Hello, my name's Mrs. Nevin, and today we're going to be looking at how we can determine a reaction equation experimentally, specifically for magnesium oxide as part of our unit on calculations involving masses.

Now you will have some experience of what we talk about in today's lesson from your previous learning, but what we do today will help us to not only answer that big question of what are substances made of, but it will also help us to bring together those three fundamental ideas of chemistry, of linking up what we do in the laboratory to how we communicate chemistry to others and how we talk about those reactions in terms of particles.

So by the end of today's lesson you should hopefully feel more comfortable describing a safe experiment that we could use to calculate the reacting molar ratios for the oxidation of magnesium.

Now throughout this lesson I will be referring to some keywords and these include limiting reactant, ratio, stoichiometry, mole and balance symbol equation.

Now the definitions for these keywords are given in sentence form on the next slide, and you may wish to pause the video here to read through them or perhaps make a note of them so you can refer back to them 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 can heat magnesium and then we'll move on to look at how we can calculate the ratios for an oxidation reaction.

So let's get started by looking at how we can heat magnesium safely.

Now one thing to remember is that oxidation reactions are those in which oxygen atoms are bonded to the elements of the reactants.

And an example here would be burning coal.

In this instance, carbon reacts with oxygen to form carbon dioxide.

And a big indicator that this is an oxidation reaction is that we have oxygen as a reactant.

Now most oxidation reactions occur when substances are either exposed to the oxygen in the air or heated in the air because air contains oxygen.

Let's stop here for a quick check.

Which of the following equations describes an oxidation reaction? Well done if you said C.

You may have been tempted by A, because oxygen is reactant here, but the product is incorrect.

So the best answer here was letter C.

Very well done if you managed to get that correct.

Great start guys.

Now when chemists are discussing reactions they'll often refer to what's known as the observed environment, where that reaction is taking place, and they'll often be described as either closed or open systems. And we have two examples here, on the left, we're heating iron and sulphur and on the right we're simply burning a match in air.

If we take a closer look at the reaction on the left, we can see that the apparatus contains a plug.

But burning a match in air does not contain a plug.

In fact, gases can freely enter or leave that observed environment and that's a key difference.

The reaction on the left of heating iron and sulphur is a closed environment because any of those substances are not able to leave or enter that environment where the reaction is taking place.

Whereas burning a match in air is referred to as an open environment because substances can freely enter or leave that observed environment.

Magnesium undergoes oxidation when it's heated in an open environment and there are some pros and cons to heating magnesium in this way.

For instance, we can see that the reaction is very, very fast and it's incredibly easy to observe what's going on in this reaction.

However, it does produce quite a bright light that can damage eyes and as we've seen here just now, we could actually lose some of the product very easily heating it in this manner.

Now, magnesium could also be heated in a closed environment that's occasionally opened and it's opened by simply lifting that lid every once in a while to allow substances to either enter or exit that environment.

But as before, there are pros and cons to this.

If we heat magnesium in a crucible, there's potentially less product that's going to be lost because it's contained within that crucible.

And it also helps to protect chemists from that bright light that's produced by burning that magnesium.

However, it's not as easily observed as it was when we simply held the magnesium over a bunsen burner.

And also there's a bit of a balance that needs to be held here because if you don't open the lid often enough, it's possible that full oxidation or the full reaction of that magnesium may not take place because not enough oxygen's actually getting into this system.

Likewise, because substances could be lost from that closed environment, if you leave the lid off for too long, some of that product could be lost from that environment.

So there's a a very clear balance about lifting the lid often enough to get enough oxygen in, but not leaving the lid off long so much that some product is lost.

So it's quite tricky thing sometimes, but the benefit of burning magnesium in an open environment simply over the bunsen burner allows us to collect some very useful qualitative observations.

By heating magnesium in a crucible then, in that closed environment where you have a little bit more control over how much is entering or leaving that environment, we're able to collect then quantitative measurements from which we can maybe do some calculations.

