Lesson video

- Hi folks.

My name's Mrs. Farrow.

For today's session, you're going to need a pen and a piece of paper, and it's a good idea if you can switch off or pause any mobile devices, as this will help to minimise distractions.

We're gonna do some really cool stuff today.

By the end of the session, you're gonna feel super smart, so pause the video now, get those things organised, and press play when you're ready to continue.

Today's session is about chemical formula, and the conservation of mass.

We're gonna define conservation of mass, and apply this to chemical equations and reactions.

Some of the keyword that you can expect to see pop up during today's lesson include, the Law of conservation of mass, which states that mass is neither created nor destroyed in chemical reactions.

A chemical formula, which is a set of chemical symbols showing the elements present in a compound, and no relative proportions.

So the type number of elements that we see, and a balanced equation.

This is one in which the number of atoms of each element in the reaction are the same for both the reactants and the products.

Today's session has been broken down into three learning cycles.

We're going to look at reading formula, and talk about the conservation of mass within chemical reactions, and then have a go at balancing some equations.

So if you're ready, we'll begin and jump in looking at reading formulae.

So what are formulae? Well, a formulae as we've already said, is a set of chemical symbols, which shows the elements that are present in a compound, in they're relative portions.

When we want to write about a compound or a reaction, not only is it quicker to use the formulae than to write out the whole name, but it actually gives us a lot more information.

It not only tells us the type of element that we have, but also the number of each of the atoms within the compound.

So for example, calcium carbonate is called CaCO3.

Now you may be able to defer from the name that it contains calcium, carbon, and oxygen, but we don't know in which proportions until we look at the formula.

We can see here that this formulae indicates there is one calcium, one carbon, and three oxygens.

When it comes to reading formulae, there is important rules to follow.

Firstly, each capital letter represents the start of a new element.

So here we have capital F, lowercase E.

This represents one iron atom.

We then have a capital S.

On the periodic table a capital S is for sulphur, and there is no numbers next to it, and therefore only one sulphur atom present.

And then a capital O means oxygen.

And here we have the small subscript four following it, which means there are four oxygen atoms in this particular compound of iron sulphate.

Small subscript numbers only affect the element that they're following.

As we can see here, there are only four oxygens, whereas there remains one iron and one sulphur because the four follows the oxygen atom only.

Let's take a look at some more examples.

This is aluminium oxide.

Can you work out how many atoms are aluminium and oxygen there are present? Well, Al2 two means there are two aluminium atoms, and then the three following the O means that there are three oxygen atoms. Well done if you've got that right.

Let's take a look at another carbonate, this time, copper carbonate, CuCO3.

This is a capital C for carbon and a capital O for oxygen, not capital C, lowercase O, which would in fact be cobalt.

Let's take a look at the number of each of the elements within this compound.

We've got one Cu, which means one copper atom, one C, which means one carbon atom, and the subscript three following the O, showing that there are three oxygen atoms present.

Large numbers in front of a formula, multiply all of the atoms in the formulae by that number.

See for example, we'll often see H2O, which is water written like this.

Sometimes, however, more than one water molecule is made.

And so instead of having two hydrogen atoms and one oxygen atom, as there is in just one molecule of water, if you have two molecules of water, you have two lots of H2, which in total gives us four hydrogen atoms. And because the two is in front of H2O, it also times as the oxygen by two as well.

We now have two complete molecules of water.

We have two lots of oxygen and four lots of hydrogen atoms, but arranged as two molecules of water, or H2O.

Some formulae contain brackets.

This has become some elements form polyatomic ions.

Here we can see we've got a capital C and a lowercase A, representing one calcium.

And then in brackets we have OH.

OH represents a hydroxide group.

Outside of the brackets we have a two, and what this means is everything within the brackets is times by that subscript number two.

And so in this instance we have two oxygens and two hydrogens within our compound.

The OH, or hydroxide iron, always stays together, and so cannot be split up, and subscript numbers cannot be added within the brackets.

You can increase the numbers of everything inside the brackets by adding subscript numbers after it.

