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Concentrations and Dilutions:
A survival guide
In your science experiments, often you need to know how much material to use or how
much of a solute is present in a solution. This applies particularly to biology and chemistry
experiments. Almost certainly you will need to do some calculations!
This survival guide should help you to do these calculations. It looks closely at how to handle
numbers correctly, how to express concentrations and how to calculate dilutions. There are
plenty of worked examples to guide you, then lots of questions (with answers) for you to
practise your skills.
The five sections in the guide are listed below:
1. Numbers in standard form
2. Relative atomic mass and molar mass
3. Amounts in solution
4. Percentage weight/volume
5. Dilutions
You may decide to work through step by step from the beginning or you may prefer to jump
to the section that will help you with your problem.
Look for red to find the questions you are encouraged to attempt.
Look for green to find the answers just click on check answer.
At the end of the guide, we hope you will have developed greater confidence with handling
numbers, concentrations and dilutions. You might even be more enthusiastic!
1. Numbers in standard form
It is important that you are able to handle, with confidence, calculations that deal with num-
bers in so-called standard form. Sometimes standard form is called scientific or index notation
but they mean the same thing.
Invariably concentrations are expressed in standard form and so this is an important skill that
you must develop. At first it may seem daunting but stick with it! In our experience the use of
standard form is a very common area where students have problems and so it will be good
practice for you to try out the calculations in the examples that follow later. You might also
find it worth looking at the additional explanations and examples given in Appendix 1.
Table 1. Some common powers of 10
Number Standard Form
100 000 000 1.0 x 10 8
10 000 000 1.0 x 10^7
1 000 000 1.0 x 10^6
100 000 1.0 x 10 5
10 000 1.0 x 10 4
1 000 1.0 x 10 3
100 1.0 x 10^2
10 1.0 x 10^1
(^1) 1.0 x 10^0
Let’s answer this by looking at an example. Suppose we wish to convert the number 512 to standard form – how would we do it? Well a number in standard form is expressed as a value between 1 and 10 multiplied by a power of 10. Let’s look at the number 512. We can say that
512 = 5.12 x 100
Now 100 = 1.0 x 10 2
(If you have not come across this way of writing numbers before don’t worry! We have included an appendix that gives a more detailed explanation. Common powers of 10 are also given in the Table below).
If we combine the two equations above we obtain
512 = 5.12 x 1.0 x 10 2 = 5.12x10^2
How do we go from a normal number to the standard from version?
The easiest way to deal with numbers in standard form is, of course, to use a scientific
calculator that will do the conversions for you. Most calculators automatically convert numbers
to standard form once they get bigger than 100 000 000 (100 million) – although this state-
ment is meant to be a rough guide only! Most calculators have a 'mode' key that allows you to
have all numbers presented in scientific format (a term often used to describe standard form).
(Ignoring a few of the decimal places) the calculator display will look something like
It is important that you realise the meaning of the number 11 which is separated from the
rest of the digits.
In fact 123456 x 4567890 = 5.63933 x 10^11
Do we need to deal with large numbers such as 6.02 x 10 23 ?
Well the answer to that is yes!
The number 6.02 x 10^23 is very important in science
- it is called Avogadro's number - we shall return to
Avogadro's number later.
Try the following calculation (using your calculator): 123456 x 4567890
and so the number 11 represents the index of the power of 10 of the number. We would
normally state verbally that the calculator reads as "5.639 times ten to the eleven".
We said before that most calculators will automatically convert numbers bigger than 10^8 into
standard form. You should try and find out whether you can convert numbers lower than this
to standard form using your calculator (you might need to read the instruction booklet that
accompanies your calculator!)
The chances are that your calculator display will look similar to one of the following:
In a similar way to the multiplication above, the -03 represents a power of 10 that is less than
zero. In this case the number can be written as:
0.009009009 or 9.