Now, the key for these quantitative results to be useful though, it's essential that the substance that's undergoing oxidation is the limiting reactant for that reaction.

So when we are oxidising magnesium in our crucible, magnesium must be the limiting reactant.

That means that I'm gonna need to make sure that that crucible lid is lifted often enough to ensure that enough oxygen is allowed into that environment, into that open system.

And by doing so, I'm ensuring that the oxygen is the excess reactant that I have too much of it.

And that ensures then that every atom of magnesium that was put into my crucible at the start of the reaction reacts with oxygen that is entered it to form the maximum amount of magnesium oxide possible.

Let's stop here for another quick check.

True or false, heating magnesium in a crucible is more dangerous than heating it in an open environment.

Well done if you chose false.

But which of these statements best justifies that answer? Well done if you chose A.

Now the eagle eye among you may have noticed that both of these statements are correct when talking about heating magnesium in an open environment.

However, the original statement was about safety, which was more dangerous.

And heating a magnesium in an open environment could potentially damage one's eyes, which makes it slightly more dangerous than heating magnesium in a crucible.

So very well done if you manage to get those correct guys.

Great job, okay, now we are moving on to the first task of today's lesson, and what I'd like you to do in this first part is to draw a circle around each of the following equations that describes an oxidation reaction.

So you may wish to pause the video here to discuss your ideas with the people nearest to you and come back when you're ready to check your answers.

Okay, let's see how you got on.

Now, if you recall, an oxidation reaction is one in which oxygen is a reactant.

So we're looking for it to be found on the left hand side of the reaction arrow.

And when we see that it's the first, fourth and last equations that were examples of oxidation reactions.

So very well done if you got those correct.

For the next part of this task, I'd like you to carry out your own oxidation reaction of magnesium and the equipment that you'll need to carry this out is shown here, including a heat proof mat and a bunsen burner.

The method I'd like you to follow is shown here.

The first thing I'll need you to do is to record the masses that's shown in the results table, which I'll show you in a moment.

Then you're going to heat that magnesium strongly as shown in the diagram.

So you'll need to lift that lid very carefully on the crucible and often to ensure that enough oxygen is entering that system.

And you're going to heat this until the mass is constant.

So you may need to stop a moment, then double check the mass of your reactance in that crucible with the lid and then perhaps continue heating until that mass stays constant.

Okay, that's going to ensure then that all the magnesium ribbon has reacted.

Once you've got that constant mass, you're going to allow the crucible to cool and then record your final mass in that results table.

Now this is the results table that I'd like you to use, and it describes exactly what you should be recording for each of those masses.

Once you have recorded the masses from your practical, I'd like you to use your results to calculate the mass, the magnesium that reacted and the mass of magnesium oxide that was produced.

You'll then go on to use your understanding of conservation of mass to calculate the mass of oxygen that reacted.

Now, if you were unable to carry out this practical in the laboratory, you can see we've got a video clip here to show you how it would've been carried out and the results that were collected during this recording.

So what I'd like you to do then is to use these results to fill in your table and to carry out the follow-up calculations.

And you may wish to pause the video here so you can make those recordings in your table, but it gives you an idea of what it would've looked like if you were able to carry it out yourself.

Okay, let's see how you got on.

Now everybody's results might be slightly different, but if you use the information that was supplied in the practical clip, your results table should look like this.

Mass one was 23.

95 grammes, mass two was 24.

21 grammes, and mass three should be 24.

38.

Now for part C, you were asked to use your results to calculate the mass of the magnesium that reacted and the massive magnesium oxide that was produced.

So using the information again provided from the practical clip, and you could follow this in your own calculations, is essentially the massive magnesium that reacted would've been mass two minus mass one.

So the information from the practical clip then told me that 0.

26 grammes of magnesium reacted.

In order to calculate the mass of the magnesium oxide produced, that would be mass three minus mass one.

And again, using the information from the practical clip earlier, I got a mass of 0.

43 grammes.

And finally then for part D, you were asked to use your understanding of conservation of mass to calculate the mass oxygen that reacted.

