Hello, my name's Mrs. Nivin.

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.

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

The definitions for these keywords are given on the next slide in sentence form.

And you may wish to pause the video here so that you can jot down these definitions for reference later on in this lesson or later on in your learning.

So today's lesson is broken into two parts.

Firstly, we'll look at how we can gather the data that we need in order to deduce the empirical formula of a substance, and then we'll move on to how we process that data and consider it.

So let's get started by looking at how we can gather data that we need 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 balance symbol equation is that no atoms have been gained or lost.

Nothing's just been (claps) 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, the 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 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 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 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 a 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 in 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 (clears throat) 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.

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 every 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 every 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.

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 teared 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 massive 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 oxygens 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 would use 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.

Now that we're feeling a little more comfortable about how we would gather data that we need in order to find an empirical formula, let's look at how we can process that data then and consider it as well.

Now, I said earlier that we could deduce a substance's empirical formula from its reactant masses.

And there are a few steps in order to do that.

The first thing we're going to do is we're going to take that reactant mass value and divide it by the elements relative atomic mass.

And when we do that, what we're finding is the number of moles present for each element in that substance.

The next thing we're going to do then is to divide each answer from step one by the smallest answer in step one.

Now, what that does then is it forms a unitary molar ratio.

And because of that we should have at least one answer that comes out with a value of one.

Now, when that happens, you might get a few answers that are very close to a whole number and then could be rounded to it.

Sometimes that doesn't happen.

And in that case, what you might need to do is multiply accordingly in order to achieve a whole number ratio, because an empirical formula is the smallest whole number ratio of atoms of each element within a substance so we're always aiming for that whole number ratio.

Now it can be sometimes a little tricky to keep track of what information you need or how to process it in steps like these.

So I'm gonna go through an example with you.

I'd like to know the empirical formula of a substance that contains 8.

57 grammes of carbon and 1.

43 grammes of hydrogen.

And I'm going to use a calculation grid to help me.

So in the far left hand corner, I have the information or the step that I need to carry out.

And then I have a separate column for each of the elements in my substance so in this case, it's carbon and hydrogen.

So the first thing I need is that reactant mass, and I'm gonna copy those straight out of the questions.

So for this, it's 8.

57 for carbon and 1.

43 for hydrogen.

Then I'm gonna grab my periodic table and copy over those relative atomic masses for each element.

So the carbon it's 12, for hydrogen its one.

When I divide those values and its reactant mass divided by relative atomic mass, I'm gonna find the number of moles.

So for carbon it's 0.

714, and for hydrogen it's 1.

43.

At this point, I need to determine which of those values, molar values is the lowest value.

And for this one, it's carbon.

So I divide both of my molar values by 0.

714, carbon becomes a one and hydrogen becomes a 2.

002.

So I'm looking for whole numbers here.

Carbon is definitely a whole number, but hydrogen is not.

However, it's very close to a whole number and therefore I can round it.

So I find a molar ratio then, one for carbon and two for hydrogen.

But we very rarely write a one in our chemical formula.

And so my final answer for the empirical formula of this substance is CH2.

What I'd like you to do now after I've gone through this example, is I'd like you to determine the empirical formula of a substance form from 6.

902 grammes of iodine and 3.

098 grammes of fluorine.

Now this is gonna take a little bit of time.

You definitely need a periodic table and a calculator.

You may wish to pause the video here then and come back when you're ready to check your answer.

Okay, let's see how you got on.

Now, if you have done all of your processing correctly, you should have got a final answer of IF3.

If you didn't get that answer, you may wish to pause the video and just take a closer look at that processing that's been laid out for you so you can identify where you've gone wrong and think about how you might fix that to avoid it in your future calculations.

But very well done if you got that correct answer.

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 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're 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 its 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 reactions 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're 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 your 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 everybody's answers are going to be slightly different based on the practical that they carried out.

So the feedback I provide here is based on the practical video that was shown earlier in today's lesson.

So using the values from that, I got a final empirical formula calculated out as MG4O3.

And you can see here I've tried to highlight where the reactant mass values for magnesium and oxygen came from from my practical earlier in the lesson.

I've divided by the relative atomic mass to find the number of moles of each.

I've created my unitary ratio by dividing by the number of moles of oxygen present.

And then I had to multiply accordingly in order to get that whole number molar ratio of four to three.

Now, some of the eagle-eyed among you might have noticed that this formula looks a little odd, and that's okay.

Because sometimes that happens with the practical values.

What we're looking here for though is for you to have processed those values correctly in order to calculate an empirical formula.

So very well done if you manage to get a final answer.

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 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 that overall charge of zero, which means the actual empirical formula for magnesium oxide should be MgO.

So then I asked you to think about comparing your answers from your calculated empirical formula to the one that we determined from the diagram.

And everybody's answers are gonna be slightly different depending on essentially your calculated value.

But the crucial thing here is to reflect on why they might not match if they don't match.

And there's a few different things that we could be considering.

First of all, maybe not enough oxygen was actually allowed to enter that crucible.

Maybe the lid wasn't lifted often enough, and that would mean then that not all of that magnesium actually was allowed to react.

So less magnesium reacted than what we thought had.

It's also possible that some of that really lightweight white magnesium oxide powder was lost to the environment when the lid was lifted.

Maybe the lid was lifted a little bit too high, and if that happens, then we have a lower mass for our magnesium oxide and therefore, a lower calculated mass for our reactant oxygen.

So there's a few different reasons why practically our values for our reactant masses could be different and result then in a calculated empirical formula that doesn't match the actual empirical formula based on ion charges.

It is possible though to have achieved results if we were to allow all of that magnesium to react and we were really careful in how we lifted that lid to make sure nothing was lost in the process to have calculated out using moles and molar ratios to get an empirical formula of MgO.

But it all comes down to that practical technique.

So very well done if you manage to get answers that match.

Very impressive.

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 of 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.

My goodness.

We have gone through a lot today.

Let's take a moment to summarise what we've learned.

Well, we've learned that we can take a really small mass of substance including things like magnesium, titanium, carbon even, and they could be really easily heated in a crucible.

And that crucible then needs to be opened periodically.

So every once in a while to let the air or the oxygen in the air to enter that environment.

But we need to do that carefully because if we lift the crucible lid too often or hold it open for too long, you could actually lose some of that product and that would lead then to a lower final mass recording.

And that kind of has a knock-on effect then for the calculations that you're going to do as a result of this practical.

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

And once we have that, we can process those reactant mass values to determine the empirical formula of a substance, including that ionic molar ratio of a metal and oxygen.

(exhales) You've done so well today.

I'm really, really impressed.

I hope you had a good time learning with me today.

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

Bye for now.