Hello, my name's Mrs. Navin and today we're going to be looking at how to determine the empirical formula of a substance experimentally as part of our topic on calculations involving masses.

Now, you may be familiar from your previous learning on some of the mathematical processing that's needed, but what we do in today's lesson will help us not only answer that big question of what are substances made of, but will help us to better link up what we see and observe during a practical to how we discuss reactions and talk about the particles involved in those reactions.

So by the end of today's lesson, you should be able to describe and carry out an experiment to help you determine the empirical formula of a simple compound.

Throughout the lesson, I'll be referring to some keywords, and these include empirical formula, relative atomic mass, and evaluate.

The definitions for these keywords are given in sentence form on the next slide and you may wish to pause the video here so that you can jot down these definitions for reference later on in the lesson or later on in your learning.

So today's lesson is broken into two parts.

Firstly, we'll look at how we gather the data that we need in order to deduce the empirical formula and then we'll move on to process that data that's been collected and consider it.

So let's get started and looking at how we can gather data that we need in order to find an empirical formula.

So first of all, we need to remind ourselves that during a chemical reaction, all of those reactant atoms are being reorganised to form the products in this chemical reaction.

So if we look at this particular example of iron added to sulphur and then they bond together to form iron sulphide, what we can see here in our balanced symbol equation is that no atoms have been gained or lost, nothing's just been (snaps fingers) suddenly created or completely destroyed and lost in the course of that chemical reaction.

Atoms have been conserved.

Now because the atoms aren't being lost or gained throughout that chemical reaction then the mass of those atoms, so again, the mass of our reactants and our products is being conserved as well.

And that means then the total mass of our reactants is equal then to the total mass of my products.

And if we go back to that reaction earlier of iron and sulphur being bonded together to make iron sulphide, if I start with five grammes of iron and two grammes of sulphur, I would expect then to form seven grammes of iron sulphide.

So what chemists can do then is use our understanding of these chemical reactions, the idea that atoms and the reactants are rearranging to form the products and that conservation of mass, so the mass of my reactants combined equals the total mass of my products combined.

And if we use that then, we could mathematically process those values to find an unknown reactant mass.

So if we look at this example here, I have 3.

28 grammes of chromium that's reacting with some chlorine, and that's forming 10 grammes of my chromium III chloride.

If I mathematically process this, so I need to find the mass of chlorine, I'm gonna take the total mass of my products, which is 10 grammes, and subtract my one known mass of my reactant for chromium 3.

28, and I get a value of 6.

72.

What that tells me then is that in this particular reaction, 6.

72 grammes of chlorine reacted.

So why do we care? Well, it's really useful because we can actually process those reactant mass values to find the empirical formula for my chromium III chloride.

Let's stop here for a quick check.

What do you reckon is the missing mass for oxygen in this reaction? You may wish to pause the video here so you can grab a calculator and come back when you're ready to check your answer.

Well done if you said, C, 2.

857 grammes of oxygen was needed, and we know that because that value has to be added to the titanium value in order to equal the mass of my product, which was 7.

143 grammes.

So very well done if you managed to get that correct.

Great start guys.

Now, when we're gathering data for a mathematical processing, it's normally data that's collected through an investigation.

And when we're discussing reactions in a practical setting, chemists often refer to an observed environment.

So these are usually described as being open or closed systems. So we have two examples here.

The first one on the left, the heating, the iron and the sulphur.

What we can see here is that the apparatus, the setup contains a plug.

So it's enclosed that actual environment.

And so we call that then a closed environment and that's where substances cannot enter or exit where that reaction is taking place.

If we look then at the reaction that's taking place on the right, this burning of a match in air, we can see that it's fully open to the surroundings, okay? And that reaction being fully open allows then gases to enter or leave that it observed environment and therefore this particular observed environment would be described as being open.

Now you may recall that oxidation reactions are those in which oxygen can bond to the elements of the reactants.

And we have an example here.

If you're burning coal, which is mostly carbon, we can see that oxygen is bonding here because we have plus oxygen in our chemical reaction and it's forming here then carbon dioxide.

Now, most oxidation reactions occur when our reactants are exposed to or heated in air because air contains oxygen.