And in order to do this, you needed information from earlier on in this task, part 2A and 2C.

From part 2A, we have our reaction equation, which is magnesium plus oxygen makes magnesium oxide.

From part 2C then, we have the massive magnesium that reacted and the mass magnesium oxide that was produced.

Now using this conservation of mass understanding, we need to then find the oxygen mass, and we know then the mass of memory reactants should add together to give me the mass of my products.

And when I rearrange that equation, I get a mass of oxygen that was reacted as not 0.

17 grammes.

Now this is to be different if you had different values.

So just make sure that you are carrying out the calculation processing correctly, and you might want this checked by the people nearest you, but very well done if you manage to carry out that processing correctly guys, great job.

Now that we're feeling a little more comfortable talking about heating magnesium, let's look at how we can determine the ratios for oxidation reactions.

Now, you may remember that during a chemical reaction, reactant atoms all rearranged to form all of the products.

And because of this, atoms and mass are conserved.

Now, this conservation of mass then allows chemists to calculate the mass of oxygen that reacts during an oxidation reaction.

And you'll recall from our oxidation of magnesium, by having the mass of the magnesium that reacted and the mass of the magnesium oxide that was produced, we are able to calculate the mass of oxygen that was able to enter that environment and react with the magnesium simply through a little bit of algebra.

And because of this, we could say for instance, in this example that 0.

65 grammes of oxygen gas was gained from the surroundings and reacted with the magnesium in our crucible.

So let's stop here for a quick check.

What do you think is the missing mass of oxygen in this reaction? Well done if you chose C, by taking the mass of the carbon dioxide that was produced and subtracting the mass of the carbon that was used, we could then find the mass of the oxygen that reacted, very well done if you managed to get that correct.

Now, once the mass of each reactant has been determined, we can calculate the reacting molar ratios or the stoichiometry for the reaction that occurred.

So for example, if I reacted 0.

55 grammes of sodium and I determined that 0.

19 grammes of oxygen reacted, and I want to find the reacting molar ratios, the first thing I'm going to do is to create a calculation grid that looks a little bit like this and I have one column for each of my reactants.

Now you'll notice that the first row says mass in grammes, and that is genuinely just the masses that you were able to measure from your experiment.

So copy those values over.

The next thing I need is the relative mass for each of these reactants.

And for this, I need my periodic table.

Then I'm going to calculate the moles or the number of particles in each of these mass samples.

And in order to do that, I'm simply going to take the mass in grammes and divide that value by the relative mass for each substance.

And that tells me the number of particles then in each of my mass samples, I'm then going to calculate the unitary ratio.

And in order to do this, I'm going to divide each of the mole's values by the lowest mole's value.

And for this reaction, it's the oxygen that's the lowest one.

And when divide those values, I get one as a value for oxygen and 4.

0284 sodium.

Now it's important that in this step at least one of the values should come out as one.

That's what makes it a unitary ratio.

The final step then to find that reacting molar ratio is you need to either round those unitary ratio values or multiply accordingly in order to achieve whole numbers.

And in this example, I'm going to round and when I get that then, the reacting molar ratio for sodium to oxygen is a four to one ratio.

Now, once the reacting ratio of moles has been calculated that rest of that balanced symbol equation can be deduced.

Now if you remember, we had a four to one ratio of sodium to oxygen, and that could be read as four times as much sodium is going to be needed to react with the oxygen in order to produce sodium oxide.

So the first thing I'm going to do then is I'm gonna write the chemical formula, including any reacting ratios out.

So I've got my reactance of sodium and oxygen, and I know I'm making sodium oxide and I know that the ratio for my sodium to oxygen is a four to one, in order to double check the stoichiometry for my products then, I need to balance the atoms of each element in the equation.

So I separate my reactants from my products and I draw a diagram to represent each of my particles of my reactants.

So I've got four sodiums and one oxygen molecule.

I need to keep track though of the atoms. So I have four sodium atoms and two oxygen atoms. I'm going to do the same then for my products and show this with my formula units of the sodium oxide.