Now, 1 ÷ 1000 can be written as 1 x 10-3^ and so 0.009009009 can be written as 9.009009 x 10-
In fact some (often older) calculators convert numbers smaller than 0.01 to standard form
automatically. Try and type in the number 0.0032. Once you have typed in the final digit press
the '=' button. Some calculators will automatically change the display to 3.2 -03, i.e. 3.2 x 10-3.
Other calculators will convert numbers to scientific notation only when you set them to do so
(check the instruction booklet). We would normally state verbally that the calculator reads as
‘3.2 times ten to the minus three’. It might be useful to look at the following table which lists
some numbers and their equivalents in standard form.
Try the following calculation (use your calculator): 5 ÷ 555
You need to be careful that you follow these instructions in the correct order and you
also need to check how your calculator works.
How do I enter numbers in standard form into my calculator?
Suppose that you wish to enter the number 3.46 x 10 12 into
your calculator - what do you do?
Each calculator is slightly different but in general you would
type in 3.46 followed by the key marked 'EXP' followed by 12.
The 'EXP' key takes the place of 'times 10 to the power of'.
When dealing with numbers less than 1, for example 0.00346,
you would type 3.46 'EXP' ± 3. The button marked ± changes
the sign of the number. On some calculators it looks like +/-.
2. Relative atomic mass and molar mass
Atoms are very small particles! This means that their masses give us figures that are difficult to
work with. The table below lists the average mass of an atom of some common elements in
grams.
To overcome the difficulty of working with such small amounts of material it is usual to work
with a relative scale for the atomic mass. The scale used is known as the carbon-12 scale. This
was adopted by IUPAC (the International Union of Pure and Applied Chemistry) in 1961 as its
standard. In this carbon-12 scale, the mass of a carbon atom of isotope number 12 (symbol -
12 C) is 1.99252 x 10 -23 g and the relative mass of all atoms is compared to 1/12 th of the mass
of 12 C. The relative mass is, therefore calculated as:
Relative mass = Average mass ÷ 1/12th^ of mass of 12 C
We can now re-write the above table to give relative masses. You can see that relative masses
numbers are much easier to handle than average masses
When we come on to looking at the concentrations of materials we will need to be able to
express the amount of a substance that is contained in a given volume of solution. The SI unit
for quantity of material is called the mole (abbreviation mol). The mole is defined as the
amount of substance that contains as many elementary particles as there are atoms in 12 g
of 12 C. This definition of a mole is important.
Element Symbol Average mass of an atom / g
Hydrogen H 1.67355 x 10 -
Helium He 6.64605 x 10 -
Lithium Li 1.15217 x 10 -
Carbon C 1.99436 x 10 -
Oxygen O 2.65659 x 10 -
Sodium Na 3.81730 x 10 -
Argon Ar 6.63310 x 10 -
Uranium U 3.95233 x 10 -
Symbol Average mass of an atom / g Relative mass
H 1.67355 x 10 -24^ 1.
He 6.64605 x 10 -24^ 4.
Li 1.15217 x 10 -23^ 6.
C 1.99436 x 10 -23^ 12.
O (^) 2.65659 x 10 -23^ 15.
Na 3.81730 x 10 -23^ 22.
Ar 6.63310 x 10 -23^ 39.
U 3.95233 x 10 -22^ 238.
You should now attempt the sample questions in SAQ3 - answers are given. You will probably
need to refer to the table of relative atomic masses of a range of elements. You may also need
to consult a suitable textbook that gives the molecular formulae for the compounds listed.
Hopefully (!) once you have completed SAQ3 you will feel more confident about how to
calculate the molar masses of substances.
It is often necessary to calculate how many moles of a substance there are in a given mass of
material. This can be done using the following expression:
Number of moles = mass present ÷ molar mass
During practical work you may come across occasions when you need to do calculations such
as these and so it is important that you are able to handle them accurately and confidently.
To help you further, we recommend that you try the examples in SAQ4.