Now if we take that idea of oxidation taking place within the laboratory setting, possibly, okay, we could potentially expose something to air by simply lifting the lid on an otherwise closed system.

So here we have a crucible that's being heated over a Bunsen burner, and if we just take the lid off, that will allow substances like air to enter that observed environment.

Now that could potentially cause an oxidation reaction to occur, in which case we might be able to then use our conservation of mass understanding to find an unknown value.

So let's imagine we have some magnesium in that crucible and we've lifted the lid on it and exposed that magnesium whilst it's being heated to the oxygen in the air that has entered into that environment.

And by doing so, it's formed some magnesium oxide.

We have some values here then where you're able to measure the mass of the magnesium at the start and the magnesium oxide at the end, and then we can process those values and find out what mass of oxygen then actually was able to enter that environment and react with that magnesium.

And we can find then that 0.

65 grammes of oxygen gas was gained through this reaction from the surroundings.

Okay, time for our first task and we're starting with a practical.

What I'd like you to do is to react some magnesium with oxygen, and by doing so, we're going to form magnesium oxide.

Now, in order to do this, you're going to need a few things.

You're gonna need your heat proof mat, a Bunsen burner, tripod, clay triangle, a crucible with a lid.

And it's crucial that that lid fits over the crucible itself very easily.

So you have a closed system to start with.

Then you'll have some magnesium ribbon that you'll put inside it, and you'll need some tongs in order to open that environment of the crucible as it's being heated.

Now what you'll be doing is you'll be recording some masses throughout we go, as we go, sorry, and I will show you those in the table on the next slide in a moment.

But ultimately this reaction is gonna require you to heat the magnesium ribbon strongly.

So we're gonna have it on its full roaring blue flame, okay? And then as that's heating, you're gonna carefully lift the crucible lid every so often.

Now, it can be a little tricky sometimes if you've not done this very often.

So I would recommend practising with the tongs, lifting the lid on the crucible a few times before you start heating it.

Also, it's really important that you don't look directly at this reaction because it could potentially hurt your eyes, okay? So maybe just look, see if it's still reacting, and then look away after you've put the lid back on.

After you've heated this for about five minutes, you're gonna allow the crucible to cool.

And then finally, record that last mass in your table.

Now if for some reason you don't have access to the equipment in order to carry out this practical, there is a video that you can click here to watch that.

Now, as I said, I would provide you with a table of the measurements I'd like you to make throughout this practical, and this is it.

So I've outlined the numbers, mass one, two, and three and referred to those in the method on the previous slide.

So you may wish to just double check that so you know when you are recording these different masses.

And then once you've finished this practical, I'd like you to process your results.

I'd like you to use them and your understanding of conservation of mass to calculate the mass of oxygen that actually reacted with your magnesium.

So this is gonna take a little bit of time, so pause the video and come back when you're ready to check your answers.

Okay, so we've got all of our equipment ready for our practical, and we moved the crucible and the lid onto our balance to get that first mass, which is 23.

94 grammes.

And next what we're going to do then is to move our magnesium strip into the crucible, and that mass then is 24.

20 grammes.

Now that magnesium strip has been curled up a little bit to make sure that it can fit into our crucible easily and that the lid will fit on it closely as well to make sure we've got a closed system here.

Now we put it in our clay triangle on top of our bun, sorry, on top of our tripod, move our Bunsen burner then to that roaring blue flame and put it underneath our crucible to let it heat quite strongly.

Now remember, we're gonna be lifting that lid ever so slightly, changing our closed system into an open system, and we can see that it's definitely reacting with that bright glow that's forming there.

And we're gonna keep doing this ever so often, double checking up, it's still glowing, which means it's still reacting.

And we're gonna close that lid again to make sure that none of the products that are forming are escaping as well.

Keep checking it now, and every time it glows, we know that it's still reacting.

Put that lid back on and we're gonna keep doing that until it stops glowing.

Now this one then has finished reacting and it has now cooled down.

So I'm removing it from the clay triangle and I'm gonna put it onto the balance once it's been tared.

So it's should say 0.

00 grammes.

Once that happens, you can now put your crucible with our product and we know it's product 'cause of that white powder we can see within it, put the lid on as well.

And we now have our final mass of 24.

33 grammes.