And again, looking at the number of atoms, I have two sodium and one oxygen, and it's clear that I need more atoms on the product side.

So I draw another whole formula unit and double check how that changes the number of atoms on that side.

And I now have four sodium and two oxygen, which matches what I have on the reactant side.

So the final step then is the number of diagrams that have been drawn for each substance is the same then as the coefficient that you would write for that balanced symbol equation.

So the balanced symbol equation for this reaction would look like this, four sodium plus one oxygen makes two sodium oxide.

Let's go through another example.

This time I want to use molar ratios to write a balanced equation for the reaction when iron reacts with oxygen to form iron oxide.

So the first thing I'm going to do is identify the reactants, which is iron and oxygen, and create a calculation grid below each of those symbols.

I then go back to my question to identify the masses that have reacted for each and copy that into the relevant place within my calculation grid.

I'm then going to grab a periodic table and copy over the relative mass for each of these substances and then divide these values, the mass divided by the relative mass to find the number of moles.

To find the unitary ratio, I identify then the lowest value of moles between my substances and divide all of those values by that lowest value.

When I do that, I get a unitary ratio.

Now, in order to find the molar ratios, I need to be able to round or multiply accordingly to get whole numbers.

And I can see clearly that the iron unitary ratio value cannot be easily rounded.

I'm going to need to multiply by three in order to achieve whole numbers that can be easily rounded.

When I do that then I can start to develop my balanced symbol equation.

So I write out the chemical formula for each substance and copy over that reacting molar ratio.

So I have 4FE plus 302 makes, I don't know how many iron oxides FE 203.

So to help me identify the coefficient that I need for the iron oxide, I'm going to create some representations for each of these particles.

So I have four iron atoms, three oxygen molecules, and I'm going to start with one formula unit of my iron oxide.

And I can quickly see that I don't have enough iron atoms between on the product side when I compare it to the reactant side.

So I'm going to draw another formula unit for my iron oxide.

And when I count up the number of atoms between my reactants and products, I can see that it's now balanced.

So my final balanced equation is going to read 4FE plus 02 makes 2FE 203.

Now what I'd like you to do is to use molar ratios to write a balanced equation for the reaction for when zinc reacts with oxygen to form zinc oxide.

Use the calculations that were shown on the left to help guide you, but you'll definitely need a periodic table and calculator pen and paper and a bit of time.

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

Okay, let's see how you got on.

So if you have calculated out your molar ratios correctly, you should have got a two to one ratio for the zinc to oxygen.

And when we start to create then our representations for each of these particles, we can see then that we needed one more zinc oxide giving us a final balanced equation of two zinc plus one oxygen makes two zinc oxides.

Now, if you didn't get that as an answer, make sure that you pause the video and go back to check your calculations to see where you may have gone wrong and then that hopefully will help you avoid that in the future.

But incredibly well done if you manage to use those molar ratios to write that balanced equation correctly guys, fantastic job.

Okay, we're ready for the next task in today's lesson.

Now, combustion is a very common example of an oxidation reaction, and what I'd like you to do is to use molar ratios to determine the balance symbol equation for each combustion reaction that's described below.

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

Okay, let's see how you got on.

So if you've done this correctly then, you should have got a final answer of C2 H5 OH plus 302 makes two CO2 plus three H2O.

Now, if you didn't get that answer, definitely make sure you've paused the video to double check your calculations to see where you may have gone wrong, but fantastic work if you managed to get that correct.

Great job, guys, and for part B then, if you've done your calculations correctly here, you should have had a final answer of 2C3 H8 plus 702 makes six CO plus eight H2O.

Very, very well done if you managed to get these correct guys, I'm so impressed.

Okay, time for the last task of today's lesson.

And what we're going to do is go back to that practical results that we did earlier in the lesson.

I'd like to use your results from task A, part two to calculate the reacting molar ratios of the magnesium and oxygen.

And then I'd like you to use that information to write a balance symbol equation.

