
by 1000 (as there are 1000 cm3 in 1 dm3).
Number of moles of H2SO4  1mol dm- 3 
Reagent
CuCO3
Relative
molecular mass
123.55
H2SO4
90 cm3
 1 mol dm-3  0.09 dm3 = 0.09 mol
1000
Amount used
Molar amount
5.00 g
0.04 mol
90 cm3 = 0.09 dm3
0.09 mol
From the equation:
CuCO3 + H2SO4 + 5H2O  CuSO4.5H2O + H2O + CO2(g)
…we can see that one mole of CuCO3 reacts with one mole of H2SO4. We are
reacting 0.04 moles of CuCO3 with 0.09 moles of H2SO4, so we can see that the acid
is in excess and the CuCO3 is the limiting reagent. The maximum yield of the
reaction is therefore equal to the number of moles of CuCO3, i.e. 0.04 mol.
Now we need to calculate the relative molecular mass (Mr) and the number of moles
of product (CuSO4.5H2O) we have obtained:
105
trans:it science
Relative atomic masses:
Cu
63.546
S
32.06
O
15.9994
H
1.008
Mr CuSO4.5H2O = 63.546 + 32.06 + (9 × 15.9994) + (10 × 1.008) = 249.6806
So the molar mass of CuSO4.5H2O is 249.68 g mol-1
Number of moles of CuSO 4 .5H2O 
9.2 g
 0.037 mol
249.68 g mol -1
The percentage yield of the reaction is therefore:
Percentage yield 
actual yield of product
0.037 mol
 100 
 100  92.5%
theoretica l yield of reagent
0.040 mol
106
trans:it science
Example Two:
Try this one for yourself then check your answers on pages 11 -12 of this Appendix.
You react 1.65 cm3 benzaldehyde with 1.0 cm3 pyrrole and obtain 0.51 g of the
desired product, tetraphenylporphyrin:
H
O
NH
H
N
N
4
4
N
benzaldehyde
HN
pyrrole
tetraphenylporphyrin
To calculate the molar amounts, you would need to look up the densities and the
molar masses of the reactants, but we have done that for you:
Reagent
Volume used
Density at 20C
Molar mass
benzaldehyde
1.65 cm3
1.044 g cm-3
106.12 g mol-1
pyrrole
1.00 cm3
0.969 g cm-3
67.09 g mol-1
1a: Using the volume and the density, calculate the mass of benzaldehyde used.
1b: Using the mass and the molar mass, calculate the number of moles of
benzaldehyde used.
2: Repeat the process for pyrrole, to work out how many moles of pyrrole were used.
Now looking at the product:
Product
Mass produced
Molar mass
tetraphenylporphyrin
0.51 g
614.74 g mol-1
3: Calculate how many moles of tetraphenylporphyrin were produced.
From the equation for the reaction, we can see that four molar equivalents of
benzaldehyde are reacted with four of pyrrole to give one mole of product.
107
trans:it science
4: Looking at the results of your calculations, work out which of the two reagents is
the limiting reagent. This is the one that should be used to calculate the yield of the
reaction.
5: What is the maximum theoretical yield of product for this reaction?
6: What is the percentage yield of tetraphenylporphyrin?
(Now check your answers on pages 11-12 of this Appendix).
Report writing

Make sure you carefully read what is required.

If things have not gone according to plan, don’t panic! Having a bad day in the
lab will occasionally happen to everyone, even the best of us! If things have
gone wrong, can you work out in which part of the experiment this happened?
Often you will be given some credit if you can objectively offer plausible
explanations as to why things did not go right.

When answering questions on the experiment, it is often helpful to use
textbooks to provide some background information.