Okay, let's see how you got on.

Now it's going to depend on what mass of magnesium you actually started with, but if we use the measurements that were available on the practical video, then mass number one of the crucible and the lid was 23.

94 grammes.

With the magnesium ribbon before it was heated was 24.

20 grammes, and after it was heated, had a value of 24.

33 grammes.

Now, I then asked you to use your understanding of conservation of mass to calculate the mass of the oxygen that actually reacted here.

Now again, this is gonna depend on your own recordings of the masses throughout your practical, but if we go through here what it would be like if you had used the practical measurements, we know that 24.

20 grammes of magnesium and the crucible at the start and then the 24.

33 grammes of the magnesium oxide with the crucible and the lid at the end.

Again, remember, those mass of the reactants totaled, has to equal the mass of the products totaled.

And when we process those numbers then, we get a value of 0.

13 grammes of oxygen then reacted.

Okay, now that we're feeling a little more comfortable about how we can collect data that we need in order to find empirical formula, let's actually look at how we can process that information and consider that practical data.

Now, I said earlier that a substance's empirical formula could be calculated if we knew the reactant masses.

And there are a few steps within this process.

The first thing we need to do is we're going to divide the reactant mass of each element by its relative atomic mass.

Then what we're going to do after that is divide each answer by the smallest answer from step one.

And what that does then is it forms a unitary ratio, meaning that at least one of the answers should becoming a one, okay? Now, some of the answers then as a result of step two will be very close to a whole number and could be rounded to it.

And if it isn't, then it might be necessary to multiply accordingly in order to achieve a whole number ratio.

So let's go through an example of how we could follow this processing procedure.

If I want to know the empirical formula of a substance that contains 8.

57 grammes of carbon and 1.

43 grammes of hydrogen, I'm going to set up a little bit of a calculation grid.

And what you can see here is I have both of the elements listed at the top in headings with spaces below to do my processing of the numbers.

And then along the far left-hand column, I am keeping track of either the information I need in order to do that processing or the actual step that I'm carrying out.

So the first thing I'm going to do is I'm going to record then the masses for each of my elements straight from the question.

So I have 8.

57 grammes of carbon, so I'll put that under the C column, and I have 1.

43 grammes of hydrogen, so that's going under the H column.

Now the first step, if you remember, is that I need to divide these answers by their relative atomic mass.

So for carbon that's 12, and for hydrogen that's one.

And when I do that, I get a value of 0.

714 for carbon and 1.

43 for hydrogen.

Now when I compare those two answers then, I can see that 0.

714 is smaller.

So I'm going to divide those answer values by 0.

714.

Carbon changes to a one and hydrogen changes to a 2.

002.

Now carbon is definitely a whole number here we can see that hydrogen is not, however, the value is very, very close to a whole number.

And because of that I can round it.

And when I do that, I get a value of one for carbon and two for hydrogen.

And we don't ever really write the ones in a chemical formula.

So my final answer here for the empirical formula of this substance would be CH2.

So I've given an example here.

What I'd like you to do now then is to determine the empirical formula of a substance that's formed from 6.

902 grammes of iodine and 3.

098 grammes of fluorine.

You're going to need a periodic table and a calculator and a little bit of time to carry out your processing.

So pause the video here and come back when you're ready to check your answer.

Okay, let's see how you got on.

Now, what I'm going to do is I'm gonna show you the final answer, and if you've got that correct, incredibly well done.

If not, all the processing is laid out very similar to what I showed you earlier.

So you can have a quick look through that and see if you can identify where you've gone wrong.

So you can avoid that or fix your processing as we go forward.

But very well done if you managed to get an answer of IF3.

Great job guys.

Now you may recall that the empirical formula for an ionic substance is the same as its chemical formula.

So we have an example here of sodium chloride.

So we have a chloride one minus ion and sodium one plus ions, and we compare the ratio of those, we can see that we have one sodium ion for every one chloride ion.

And because each of those charges on our ions, when we have a one-to-one ratio balance out to an overall formula unit charge of zero, then its chemical formula for sodium chloride is NaCl.

But when we look at another example, this time of sodium oxide, our oxide ion has a charge of two minus, and our sodium has a charge of one plus still.

