Lesson video

Hello, my name's Mrs. Navin.

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 talked 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 start off by looking at balanced masses in a chemical reaction, and then we'll look at how we can use a balanced symbol equation to calculate those unknown masses.

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 in the 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 make 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 and 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 2XY 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 symbol 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 0.

505 grammes, I can actually use that information to calculate the mass of any number of similar-sized ball bearings, because all I need to do is know how many ball bearings I have and I'll multiply that by the average mass of 1, and I can get a basic idea of how heavy that's gonna be.

Now, the same is true for a chemical reaction.

I can find the mass of one particle because that is essentially its 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 those 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 1.

What this means then is that I can calculate the relative masses of the reactants and the products simply by using the relative formula mass of that substance based on its chemical formula, and the coefficient from that balance symbol equation.

So if I look at this reaction of magnesium reacting with oxygen to form magnesium oxide, I can find the relative formula mass, the MR, for each particle, simply by using a periodic table and its chemical formula.

And when I do that, I see that magnesium is 24, oxygen is 32, and magnesium oxide then is gonna be 40.

The coefficient then tells me how many particles I'll have.

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

To find the relative mass then for each of these substances in my reaction, all I'm going to do is take that relative formula mass and multiply it by the number of particles.

So magnesium's relative mass in this reaction is 48, oxygen is 32, and the relative mass then for my product of magnesium oxide is going to 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 to that magnesium that reacted and the oxygen that reacted together, I get a mass of 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.

If I take that conservation of mass idea, the total mass of the reactants is equal to the total mass of the products.

I could alter that ever so slightly and talk about 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.

Remember, they're simply reorganising from reactants to products.

So, if I look at that balance symbol equation before of that magnesium oxide, and I remember those relative formula masses for each of the particles for those substances, and the coefficient tells me how many particles I have, those relative masses for my reactants of magnesium would be 48, and the oxygen was 32.

And when I add those together, that relative mass of my reactant magnesium oxide is equal to 80.

Crucially, that is the relative mass of my products, the magnesium oxide.

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 symbol 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 keywords 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 coefficient 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 my first reaction here of hydrogen reacting with oxygen to make water, that I have three different substances, so I should have three different relative masses.

So the hydrogen should be 4, the oxygen is 32, and the water then is 36, not forgetting that we need to multiply that relative formula mass by the coefficient in each of these.

So well done if you got those three correct.

So for B, I'm gonna follow that same technique.

So the nitrogen relative mass should be 28, hydrogen is 6, and the ammonia or NH3, has a relative mass of 34.

Well done if you got that correct.

So C was a little bit more complicated, but you're gonna use the exact same strategy.

I have four different substances, so I'm gonna have four different relative masses.

The first one of silver nitrate should be 170.

The sodium chloride should be 58.

5.

Sodium nitrate then is 85.

And the silver chloride should be 143.

5.

Now, I should mention here, you can see I've put all of my working out so you can keep track of what I've done.

If you have got any of these answers incorrect, it's worth going back through your working out to see where you've gone wrong so we can correct them going forward.

If you haven't written out your working out, please make sure you're doing so.

So just for that reason, so we can find any errors and correct them if needed.

But incredibly well done if you've got those correct, guys.

What a great job.

Okay, for the next part of this task, we're gonna be helping Andeep.

He's been asked to prove that mass has been conserved in this reaction, and he's done his calculations, but it's clear he's gone wrong and he can't find out where.

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

You may wish to discuss this with the people nearest you.

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

Okay, let's see how you got on.

Now, the eagle-eyed among you may have noticed what Andeep has forgotten to do is multiply that relative formula mass by the coefficients for each of those substances.

So if you go back and complete those, you should now find that the mass of the reactants is 356, and therefore the products is also 356.

And we have proven that mass has been conserved in this reaction.

So really well done if you found and corrected that error correctly.

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 an 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 with that 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 from magnesium changes from a 2 to a 4, 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 1 to a 2, and the magnesium oxide goes to a 2 to a 4.

The same thing happens if that coefficient, let's say, for the magnesium this time changes from a 2 to a 12.

It has increased six full, 2 times 6 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 1 to a 6, and the magnesium oxides coefficient moves from a 2 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 N2 plus 3H2, makes 2NH3.

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 NH3, 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 NH3 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 symbol 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're making assumptions when we're making these calculations.

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

Okay? 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 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.

