So, in this series I want to go over what intermolecular forces are, the three types to consider and what sets each type apart from the other.

Firstly, what do we mean by intermolecular forces? The inter part essentially means between, as in international travel means travel between nations. So, intermolecular forces means forces between molecules, not forces within molecules. That is quite important. Forces within molecules, in other words The regular kind bonding such as covalent, ionic and metallic, is referred to as intramolecular forces. We are looking at the forces between one molecule and another nearby molecule. These forces are much weaker than the bonds within a molecule.

The three types of intermolecular forces

So, the three types we shall look at are:

1. Van der Waals (often referred to as vdw)
2. Dipole-dipole
3. Hydrogen bonding.

First thing to note here is that I have arranged them in order of strength, with the weakest being at the top with Van der Waals, going down to the strongest at the bottom with hydrogen bonding.

THOUGHTS THAT CAN STOP YOU STUDYING

The weakest of the three: Van der Waals

Let’s imagine a molecule, for example chlorine, Cl2. The two atoms are bonded by a single covalent bond, but the electrons are constantly moving around and at any one time, they could be more around one atom than the other. We could say that the electron density is higher around one atom than the other. The diagram below shows the electron density is higher around the right hand atom within the molecule.

As a result, the part of the molecule that has a lower electron density will have a slightly positive charge. The other side will have a slightly positive charge.

We refer to this mis-distribution of charge as a dipole. In this case, it is a temporary, or induced dipole because no sooner do we have an arrangement like the one above, the distribution could change and so the dipole would reverse. Of course, there would be moments of no dipole and the strength of the dipole will fluctuate also.

So, two neighbouring chlorine molecules may feel a slight attraction to each other as the slightly positive area of one molecule is attracted to the slightly negative area of another.

Something which is often not explained here, is that a randomly occurring dipole in one molecule causes a dipole in the other molecule, hence the expression induced dipole. If a positive charge from the first molecule approaches the electron cloud of the second molecule it causes the electrons in the second molecule to be attracted to it which causes that part to be negatively charged. The opposite happens if the first molecule’s negative side approaches the second molecule. A similar thing occurs when you charge a balloon by rubbing it on your head and it sticks to a neutral wall. Click here for a simulation of this.

RADIOACTIVITY, NUCLEAR DECAY, RADIATION… WHAT’S THE DIFFERENCE?

How molecular size affects things

It must be no surprise that molecules with more atoms in them would have stronger van der Waals forces, as they have more surface area. However, less obvious is that molecules with larger atoms in them also have a stronger van der Waals. This is because more electrons are able to move around the molecule and lead to a stronger dipole

Saturated and unsaturated fats

Another interesting case study is saturated and unsaturated fats. Saturated fats include most animal fats and occur in butter and lard. They also occur in coconut oil You may have noticed that they are always solid at room temperature. Whereas unsaturated fats, such as olive oil and other vegetable oils are liquid at room temperature. This must be because room temperature does not produce enough energy to melt saturated fats, but does to melt unsaturated fats. Why is this?

Unsaturated fats, due to the double bonds, tend to have kinks in them and not lie flat next to each other, whereas saturated fats are more straight which means they can lie very close to one another. Because of this, the saturated fats get closer and can produce larger van der Waals forces. See the diagram below.

As you can see, the double bond bends the molecule and they are less able to lie close together.

Summary

So, I have gone over the first of the three intermolecular forces, which is also the weakest. The next one is usually called dipole to dipole, but more specifically involves permanent dipoles. In van der Waals, it is technically a dipole to dipole also, but in this case they are randomly occurring dipoles, or more accurately: One randomly occurring dipole inducing a dipole in another molecule leading to an attractive force.

I hope that cleared up a little bit of confusion on the topic. I shall see you in the next post!