Again, here we have another example.

This is copper nitrate.

We can see that at we have one Cu representing one copper atom.

Inside the brackets we have NO3.

This means there's one nitrogen atom and three oxygen atoms, moving around together.

Outside the brackets, however, there's a two, and so everything within the brackets is now times by two.

So we have two lots of nitrogen atoms, and then the oxygen of which we already have three is times by two.

And three times two gives us six.

So we have six atoms of oxygen in total within this compound.

You cannot add numbers inside the brackets.

This would change the polyatomic ion, and affect the compounds that you have formed.

Let's do a quick knowledge check and see what you can recall.

What's the chemical symbol for carbon? Is it A, Cu? B, Ca? C, C? Or D, Co? The correct answer is C.

Carbon atoms are represented by a single uppercase C.

What is the correct formula for a compound which contains two sodium atoms and one oxygen atom? Is it A, 2SO? B, NaO2? C, S2O? Or D, Na2O? Though it's important to have a quick look at the periodic table, if you're not familiar with the symbols of these elements.

Sodium has a symbol of Na, so the answer has to be either B or D.

It says that there are two sodium atoms presence, but only one oxygen.

This means the correct answer is D, Na2, so two lots of Na, and only one oxygen.

Well done if you've got that one right.

Next question asks, what is the correct formula for a compound with one calcium, one carbon, and three oxygen atoms? Is it A, 3CACO? Is it B, CaCO3? C, 3CaCo? Or D, CaC3O? Remember we're looking for one with only one calcium atom and one carbon atom.

So the formula which only have one calcium in are B and D, but we can see that in formula D there are three carbon atoms. Our question says that there is only one and three oxygens, so the correct answer must be CaCO3.

How many hydrogen atoms are there in three molecules of this compound? So let's take a look.

We've got NH4, two lots of it.

CrO4.

Well we could ignore the CrO4.

What we need to do is concentrate on the number of hydrogen atoms, which is represented by capital H in the formula.

There are four hydrogen atoms inside the brackets, but we need two lots of it.

Four times two is eight.

But earlier in the question it said that we have three lots of this molecule, three times eight by three.

The correct answer would be D, 24.

There are 24 hydrogen atoms in three molecules of this compound.

Now it's time for you guys to have a go at practise task, and see how well you've got on reading formula.

I'd like you to write out the name and number of atoms of each element in each of the formula that follow.

You want to pause the video now, and it's a good idea to use a periodic table, if you're not familiar with their symbols.

When you're ready to hear the answers, press play to continue.

How'd you find it? Let's take a look and see how you got on.

So A is CO2.

Many of you may recognise this compound as it's very common, it's carbon dioxide.

We should have written that it contains one carbon atom and two oxygen atoms. Moving on to B, we've got NaCl.

Notice that the N is a capital letter, but the A is lowercase.

So this is one element.

We then have a capital C followed by a lowercase L.

So again, this is another element.

Na represents sodium and Cl chlorine.

So we have one Na and one chlorine in this compound.

C is H2SO4.

For this answer you should have got two hydrogens, one sulphur, and four oxygen atoms. Followed on to D, we have CaCO3.

This is an example we looked at in the first section of the video.

Hopefully you got this right.

One calcium, one carbon, and three oxidants.

Now you're gonna have a go at a few that are a little trickier, and contain those brackets, due to the polyatomic ions.

We're going to do exactly the same, writing out the type of number of each of the elements within the compound.

Pause the video now and press play when you're ready to hear the answers.

How do you find these? A little trickier but hopefully still manageable.

He has one calcium outside of the brackets, and then in the brackets we have O, oxygen, and H for hydrogen.

There's a two.

This means we should have one calcium, two oxygen, and two hydrogen.

F shows that there are two ion atoms, and four oxygen atoms. G, Al2, SO4 in brackets, three.

This means there are three lots of the SO4 in brackets, and the three does not affect the Al2, so we can say that there are two aluminiums. There will be three sulphur, and four times three lots of oxygen, which is 12.

And the last one by far the trickiest.

