Maths and Chemistry for
Biologists
Chemistry 2
Solutions, Moles and Concentrations
This section of the course covers –
• why some molecules dissolve in water to form
solutions
• how we measure the amounts of atoms and
molecules - the mole
• The relative molar mass of molecules
• how the concentrations of solutions are
described and how solutions of a given
concentration are made
The bonds in some molecules are polarised
C–H
N–H
O–H
Cl – H
increasing electronegativity of the heavy atom
As electronegativity increases the pair of
electrons in the bond is displaced towards the
heavy atom and hence the bond is polarised
Electrons spend a larger proportion of their
time near the electronegative atom. Shown by
e.g. - O – H + The  indicates a fraction of a
charge
Water is a bent, polar molecule
The oxygen in a water molecule has two
lone pairs of electrons which take up
space. The two bonds and two lone
pairs point to the corners of a tetrahedron
– hence molecule is bent. The bonds are
polar so the molecule can be drawn as
In water the value of  is about 0.3
O
H
H

O


H H
Consequences of the structure of water (1)
It forms hydrogen bonds – electrostatic interaction
between + on H and - on O of adjacent molecules
O – H+ -O – H
hydrogen bond
H-bonds are weak (about 1/20th strength of covalent
O – H bond) and long (about 0.176 nm compared
with O – H bond) but are enormously important for
the properties of water.
H-bonds also occur in compounds containing N – H
Consequences of the structure of water (2)
Water dissolves ionic
compounds such as NaCl
because it interacts
electrostatically with the ions
Water dissolves polar
molecules such as ethanol
because it interacts
electrostatically with the
partial charges
+
-
CH3CH2O
-
H O
+ -
H
H
H O
H
+
Consequences of the structure of water (3)
In liquid water molecules stick together to give a
partially ordered structure
The need to “weigh” atoms and molecules
Consider the reaction Na + F  NaF
One meaning of this is that 1 atom of sodium reacts
with 1 atom of fluorine to give 1 molecule of NaF
Can’t weigh atoms so how can we know how much
sodium to take to react exactly with a given amount
of fluorine?
Weighing atoms continued
Na has a mass number of 23 (it contains 11
protons and 12 neutrons) so its relative atomic
mass is 23
F has a mass number of 19 (9 protons and 10
neutrons) so its relative atomic mass is 19
So 23 g of sodium will react exactly with 19 g of
fluorine to give 42 g of sodium fluoride
The mole
23 g of Na and 19 g of F contain the same numbers
of atoms
23 g of Na is a mole of sodium
19 g of F is a mole of fluorine
Definition:- a mole is the amount of a substance
that contains the same number of particles as
there are atom in 12 g of the isotope 12C
This is the Avogadro constant equal to 6.022 x 1023
The mole continued
Na + F  NaF
Also means that 1 mol of Na reacts with 1 mol of F
to give 1 mol of NaF
(mol is the abbreviation of mole)
What the mol gives us the weights of atoms that
will exactly react together to give molecules
One for you to do
Relative atomic mass of fluorine = 19
What mass of F is 0.1 mol?
How many mol of F is 38 g?
Answers
1 mol = 19 g  0.1 mol = 1.9 g
19 g = 1 mol  38 g = 2 mol
Relative molar mass (Mr) of compounds
Calculate these by adding the relative atomic
masses (Ar) of the constituent atoms
H2O: 1 x 2 + 16 = 18
Urea (NH2CONH2): (2 x 14) + (4 x 1) + 12 + 16 = 60
Hence 18 g of water is 1 mol
1.8 g of water is 0.1 mol
18 g of water is 18 mol
60 g of urea is 1 mol
Molecular weight
You will see this term used sometimes. It is
numerically equal to the relative molar mass but
it needs units. The correct unit is the Dalton (Da)
which is 1/12th of the mass of an atom of 12C
Better to stick to relative molar mass
One for you to do
C + O2  CO2
(means that 1 mol of carbon reacts with 1 mol of
oxygen molecules to give 1 mol of carbon dioxide)
What weight of O2 is required to react completely
with 1.2 g of C?
Answer
1.2 g of C is 0.1 mol.
Hence we require 0.1 mol of O2
Mr of 02 is 2 x 16 = 32
Hence we need 3.2 g of O2
Product is 4.4 g (0.1 mol) of CO2
Concentrations of solutions
Expressed in terms of MOLARITY – the number of
moles per litre
A 1 MOLAR solution contains 1 MOLE in 1 LITRE
or a 1 M solution contains 1 mol/L
NB the abbreviations for mole (mol) and molar (M)
– NEVER confuse them. One is an amount of
substance (mol) the other is a concentration (M)
Concentrations of solutions (contd)
A 1 MOLAR solution contains 1 MOLE in 1 LITRE
or 1 mol in 1 L gives a 1 M solution
1 mol/L solution is 1M
0.1 mol/L solution is 0.1 M
0.1 mol/100 ml solution is 1 M
1 mol/100ml solution is 10 M
1 mmol/ml solution is 1M
Examples of how to make solutions
How much NaOH is needed to make 500 ml of a
0.2 M solution?
Mr of NaOH is 23 + 16 + 1 =40. 1L of a 0.2 M
solution contains 0.2 mol. Hence 500 ml
contains 0.1 mol. 0.1 mol of NaOH = 4 g
How much KCl is there in 100 ml of a 2 M soln?
Mr of KCl is 39 + 35.5 = 74.5. 1L of 2M soln
contains 2 mol hence 100 mol contains 0.2 mol.
0.2 mol of KCl is 0.2 x 74.5 = 14.9 g
More examples
What is the conc of a solution of NaH2PO4
containing 2g/L?
Mr is 23 + 2 + 31 + 64 = 120. Hence 2 g is 2/120 of
a mol or 0.0167 mol. This is contained in 1 L so
conc = 0.0167 M (or 1.67 x 10-2 M or 16.7 mM)
How many mol is 1 mg of a substance with Mr =
250?
1 mg = 10-3 g. Hence 10-3/250 = 4 x 10-6 mol or
4 mol
One for you to do
What is the concentration of a solution of NaOH
containing 20 g in 500 ml? (Mr = 40)
How much NaOH is required to make 10 ml of a
1M solution?
Answer
Solution has 20 g in 500 ml and hence 40 g in 1 L.
40 g is 1 mol and hence conc = 1 M
For 1 M need 40 g / L
10 ml = 0.01 L. Hence need 0.01 x 40 g = 0.4 g
(or 400 mg)
Dilutions
If you dilute a solution (add more solvent) the
amount of solute remains the same but the
concentration decreases
Take 250 ml of a 0.1 M solution and dilute to 1L
250
New conc = 0.1 x
= 0.025 M
1000
Amount of solute remains as 0.025 mol
Example
How would you make 50 ml of a 10 mM solution by
dilution of a 1M stock solution?
New solution has conc 1/100 that of stock. So
need 50/100 = 0.5 ml of stock solution plus
49.5 ml water
or 1 L of 10mM solution contains 10 mmol of solute
so 50 ml (1/20 L) contains 0.5 mmol.
Stock has 1 mol/L or 1 mmol/ml. So need 0.5 ml
of stock plus 49.5 ml water as above.