A bit about the author, Paul H:

Paul is a qualified and experienced Physics, Maths, and Science teacher, now working as a full-time tutor, providing online tuition using a variety of hi-tech resources to provide engaging and interesting lessons.  He covers Physics, Chemistry, Biology, and Science from Prep and Key Stage 3 through to GCSE and IGCSE. He also teaches Physics, Maths, and Chemistry to A-Level across all the major Exam Boards.

You can enquire about tutoring with Paul here

The post Intermolecular forces 1: Van der Waals appeared first on The Tutor Team.

The Kekule structure of Benzene

So, this is how we once thought Benzene looked. It would be called cyclohexa-tri-ene. We knew it was a ring of 6 carbon atoms with a hydrogen on each one, so this is how we theorised it must be. This structure is called the Kekule structure of Benzene, after August Kekulé. The model is wrong, and you’ll see why soon, but you can see why we used to think this.

Each carbon has 4 bonds, each carbon has one hydrogen. Everything seems fine, but even here there is something a little strange… The double bonds could, in theory, actually be in two different positions, and why be in one, rather than the other? See below for a diagrammatic explanation:

Incidentally, the name of the type of chemistry which looks at chains of atoms, as in most of organic chemistry, is called aliphatic, in case you start to see that word anywhere when learning about this.

ARRHENIUS EQUATION

Bond lengths

The bond lengths of C-C bond and of a C=C bond are different. C=C bonds are shorter, but when measuring the bond lengths of benzene, they were found to all be the same length. This is our first big clue that the Kekule structure is incorrect. If the Kekule structure was correct, we should have seen two distinct bond lengths within the ring.

Hydrogenation of Benzene

The final nail in the coffin for the Kekule structure is the hydrogenation of Benzene. Hopefully, you know that hydrogenation simply means to add more hydrogen to a compound. We will imagine this happening with the Kekule structure. If we break all of the double bonds, each carbon atom will be able to hold another hydrogen atom, and so produce cyclohexane. We could do partial hydrogenation, and just break one double bond, but let’s do full hydrogenation for simplicity.

It is important to remember that, when we do this, we break bonds, which requires energy, and then make bonds, which gives out energy. You can think of breaking bonds as an endothermic process and making bonds as an exothermic process. So if it requires more energy to break the bonds, than the energy released to make the bonds, the reaction will be endothermic overall and vice versa for exothermic reactions. Doing a bit of maths, we find that the enthalpy change of adding one molecule of hydrogen to our Kekule structure of Benzene should be -120kJmol-1, and so for 3 molecules as shown above, it should be -360kJmol-1. A negative enthalpy change means an exothermic reaction.

But…

When we measure the enthalpy of the hydrogenation of benzene, it is actually -208kJmol-1. Now, importantly, this tells us two things. Firstly, the structure of benzene needs to be remodelled. Secondly, the actual structure of benzene as more stable than the Kekule structure of benzene, which of course is why it is that structure. Being given the choice, nature would always pick the most stable structure. Why does this finding mean that? Because the reaction is less exothermic. Meaning Benzene is more stable than we thought it was. It must have required more energy to break the bonds within Benzene to then add the hydrogens to make cyclohexane.

CHEMICAL BONDING AND ELECTRONEGATIVITY

So, what is the structure of benzene??

So, it is not single bond, double bond, single bond, double bond all the way around. If we first start off making everything a single bond then each carbon has a free (delocalised) electron which could bond in either direction around the ring. I’ve shown these electrons as dots on the diagram below:

These delocalised electrons form a ring. Actually, they form two rings, one above and one below. See below:

Another go at our first question

So aromatic chemistry is really referring to the study of molecules with this delocalised ring of electrons. Benzene is the most simple molecule with this ring, but aromatic chemistry includes other molecules which are derivatives of benzene and as such still contain the delocalised ring of electrons.

Ok. I hope that cleared up some understanding and I may return to this topic in future posts. Thanks for reading!