Let's take a look.

We've got one copper of four nitrogens.

Hydrogen is particularly difficult, as there is three times four, which is 12 in the first set of brackets, and then two times two, which is four, on the outside of the brackets.

12 plus four gives us 16 hydrogens, and then we have two lots of oxygen.

This was a really tricky one, so well done if you got this one right.

Hopefully, you're now beginning to feel a little bit more confident at reading formulae.

If you feel like you need to practise reading formulae a bit more, you can always go back and rewatch that section of the video again.

Next, we're gonna take a look at conservation of mass.

It's really important to recognise that atoms are not created or destroyed during chemical reactions.

We only rearrange them.

So the atoms that go into a chemical equation must come out of a chemical reaction.

They are just simply in a slightly different order or arrangement.

This means that the total mass of the products of a reaction will be equal to the total mass of the reactants in a reaction.

If we take a look at a simple chemical reaction, such as the one below, we've got lithium metal, one of the alkaline metals in group one, reacts with oxygen in the air, the form lithium oxide.

If we use six grammes of lithium and react it with four grammes of oxygen, the total on the left hand side of the equation is 10.

This means that the total amount of lithium oxide made must also be 10 grammes because we are not losing or creating any atoms within the reaction.

This is an example of a displacement reaction.

Silver nitrate will react with sodium chloride to form sodium nitrate and silver chloride.

Notice that both the reactants on the left hand side and the products on the right hand side both have a mass of 80 grammes.

This is because no material or atoms have been lost or gained during the chemical reaction.

The mass of the reactants before the reaction is equal to the mass of the products, after the reaction has occurred.

Atoms are not being lost.

They are simply being rearranged to form the products.

Let's see if we can calculate the missing mass.

We're going to select one of the options A, B, or C to answer the question.

So how much calcium carbonate would break down to form calcium oxide and carbon dioxide, if you get 44 grammes of calcium oxide and 56 grammes of carbon dioxide? So this is very much a mass question.

We just need to work the equation backwards.

So we'll add together the two masses of the products that have been made to give us the answer.

So will you choose A, 100 grammes, B, 12 grammes, or C, 38 grammes? The correct answer is A, 100 grammes.

Well done.

Let's do another one.

Magnesium burns with a bright white light in oxygen to form magnesium oxide.

If we react 24 grammes of magnesium and produce 40 grammes of magnesium oxide, how much oxygen was used? Is it A, 64 grammes, B, 68 grammes, or C, 16 grammes? The correct answer is C, 16 grammes.

Onto our next question.

At the end of a chemical reaction, how does the total mass of products compare to the total mass of reactants? Is it A, the mass is doubled, B, the mass is halved, C, the mass is the same, or D, the mass depends on the reaction happening? The correct answer is C.

The mass is the same.

It's the Law of Conservation of mass.

And to conserve something means to keep it the same.

Now, some reactions can't, sometimes make us think a little, and these are reactions which involve gases.

If a reaction happens in an open system and a gas is made, the gas is able to escape.

So for example, when mercury oxide is broken down into its component elements of mercury and oxygen, it can appear as if the masses has changed between the reactants and the products.

This isn't however the case.

It's just that the oxygen gas is escaping into the surroundings, rather than being captured and able to be weighed.

So we need to watch out for this during chemical reactions.

Gases can also be tricky when they're part of the reactants as well as products.

We've already seen that magnesium burns and reacts with oxygen in the air, and it forms magnesium oxide.

We can see here that the number of magnesium atoms and oxygen atoms, going into the reaction on the left-hand side of the equation is equal to the number of atoms of, that are contained within magnesium oxide on the right-hand side of the equation.

It will appear when doing the experiment that mass has increased during the reaction, that no atoms were created.

It's only that the oxygen atoms could not be weighed at the start of the experiment because they've come from the air in the surroundings around the magnesium.

It's only once the auction has become part of the solid product that you can actually weigh it and find out its mass.

Let's do a quick progress check and see how you're getting along.

True or false, mass can be lost or gained in chemical reactions involving gases.