In order for our formula unit to have an overall charge of zero, the ratio of sodium to oxide ions is going to be a two-to-one.

And that means then the chemical formula, and therefore the empirical formula of this ionic substance will be Na2O.

So why do we care? Well, what chemists can do is they can use this understanding of ionic substances, their chemical formulas, empirical formulas to evaluate then the quality of the data that was collected to determine the empirical formula of an ionic compound.

So we might then compare our calculated value against our ionic prediction.

Now, every good scientist is going to evaluate the quality of their data by both comparison and reflection.

They might compare their data in order to ensure the method was repeatable.

So that means that they're producing reliable or very similar results themselves, and they might also compare their data to see if it was reproducible.

Are other people able to achieve similar results following the same method that they've used? They might also evaluate the quality of their data by reflecting, thinking about how that data was collected.

Could it be improved either by changing the equipment that they're using or considering the steps and how they were carried out within the method itself? When we are considering the equipment that's been chosen, you might wanna think about trying something else.

Is there a different piece of apparatus that would provide more precise measurements to be taken? Because more precise measurements have lower uncertainty.

So a smaller range within the accurate value can be found, and that means that you would have better quality data that's been collected.

So I have here an example of two balances.

They both can measure the mass of a substance, one will measure it to one decimal place, and the one on the right will do it to two decimal places.

When we compare the uncertainty, the one on the left has an uncertainty of 0.

05 grammes, but the one on the right has an uncertainty of 0.

005 grammes.

So it's uncertainty range is a lot smaller.

And that means then, that it's going to provide a better quality data because the correct value would be within a smaller range of values, that smaller plus or minus range.

When chemists are reflecting on the method they've used, they might consider when they are looking at a closed system, okay? Especially if there is a step that includes opening that closed system, okay? Because remember opening a closed system, so removing that lid, allows substances to both enter and exit that environment.

And that means really any gases or any lightweight products could leave that environment if the system is open for too long.

Moreover, if the reaction involves a gas, a chemist might think about keeping track of the reaction's mass until it doesn't change any longer.

For instance, if we were reacting some magnesium in acid and we're keeping track of its mass as it changes, if it's after 30 seconds after the start, you might see a mass and you could take that mass immediately.

But is it an accurate mass? Possibly not.

If you leave that reaction going until it no longer changes, the mass that you then record at the end, that final mass is more likely to be correct because you're allowing the full necessary volume of gas to react in the first place or for all of that gaseous product to form before you take that final mass recording.

Let's stop here for another quick check.

Which apparatus choice listed below would provide the highest quality data when measuring mass? And I'd like you to explain your choice.

Well done if you said D, if a balance has more decimal places, it's more likely to have a smaller value range for the uncertainty, and therefore you are going to have higher quality data.

So very well done if you chose D.

Okay, moving on to the last task for today's lesson.

For this first part, what I'd like you to do is to use your results from Task A, parts B and C to calculate the empirical formula for the magnesium oxide.

And I'd like you to show you working out please.

So pause the video here and come back when you're ready to check your answer.

Okay, let's see how you got on.

Now your results are all gonna be different depending on the practical that you carried out.

So the value I'm going to show you is based on the practical video that was shown earlier in the lesson.

So when I've processed the values from that particular practical, I got a final empirical formula of Mg4O3 and I've shown where we got the different values for the magnesium and the oxygen as those reacted masses, divided by the relative atomic mass to find an answer.

And then in this case, because I didn't get whole number ratios or ones that needed to be easily rounded, I had to multiply accordingly to get that whole number ratio at the end.

Now, some of you eagle eyed out there might be saying that formula looks a little odd and it might do to you, but ultimately what we're looking for here is have you processed your data correctly in order to find an empirical formula? And if you have and you followed these ones here, incredibly well done.

So for the next part of this task, what I'd like you to do is to use the diagram provided to determine the actual empirical formula for magnesium oxide.

And then once you have that, I'd like you to compare your answers to Task B part two A and Task B part one.

So pause the video here.

You may wish to make some discussions with the people around you and then come back when you're ready to check your answer.

Okay, let's see how you got on.

So for this first part, you needed to use the diagram to determine the actual empirical formula for magnesium oxide.