If you have to do a full experimental or project report, taking time to plan it first
will help you to write a focused and well structured piece of work.
The full report (stages):
Abstract: A short summary of what was done and the main results.
For example:
Mono- and bis-hydride ruthenium(II) complexes containing triphenylphosphine
were prepared in 58% and 42% yields respectively, and were characterised
using 1H and 31P NMR and infrared spectroscopy.
Introduction: Introduce background theory to put the science of the experiment into
context.
108
trans:it science
Experimental: This is where you set out exactly what you did, the techniques used,
the amounts of reagents etc.
Results and Analysis: Clearly present all your experimental data in an appropriate
format, such as tables and properly labelled figures.
Discussion: Relate your experimental results to theory; explain the science behind
your results using academic paper references and text books. Where possible,
compare your results to similar results that have already been published. Where
appropriate, comment on how reliable your data is.
Conclusions: Clearly restate the significant conclusions you have drawn from your
experimental data including qualitative (e.g. state or colour changes) and quantitative
results, such as reaction yields and other calculated or measured values.
References: You need to clearly state what sources you have used for any particular
pieces of information. Sources can include textbooks, journals and sometimes
websites (but be very careful about the reliability of information on websites).
Useful books to help with your Chemistry degree
Usually you can find out which textbooks are recommended before or when you start
your degree, but if not you might find the following helpful:
1. Keynotes in Organic Chemistry (Andrew F. Parsons, Blackwell Publishing).
This book is very useful when revising organic chemistry, but also to use in
conjunction with a larger organic textbook.
2. Chemistry3: introducing inorganic, organic and physical chemistry. (Andrew
Burrows, John Holman, Andrew Parsons, Gwen Pilling and Gareth Price,
Oxford University Press). A text book covering the principles from the three
main areas of Chemistry that you will cover within the first year of a degree
course.
109
trans:it science
3. Maths for chemists (Martin C.R. Cockett, Graham Doggett, Royal Society of
Chemistry). Volume one covers numbers, functions and calculus whilst
volume two covers power series, complex numbers and linear algebra.
4. Beginning Mathematics for Chemists. (Stephen K. Scott, Oxford University
Press). A workbook designed to help with mathematical skills by providing
practice through working through problems.
5. Oxford University Press has published an extensive series of Chemistry
Primers; these are short textbooks on a wide range of chemistry areas such
as aromatic chemistry, chemical bonding, and first-row transition metal
chemistry. Details available:
http://ukcatalogue.oup.com/nav/p/category/academic/series/chemistry/ocp.do
?sortby=bookTitleAscend&thumbby_crawl=10&thumbby=50
Useful equations
(You will probably have come across some of these before)

Gradient of a straight line: y = mx + c
y = value on the y axis
x = value on the x axis
m = gradient
c = the intercept on the y axis

Ideal gas equation: pV = nRT
p = pressure in Pa
V = volume in dm3
n = molar amount in mol
R = gas constant = 8.314 J K-1 mol-1
T = temperature in K

Calculating the enthalpy change of a reaction:
ΔH = Hproducts – Hreactants

Calculating the entropy change of a reaction:
ΔS = Sproducts – Sreactants
110
trans:it science

Relationship between free energy change, enthalpy change,
entropy change and temperature: ΔH = ΔG -TΔS
ΔH = change in enthalpy
ΔG = change in free energy
T = temperature
ΔS = change in entropy

Relationship between pH and proton concentration: pH = -log10[H+]
[H+] = concentration of hydrogen ions (protons)

Calculation of the acid dissociation constant, Ka, for a particular
acid:
Ka 
HA
acid
[ A - ][H ]
[HA]
H+ +
proton
Aconjugate base
For example:
CH3CO2H
acid
H+ +
proton
CH3CO2conjugate base
[A-] = the concentration of the conjugate base
[H+] = proton concentration
[HA] = concentration of acid