The correct answer is false.

As we've just said, it may appear as though mass is lost or gained.

However, mass is never lost or gained in chemical reactions.

It is conserved.

Now it's time for you to have a go at a practise task.

In task A, we'd like you to answer the questions using the equation and the masses provided.

The chemical equation is that of calcium carbonate breaking down to form calcium oxide and carbon dioxide.

The reactants, as shown, is calcium carbonate of which you have 200 grammes.

And the products of the reaction are calcium oxide at 112 grammes and carbon dioxide at 88 grammes.

Use this information to answer the four following questions, and when you're ready to hear the answers, press play to continue.

Question one asks, what is the mass of the reactant? So we look to our equation and we're able to read off that the mass of calcium carbonate, the reactant in this equation, is 200 grammes.

What is the mass of each of the products? There are two products in this reaction, calcium oxide and carbon dioxide, and the masses are 112 grammes and 88 grammes.

What is the total mass of the products? In order to answer this question, we need to add together 112 and 88, which gives the answer 200 grammes.

Does this experiment data support the Law of Conservation of mass, and can you explain why? Well, let's look.

We had 200 grammes of calcium carbonate entering the reaction as a reactant.

We then had a combined total mass of products of 200 grammes.

So we can see that no mass has been lost or gained between the reactants and the products during this chemical reaction, so yes, it does follow the Law of Conservation of mass.

Your answer doesn't need to be word for word, the same as this one, but should follow this general theme.

Yes, the total mass of the reactant is equal to the total mass of the products.

No mass has been lost or gained.

Once the second part of this task, we'd now like you to calculate the missing masses for each of these reactions.

You'll need to pause the video and you're quite welcome to use a calculator should you choose to.

When you're ready to see the answers, press play to continue.

Hopefully you found these quite easy.

Let's see then how you got off.

So if we make 80.

6 grammes of magnesium oxide in the first reaction, and 32 grammes of that is oxygen.

If we take 80.

6 and minus 32, we should get an answer of 48.

6 grammes.

Well done if you've got that one right.

The second question, we are reacting lithium with oxygen to make lithium oxide.

We've got 21.

9 grammes of lithium and eight grammes of oxygen.

So in order to work out the mass of lithium oxide, we need to add those two products together.

So 21.

9 add eight grammes should give you an answer of 29.

9 grammes.

And then lastly, sodium reacts with oxygen to form sodium oxide, and in this instance we have an excess of oxygen available.

If we react 46 grammes of sodium with 32 grammes of oxygen, and this produces 62 grammes of sodium oxide.

What is the oxygen that's remaining? So to find out the answer to this, we need to do two sums. First of all, we need to add 46 to 32, in order to find out the total mass of the reactants.

We then need to subtract 62 from this figure, and it would give us the answer of 16 grammes.

Well done if you got all of those right.

Hopefully looking at the equations like this makes it easier for you to be able to see what it is you're trying to work out, and where the values are coming from.

Unfortunately, sometimes in exam questions, they're not quite so friendly.

And so, for the third and final part of this task, we've given you some written examples of questions.

You are calculating the missing masses for the reactions, just as you have done in each of the cases before.

But now you're going to have to read the information to work out which are the reactions and products, and calculate the amounts that is missing.

I suggest pausing the video now, and you may find a pen and paper handy to write down the equations, as you read through each of the sentences.

When you're ready to hear the the correct answers, press play to continue.

How did you find it? It's easier when you've got a pen and paper, I'll admit.

So let's see if we can see the wood for the trees, in these questions.

Question one says, we have four grammes of hydrogen, which reacts with oxygen to make 36 grammes of water.

So our reactants are hydrogen and oxygen.

We know that we have four grammes of hydrogen, but we haven't been told the mass of oxygen.

We know that we're going to make 36 grammes of water product.

So in order to calculate the amount of oxygen used, we would need to say 36 minus four equals 32 grammes.

32 grammes of oxygen was needed in order to produce the 36 grammes of water.

Well done if you got that one right.