Self-Assessment Questions 3
Calculate the molar masses (the mass of 1 mol) of the following molecules:
(i) water (H 2 O) (ii) carbon dioxide (CO 2 ) (iii) oxygen (O 2 ) (iv) nitrogen (N 2 ) (v) ammonia (NH 3 ) (vi) glucose (C 6 H 12 O 6 ) (vii) sucrose (C 12 H 22 O 11 )
Check answers
Worked examples
(i) Calculate the number of moles of material present in 6.3 g of carbon dioxide.
Answer:
Number of moles = mass present ÷ molar mass
= 6.3 ÷ 44 = 0.143 mol carbon dioxide
(ii) Calculate the number of moles of material present in 12.5 g of glucose.
Answer:
Number of moles = mass present ÷ molar mass
= 12.5 ÷ 180 = 0.069 mol glucose
Self-Assessment Questions 4
Calculate the number of moles of material present in each of the following:
(i) 0.5 g of glycine (molecular formula C 2 H 5 NO 2 )
(ii) 10 g of urea (molecular formula CN 2 H 4 O)
(iii) 25 g of disodium hydrogen phosphate (molecular formula Na 2 HPO 4 )
(iv) 250 g of fructose
(v) 0.0028 g of ribose (molecular formula = C 5 H 10 O 5 )
(vi) 12 g of sodium hydroxide
(vii) 15.8 g of glucose
Check answers
The concentration of a substance is defined as the number of moles of the substance that is
dissolved in 1 dm 3 of solvent. So the concentration of a substance is measured in mol dm-
(verbally stated as moles per decimetre cubed or moles per cubic decimetre). You will occasion-
ally see the symbol M or the term molarity both of which are used as alternatives to mol dm-^.
The use of both these terms is discouraged and we recommend that you adopt mol dm -^.
It is important that you practise calculations such as these since they form an integral part of
any advanced course in biology and related areas.
Worked example
A solution of sodium chloride in water is prepared by dissolving 11.7 g of solid in 1.0 dm 3 of water. What is the concentration of the sodium chloride in mol dm -3^?
Answer:
The mass of one mole of NaCl may be calculated as 23 + 35.5 g = 58.5 g
So, 11.7 g of NaCl represents 11.7 ÷ 58.5 mol NaCl = 0.2 mol
The solution contains 0.2 mole NaCl dissolved in 1 dm 3 of water.
So, the concentration of the solution is:
0.2 ÷ 1 mol dm-3^ = 0.2 mol dm-
Worked example
A solution of sodium chloride in water is prepared by dissolving 6.5 g of solid 250 cm 3 of water. What is the concentration of the sodium chloride in mol dm -3^?
Answer:
The mass of one mole of NaCl may be calculated as 23 + 35.5 g = 58.5 g
So, 6.5 g represents 6.5 ÷ 58.5 mol NaCl = 0.111 mol
The solution contains 0.111 mol NaCl dissolved in 250 cm 3 of water.
Now 250 cm^3 represents 250/1000 dm^3 = 0.25 dm^3.
So, the concentration of the solution is:
0.111 ÷ 0.25 mol dm-3^ = 0.444 mol dm-
Why don’t you try the following examples!
Self-Assessment Questions 5
(i) A solution of disodium hydrogen phosphate (Na 2 HPO 4 ) is prepared. If the solution contained 71.0 g of Na 2 HPO 4 in 1 dm 3 of water, what is the concentration in mol dm -3^?
(ii) A solution is made containing 2.38 g of magnesium chloride (MgCl 2 ) in 500 cm^3 of water. What is the concentration of MgCl 2 in this solution?
(iii) The table below indicates the masses of various compounds that were used to prepare solutions of the stated volumes. Calculate the concentrations of these solutions.
