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Hello, my name's Mrs. Niven and today we're going to be talking about masses in a chemical reaction as part of our unit on calculations involving masses.

Now, you may have some experience of what we talk about today from your previous learning, but the crux of today's lesson is really about seeing how we can make that link between what we do in practical work, the balanced equations that we use to represent the reactions that are taking place in that practical work, and really digging down into understanding the particles and how they change from reactants to products throughout those reactions.

And it's all having to do with those mathematical relationships and how we use them.

So by the end of today's lesson, you should be able to use conservation of mass and our understanding of relative formula masses and balanced equations to predict the mass of an unknown product or reactant.

Now throughout today's lesson, I'll be using some key terms and they include balanced symbol equation, coefficient, relative formula mass, and conservation of mass.

Now the definitions for these key terms are given on the next slide in sentence form, and you may wish to pause the video here so that you can jot down a little note of what each means so that you can refer back to them later in your lesson or later on in your learning.

Now, today's lesson is broken into two parts.

We'll look at how masses balance in a chemical reaction, and then we'll move on to look at how we can use a balanced symbol equation to calculate an unknown mass.

So let's get started by looking at balanced masses.

Now, one of the main things we need to keep in the back of our head when we're talking about balanced masses is the idea that atoms do not change into different atoms during a chemical reaction.

What's really happening is that the atoms and our reactants are reorganising to form those products.

So if we look at this reaction of calcium carbonate that's being heated and decomposing into calcium oxide and carbon dioxide, I can take my simple equation and simply represent each individual atom from that formula underneath.

And when I take a closer look, I can see that the calcium and an oxygen atom have combined differently in the product to fake calcium oxide.

And then the same thing happens with a carbon and two oxygens to form the carbon dioxide of my product.

But what this shows me then is that the number of atoms that are involved in a reaction are conserved.

That means they don't change throughout that reaction.

The number of atoms I start with is the same number of atoms that I finish with.

The only thing that's happened is they've reorganised themselves into those products.

What's great is that a balanced symbol equation actually shows us this conservation of atoms in a chemical reaction by the use of coefficients, and that's because a coefficient simply multiplies the atoms in the formula.

So if we look here at this reaction, we've got magnesium reacting with oxygen to form magnesium oxide, and what I've done is I've highlighted the coefficients.

What that tells me then is I'm gonna have two particles of magnesium that reacts with one particle of oxygen to form those two particles of magnesium oxide, simply multiplying what I have in my atoms by the coefficient shown in front of them in that balance symbol equation.

Let's stop here to have a quick check.

Which particle diagram correctly represents the balance symbol equation shown, and you have a key here, so you know which is representing the X atoms in which represents the Y atoms. This may take a little discussion, so pause the video here and come back when you're ready to check your answer.

Well done if you said D, very well done.

That is the correct answer.

Now, A unfortunately is incorrect because it's showing two atoms of X and two of Y in the reactants, and they should be bonded together and they're showing separately.

B is incorrect because the coefficient of two XY being our products means I should have two lots of this formula, unit of X and Y.

And in fact, they've been bonded together.

That's why they're overlapping slightly.

C is completely unbalanced, so that's why it is incorrect as well.

So very well done if you chose D.

Good job.

Okay, now we understand that atoms are conserved within a chemical reaction and those atoms are represented in a balanced simple equation with those coefficients.

I wanna just move a little bit now to talk about how we can link that idea then to this understanding of masses.

So let's use an example here.

If I know that one ball bearing has an average mass of about not 0.

505 grammes, I can actually use that information to calculate the mass of any number of similar size ball bearings because all I need to do is know how many ball bearings I have and I'll multiply, but that by the average mass of one and I can get a a basic idea of how heavy that's gonna be.

Now that's the same as true for a chemical reaction.

I can find the mass of one particle because that is essentially it's relative formula mass.

Now sometimes you can see relative formula mass represented as RFM, and sometimes it is represented as MR. Essentially it's the same process to find that relative formula mass.

The number of particles then that I have is indicated by the coefficients in a balanced symbol equation.

Now you can see here in the example I've given of calcium carbonate making calcium oxide and carbon dioxide, those coefficients, those numbers weren't there.

And the thing to remember is if there's no number in front of that formula, it's coefficient is a one.

Now, what that means then is I can calculate the relative masses of the reactants and the products simply by using the substances relative formula mass and the coefficient that I'll get from its balanced symbol equation.

So if I look at this reaction of magnesium reacting with oxygen to form magnesium oxide, I can simply use the chemical formula for each of those substances to find the relative formula mass, the MR of each particle.