Question two, in a chemical reaction, 150 grammes of sodium bicarbonate and vinegar on heating produces 87 grammes of carbon dioxide gas.

What mass of solid residue is left? So we're starting with sodium bicarbonate and vinegar, and we're heating it.

The 150 grammes of sodium bicarbonate and vinegar loses 87 grammes of carbon dioxide gas's product.

So how much will be left? In order to calculate this, we need to say 150 grammes, subtract 87 grammes, which gives us the solid residue left behind of 63 grammes.

And lastly, when 0.

0976 grammes of magnesium is heated in air, 0.

1618 grammes of magnesium oxide is produced.

What mass of oxygen is needed to produce the 0.

1618 grammes of magnesium oxide.

So we've been given the mass of the product, and we've been given the mass of one of the reactants.

So we're going to take the mass of the product, 0.

1618 grammes and deduct the mass of the reactant, the known reactants, magnesium, 0.

0976, which equals 0.

0642 grammes of oxygen, would've been needed in order to create this mass of magnesium oxide.

You've done so well if you've managed to get all of those right.

It's worth remembering a highlighter for your exams to help pick out the important information in the questions, and this will help you to see what you need in order to do what are actually quite simple mathematical subs.

Don't forget you're also allowed a calculator in your exams too.

So we've now taken a look at how to read formulae, and we understand that mass is conserved in chemical reactions.

Bet you're already feeling pretty smart with all the stuff we've done so far.

Well, now we're gonna take a look at balancing equations, notoriously something that lots of students find tricky, but hopefully by the end of this cycle you're going to be a whiz at it.

So let's jump in and take a look.

Balancing equation means getting out what you put into the equation.

This is the same as conserving mass.

Atoms have mass, we know this.

Atoms contain protons and neutrons within their nucleus, and the proton and the neutron each have a relative mass of one.

And so in the same way that we can't lose or gain massing grammes, that means we can't lose atoms. If we look here at our scales, we've got oxygen and lithium on the left-hand side, which then react together to form lithium oxide on the right-hand side.

Products of, the reactants, sorry, of oxygen and lithium rearranged to form lithium oxide, but the mass doesn't change.

There is an equal amount of mass in the reactants and the products, and there's an equal number of atoms in the reactants and the products.

One of the important things that we really need to remember when balancing equations is that you can't change the formula.

So think back to the beginning of our session, we talked about the idea of water having a formula of H2O.

Water, which is H2O we know is safe to drink.

What we can't then do is add small subscript numbers to formula.

If you were to do this to water and form H2O2, this is actually hydrogen peroxide which is found in bleach.

Very much not safe to drink.

I guess the message is don't add small subscript numbers to formula, you could kill somebody.

So, we can only add numbers in front.

If we wanted to increase the number of oxygens there by two, or two two, then we would need to balance the equation by adding a number in front of the elemental compound, not after.

However, we will need to consider the fact that putting a big number at the front multiplies everything that follows it, not necessarily just the atom that we want to increase the number of.

So let's take a look at a few examples.

Hydrochloric acid reacts with sodium hydroxide in a neutralisation reaction to make sodium chloride and water.

So we have HCl, H is hydrogen and Cl is chlorine.

As represented below, you can see by the little coloured atoms. We react that with NaOH or sodium hydroxide.

Sodium hydroxide is made of one Na, one oxygen, and one hydrogen.

This is why the reading formulae section of our learning today was so important.

It's impossible to balance an equation correctly, if you can't read the formula.

Now these react, and in reacting they rearrange to form the products NaCl and water.

Now, in order to make sure that this reaction is balanced, we need to make sure that we have the same number and type of element on each side of the reaction, both the reactants and the products.

This has been made easier in this particular image because you can count the coloured circles.

We can see that on the reactants side of the equation we have two hydrogens, as shown by white circles with Hs in.

On the product side we also have two white hydrogens with Hs in.

So we know that the number of hydrogen atoms going in is the same as the number of hydrogen atoms coming out.

We can see that one purplely-colored sodium atom has gone into the equation, and we've got that same sodium atom out in the products.

