An equation for a chemical reaction can provide us with a lot of useful information. It tells us what the reactants and the products are in the reaction, and it also tells us the ratio in which the reactants combine to form products. Look at the equation below:

Fe + S → FeS Fe + S → FeS

In this reaction, every atom of iron (FeFe) will react with a single atom of sulphur (SS) to form one molecule of iron sulphide (FeSFeS). However, what the equation doesn't tell us, is the quantities or the amount of each substance that is involved. You may for example be given a small sample of iron for the reaction. How will you know how many atoms of iron are in this sample? And how many atoms of sulphur will you need for the reaction to use up all the iron you have? Is there a way of knowing what mass of iron sulphide will be produced at the end of the reaction? These are all very important questions, especially when the reaction is an industrial one, where it is important to know the quantities of reactants that are needed, and the quantity of product that will be formed. This chapter will look at how to quantify the changes that take place in chemical reactions.

Sometimes it is important to know exactly how many particles (e.g. atoms or molecules) are in a sample of a substance, or what quantity of a substance is needed for a chemical reaction to take place.

You will remember from Relative atomic mass that the relative atomic mass of an element, describes the mass of an atom of that element relative to the mass of an atom of carbon-12. So the mass of an atom of carbon (relative atomic mass is 12u12u) for example, is twelve times greater than the mass of an atom of hydrogen, which has a relative atomic mass of 1u1u. How can this information be used to help us to know what mass of each element will be needed if we want to end up with the same number of atoms of carbon and hydrogen?

Let's say for example, that we have a sample of 12g12g carbon. What mass of hydrogen will contain the same number of atoms as 12g12g carbon? We know that each atom of carbon weighs twelve times more than an atom of hydrogen. Surely then, we will only need 1g1g of hydrogen for the number of atoms in the two samples to be the same? You will notice that the number of particles (in this case, atoms) in the two substances is the same when the ratio of their sample masses (12 g carbon: 1g hydrogen = 12:1) is the same as the ratio of their relative atomic masses (12 u: 1 u = 12:1).

To take this a step further, if you were to weigh out samples of a number of elements so that the mass of the sample was the same as the relative atomic mass of that element, you would find that the number of particles in each sample is 6,022× 10236,022×1023. These results are shown in Table 1 below for a number of different elements. So, 24,31g24,31g of magnesium (relative atomic mass=24,31urelative atomic mass=24,31u) for example, has the same number of atoms as 40,08g40,08g of calcium (relative atomic mass=40,08urelative atomic mass=40,08u).

This result is so important that scientists decided to use a special unit of measurement to define this quantity: the mole or 'mol'. A mole is defined as being an amount of a substance which contains the same number of particles as there are atoms in 12 g of carbon. In the examples that were used earlier, 24,31g24,31g magnesium is one mole of magnesium, while 40,08g40,08g of calcium is one mole of calcium. A mole of any substance always contains the same number of particles.

Definition 1: Mole

The mole (abbreviation 'n') is the SI (Standard International) unit for 'amount of substance'. It is defined as an amount of substance that contains the same number of particles (atoms, molecules or other particle units) as there are atoms in 12g12g carbon.

In one mole of any substance, there are 6,022× 10236,022×1023 particles.

Definition 2: Avogadro's number

The number of particles in a mole, equal to 6,022× 10236,022×1023. It is also sometimes referred to as the number of atoms in 12g12g of carbon-12.

If we were to write out Avogadro's number then it would look like: 602200000000000000000000602200000000000000000000. This is a very large number. If we had this number of cold drink cans, then we could cover the surface of the earth to a depth of over 300km300km! If you could count atoms at a rate of 10 million per second, then it would take you 2 billion years to count the atoms in one mole!

We can build up to the idea of Avogadro's number. For example, if you have 12 eggs then you have a dozen eggs. After this number we get a gross of eggs, which is 144 eggs. Finally if we wanted one mole of eggs this would be 6,022× 10236,022×1023. That is a lot of eggs!

Definition 3: Molar mass

Molar mass (M) is the mass of 1 mole of a chemical substance. The unit for molar mass is grams per mole or g·mol-1g·mol-1.

Refer to Table 1. You will remember that when the mass, in grams, of an element is equal to its relative atomic mass, the sample contains one mole of that element. This mass is called the molar mass of that element.

You may sometimes see the molar mass written as MmMm. We will use MM in this book, but you should be aware of the alternate notation.

