How to tackle the more challenging GCSE Chemistry Topics?

Some topics in GCSE and IGCSE chemistry students find more challenging than others. Below we look at three tricky topics: calculations using the mole, structure and bonding, and equilibria and Le Chatelier’s principle.

Philippa Platt

Chemistry

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10 minutes

As a chemistry teacher and tutor with over 15 years’ experience, I have found that there are some topics in GCSE and IGCSE chemistry that students find more challenging than others. For example, questions about atomic structure are generally well answered in exams, whereas equilibria seems to involve far more in-depth thought and contain taxing aspects where students misinterpret the question or miss out on marks.

I have highlighted three topics where, as a teacher and examiner, I often see students make mistakes or where there are misconceptions.

I like to take time teaching these so my students are able to tackle them by themselves when revising and in their exams. 

Let’s look at them one by one.

Calculations using the mole

Some calculations in GCSE and iGCSE chemistry are certainly more challenging than others. Examiner reports often comment that for the more challenging calculations, a low percentage of students gain full marks. 

It is not necessarily the whole topic that my students would like to revise, but generally more niggly areas which are preventing students from really mastering the layout of the answers they need to write in the exam which will hit all the marking points. To help them there are a number of teaching points that I use which can be used when constructing answers. 

The biggest confusion I find to begin with is the question of what the mole actually is! It seems an alien concept as it is not a unit of measurement that is often visited in any other course. The main thing we need to get our head around is that it is just a different way of measuring the amount of something. It makes it easier for us, as chemists, to determine how much of something we have. An equation that is often used in chemistry lessons is:

N u m b e r space o f space m o l e s space equals space fraction numerator m a s s space open parentheses g close parentheses over denominator M subscript r end fraction

This is also only true of solid substances and it is important to understand initially what makes up a mole.

One mole contains 6.02 x 1023 particles (e.g. atoms, ions, molecules); this number is known as the Avogadro Constant. This is an incredibly large number which is why we need to use easier numbers using the equations that we need to learn. So, one mole of magnesium atoms contains 6.02 x 1023. Or one mole of oxygen molecules, O2, contains 6.02 x 1023 molecules.

We also have to work out the concentration of solutions, number of particles and volume of gas. These concepts don’t seem to link together and this can be the sticking point when solving a calculation. The concepts are all linked together by moles, however, as shown in this great tool, the mole roundabout:

 

The mole roundabout

mole-roundabout

If you can get to moles you can go anywhere from here!

I have found that students struggle with working through calculations that require a conversion of one form to another; for example, mass of a substance to a volume of gas. This is where the mole roundabout is useful. The most important tip I give my students is to calculate the number of moles first. This is the centre of your roundabout, and you can go anywhere from here: mass (in grams), gas (volume in cm3 dm3), number of particles, solutions (volume in cm3 or dm3 or concentration mol / dm3).

The second tip I tell my students is to check the number of moles that you have calculated. If the number is very large the likelihood is that the equation has been rearranged incorrectly

Here is a worked example that I use to help demonstrate:

3 g of magnesium ribbon is used to neutralise 125 cm3 of sulfuric acid. Using the following equation, calculate the concentration of sulfuric acid. (Ar of Mg = 24)

H2SO4 + Mg → MgSO4 + H2

This question requires you to firstly calculate the number of moles, then the number of moles of acid which can then be converted to concentration.

  • Number of moles of magnesium = 3/24 = 0.125 mol

  • Number of moles of acid = 1:1 ratio = 0.125 mol

  • Concentration of acid =  moles of acid / volume of acid (dm3) = 0.125/0.125 = 1.00 mol / dm3

You could also be asked to calculate the volume of gas produced as well. Provided you know the moles you can work out most chemical reactions. A further possibility examiners could ask would be to calculate the number of magnesium atoms used in the equation.

Top tips

  • If given an equation, write down the information you are given.

  • Set out your working clearly.

  • Use the correct symbols to help avoid confusion: m for mass and n for moles.

  • Work out the moles (n) of what you know the most about first.

  • Be careful with unit conversion of cm3 to dm3 and vice versa.

unit-conversion

Structure and Bonding

As an examiner, I read many answers that contain misconceptions about structure and bonding. The wording in mark schemes is very specific and therefore students often miss marks by stating too many types of bonding.

A common question asks why a substance such as ammonia, NH3, has a low boiling point. There are specific things an examiner is looking for in the answer. Setting this out clearly helps examiners like me award marks. When asked about a property (such as boiling point, melting point, electrical conductivity) it is always a good idea to state the type of substance and bonding that is present. Looking at the question of why ammonia has a low boiling point, you should state for the first mark that it is a simple covalent molecule or small molecule.