Okay, I'm gonna use an example question to go through this step-by-step strategy.

Now, there are five steps in this strategy, and some previous students have preferred to use some coloured pencils where they'd write out the step in words in one colour, and use that same colour to show that step being used in their actual calculation.

So if you'd like to do that as well, you may wish to get some coloured pencils.

So let's get started.

The first step that we need to do is identify the known and the unknown substance in your balanced symbol equation from your question.

And the question we have is "What mass of magnesium oxide is produced when 45 grammes of magnesium is burning in oxygen?" And we've been given that balanced symbol equation as well.

Now, the known then from your question, is the one that has the mass provided.

So we've got 45 grammes of magnesium, therefore it's the magnesium in my balanced symbol equation, that is known.

And I'm gonna indicate that just by putting a little tick above it, 'cause that's the stuff I have information about.

The unknown then is the question that you're being asked, okay? I want to know what mass of magnesium oxide is.

So I need to find in my formula, or I'm sorry, my balanced symbol equation, the formula for magnesium oxide.

And what I'm gonna do is above that formula, I'm gonna put a little question mark because that's my unknown.

Now, the reason I'm doing that is because I'm going to refer back to the answers for known and unknown later on in my strategy.

So if you're unsure, you can always just go back and double check those little notations that you've put above the formula in your balanced symbol equation to keep things clear.

The next thing you're gonna need to do then is you need to calculate the relative formula mass for both the known and unknown substances.

Now, if there's other substances within that balanced symbol equation, you do not need to find the relative formula mass for that.

That is just wasting time, and it could potentially confuse things later 'cause you have extra numbers getting in the way of things.

So keep it nice and clear.

Just find the relative formula mass for your known and unknown substance.

And when you do that, you have 24 for magnesium and 40 for the magnesium oxide.

So the next thing you're going to do then is you're gonna multiply that relative formula mass by its coefficient.

You're effectively finding those relative masses for your known and your unknown substances.

So when I do that then, I get 48 for the magnesium and 80 for the magnesium oxide.

The next step then is the one that most people forget to do, okay? You need to divide each of your answers to step three, by the answer you have for your known substance.

And what you're doing here is you're making a ratio of masses, ratio of your relative masses.

So my known answer was 48 for the magnesium.

So when I divide that, I get 1, and when I divide the magnesium oxide by 48, I get 1.

667.

Now, the key here to remember is that in step four, when you're dividing an answer by a known, you should always have at least one answer that comes out as 1, okay? That's your ratio value.

And it should always be the known substance that comes out as 1.

So, this is a really good quick and easy check to make sure that you're on the right track with your processing, and it might save you some grief later on to try and find out where you've gone wrong.

So, you should always have, step four, at least one answer that's 1, and that answer should be for your known substance.

So the last step then is you're going to then multiply the answer to number four for only the unknown substance by the mass that we have from our known substance.

So that's the 45 grammes from our question that we had for magnesium.

So 45 grammes of magnesium is burning, I'm gonna multiply that value by my answer to step four, which was my ratio of relative masses, and I get a value of 75.

What that means is my answer to this question then is that 75 grammes of magnesium oxide will be produced when 45 grammes of magnesium burns in oxygen.

Okay, we're gonna go through another example again, but this time I'm not gonna show you the different steps and the strategy.

I'll just talk you through it, okay? I wanna know what mass of hydrogen is needed to make 20 grammes of ammonia, and to give my answer to three significant figures, and I've been given the balanced symbol equation.

More often than not, you are provided this balanced symbol equation.

It's more about how we process the numbers to get that final answer than it is for you to balance that equation, okay? That's what we care about when these types of questions.

So, the first thing I need to do is identify my known and unknown.

So my hydrogen is unknown, so I'm going to put a question mark above that.

I've been given the mass for ammonia, so I'm going to put a tick above that.

Now, I know that NH3 is the ammonia formula here, because the only other formula in my balanced symbol equation is N2, and that's nitrogen.

So process of elimination, NH3 is the ammonia formula.

So a tick goes above that for my known.

So first step done.

Second step, I need to find the relative formula mass for my known and unknown.

So hydrogen is 2, and ammonia is 17.

I need to find these relative masses next.

So I'm going to multiply the relative formula mass by the coefficients.

So hydrogen then becomes 6, and the ammonia becomes 34.

The next step then is to find your relative mass ratios.

So that's dividing these most previous answers by the known answers.

