Maths and Chemistry for Biologists. Chemistry 2 Solutions, Moles and Concentrations This section of...

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

CH3CH2OH O

H

HH 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 23F 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 mol1.8 g of water is 0.1 mol

18 g of water is 18 mol60 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 LITREor 1 mol in 1 L gives a 1 M solution

1 mol/L solution is 1M0.1 mol/L solution is 0.1 M

0.1 mol/100 ml solution is 1 M1 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

New conc = 0.1 x = 0.025 M

Amount of solute remains as 0.025 mol

1000

250

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 wateror 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.