Next, I would be looking for a statement that tells me about the characteristic of simple molecules that links to the property stated in the question. In this situation, it would be intermolecular forces. These are weak forces of attraction that occur between molecules. A common misconception is that weak intermolecular forces are between atoms. They are not, they are between the molecules.

Weak intermolecular forces and bonds

Intermolecular Forces Vs Covalent Bonds

The intermolecular forces are between the molecules and bonds are between atoms

Having taught, tutored and examined this topic I have come across this misconception many times and advise my students to take time learning the difference between bonds and forces. For simple covalent molecules or simple compounds, it is not the bonds breaking that cause either boiling or melting, it is the dispersion or spreading out of the molecules themselves. This is why giant structures such as diamond have very high melting points. In order to melt diamond, the strong covalent bonds (there are four per each carbon atom) in the structure have to be broken, which is different from simple covalent structures.

Practise some bonding questions and look at the commentaries for each type of structure.

Top tips

  • State the type of bonding and structure of the substance in the question.

  • Remember that bonds are between atoms and intermolecular forces are between molecules.

  • Link the characteristic of the type of bonding to the property being asked about:

    • For example, ionic lattices generally have high melting points. Ionic bonds are strong forces of attraction between positively and negatively charged ions which take a lot of energy to overcome.

Equilibria and Le Chatelier’s principle

When I have tutored this topic I’ve found that students have a solid understanding of reversible reactions and equilibria but find it difficult to word their answers. A great tip for tackling the more complex aspects of chemistry is always go back to your first principles ​– to what the words really mean. In equilibria topics, we talk about reversible reactions which occur when reactants can turn into products and products can also turn into reactants. One pathway will be exothermic (energy released to surroundings so temperature increases) and the other endothermic (energy taken in so temperature decreases). This can change depending on the reaction. It is important to remember that Le Chatelier’s Principle states:

‘When a change is made to the conditions of a system at equilibrium, the system automatically moves to oppose the change.’

The Haber process is the production of ammonia from nitrogen and hydrogen and is commonly referred to when discussing Le Chatelier’s Principle.

3H2 (g) + N2 (g) ⇌ 2NH3 (g)    ΔH = -46 kJ / mol

The negative energy change shows that the forward reaction is exothermic. If the temperature of the reaction is increased, the system will work to restore the equilibrium. It will do this by picking a pathway to follow. In this case it will be the endothermic pathway as this will reduce the temperature. :

Another example is the reaction between ICl and Cl2:

ICl + Cl2 ⇌ ICl3                               ΔH = -106 kJ / mol

Even though this may be a more unfamiliar reaction, the same principle applies if the temperature is changed. I like to walk through the first principles as shown in the notes below:

 

Consider: What needs to happen to restore the equilibrium?

 

 

exo

 

 

COLD

ICl + Cl2

ICl3       

HOT    

 

 

endo

 

 

ΔH = -106 kJ / mol is negative and therefore the forward reaction is exothermic

  • Write exo and endo on the correct arrows

  • Write HOT and COLD on correct sides

  • Temperature change

    • Increase; to restore the equilibrium, the endothermic reaction / backward reaction is favoured.

    • Decrease; to restore the equilibrium, the exothermic reaction / forward reaction is favoured.

exothermic-and-endothermic-article-printed-notes

The interesting thing about this reaction is that we can see a colour change. The reactant side is a dark brown colour and the product is a yellow colour. So if the forward reaction is favoured, the yellow colour is seen; if the backward reaction is favoured, the brown colour is seen.

Top tips

  • Write the correct information from the question onto the equation.

    • You can also include the number of molecules.

    • Include hot and cold! (I’ve found this really helps students.)

  • Always consider ‘what needs to happen to restore the equilibrium’.

  • Remember that if a change is made to a reversible reaction, the (closed) system will restore the equilibrium by opposing the change.

 

Conclusion

Hopefully these top tips will help you, but it is also essential that you do the background learning that will enable you to tackle the questions containing these topics in the exam. For example, make sure you know how to convert moles to mass (in grams), gas (volume in cm3 dm3), number of particles, solutions (volume in cm3 or dm3 or concentration mol / dm3).

In the exam, set out your answer so the examiner can see a clear thought process. When examining, I find that students who set out their calculations clearly score more marks, even if they make a mistake, as I can see where they went wrong and award credit for their working.

For longer answers that require a deeper explanation, make sure you have information to work from and if you get muddled go back to your first principles. For example, write down if a reaction is endothermic or exothermic, or an equation to calculate the number of moles from mass and relative molecular mass.

There may be other areas that you find challenging so use the resources you have on offer. One watch of a video along with a conversation with your teacher or tutor may be all you need to crack it.

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Written by Philippa Platt

Chemistry 2 articles

Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener

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