So magnesiums would be 24.

3, oxygen is 32.

0, and magnesium oxide is 40.

The coefficients then tell me how many particles I'll have of each of these substances.

So I'll have two magnesium atoms, one oxygen molecule, and two magnesium oxide formula units.

In order to find the relative masses of each of these, then I'm simply gonna take the relative formula mass and multiply that by the coefficient.

So magnesium's relative mass in this reaction will be 48.

6, oxygens will be 32.

0, and the relative mass of my product, magnesium oxide will be 80.

Now, if you cast your minds back, you might remember that conservation of mass means that the total mass of the reactants is equal to the total mass of our products.

So if we go back to this reaction of burning magnesium and oxygen to make that magnesium oxide, if I was to actually measure out that magnesium before it was burned and the magnesium oxide, then that is produced, I could actually calculate the amount of oxygen that reacted with that magnesium.

And what's crucial here is that if I add together the masses of my reactant, so that magnesium that reacted and the oxygen that reacted together, I get a mass of not 0.

40 grammes, and that actually equals then the mass of my products.

Now, the whole point of this is that I had to use a scale, I had to use a balance of some sort to measure out the mass of these reactants.

But crucially, we can use this understanding conservation of mass without measuring masses on a balance.

So remember that conservation of mass ideas, the total mass of the reactants is equal to the total mass of the products.

I could alter that ever so slightly to say the total relative mass of the reactants is equal to the total relative mass of the products.

And that's because the atoms aren't changing.

They are simply reorganising from the reactants to products.

And remember, I have as many atoms that I start with and I end with is exactly the same.

So if I have this balance simple equation again of my magnesium and oxygen and remind myself the relative formula mass for those substances is as shown, and that those coefficients tell me how many particles I have, those relative masses, when I add them together for my reactants of magnesium and oxygen, those two relative masses add together to form 80.

6 crucially the relative mass of my product, the magnesium oxide.

So I've been able to prove conservation of mass using relative masses rather than measured masses on a scale here.

Okay, time for our first task, what I'd like you to do is to match each key term to the correct description.

So you may wish to discuss these, maybe look for those keywords in the descriptions before you decide which word matches it best.

But pause the video then and come back when you're ready to check your answers.

Okay, let's see how you got on.

So the relative formula mass matches the sum of the relative atomic masses of all the atoms in a substance.

So the definition that was second from the bottom.

Now a balanced simply equation then is the top description.

It's representing that chemical reaction using the chemical formula and ensures we have a conservation of atoms using those coefficients.

The coefficient then is the second description down that's indicating the number of particles of each substance we have throughout a reaction.

The chemical formula then indicates the ratio of atoms in a substance and conservation of mass tells me that the mass of my reactants is equal to the mass of those products.

Now what I might recommend is when you have some of these definitions, it can be a little convoluted sometimes.

So it might be worth taking a moment to just highlight, underline, or circle some of the key words in these definitions that you think might have helped you to remember that description for matching up to that keyword.

But very well done if you managed to get those correct.

Great job guys.

For this next task, what I'd like you to do is calculate the relative masses of each substance in the reactions described below.

So as a reminder, you need to find the relative formula mass for each substance, and then use the coefficients to find the relative mass of each substance.

So you're going to need your periodic table and a calculator.

And whenever doing maths in chemistry, I would always recommend maybe checking your answers with someone near you in case you make a mistake when you're popping that into the calculator.

But it's gonna take a little time.

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

Okay, let's see how you got on.

So I can see in this first reaction of hydrogen reacting with oxygen to make water that I have three different chemical formula, and therefore I should have three different relative masses.

So the hydrogen should be 4.

The oxygen should be 32.

0, and the water should be 36.

Now don't forget, once you've found that relative formula mass, you need to multiply it by the coefficient, which is that large number before the chemical formula in to get that relative mass.

Very well done if you got those correct.

So for B, I'm gonna use the same strategy.

The nitrogen then should be 28.

Hydrogen was 6.

0, and the ammonia or NH3 should be 34.

Well done, guys.

You're doing great.

Now C looks a little bit more complicated, but we're gonna use the same strategy in order to find those relative masses.

Now I can see four chemical formulas, so I should have four relative masses.

The silver nitrate was 169.

9, the sodium chloride was 85, sorry, 58.

The sodium nitrate was 85.

0, and the silver chloride was 143.

Now you see that for all of these, I have shown my working out.

So in case we got the wrong answer, we could go back and see how we got to those answers.

So if you did get any of these incorrect, make sure you're going back through that working out to find out where you've gone wrong.

So we can easily fix that going forward.