A bit about the author, Paul H:

Paul is a qualified and experienced Physics, Maths, and Science teacher, now working as a full-time tutor, providing online tuition using a variety of hi-tech resources to provide engaging and interesting lessons.  He covers Physics, Chemistry, Biology, and Science from Prep and Key Stage 3 through to GCSE and IGCSE. He also teaches Physics, Maths, and Chemistry to A-Level across all the major Exam Boards.

You can enquire about tutoring with Paul here

The post What is aromatic Chemistry? (A-level Chemistry) appeared first on The Tutor Team.

This is quite a mathematical area of the Chemistry A-level, but we can start off with a brief recap of some fundamental principles within Chemistry, which are not really mathematical

The Arrhenius equation (above) relates to the rate of reaction and contains the following variables within the equation:

k = the chemical reaction rate (variable units)

A = the pre exponential factor (same units as k)

Ea = the activation energy (J mol-1)

R = the gas constant (J K-1 mol-1 

T = temperature (K)

How the rate of reaction might change?

Let’s think about how these variables should impact the rate of reaction. 

Temperature

It stands to reason that a higher temperature should lead to a higher rate of reaction. The molecules are moving more quickly and so there will be more collisions per second and each collusion is more likely to be successful. So presumably, the equation above should show that…

 …a higher temperature leads to a higher rate of reaction.

Activation energy

Next is activation energy. Remember that the activation energy is the energy required for the reaction to occur. If the activation energy is low, then the collisions are more likely to be successful. So the equation above should show that…

…a higher activation energy leads to a lower rate of reaction.

WHAT IS A MOLE?

And now the maths…

So let’s see if the equation does what it is supposed to do.

Let’s rewrite the equation as…

k = A e-x

Where     

Hopefully you can see that A is directly proportional to k from the top equation. 

We know from our GCSE Maths lessons that…

a-b = 1/ab

So,

e-x = 1/ex

If x is large, then ex would be large. This would make e-x small.

So, A large x value means a small k value and a small k value means a low rate of reaction. In other words, a large x value means a low rate of reaction.

As you can see, x is directly proportional to the activation energy, so a large activation energy, means a large x value, which would mean a low rate of reaction.

We can also see that x is inversely proportional to the temperature. So, a high temperature would mean a low x value, which would mean a high rate of reaction.

So, it all checks out. The equation does what it is supposed to do. Awesome. Truly awesome.

CHEMICAL BONDING AND ELECTRONEGATIVITY

The other Arrhenius equation…

You may have come across this other equation which looks similar, but not the same. To understand where this came from, we need to understand a very useful mathematical tool called logarithms, or log for short.

What are logs?

So, logs are the final piece of the jigsaw puzzle when it comes to an equation like the one below:

If we want to rearrange the equation to make b the subject, we use the mathematical operation known as a root:

But how we can we make c the subject? How can find the power in the first equation if we knew a and b? The answer is use logs! In bc, we would say c is the power and b is the base. So when we use a log, we need to include in it what base to use.

On your calculator, you have a few log buttons that can help you with your Arrhenius equation:

In the blue circle at the top, you have a log button where you can choose the base.

Eg. Log2(8) = 3, because 23=8. Remember, logs help you to find the power.

In the red circle on the left, you have a log button which does not seem to have a base. It does, but is often not written. The base is 10.

Eg. Log (10,000) = 4 , because 104 = 10,000.

In the green circle on the right, you have ln. (Lower case L, not capital I) This is often pronounced “Lun”.

This last one means natural log it is a log with the base e. e represents a mathematical constant called Euler’s number. We won’t worry much about the nature of this constant, but if:

Then,

LEARNING SCIENCE – CAN IT MAKE YOU HAPPY?

Deriving the other Arrhenius equation

So, we can take the equation:

And natural log both sides of the equation:

In logarithms, there are some log rules. See below:

I’m going to apply the product rule to our equation so far:

On the second term, the natural log cancels out the e, so now we have:

And finally, with a little bit of moving terms around we have…

If you have followed me up to this point, and you have only just today heard of logs, well done. If you have been learning this in A-level maths then some of that might have been fairly simple, but if you are struggling with logs in general I shall be writing a post on that topic at some point in the future.