Relationship between pKa and Ka:
pKa = -log10Ka

Relationship between pH and pKa:
 [A - ] 

pH  pKa  log10 
 [HA] 
The laws of thermodynamics:
1. Energy can neither be created nor destroyed, only interconverted between
forms.
2. The entropy within a closed system increases i.e. becomes more disordered.
3. Zero entropy can only be achieved in a perfect crystal at absolute zero i.e.
when the temperature is 0 K.
111
trans:it science
Answers to Example 2
Determining the yield of a reaction
Example Two: The answers
You react 1.65 cm3 benzaldehyde with 1.0 cm3 pyrrole and obtain 0.51 g of the
desired product, tetraphenylporphyrin:
H
O
4
benzaldehyde
NH
H
N
N
4
N
HN
pyrrole
tetraphenylporphyrin
To calculate the molar amounts, you would need to look up the densities and the
molar masses of the reactants, but we have done that for you:
Reagent
Volume used
Density at 20C
Molar mass
benzaldehyde
1.65 cm3
1.044 g cm-3
106.12 g mol-1
pyrrole
1.00 cm3
0.969 g cm-3
67.09 g mol-1
1a: Using the volume and the density, calculate the mass of benzaldehyde used.
Mass of benzaldehyde used = 1.65 cm3  1.044 g cm-3 = 1.72 g
Note that the answer is given to 2 decimal places, in line with the mass that
was measured. We should not imply a greater level of accuracy than this.
1b: Using the mass and the molar mass, calculate the number of moles of
benzaldehyde used.
Number of moles benzaldehyde used =
1.72 g
= 0.016 mol
106.12 g mol -1
Here the answer is given to 2 significant figures.
112
trans:it science
2: Repeat the process for pyrrole, to work out how many moles of pyrrole were used.
Mass of pyrrole used = 1.00 cm3  0.969 g cm-3 = 0.97 g
Number of moles of pyrrole used =
0.97 g
= 0.014 mol
67.09 g mol -1
Now looking at the product:
Product
Mass produced
Molar mass
tetraphenylporphyrin
0.51 g
614.74 g mol-1
3: Calculate how many moles of tetraphenylporphyrin were produced.
Number of moles of product obtained =
0.51 g
= 0.00083 mol
614.74 g mol -1
Answer given to 2 significant figures.
From the equation for the reaction, we can see that four molar equivalents of
benzaldehyde are reacted with four of pyrrole to give one mole of product.
4: Looking at the results of your calculations, work out which of the two reagents is
the limiting reagent. This is the one that should be used to calculate the yield of the
reaction.
From the molar calculations, we can see that the benzaldehyde (at 0.016 mol)
is in slight excess compared to the pyrrole (0.014 mol), so pyrrole is the
limiting reagent for the reaction and therefore will be used to calculate the
yield.
113
trans:it science
5: What is the maximum theoretical yield of product for this reaction?
Maximum theoretical yield of product
=
no. of moles pyrrole used
4
=
0.014
= 0.0035 mol
4
6: What is the percentage yield of tetraphenylporphyrin?
Percentage yield 
actual yield of product
0.00083 mol
 100 
 100  23.7%
theoretica l yield of product
0.0035 mol
CREATIVE COMMONS LICENSING:
You are free:

to Share — to copy, distribute and transmit the work

to Remix — to adapt the work
Under the following conditions:

Attribution — You must attribute the work in the manner specified by the author or
licensor (but not in any way that suggests that they endorse you or your use of the
work).

Noncommercial — You may not use this work for commercial purposes.

Share Alike — If you alter, transform, or build upon this work, you may distribute the
resulting work only under the same or similar license to this one.
With the understanding that:

Waiver — Any of the above conditions can be waived if you get permission from the
copyright holder.

Public Domain — Where the work or any of its elements is in the public domain under
applicable law, that status is in no way affected by the license.

Other Rights — In no way are any of the following rights affected by the license:
o
Your fair dealing or fair use rights, or other applicable copyright exceptions and
limitations;
o
The author's moral rights;
o
Rights other persons may have either in the work itself or in how the work is
used, such as publicity or privacy rights.
Notice — For any reuse or distribution, you must make clear to others the license terms of this
work
114