One chlorine as shown by a green circle here goes into the equation and one comes out.

And lastly, one oxygen atom is part of the reactants and one oxygen atom is part of the products.

So what we can see is that the same number and type of atoms have gone into the equation as have come out.

This equation is what we call balanced.

Let's take a look at another reaction.

Hydrogen reacts with chlorine to form hydrogen chloride.

H2 plus Cl2 gives HCl.

So in this instance we've got two hydrogens reacting with two chlorines to us HCl.

Hopefully you can spot that there's a problem here, just from looking at the images of the coloured circles.

We've put two hydrogens in and two chlorines in, but we've only got one of each out.

This equation was suggests that we've destroyed atoms, and this isn't possible.

The only way to fix this is to increase the number of HCls that we've made on the product side of the reaction.

Now remember, we can't change the formula, and so we are not going to add any small subscript numbers.

We must in fact put a large number in front of the HCl.

This doubles the number of HCls that we have, and so now we have two hydrogen atoms on the reactant side of the equation, and two on the products, and two chlorines on the reactant side, and two chlorines on the products.

Hopefully you can see how we've balanced this equation.

Here's another example.

Methane is natural gas and burns in oxygen to reduce carbon dioxide and water.

This reaction is shown by the similar equation, CH4 plus O2 gives CO2 plus H2O.

If we read the formulae CH4, we know that it has one carbon and four hydrogen atoms. This reacts with O2, which has two oxygen atoms. You can see how important our formula reading is.

We then make carbon dioxide, which is one carbon with two oxygen atoms, and we make water which is hydrogen with one oxygen atom.

Hopefully by looking at the different coloured atoms that's represented in the image, we can see that there is a disbalance in both the number of hydrogens and oxygens that we find in the reactants and the products.

We have two oxygens in the reactants, but three in the products.

And we have four hydrogens in the reactants, but only two in the products.

I'm gonna start by sorting out the hydrogens because I can see that at an easy fix for hydrogen to make go from two to four would be to times it by two.

So two times H2O would now give me the four hydrogens that I require.

In addition, this means on this side of the equation now, the products, I now have four pinky coloured oxygen atoms in this image, I'm only putting two auction atoms in however.

This again would now suggest I'm creating atoms, which can't be done, and so I must increase the number of oxygen atoms that have been put in.

If I times the number of oxygen atoms that have gone in by two, I then have four oxygen atoms. If we now count rounds the different atoms on the reactants and the products, we can see that they balance.

We have the same number of hydrogens going in as out.

We have the same number of oxygens going into the equation as out, and carbon has remained the same throughout.

This is another example of how equations are balanced.

Balance the equations basically means making sure that they follow the Law of Conservation of mass.

Let's do one more together.

Lithium reacts with oxygen air to form lithium oxide.

Li plus O2 gives Li2O.

So we've got one lithium atom going in and two oxygen atoms going in.

But in the compound we've made, we've got two lithium atoms and one oxygen.

In order to use up the other oxygen atom that we've put in, I'm going to have to make an extra lithium oxide compound.

I can't just put a small subscript number after that oxygen.

As much as you may want to put a small tube there that's changing the formulae, and we don't do that because like in the case of water, it kills people.

So, we need to say we've got two lots of Li2O.

This would look something like this.

We've now sorted out our oxygen problem, but we now have four lithium atoms in our products.

We've only put one lithium atom in as a reactant.

And so to balance this equation and make sure it adheres to the Law of Conservation of mass, we need to increase the number of lithiums to four.

Hopefully this is starting to make a bit more sense.

Let's have a look at this one.

Do you think it's balanced or not? Magnesium plus oxygen gives magnesium oxide.

The reaction for this is Mg plus O2 gives MgO.

Taking a look at the diagram, hopefully you can see that this equation is not balanced, so let's balance it.

What are we going to need to do? Well, we're going to need to increase the amount of magnesium oxide made.

In doing that, we now need to increase the amount of magnesium that we're putting in.