I hope that was useful and you now understand the Arrhenius equation a bit better. I’ll see you again in the next post!

A bit about the author, Paul H:

Paul is a qualified and experienced Physics, Maths, and Science teacher, now working as a full-time tutor, providing online tuition using a variety of hi-tech resources to provide engaging and interesting lessons.  He covers Physics, Chemistry, Biology, and Science from Prep and Key Stage 3 through to GCSE and IGCSE. He also teaches Physics, Maths, and Chemistry to A-Level across all the major Exam Boards.

You can enquire about tutoring with Paul here

The post Arrhenius equation appeared first on The Tutor Team.

At GCSE we learned that the three types of chemical bonds:

1. Covalent bonding: The electrons are shared between the atoms. The positively charged nuclei of each atom are attracted to theshared electrons. Covalent bonding usually lead to molecules being formed, which are groups of atoms bonded together in a particular way. Molecules are attracted to each other, but not so much, compared with the attraction between atoms within each molecule. However, there are a few exceptions in which covalent bonds lead to a giant covalent structure, such as graphite, diamond or silicon dioxide. In these cases, the group of atoms is never complete as such, as there are always atoms which have spare bonds and as such would happily bond with another atom, if it were to come along.

2. Ionic bonding: One atom donates electrons to another. This causes one atom to be positively charged and the other negatively charged. As a result, the two ions are attracted to one another. The ions form an ionic lattice, which like a giant covalent structure, has no size limit. No matter how large the lattice is, there is always room for more ions to join on to it. 

3. Metallic bonding: When metal atoms join together, the outer electrons are free or  delocalised and form what is called a referred to as a sea of electrons. This ‘sea’ is of course negatively charged and the positively charged metallic ions are attracted to it, which is what forms the bond.

Introducing electronegativity

At A-level we learn that the difference between ionic and covalent bonds is actually not all that clear cut. There’s covalent, ionic and er… sort of covalent! This last category is more commonly referred to as polar bonds.

So, what is electronegativity? Essentially, it is how much an atom attracts the bonding pair of electrons within the covalent bond. At GCSE we tend to think of the atoms being shared equally, but that is often not the case. For example, a molecule of Hydrogen Fluoride the bonding pair of electrons are much more attracted to the Fluorine atom and as such you are more likely to find them there, than around the hydrogen atom.

Below is what looks like the periodic table, but the numbers you see are referring to the electronegativity of the element. The higher the number, the more attractive bonding electrons will find that atom. These numbers are determined using the Pauling scale of electronegativity invented by Linus Pauling.

WHAT IS A MOLE? A 3 part series

Trends in electronegativity

You will notice a couple things in this table:

1. Some noble gases do not seem to have any electronegativity.
2. If you stay within the same period (row) the electronegativity increases from left to right, apart from some noble gases.
3. If you stay within the same group (column), the electronegativity increases as you go upwards.
4. The most electronegative element is fluorine, in the top right (excluding noble gases)
5. The least electronegative elements we can see are in the bottom left: Caesium and Francium.

Let us try to understand these observations:

Observation 1: Noble gases have a full valence shell of electrons and as such do not have any need for attracting electrons. So, what about krypton and Xenon?? It was discovered in the 1960s that these two elements can react with very electronegative elements such as fluorine and oxygen.

Observation 2 to 5: The rest of the observations can be explained together and it helps here to think of how we understood reactivity at GCSE. As you go down a group, you are adding another shell of electrons. This means that bonding electrons are further away from the positively charged nucleus and as such will not feel as much of an attractive force. As you move across the period, you are adding extra protons to the nucleus. This means that the nucleus will become more positively charged and so would attract the bonding electrons more strongly. So, the element with the most positively charged nucleus but with still a small number of shells is fluorine; whilst the opposite is true for Caesium and Francium.