And if you haven't been writing out your working out, please make sure you're doing so.

It's very easy to go wrong.

It's very simple maths, and I would hate for you to miss out on marks purely because you haven't shown that processing.

We want to identify those errors now so we can reduce them going forward.

What a great start though, guys.

I'm really pleased.

Keep it up.

Okay, for the last part of this task, we're gonna help Andeep.

He's been asked to prove that mass has been conserved in this reaction, and he is done his calculations, and it's clear he is gone wrong somewhere, but he's not sure where.

So what I'd like you to do is to identify and correct Andeep's error.

You may wish to converse and discuss ideas with the people nearest you.

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

Okay, let's see how you guys got on.

So the eagle line among you may have noticed that Andeep has forgotten to multiply the relative formula mass by their coefficients.

So if you go back and calculate those out, these are the values you get.

And then when you add those relative masses together for the reactants, you get a value of 355.

And when you add them together for the products, you get a value that's also 355.

2, and therefore proving that mass has been conserved in this reaction using relative masses.

Very well done.

Great job, guys.

Okay, now that we're feeling a little more comfortable talking about balanced masses, let's look at how we can use a balanced symbol equation to calculate at unknown mass.

Now, we said earlier that a coefficient multiplies the number of particles within a balanced symbol equation, but what it also does is it indicates the ratio of each substance that's needed or produced in a reaction.

So if we go back to that idea of the magnesium reacting with oxygen to make magnesium oxide, what I could read from looking at the coefficients in this balanced symbol equation is that I'm gonna need twice as much magnesium to react without oxygen in order to make the magnesium oxide.

And what's really crucial here is that if the coefficient is changed on any of these substances, the ratio has to remain.

So I will always have two magnesium to one oxygen particle to form two magnesium oxide formula units.

So if my coefficient for magnesium changes from a two to a four, I can see that it has doubled.

Therefore, the coefficients for the other substances in my balance symbol equation would also need to double.

So the oxygen goes from a one to a two, and the magnesium oxide goes to a two to a four.

The same thing happens if that coefficient, let's say for the magnesium this time changes from a two to a 12, it is increased six fold, two times six equals 12.

So I'm gonna have to do the same thing for those other coefficients to maintain that ratio.

The oxygen then goes from a one to a six, and the magnesium oxides coefficient moves from a two to a 12.

So if one coefficient is changed, the others must also in order to maintain that ratio.

Let's stop here for a quick check.

What I'd like you to do is to decide which chemical equation or equations shows a similar ratio to the reaction of N two plus three H two makes two NH3.

Now, you may wish to take a little bit of time to really dig down into the differences and similarities between these chemical equations.

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

Well done if you said C and D.

We can see that just looking down the nitrogens on each of these reactions that the nitrogen has changed.

So if you focus in on that, that A is incorrect because whilst nitrogen has doubled, and so is the NH three, the hydrogen hasn't doubled.

So that hasn't worked.

The ratio hasn't stayed.

For B, we can see that the nitrogen and hydrogen is both doubled, but the NH three has not.

So again, that ratio has not been maintained and therefore it's incorrect.

But for C and D, the C, the ratio has been doubled.

And in D, the ratio has been tripled.

So those two are correct.

Well done if you got at least one of those and very well done if you managed to choose both of the equations that showed the same ratio as the reaction.

Great job, guys.

Okay, what we're gonna move on to now then is to look at how chemists exploit the ratios that are shown in a balanced simply equation in order to calculate an unknown mass.

Now, these calculations assume that every single atom has reacted.

It doesn't always happen because some atoms are lost in the transfer of things, or maybe a reaction is reversible.

There are different things, but ultimately we are making assumptions when we're making these calculations.

And because of that, these calculated masses are sometimes referred to as a theoretical mass.

Now, in order to do this, you're gonna need a few things, including a calculator, periodic table.

You're gonna need a balanced symbol equation, and you're gonna need a step-by-step strategy.

So if you don't have your calculator or periodic table to hand, I would recommend you pause the video and grab it.

And then we will come back to look at that step-by-step strategy using a balanced symbol equation that I'm going to provide.

I'm gonna go through this strategy using an example question just to show you exactly how it would work.

Now, some of my previous students have preferred to show a strategy using a colour coding.

So they'd write out one step in a colour and use that same colour to show the calculation step on the side.

So if you'd like to do that, there are five steps in this strategy, and you may wish to use five different colours then, does it matter, it's completely up to you how you record how to carry out this procedure.

But let's go through this strategy.

I want to know what massive magnesium oxide is produced when 45 grammes of magnesium reacts in burns in oxygen, and I've got the balance symbol equation provided.