Now we have the same number of atoms of each element on both the reactant side and the product side of this equation.

Time for a quick knowledge check.

True or false? This equation is balanced.

N2 plus H2 gives NH4? True or false? The correct answer is false.

The number of atoms of each element changes between the reactant side of the equation, and the product side of the equation.

Two atoms of nitrogen are going in, but only one atom of nitrogen is going out, so this is an unbalanced equation.

Well done if you got that one right.

These are quite tricky to do particularly at speed.

Let's have another look at another one.

True or false? This equation is balanced.

2Al plus 3B2 reacts to give 2AlBr3.

Okay, what do we think? Looking at this equation I can see I've got two aluminiums going in, and I've got two aluminiums coming out.

I've got three times two bromine on the reactants side.

So three times two gives me six.

On the reactants, or on the product side, I've also got three times two because the two is at the big start of the formula.

So the two at the start times by the three R that follows the bromine gives me six bromines.

This means that this equation is balanced, and therefore it's true.

The number of atoms of each element is the same on both sides of the equation.

Let's have a look at the last one.

True or false, this equation is balanced.

H2 plus O2 gives H2O.

Hopefully it's easy to spot that this equation is not balanced, and so the answer is false.

The number of oxygen atoms on the left is greater than the number of oxygen atoms on the right, and so this equation has not been balanced.

Now it's time for you to have a go at a practise task.

You are going to be given a series of five equations that you are going to have a go at balancing.

Remember, you mustn't change the formula, so no adding small subscript numbers after any of the elements.

You can add large numbers in front of the elements or compounds, in order to increase their number.

Give it a go.

If you find it easier to draw them out on paper and use coloured pens or symbols to represent the atoms, then do so.

When you're ready to see the answers, then press play, and we'll go through them together.

How you found it? They do get progressively more difficult, as you go down through the list.

So don't worry if you've only managed the first two or three.

It's important that you keep practising , and that you begin to your skills in balancing equations.

Let's take a look at these answers.

So for equation one, Zn plus O2 gives ZnO.

This is currently unbalanced.

In order to balance this equation, you need to increase the number of ZnO by two, and then add extra zinc to the reaction.

So we need 2Zn plus O2 gives us 2ZnO.

In equation two, we have Cl2 plus Al gives AlCl3.

This equation is unbalanced.

In order to make it correct, we need to place a three in front of the Cl2, in order to make six.

We then place a two in front of the AlCl3.

This gives us the six chlorines, but also increases the number of aluminium by two.

And so we place a two in front of the aluminium on the reactants side.

Giving us 3Cl2 plus 2Al reacts to form 2AlCl3.

Equation three, Na plus O2 gives Na2O.

In order to balance this equation, you needed to increase the number of sodiums going into the reaction by four.

This is because we need two oxygens on the product side, and so place a large two in front of the equation.

Place a four in front of the sodium.

Equation four, Mg plus HCl gives MgCl2 plus H2.

In order to balance this equation, we need to increase the number of HCls by two.

And then lastly, we have Fe2O3 plus 2Al, reacts to form 2Fe plus Al2O3.

Really well done if you've been able to get all of those right.

This is not easy, but you will get there with continued practise, if you've not already done so.

Don't forget you can always go back and watch any sections of the video that you want to again, in order to practise your skills of reading formulae, identifying that masses can serve within reactions, or balancing equations.

We've reached the end of our learning cycle for today on chemical formulae and the conservation of mass.

So what are our takeaway points? Well, a formulae is chemical symbols showing the elements in compounds, and the number of atoms of each element that are present.

The Law of Conservation of mass says that no mass can be lost or gained during a chemical reaction.

The total mass of the reactants and the products must be equal.

And balancing equation involves ensuring the number of atoms of each element are the same for the reactants and the products.

If you're feeling confident, why not have a go at our exit quiz before you sign out? I really hope you've enjoyed today's lesson, and got lots from it.

I always felt really clever after completing something really challenging like this.

We'd like to thank you for choosing to use Oak National Academy today, and we hope to see you again soon.

Bye for now.