RATIONALISING THE DENOMINATOR

Polar bonds

Let’s take the compound Hydrogen Chloride:

Chlorine has a higher electronegativity than hydrogen and so the electrons are going to be more strongly attracted to the chlorine than the hydrogen. As such, the electrons will be closer to the chlorine than the hydrogen. This can be shown in a second diagram on the right.

As a result the chlorine will be slightly more negatively charged than the hydrogen and this charge difference is called a dipole. This is represented by the small delta positive and negative symbols on the diagram. This slight charge difference is always present and so we can call it a permanent dipole. There are such things as temporary dipoles which you will come across during your A-level.

So, there is more to learn along this topic, but I hope for now this should suffice as a brief introduction to electronegativity

A bit about the author, Paul H:

Paul is a qualified and experienced Physics, Maths, and Science teacher, now working as a full-time tutor, providing online tuition using a variety of hi-tech resources to provide engaging and interesting lessons.  He covers Physics, Chemistry, Biology, and Science from Prep and Key Stage 3 through to GCSE and IGCSE. He also teaches Physics, Maths, and Chemistry to A-Level across all the major Exam Boards.

You can enquire about tutoring with Paul here

The post Chemical bonding and electronegativity appeared first on The Tutor Team.

This is the third post in my series on moles. This one covers performing a titration calculation. You can click here to see part one and part two.

In this post, we look at how to use a titration result to calculate the concentration of an acid or alkali, as in this question:

25.0 cm³ of 0.200 mol/dm³ sodium hydroxide solution reacted with 28.7 cm³ sulphuric acid. Calculate the concentration of the sulphuric acid in mol/dm³ .

2 NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2 H₂O(l)

What is a titration?

That’s the first question we need to answer.

Basically, it is a neutralisation reaction used to identify the concentration of an acid or alkali. We carefully drip one of the reactants through a burette into another which is classically held inside a conical flask. Using universal indicators and constantly stirring we note from the burette the volume required to neutralise the acid or alkali.

So we have substance A of known concentration and known volume and substance B will have an unknown concentration but a known volume by the end of the experiment. We can refer to substance A as the ‘known’ substance and substance B is the ‘target’ substance, as that is the substance we are trying to find out about.

A note about concentrations, volumes and moles

If the concentration of something is 0.200 mol/dm³ of sodium hydroxide, that would mean if we had 1dm3 of the solution then inside the solution would be 0.2 moles. So if we had 2dm3 then we’d have 0.4 moles. Using this thought process and understanding we can see the formula that relates moles, concentration and volume is as follows:

Number of Moles = concentration X volume

This could then be put into a formula triangle:

A dm³ seems an odd unit of volume, especially when the volumes we’ll most likely be measuring will be in cm3. Without going into too much detail I’ll just say that:

• • 1cm³ = 1ml and 1000cm³ = 1dm³
  • Also, 1000ml = 1 litre.
  • So, 1dm³ = 1 litre.

Solving the problem

Let’s return to the problem and solve it now with our improved understanding.

25.0 cm³ of 0.200 mol/dm³ sodium hydroxide solution reacted with 28.7 cm³ sulphuric acid. Calculate the concentration of sulphuric acid in mol/dm³.

2 NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2 H₂O(l)

Step 1 – identify the known and target substance

We know the concentration and volume of the sodium hydroxide so that is our known substance. The target substance is the sulphuric acid. We are to find out the concentration of this. I’ve colour coded this to help follow along.

Step 2 – Convert the volume(s) into dm³

The volumes will usually be in cm³, but the concentrations are in mol/dm³, so in order for us to use the formula I introduced earlier we need them in consistent units. There are 1000cm³ in 1dm³, so to convert cm3 to dm³ you need to divide the volumes by 1000.