Once you've done that, I'd like you to write a balance symbol equation for the reaction between magnesium and oxygen by using chemical formula to balance the atoms. Okay, and then I want you to compare your answers for task B part two A and two B.

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

Okay, let's see how you got on.

So the first thing you were asked to do was use your results from task A part two to write a balanced symbol equation.

So if I go to part B and part C of that task, I then have the mass of the magnesium and the mass of oxygen that reacted.

Now these values are gonna be different for each person depending on whether or not you used the practical video clip that was used earlier on in this lesson.

Or if you did the practical yourself, regardless, the first thing you're gonna do then is put the masses that you measured or calculated into your calculation grid.

Now, when I used the values from the practical clip that was shown earlier, I got a molar of two to one for the magnesium to oxygen.

Then when I wrote my balance symbol equation, I got a final answer of two MG plus O2 makes two MGO.

Now, do not worry if you did not get the same answer because everybody's values are gonna depend on what you manage to collect from your experiment.

What I'm looking for here is that you've managed to process those values correctly.

Did you put the correct masses that were measured into your calculation grid? Did you divide those by the appropriate relative mass to get the number of moles in each sample? Did you then divide each of those mill values by the lowest value to get your unitary ratio? And then did you either round or multiply appropriately to get that whole number molar ratio? So don't worry if your balanced symbol equation doesn't match, have you processed those values correctly? 'Cause sometimes it doesn't work out like you expected.

Maybe something went wrong with the practical and that's okay.

But what we're looking for here is that you have processed those values correctly and very well done if you have.

For part 2B then, you were asked to use chemical formula for each of the substances and balance that reaction that way.

And when you do it that way, you drawing out your representations for the magnesium atoms, the oxygen molecules, and the magnesium oxide formula unit, you should have hopefully got a final balance symbol equation that reads two MG plus O2 makes two MGO.

Now, the last thing you were asked to do then was to compare your answers.

So comparing those balance symbol equations that you calculated from your experimental results versus balancing using just the formula for each of those substances.

Now everybody's answers are going to be slightly different because of the different practical values that you collected.

But if those answers didn't match, there might be a few reasons why.

For instance, perhaps you didn't lift the lid on the crucible often enough to allow a sufficient amount of oxygen to enter that reaction environment to ensure that the magnesium was that limiting reactant.

So that may mean that not all of the magnesium actually reacted.

There's also the possibility that actually you did lift the lid often enough, but maybe you lifted it a little too high and because of that, some of the magnesium oxide was actually lost out of that crucible and into the environment, then that would lead to a lower calculated mass for the reactant oxygen.

Now, ultimately, it doesn't matter at this stage whether or not your calculated balanced symbol equation matches the theoretical balanced symbol equation, the chemical formula.

What we're looking for here is have you processed the values correctly and have you been able to reflect on whether or not those match appropriately? If they don't match, can you reflect back on your technique to think about what could have caused the differences and how you might change that going forward when you carry out a similar technique in the future? That's part of the comparison and reflective nature of science when we get to the end is what did we expect? What did maybe we do that we could improve on in going forward? That's what science is all about.

How can I get better going forward? So very well done if you managed to reflect appropriately on your values and incredibly well done if you managed to get the technique so well done that you managed to get those equations to match guys, superbly done.

Let's take a moment now to summarise what we've managed to do in today's lesson.

Well, we've learned that we can heat magnesium in a crucible and it will react with oxygen.

Before the oxygen to react, the crucible needs to be opened periodically so that we're changing our closed system into an open system, and we need to do it quite regularly to ensure that that magnesium completely oxidises so that magnesium is the limiting reactant.

But we also need to do so carefully because if we open the crucible too much or too high, some of that product that's formed could be lost and that would then affect my experimental data that is collected during that practical.

But we also need to make sure we're carrying out this practical safely to ensure that that bright light that's produced when magnesium oxidises doesn't damage anybody's eyes.

But ultimately, we've found that we can calculate the stoichiometry or the reacting molar ratios for reaction given that experimental data.

I hope you've had a good time learning with me.

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

Bye for now.