So the first thing I need to do is use the question to identify the known and unknown substance in my balance symbol equation.

Now, the known substance then is that which has been provided a mass for.

So I have 45 grammes of magnesium.

So the magnesium is my known substance.

I'm gonna put a little tick above that because that's the one I have information for.

The unknown substance then is what I'm being asked to find.

It's what mass of magnesium oxide.

So I look at my balance simple equation for the formula for magnesium oxide, and above that, I'm gonna put a little question mark, that's my unknown.

So why do we care? Why are we doing this step? Because later on in the other steps, I'll refer to a known or unknown sample.

And if you're not sure, having these little marks above the correct formula helps it to keep clear what you're doing in each step.

So once you've done that, what you're gonna do is find the relative formula mass for the known and unknown substances, the other substances in that chemical reaction we don't care about mathematically, you are not gonna lose anything by finding out the relative formula mass for this except time.

And it could potentially make things a little bit unclear depending on your penmanship and how you lay things out.

So I would recommend avoiding all other substances except for your known and unknown substance, to find the relative formula mass.

And when we do that, then we have magnesium is 24.

3, and the magnesium oxide then is 40.

Now, the next thing you're going to then is you're going to multiply the MR or the relative formula mass for our known and unknown substance by their coefficients.

Effectively, you're finding those relative masses.

So magnesium then is going to be multiplied by two, and we get a value of 48.

6, and the magnesium oxide then becomes 80.

Now it's this next step, step four, that people tend to confuse and or forget to do.

So let's slow down, really get our head wrapped around it.

We're going to take these relative masses that we've just calculated and turn it into a relative mass ratio.

And we're gonna do that by dividing each each answer from step three by the answer to our known substance.

Now, using my little symbols at the top, I know that the magnesium was my known substance, so I'm gonna divide both of my answers from step three by 48.

So magnesium becomes one, and the magnesium oxide becomes 1.

6584.

Now, crucially, what we've done here is we have a ratio where at least one substance becomes one.

You should always have a substance that has an answer of one for step four, and an extra double check is that substance should be your known substance.

So a partway check to double check that you have at this point, at least one answer that has the value of one, helps to avoid any errors going forward.

So the final step in my strategy then is to multiply the unknowns answer to step four by the mass of the known substance.

By the mass, I mean, you're gonna have to go back to that original question, and we were given the mass of 45 grammes of magnesium.

So I'm going to multiply 1.

6584 by 45, and I get an answer of 74.

63, which tells me to three significant figures, 74.

6 grammes of magnesium oxide can be produced when we burn 45 grammes of magnesium.

I've been able to use a balanced symbol equation and mathematical processing to calculate a theoretical mass for an unknown substance.

Let's go through another example.

Now, before I actually wrote out the different step strategies.

This time I'm just gonna talk you through them.

But I want to know in this example, what massive hydrogen is needed to make 20 grammes of ammonia.

And I'm gonna give that answer to three significant figures.

And once again, I've been provided that balance simply equation.

So first things first, I need to identify my known and unknown substance.

So the hydrogen is what I'm being asked to find.

So that's my unknown.

So I'm gonna put a question mark above that, and my ammonia then, because it's been given 20 grammes of ammonia, that is my known substance, and I know that NH3 is the correct formula to put the tick above because the only other formula in my balance simple equation is N2.

That's nitrogen.

It's a different substance.

So now I have my known and unknown substances.

I'm going to find the relative formula mass for each.

So hydrogen is 2.

0 and ammonia is 17.

I then find their relative masses by multiplying those values by their coefficients.

So hydrogen becomes 6.

0 and ammonia becomes 34.

At this point, I take the relative masses and turn it into a ratio by dividing by the value that I got for the answer for ammonia, which is 34.

So ammonia then becomes one, and the hydrogen becomes 0.

1764.

At this point, I have tunnel vision to the hydrogen to my unknown because that's the question I'm trying to answer.

What mass is needed? So I'm going to multiply my ratio that was calculated by the mass of ammonia that I need to find.

And when I get that, I get an answer of 3.

5294, but I've been instructed to give my answer to three significant figures.

So my final answer is 3.

53 grammes of hydrogen is needed to make 20 grammes of ammonia.

So I've now gone through two examples with you, one where we've written out each of the steps and the second where I've walked you through them verbally, what I'd like you to do now is to use this strategy to calculate the mass of aluminium that can be extracted or removed from 350 grammes of aluminium oxide.

I'd also like you to give your answer to three significant figures, and I've provided that balance symbol equation for you.

This may take a little bit of time.