Volume of sodium hydroxide

25.0cm³ = 0.025dm³

Volume of sulphuric acid

28.7 cm³ = 0.0287dm³

Step 3 – calculate the number of moles of known substance

Being as we know the concentration and volume we can use the formula triangle from earlier to calculate the moles.

Moles of sodium hydroxide

= concentration X volume

= 0.200 mol/dm³ X  0.025dm³

= 0.005 moles

Step 4 – Use the balanced equation to work out the number of moles of the target substance

We can see from the equation that if we had 2 moles of sodium hydroxide, we could react it with 1 mole of sulphuric acid. So it’s a 2:1 ratio. We have 0.005 moles of sodium hydroxide so it could react with 0.0025 moles of sulphuric acid.

Step 5 – calculate the concentration of the target substance

We know that 0.0287dm3 of sulphuric acid was needed to complete the titration and so within that volume there must be 0.0025 moles of sulphuric acid. By using the formula triangle from earlier again:

Concentration of sulphuric acid = Moles ÷ volume

= 0.0025 moles ÷ 0.0287dm³

= 0.0871 mol/dm³ (3sf)

So, there you have it. Now, questions may vary slightly in that they may want the volume instead of the concentration, but in that case, they would give the concentration or some other means of getting it.

Notice also that with titrations there is no need to consult the periodic table to find the atomic mass or anything. However that does not mean that a sneaky follow-up question asking you to convert the moles into mass may not be included, but that is unusual.

That concludes the 3 part series on moles. The next post shall be a little different and then I may do another couple of guides on another commonly misunderstood topic within science.

Thanks for reading and I shall see you in the next post!

A bit about the author, Paul H:

Paul is a qualified and experienced Physics, Maths, and Science teacher, now working as a full-time tutor, providing online tuition using a variety of hi-tech resources to provide engaging and interesting lessons.  He covers Physics, Chemistry, Biology, and Science from Prep and Key Stage 3 through to GCSE and IGCSE, plus teaches Physics, Maths, and Chemistry to A-Level across all the major Exam Boards.

You can enquire about tutoring with Paul here

The post Performing a Titration Calculation (part 3 of 3) appeared first on The Tutor Team.

Today’s blog post will be the second part of a three-part series on a Chemistry topic called moles. I wrote the first part called ‘What is a Mole?’ recently and I would suggest reading that first, but essentially I explained that a mole is a unit of matter in which a known number of atoms or molecules is thought to be contained (specifically, 6.02×10²³ atoms/molecules which is also known as Avogadro’s number).

This number was chosen because a mole of any substance will end up having the same mass as its mass number, which is also the number of protons and neutrons in the nucleus. So, you can find the molar mass of any element by looking up the mass number in the periodic table.

In this post I will be focussing on how to apply this concept to complete calculate expected masses from chemical reactions as in the question below:

Aluminium is extracted from aluminium oxide as shown. Calculate the mass of aluminium that can be formed from 1020 g of aluminium oxide.

2 Al₂O₃ → 4 Al + 3 O₂

Calculating an expected mass

So let’s start with the first question and try to understand what information we have and what we are trying to find out.

Aluminium is extracted from aluminium oxide as shown. Calculate the mass of aluminium that can be formed from 1020 g of aluminium oxide.

2 Al₂O₃ → 4 Al + 3 O₂

Step 1: Identify the known and target substance

I’m referring to the known substance as the one we know the information the mass of. We have been given the mass of the aluminium oxide as 1020g. I’ve coloured it pink in the equation and in the question

This is not the molar mass!!

 So often people confuse this. The molar mass is the mass that one mole would have. This is just the mass we have. We don’t know how many moles there are yet, but we are going to find out.

The target substance is the substance we are being asked to calculate the expected mass of. In this case, this is aluminium. I have highlighted this in green.

