# Moles

### Basic definitions:

Mole

The amount of a substance which contains as many particles as there are
in 12g of Carbon-12.

Avagadro’s Constant = 6.02 x 1023

Number of atoms in 1 mole.

Molar Mass

The mass of a substance which contains 1 mole.

E.g. CH_{4}
= (12 x 1) + (1 x 4) = 16.0 gmol-1

Avagadro’s law

Equal volumes of any gases at the same temperature and pressure contain
an equal number of particles.

Molar volume

1 Mole of gas always occupies a volume of 24 dm^{3}
or 24 000 cm^{3}
at room temperature (25°C or 298 K) and pressure (1 atm or 100 kPa).

Moles questions will always involve one of four equations:

### Formula 1

Empirical formula

Definition: Simplest whole number ratio of the atoms of each element present
in a compound.

This first formula is often used when calculating the empirical formula
of a compound when certain data is given.

Example: Calculate the empirical formula of compound with 37.5% C, 12.5% H, 50% O and with a molar mass of 64 gmol-1

....%...Molar
Mass..Moles

C 37.5 ..12..........3.125

H 12.5 ..1 ...........12.5

O 50 ....16...........3.125

3.125 : 12.5 : 3.125

1 : 4 : 1

Note: For this question you would use the % to act as the mass in order for you to calculate the moles of each substance.

Dividing by the smallest gives the ratio gives

1:4:1 gives the empirical formula CH_{4}O.
However the questions specifies that the molar mass = 64 gmol-1

Molar mass of CH_{4}O
= 12 + (4 x 1) + 16 = 32 gmol-1

64/32 = 2

Therefore molecular formula is (CH_{4}O)
x 2 = C_{2}H_{8}O_{2}

Example of a basic moles question:

Calculate the number of moles in 15g of Pure Carbon.

Molar mass = 12 + (16 x 2) = 44 gmol-1

Moles = Mass/Molar Mass = 15/44 = 0.3409 mol

Formula 2

Example:

Calculate the number of molecules in 15g of Oxygen:

Molar mass of O_{2}
= (16 x 2) = 32 gmol-1

Moles = 15/32 = 0.46875 mol

Number of molecules = (6.02 x 1023) x 0.46875

= 2.82188 x 1023 particles

Note: When calculating the molar mass you use the molecular formula e.g.
O_{2}
not O

### Formula 3

Example:

Calculate the volume of 0.4 mole O_{2}
at room temperature and pressure.

Volume = Moles x Molar Volume

0.4 x 24 000 = 9600 cm^{3}

Formula 4

For these calculations you will need to deal with volume, moles and concentration
of a solution through the use of titrations.

The formula used in titrations is:

Note:

mol dm^{-3}
can also be written as just ‘M’. This refers to moles per
cubic decimetre. Also remember when calculating concentration, you must
always convert cm^{3}
into dm^{3}

To achieve dm^{3}
from cm^{3}
you can divide cm^{3}
by 1000 because 1000 cm^{3}
= 1 dm^{3}

Exam style questions

The first thing you must understand about moles calculations in the exam
is that you are almost always follow the same basic routine:

1. You are either told or must calculate the chemical equation and this
must be balanced.

2. You must then calculate something within the equation that is ‘know’
(something that you are already given some information about in the question
e.g. a compounds mass)

3. You then use this with the stochiometry of the equation to calculate
something of an unknown compound/element.

4. You then calculate the unknown.

This may sound confusing but it is actually quite simple. Let’s have a look at some examples:

Example 1

1. 7.9g of magnesium oxide is formed when O_{2}
reacts with excess magnesium with the equation:

O_{2}
+ 2Mg 2MgO

a) Calculate the volume of oxygen required

Right you already have a balanced equation and you ‘know’ the amount of magnesium oxide so you must first calculate the moles of MgO:

Moles of MgO = 7.9 / (16 + 24.3) = 0.1960298 mol

Now you have to look at the stochiometry of the equation – this
is basically the number in front of the chemicals – so O_{2}
has a stochiometry of 1 and MgO and Mg have a stochimonetry of 2. Because
none of the chemicals have the same number they are all in different ratios.

From stochiometry of equation 1:2 so moles of O_{2}
= 0.1960298 / 2 = 0.0980149 mol

Now you know the moles of O_{2}
you can calculate its volume.

Volume = 0.0980149 x 24000 = 2352.36 cm^{3}

Example 2

What is the concentration of 4.95g of potassium manganate(VIII), dissolved
to make 310cm^{3}
of solution?

1.) Divide 310cm^{3}
by 1000 to get your volume in dm^{3}

310/1000 = 0.31 dm^{3}

2.) Calculate moles of potassium manganate(VII)

K2MnO4 = 197.13 gmol-1

4.95 / 197.1 = 0.025114155 mol

3.) Divide moles by volume to get the concentration.

0.025114155 / 0.31 = 0.081013404 mol dm^{-3}

= 0.0810 mol dm^{-3}
(3 s.f.)

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**Module 1**

Useful Chemistry Resources

Atoms and Basic Atomic Structure

Mass Spectrometer

Moles

Covalent Bonding and Atomic Shapes

Electronegativity

Intermolecular Forces and Covalent Structure

Electronic Configuration

Ionisation Energy and Period 3 Trends

Oxidation Numbers and Ionic Equations

Chemical Reactions

Group 7 - Halides and Halogens