Compound Molecular Mass / g Volume water / cm 3 formula
(a) Glucose C 6 H 12 O 6 8.5 1000 (b) Ribose C 5 H 10 O 5 10.7 500 (c) Glycine C 2 H 5 NO 2 25.7 2000 (d) Sucrose C 12 H 22 O 11 18.5 375
(iv) You are asked to prepare a 100 cm^3 of a solution of ribose (molecular formula, C 5 H 10 O 5 ) at a concentration of 1.0 x 10-4^ mol dm-3^. How much ribose would you need?
(v) How much disodium hydrogen phosphate (Na 2 HPO 4 ) is needed to prepare 5 dm 3 of solution with a concentration of 1.8 x 10 -2^ mol dm-3^?
Check answers
Some molecules have very high molar masses.
This is particularly true for some large biological
molecules such as proteins. For example, albumin,
a protein, has a molar mass of 68 000 g mol -3^!
Some molecules may have molar masses that are
greater than 1 000 000 (or 1 x 10 6 ) g mol-3^!
What about molecules with large molar masses?
A couple of questions. Is it true that the molar masses of some molecules are unknown? If so, how do we write down values for concentrations of solutions of such molecules?
Excellent questions!
There are indeed some molecules for which the molar mass is
unknown. Typically such molecules are large (it is more difficult
to determine the accurate molar mass of a large molecule) and
have molecular structures that are not well defined. Of course
it is still useful to be able to give an indication of the amount
of material in solution and to do this we often use a so-called
percent weight / volume (% w / v) method of expressing
concentration. The % w / v of a solution is defined as the
number of g of material dissolved in 100 cm 3 of solvent. We
have given a few examples below to show you how it works.
Worked examples
(i) A solution containing 1 g of starch dissolved in 100 cm^3 of water is prepared. What is its concentration?
Answer:
Starch is a polymer made up of glucose molecules. The molar mass of starch varies according to its source and method of extraction. In order to give an indication of the concentration we can express it in terms of percentage weight / volume (often abbreviated to % w / v). We have 1 g of starch dissolved in 100 cm^3 of solvent and so the concentration is 1% w / v.
(ii) A small quantity (0.005 g) protein of unknown molar mass was dissolved in 10 cm 3 of water. What is the protein concentration in terms of % w / v?
Answer:
In this case we have 0.005 g of the protein in 10 cm 3 of water – this is equivalent to 0.05 g of protein in 100 cm^3 of water. Thus the concentration of the protein solution is 0.05% w / v.
4. Percentage weight/volume
Self-Assessment Questions 7
Express the following solutions in terms of % w / v
(i) 0.5 g of DNA in 100 cm 3 water (ii) 0.013 g of RNA in 10 cm 3 of water (iii) 0.02 g of starch in 500 cm^3 of water (iv) 0.007 g of DNA in 12.5 cm 3 of buffer
Check answers
Suppose that you are given 500 cm^3 of a stock solution of sodium chloride. The concentration of the stock solution is 0.5 mol dm -3^. You are asked to use the stock solution to prepare new solutions (to have a final volume of 50 cm 3 ) of each of the following concentrations:
0.1 mol dm-3^ ; 0.2 mol dm-3^ ; 0.3 mol dm-3^ ; 0.4 mol dm-^.
How do you calculate what is needed? Well, there are a number of ways you could do this calculation.
Method 1 Let’s start by thinking how much of the solution you wish to prepare. The final volume of each sample that we require is 50 cm 3.
If we took 1 cm^3 of the stock solution and added 49 cm^3 of water we would produce a solution whose concentration was 1/50th^ of the original (remember the concentration of the stock solution is 0.5 mol dm - and so 1/50 th^ of that is 0.01 mol dm -3^ ). Similarly if we took 2 cm 3 of the stock solution and added 48 cm^3 of water we would produce a solution whose concentration was 2/50 th^ (or 1/25th^ ) of the original and so on...
You could build up a table:
OK, I have this solution and I need to make a series of weaker solutions. How do I work out what I need to do?