Step 2: work out the molar mass of the target and known substance

Using the periodic table and the chemical formulae of the target and known substances we can calculate how many grams one mole would be. Ignore the big numbers on the left of the formulae in the balanced equation. Just find the mass of one mole of each substance. Eg. Al not 4Al.

We can see from the mass number that the molar mass of aluminium (Al) is 27g/mol.

For aluminium oxide (Al₂O₃), we can see it is:

(2 x 27g) + (3 x 16g) = 54g + 48g = 102g/mol

If I had 102g of aluminium, I would have one mole.

Step 3: Calculate the number of moles we have of the known substance

They have been nice here. We have 1020g of aluminium oxide. We know that one mole is 102g, so clearly we have 10 moles. However, sometimes they are not so nice, so it is important to know how to calculate the number of moles if you were given awkward numbers. We easily can see here that this is 10 moles, but let’s see if we can work out what sum we did.

10mol = 1020g ÷ 102g/mol

Or Moles = mass ÷ molar mass

We can use this to make a formula triangle. The right hand side of the equation shows us that mass would go at the top and molar mass at the bottom. The left hand side of the equation shows us that moles goes in the gap left behind at the bottom.

Step 4: Use the balanced equation to work out the number of moles expected of the target substance

The equation below shows us that if we had 2 moles of aluminium oxide then we could expect to get 4 moles of aluminium.

2 Al₂O₃ → 4 Al + 3 O₂

I need to make two things clear here

1. In reality we will not get this much for a couple of reasons. Some of the products could get lost in the process or maybe not all of the reactants managed to react. We should really call the expected number of moles the theoretical yield.
2. This is if we had 2 moles of aluminium oxide. So often students here record we have 4 moles of the target substance and complete the calculation from that. You need to keep the word ‘if’ needs in your mind when looking at moles.

Now… we know from earlier that we have 10 moles of aluminium oxide, so if every 2 moles of aluminium oxide produces 4 moles of aluminium, we can see the ratio is 1:2 and so we can expect 20 moles of aluminium in this case.

Step 5: Converting the moles of the target substance into a mass. THE FINAL STEP!

We’ve nearly made it! Going back to the previous formula triangle, we can calculate the mass of aluminium as follows:

mass    = moles x molar mass

= 20 moles x 27g = 540g

So, there you are. The maximum theoretical yield of aluminium expected in this example is 540g.

I hope you found that helpful and look out for the next blog post on performing a titration calculation. See you there!

9  THINGS TO ASK A PRIVATE TUTOR – BEFORE YOU BOOK THEM

A bit about the author, Paul H:

Paul is a qualified and experienced Physics, Maths, and Science teacher, now working as a full-time tutor, providing online tuition using a variety of hi-tech resources to provide engaging and interesting lessons.  He covers Physics, Chemistry, Biology, and Science from Prep and Key Stage 3 through to GCSE and IGCSE, plus teaches Physics, Maths, and Chemistry to A-Level across all the major Exam Boards.

You can enquire about tutoring with Paul here

The post More on Moles (part 2 of 3) appeared first on The Tutor Team.

This post is most suited to those of you studying GCSE Chemistry and confused about the topic of moles, which is within the area of Chemistry known as quantitative chemistry.

As a private tutor, I have become more familiar with the trickier topics within science and maths and occasionally have to brush up on the easier topics which is a little topsy-turvy, but this is because I help students with the topics they need help with and as such I get a lot of practice teaching the harder topics and the easier topics get neglected a little.

One of these topics which I have become extremely familiar with is moles. It has caused so many students so much difficulty that I have now taught it countless times, and so not only do I understand it very well myself, but I also have started to notice some of the reasons why it is so confusing.

LEARNING SCIENCE – IS THERE A SCIENCE TO IT?

Balanced equations

Let us start with a simple balanced formula equation for aluminium reacting with oxygen to form aluminium. If you want a real challenge cover the equation below and see if you can write it out and balance it, otherwise take a peek.

