For the GCSE Chemistry exam, you will be tested on calculations. In chemistry, we need to know how much of each reactant to use in order to make the right amount of product. Atoms are too small to count individually, so we weigh them instead.
Calculating the number of moles of a substance allows us to get the right ratio of our reactants for our reaction to work successfully. We also use calculations to help us decide how efficient our reactions are.
Atom economy is a way of quantifying how much waste there is associated with a particular reaction. Percentage yield is all important too. Chemists spend a long time working out the best conditions in order to maximise their yield.














M_{r} (which stands for Relative Formula Mass) is calculated by adding together all the atomic masses of the elements in the formula. In this case, rubidium is the heaviest group one atom mentioned so rubidium chloride will have the highest M_{r}. Which compound will have the lowest M_{r}?

When a formula contains brackets, anything inside the bracket needs to be multiplied by the number outside the bracket. So the formula for aluminium nitrate contains 1 aluminium atom, 3 nitrogen atoms and 9 oxygen atoms


Atom economy is a measure of the environmental impact of a reaction. In an ideal world all or most of the reactants will end up in the product that you want. By definition, addition reactions like A have 100% atom economy because there is only one product. In industry, the atom economy of a reaction can be improved if a waste product can also be collected and sold. For example, the carbon dioxide made as waste in reactions B and C could be collected and sold to a fizzy drinks manufacturer

One mole of any gas will occupy the same volume at a given temperature and pressure, regardless of what the gas is. In the above reaction 1 mole of methane reacts to form 1 mole of carbon dioxide, so the volume of carbon dioxide made will be the same as the volume of methane used


The key to this question is not to over complicate it! It’s simply a case of converting a number of moles into a mass. Since mass = moles x M_{r}  the sum you need to do is 0.2 x M_{r} of NaOH

This is a reacting mass calculation. If you start with a known mass of a chemical, by using its M_{r} you can work out how many moles of it you have. In this case we have 0.25mol of methane. You can then use the stoichiometry of the reaction (aka the big balancing numbers!) to show you that every one mole of methane produces 2 moles of water. So we deduce that 0.5mol of water will be made. Then you simply need to use mass = moles x M_{r} to calculate the mass of water produced


When doing this type of calculation, always start with the solution that you know most about – in this case the sodium hydroxide solution. Since you know both the volume and the concentration you can calculate how many moles of sodium hydroxide you have. Next look at the equation for the reaction; one mole of sodium hydroxide reacts exactly with one mole of hydrochloric acid. You therefore know how many moles of HCl there must have been in 21cm^{3} of solution. Since you know moles and volume, you can calculate the concentration

Top tip: when calculating a percentage, if you get an answer greater than 100%, you definitely need to think again about your calculation. Any percentage is calculated by multiplying a fraction by 100. In this case you are finding what fraction of your expected yield you actually made, so the sum you need is 1.8/3.2 x 100


The first thing you need to do in this question is convert the volume in cm^{3} into dm^{3}. Since there are 1000cm^{3} in a dm^{3}, 100cm^{3} is 0.1 dm^{3}. If you can’t remember how to calculate concentration, use the units to help you; grams per decimetre cubed means g divided by dm^{3}, so the sum you need is 10 / 0.1. Then simply find out how many moles there are in 100g of NaOH. The RFM is 40 so the sum you need is 100/40

The key fact to remember here is that one mole of any gas occupies 24 dm^{3} at RTP. I have 1/12 of this volume, so I must have 1/12 of a mole