Students often have problems in calculating the effect of
diluting solutions. You may be given a solution (the 'stock'
solution) and asked to prepare from it a range of solutions
of different concentrations.
In this section we will look at how to calculate the effect of
diluting solutions.
Volume of stock taken / cm 3
Volume of water taken / cm 3
Dilution Factor (= 50 ÷ volume taken)
Final concentration / mol dm^3 1 49 50.000 0. 2 48 25.000 0. 4 46 12.500 0. 6 44 8.333 0. 8 42 6.250 0. 10 40 5.000 0. 15 35 3.333 0. 20 30 2.500 0. 24 26 2.083 0. 28 22 1.786 0. 30 20 1.667 0. 36 24 1.389 0. 40 10 1.250 0. 44 6 1.136 0. 48 2 1.042 0.
5. Dilutions
During your science studies when you might have to do some calculations involving so-called
serial dilutions. Students often find such calculations challenging and so this section has been
written to help give you confidence in dealing with such problems.
Let’s work through an example. Suppose we are given a solution (the stock solution) of a dye
that is occasionally used to stain DNA. We are told that the concentration of the stock solution
is 2.5 x 10-1^ mol dm-3^. For use as a DNA stain, the dye needs to be present in solution in the
concentration range 1.0 – 4.0 x 10 -4^ mol dm-3^. How could we prepare a sample (5 cm^3 ) of the
dye from the stock solution that falls inside this range? A reliable method to use would be to
create a series of serial dilutions as shown in the diagram below.
Sounds pretty straightforward! Should I try some examples?
Self-Assessment Questions 8
Methylene blue is an intensely coloured dye molecule used for staining nucleic acids. A stock solution of methylene blue was prepared in water with the concentration of the methylene blue being 5 x 10 -5^ mol dm-^.
(i) 4 volumetric flasks (25 cm^3 ) were taken. Different volumes of the methylene blue stock were placed in the flasks such that the first contained 1 cm^3 , the second contained 2 cm^3 , the third contained 3 cm^3 , and the fourth contained 4 cm^3. Each flask was made up to the mark with distilled water. What are the final concentrations of methylene blue in each of the four volumetric flasks?
(ii) How much methylene blue stock solution would need to be placed in a flask (25 cm^3 ) in order to achieve a final concentration of 2.4 x 10 -6^ mol dm-^3 when the flask is made up to the mark with distilled water?
Check answers
What are serial dilutions and why are they used?
There are occasions when you are given a stock solution (for
example a concentrated dye solution or a suspension of bacteria)
and you need to measure the concentration. The device you use
to measure the concentration may only be accurate over a certain
concentration range and so you may need to dilute your stock
solution to ensure that you can work in this range. Often the best
way to approach this is to prepare a range of dilutions from the
stock and a common approach is to use a method called serial
dilution.
Starting with the stock solution you would take 0.5 cm 3 and add 4.5 cm 3 of water. This would
give you a solution of concentration 2.5 x 10-2^ mol dm-3^. You could then repeat this process
until you have a solution that lies in the correct concentration range.
The above case is an example where the serial dilutions differ by an order of magnitude from
one to the other. You could of course change the concentration by any suitable factor you
wished.
6. Acknowledgements
I am grateful to those individuals who have read early drafts of this guide and given freely of
their time, advice and expertise. Particular thanks go to John Adds (Abbey Tutorial College),
Jaquie Burt (Portobello High School, Edinburgh), Erica Clark (SAPS), Kath Crawford (Scottish
Schools Equipment Research Centre (SSERC) and SAPS Biotechnology Scotland Project), Debbie
Eldridge (King Ecgbert School, Sheffield), John Gray (University of Cambridge) and Stephen
Tomkins (University of Cambridge). Any errors or omissions that remain are, of course, the sole
responsibility of the author.
It is our intention to issue updates to this Guide on a regular basis and so further comments on
content and other areas that might be covered are welcomed. Please send your comments and