It is worth remembering the following: On the periodic table, the relative atomic mass that is shown can be interpreted in two ways.

The calculations that have been used so far, can be made much simpler by using the following equation:

The equation can also be used to calculate mass and molar mass, using the following equations:

and

The following diagram may help to remember the relationship between these three variables. You need to imagine that the horizontal line is like a 'division' sign and that the vertical line is like a 'multiplication' sign. So, for example, if you want to calculate 'M', then the remaining two letters in the triangle are 'm' and 'n' and 'm' is above 'n' with a division sign between them. In your calculation then, 'm' will be the numerator and 'n' will be the denominator.

Figure 1

So far, we have only discussed moles, mass and molar mass in relation to elements. But what happens if we are dealing with a molecule or some other chemical compound? Do the same concepts and rules apply? The answer is 'yes'. However, you need to remember that all your calculations will apply to the whole molecule. So, when you calculate the molar mass of a molecule, you will need to add the molar mass of each atom in that compound. Also, the number of moles will also apply to the whole molecule. For example, if you have one mole of nitric acid (HNO3HNO3), it means you have 6,022× 10236,022×1023 molecules of nitric acid in the sample. This also means that there are 6,022× 10236,022×1023 atoms of hydrogen, 6,022× 10236,022×1023 atoms of nitrogen and (3× 6,022× 10233×6,022×1023) atoms of oxygen in the sample.

In a balanced chemical equation, the number that is written in front of the element or compound, shows the mole ratio in which the reactants combine to form a product. If there are no numbers in front of the element symbol, this means the number is '1'.

e.g. N2+3H2→2NH3N2+3H2→2NH3

In this reaction, 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.

The empirical formula of a chemical compound is a simple expression of the relative number of each type of atom in that compound. In contrast, the molecular formula of a chemical compound gives the actual number of atoms of each element found in a molecule of that compound.

Definition 4: Empirical formula

The empirical formula of a chemical compound gives the relative number of each type of atom in that compound.

Definition 5: Molecular formula

The molecular formula of a chemical compound gives the exact number of atoms of each element in one molecule of that compound.

The compound ethanoic acid for example, has the molecular formula CH3COOHCH3COOH or simply C2H4O2C2H4O2. In one molecule of this acid, there are two carbon atoms, four hydrogen atoms and two oxygen atoms. The ratio of atoms in the compound is 2:4:2, which can be simplified to 1:2:1. Therefore, the empirical formula for this compound is CH2OCH2O. The empirical formula contains the smallest whole number ratio of the elements that make up a compound.

Knowing either the empirical or molecular formula of a compound, can help to determine its composition in more detail. The opposite is also true. Knowing the composition of a substance can help you to determine its formula. There are four different types of composition problems that you might come across:

It is possible to calculate the volume of one mole of gas at STP using what we know about gases.

A typical solution is made by dissolving some solid substance in a liquid. The amount of substance that is dissolved in a given volume of liquid is known as the concentration of the liquid. Mathematically, concentration (C) is defined as moles of solute (n) per unit volume (V) of solution.

For this equation, the units for volume are dm3dm3. Therefore, the unit of concentration is mol· dm -3mol· dm -3. When concentration is expressed in mol· dm -3mol· dm -3 it is known as the molarity (M) of the solution. Molarity is the most common expression for concentration.

Definition 6: Concentration

Concentration is a measure of the amount of solute that is dissolved in a given volume of liquid. It is measured in mol· dm -3mol· dm -3. Another term that is used for concentration is molarity (M)

Stoichiometry is the calculation of the quantities of reactants and products in chemical reactions. It is also the numerical relationship between reactants and products. In representing chemical change showed how to write balanced chemical equations. By knowing the ratios of substances in a reaction, it is possible to use stoichiometry to calculate the amount of either reactants or products that are involved in the reaction. The examples shown below will make this concept clearer.

When we are given a known mass of a reactant and are asked to work out how much product is formed, we are working out the theoretical yield of the reaction. In the laboratory chemists never get this amount of product. In each step of a reaction a small amount of product and reactants is 'lost' either because a reactant did not completely react or some of the product was left behind in the original container. Think about this. When you make your lunch or supper, you might be a bit hungry, so you eat some of the food that you are preparing. So instead of getting the full amount of food out (theoretical yield) that you started preparing, you lose some along the way.