 4 Al + 3 O₂ → 2 Al₂O₃

Firstly, it is worth mentioning that the following sentence is not true:

“4 atoms of aluminium reacts with 3 molecules of oxygen to form 2 molecules of aluminium oxide.”

Why not? Seems right, doesn’t it? Iron chloride is an ionic compound and ionic compounds form ionic lattices, not molecules.

However, we can rewrite the sentence in terms of moles:

“4 moles of aluminium reacts with 3 moles of oxygen to form 2 moles of aluminium oxide.”

This is fine. However, we are getting ahead of ourselves now. I haven’t answered the question at the top of this blog yet! What even is a mole??

Enter Avogadro

Lorenzo Romano Amedeo Carlo Avagadro (yes, that is one person’s name!) came up with a crazy idea that you could weigh a bunch of stuff and calculate how many atoms or molecules were in that bunch of stuff.

After doing a bit of science… he, and some accomplices, came up with a very special number which became known as Avagadro’s Number. This number is 6.02×10²³ or 602,252,000,000,000,000,000,000. Essentially, a mole of something contains this many atoms/molecules.

This is worth repeating.

Carbon – A mole of carbon is 6.02×10²³ carbon atoms.

Iron – A mole of iron is 6.02×10²³ iron atoms.

Water – A mole of water is 6.02×10²³ water molecules.

Hydrogen – A mole of hydrogen is 6.02×10²³ hydrogen molecules.

The last one catches people out a lot. People often reason that because hydrogen is an element you treat like the first two examples and count atoms rather than molecules. Water is a covalent compound and so of course you count molecules, but hydrogen also exists as molecules, not atoms. This is the same for most non-metal elements excluding noble gases. So, when you’re considering what a mole of X is, you need to consider how X exists before assuming it is a mole of X is 6.02×10²³ atoms of X.

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Why this number though??

Basically, the number Avagadro came up with is very important because a mole of carbon weighs 12g and strangely enough 12 is also carbon’s atomic mass on the periodic table. This is no coincidence. This is the same for any element. The mass of one mole, known as the molar mass can be easily found by simply looking up the mass number of that element. To check you understand this concept, have a go at these 3 questions and cover up the answers after the questions:

1. 1. What is the molar mass of potassium?
   2. What’s the molar mass of aluminium?
   3. What is the molar mass of oxygen?

You should have got 39g, 27g and… 32g! Remember what I said about molecules? Oxygen exists as molecules and so when I say a mole of oxygen I mean a mole of oxygen molecules.

The mass number of oxygen is 16, so a mole of oxygen atoms would be 16g, but because the formula for oxygen is O₂ like most non-metal gases (other than noble gases) then a mole of oxygen molecules would be the equivalent to two moles of oxygen atoms.

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Calculating the number of moles

Let’s return back to that equation:

4 Al + 3 O₂ → 2 Al₂O₃

“2 moles of iron reacts with 3 moles of chlorine to form 2 moles of iron chloride.”

The molar mass of aluminium is 27g, so if we weighed out 27g, we would have 1 mole, right? And if had 54g, that would therefore be 2 moles, right? So, we can say following

number of moles = mass ÷ molar mass

This can be used to construct a formula triangle as below:

In a future blog, I plan on writing a guide showing how to complete mole calculations such as the one below:

“Calculate the mass of aluminium oxide formed when 135g of aluminium reacts with oxygen.”

For now, I shall tell you that the answer is 255g. Catch the next blog to find out why.

A bit about the author, Paul H:

Paul is a qualified and experienced Physics, Maths, and Science teacher, now working as a full-time tutor, providing online tuition using a variety of hi-tech resources to provide engaging and interesting lessons.  He covers Physics, Chemistry, Biology, and Science from Prep and Key Stage 3 through to GCSE and IGCSE, plus teaches Physics, Maths, and Chemistry to A-Level across all the major Exam Boards.

You can enquire about tutoring with Paul here

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