You might wanna work with a person nearest to you.

So definitely pause this video and come back when you're ready to check your answers.

Okay, let's see how you got on.

So our known was our aluminium oxide AL 203, and our unknown was aluminium, which is just AL.

You find the relative formula masses for each, and then calculate the relative masses by multiplying by those coefficients.

At this point, we turn that relative mass into a ratio by dividing those answers by the answer for our known substance.

And when we get that, we get a value of 0.

52941 for our aluminium.

We're gonna take that answer and multiply it by the mass of aluminium oxide that we started, which was 350 grammes.

That gives us a value of 185.

29.

But our final answer is 185 grammes of aluminium can be extracted from 350 grammes of aluminium oxide.

So if you got final answer of 185 grammes, incredibly well done.

Okay, there are quite a lot of processing steps involved here.

If you didn't get that answer, it's really important that you're going back through that working out to find out where you've gone wrong so that you can fix it going forward.

And if you haven't been showing your working out, it's absolutely vital that you are, because sometimes you can get a mark for one of these steps, you get a mark sometimes just for finding that relative formula mass, or just for finding those relative masses by multiplying by those coefficients.

So even if you didn't get that final answer, how many marks could you get by making an effort towards that final answer? Okay, it will come through practise, but I'm really pleased with the progress you guys are making.

Keep it up.

Incredibly well done.

Okay, time for the next task in today's lesson.

What I'd like you to do is to put these steps into the correct order to let other people know how to calculate an unknown mass in a reaction.

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

Okay, let's see how you got on.

So the first step was C, identifying your known and unknown substances in the balance symbol equation.

Then E, calculating the relative formula mass for both of them.

You then need to multiply the relative formula mass by their coefficient, which was letter A.

Then you're going to do letter D, which was dividing those relative masses by the mass of the known substance, the relative mass that is, to create the ratio.

And then finally, multiply your unknown ratio mass by the mass of your known substance.

And that came from the question.

So very well done if you managed to get those in the correct order guys, great job.

Okay, for the last part of today's lesson, then I want you to use the strategy that we've outlined in Task B, part one, and that we talked through earlier in this learning cycle to answer those questions below.

For each of these answers, I'd like you to be giving them to two significant figures.

Now, some people work faster than others.

That's absolutely fine.

Go at your own pace, okay? That's step number one.

If you get only to answer B in the next few minutes, that's absolutely fine.

If you get all the way to D, maybe ask for some extra work.

But take your time, slow and steady.

Use that strategy, pause the video, come back when you're ready to check your answers.

Okay, let's see how you got on.

Now, what I'm gonna do in these feedback sessions is I'll give you the final answer, and I do have the working out that will come up and be shown to you.

And if you'd like to take a closer look at that, maybe just pause the video before moving forward so that you can double check your working out matches that what I show you otherwise.

Let's crack one.

For part A, you should have had a mass of 3.

2 grammes of oxygen.

B, we needed to find the mass of carbon dioxide formed when zinc carbonate decomposers, and that should be 42 grammes of carbon dioxide.

For C, we were looking for the mass of rocket fuel hydrazine that can be made from 500 grammes of ammonia, and that should be 470 grammes of N2H4.

And finally then for D, the mass of oxygen needed to react with some butane.

We should have had 7.

2 grammes of oxygen.

I am so proud of the progress that you have made in today's lesson.

It's not an easy lesson to get stuck into, but you have persevered and just very, very well done.

I mean, I absolutely love this topic because it brings together that idea of the practical work, those balanced equations, the particles, the maths, everything brings everything.

I just love it.

It really brings chemistry alive for me.

But having said that, we've gone through a lot in today's lessons.

So let's just take a moment and summarise what we've learned.

Well, we've reminded ourselves that the total mass of the reactants equals the total mass of the products, because atoms in a chemical reaction don't change into different atoms. They simply rearrange from the reactants into the products.

And that conservation of atoms translates into conservation of mass, but also conservation of relative masses so that the sum of those relative formula masses of our reactants is equal to the sum of the relative formula masses in our products, as long as those relative formula masses have been multiplied by the coefficient in that balance symbol equation.

But really, the crux of today's lesson is this last point.

The idea that coefficients in a balanced symbol equation indicate ratios of substances in that chemical reaction, and that we can exploit that, we can use these coefficients, these ratios along with those relative formula masses to mathematically process our understandings to calculate an unknown mass in a substance, in a chemical reaction.

I mean, did you think that we would ever get to this point? Possibly, possibly not.

Regardless, I had a really good time learning with you today.

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

Bye for now.

I've finished the video