And reminder, we needed to find the chemical formula is the same as the empirical formula for this ionic substance and that the ion charges need to add up for an overall charge of zero for that formula unit.

And we compare then the ion charges for each of our elements, we can see that the ratio is one-to-one to get to that overall charge of zero, which means the actual empirical formula for magnesium oxide should be MgO.

And then finally, I asked you to compare your answers then.

So that's comparing the calculated empirical formula versus the one that we found from the diagram, and it's going to depend on your individual answers, okay? Now crucially here, what we're looking for is if your answers don't match, we want to reflect and think about why.

Why might those answers from our experimental empirical formula not match that for the diagram of magnesium oxide? Well, it's quite possible that in some cases not enough oxygen was actually allowed to enter that crucible.

Maybe the lid wasn't lifted often enough, and that would mean that not all of that magnesium actually reacted.

Alternatively, it could be that some of that really light weight powder of magnesium oxide was actually lost to the environment when the lid was lifted, maybe the lid was lifted a little too high, and that would actually result in a lower calculated mass then for the reactant oxygen that we used in our calculations.

Some exemplar results than here, that if we had slightly different values for our magnesium and oxygen, if we still were able to process them to find that empirical formula we could have achieved a final answer of MgO.

Now, one thing about practical experiments is be able to reflect on what we've learned from an experience and think about ways that we can improve for another practical.

So I have an example here.

You're not actually gonna carry this one out, but we want to consider what have we learned from our practical, making the magnesium oxide that we could carry through to this practical possibility of reacting titanium in a similar manner.

So I'd like you to pause the video here and have a go at answering these questions and then come back when you're ready to check your work.

Okay, let's see how you got on.

So for the first part, I wanted you to write the word equation for the reaction that takes place in the crucible.

And we have started with titanium.

We need to add some oxygen to it in order to make that product of titanium oxide.

So we get titanium plus oxygen, arrow, titanium oxide, well done if you got that correct.

For B, we needed to determine the measurements that might need to be recorded in order to eventually determine the empirical formula of that titanium oxide that forms. And what you could have done is gone back to your own measurements and you'd need something exactly the same.

So we need the crucible and the lid on its own, the crucible lid and our titanium metal at the start before it's heated, and then the crucible, the lid and the titanium products that were formed at the end.

So three main measurements that were needed throughout this practical.

You were then asked to consider why it's actually necessary to lift that crucible lid while heating the titanium.

And the key here is that you need to open that closed system because if you don't do that, you're not gonna get any oxygen into the system in order to react the titanium.

So by opening the lid, opens the system, and it allows the air, which includes oxygen, to enter that observed environment and react with that titanium.

So it allows the oxygen in.

Well done if you got that correct.

And finally then you were asked to consider what could cause a loss of product during the practical and then how it could be avoided.

Now, the key here was about the product.

So it was really only one answer here, and it comes back to that lid.

If the lid is left off for too long, some powder product could accidentally blow out of that crucible.

So what we wanna do to avoid that then is to make sure that we're opening the lid only a little bit, but doing it quite often.

And alternatively, you could simply just continue heating it until you get a constant final mass observed.

So repeatedly heating it and measuring it until the mass no longer changes.

Very well done if you managed to get that guys, what a fantastic job you've done.

Really impressed.

Wow, we have gone through a lot in today's lesson.

So let's take a moment now to summarise what we've learned.

Well, we've learned that we can take a very small mass of a substance, including things like magnesium, titanium, carbon even, and you can really easily heat them in a crucible.

And that the crucible needs to be opened periodically, every once in a while to allow that air, the oxygen in the air to enter that environment.

And that if we lift that lid a little too often or for too long, it could actually lead to a loss of some of the product that's forming and that could then cause a lower final mass recorded, and it has a knock on effect for any of the other calculations we carry on as a result of that practical.

Having said that, we can actually use practical data and our conservation of mass to calculate an unknown reactant mass.

And when we have that, we can then process that, those values, sorry, to determine that empirical formula of a substance, including what we did today, which was to find the ionic ratio of a metal and oxygen.

Wow, you guys did such a fantastic job today.

I'm really impressed.

I hope you had a good time learning with me.

I certainly had a good time learning with you, and I hope to see you again soon.

Bye for now.