So that's 34 in this case, 'cause that's my known substance of ammonia.

So, doing that, I get a value of 1 for ammonia and 0.

1765 for my hydrogen.

This point, I have tunnel vision just to my hydrogen, because that's what I'm trying to find.

It's my unknown.

So I'm gonna multiply that value by 20 grammes, and I get an answer then of 3.

5294.

But I've been instructed to give my answer to three significant figures.

So my final answer for this question is going to be 3.

53 grammes of hydrogen that's needed to make 20 grammes of ammonia.

So we've now gone through two different examples.

One in which I had the step-by-step strategy written out for you, and one where I've talked you through it.

What I'd like you to do now is to have a go at following this strategy yourself to tell me what mass of aluminium can be extracted or removed from 350 grammes of aluminium oxide.

I'd like you to give your answer to three significant figures, and I've provided you with that balanced symbol equation to use as well.

Now, you may wish to go back through the strategies from the previous slides, or work with people nearest you to try and come to this answer.

What I will ask you to do, please, is to show you're working out as you go through it.

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

Okay, let's see how you got on.

So first things first, you needed to identify your known and unknown substances, and to find the relative formula mass for each.

Once you've done that, you needed to find their relative masses.

So that's multiplying the relative formula mass by the coefficients, and then dividing those answers by the answer for the known substance, which was the aluminium oxide.

When we do that, we get a value of 0.

5294 as our relative mass ratio for the aluminium, and we multiply that by 350 grammes to get an answer of 185.

29.

But a final answer was to be to three significant figures.

So, give yourself full marks if you got an answer of 185 grammes of aluminium extracted from 350 grammes of aluminium oxide.

Now, what you can see here is, I have actually written out the different steps here, and it's almost as if you can get one mark per line, essentially in your working out.

So, if you didn't get that final answer of 185 grammes, then go back to find out where you've gone wrong, and also give yourself marks as you're going through.

Maybe you managed to at least get the relative ratios of 204 and 108, but forgot to divide an answer.

That happens sometimes.

But you'd at least get two out of three marks, okay? So make sure you're being kind to yourself as you go back through these answers.

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 balanced 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 1, and that we talk 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 to 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 guys got on.

So what I'm gonna do in each of these feedback sessions then is I will give you that final answer, but I will show the working out.

And if you'd like to go back through that working out, you can pause the video and just take a double check of that, okay? But for Part 2A, where a calcium reacting with oxygen to form calcium oxide, you needed 3.

2 grammes of oxygen.

For B, we were trying to find the amount of carbon dioxide that's produced when zinc carbonate decomposes, and that should be 42 grammes of carbon dioxide.

For C, the mass of rocket fuel hydrazine that can be made is going to be 470 grammes of N2H4.

For D then, we needed to know the mass of the oxygen needed to react with the butane, and you should have had 7.

2 grammes of oxygen.

I am so proud of what you've managed to accomplish in today's lesson.

There is a lot of maths.

It's really easy to get confused, and you really persevered.

I personally absolutely love this topic because it brings everything together, the practical work, the balanced equations, the particles, the maths, the numbers, the periodic table, everything comes together, but it's very easy to get confused.

So it's important that we're keeping things nice and clear.

Maybe go make sure that you've got these notes clearly organised, put them somewhere safe so that you can refer back to them in different lessons if you wanna just refresh your memory about something.

Having said that, we have gone through quite a lot in today's lesson.

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

Well, we've learned that the total mass of the reactants is equal to the total mass of the products.

Because the atoms don't change into different atoms during reaction.

They simply rearrange themselves from our reactants into our products.

And this conservation of atoms and conservation of mass translates into this conservation of relative masses, so that the sum of those relative formula masses of our substances equals the relative formula masses of our products, as long as those relative formula masses for our substances have been multiplied by those coefficients in our balanced symbol equation.

But crucially, the crux of today's lesson is this final point.

The idea that coefficients in a balanced symbol equation tells us the ratios of the substances in a chemical reaction, and we can use those values for the coefficients with a relative formula mass to mathematically process information about a reaction to calculate an unknown mass of a substance in a reaction.

I had a really good time learning with you guys today.

I hope you did with me.

That you've found a strategy that works for you.

If you've got a different way of mathematically processing this information, that's absolutely fine.

Go ahead and use it as long as it works consistently correctly for you.

I hope my process has helped you a little bit as a starting point.

But I hope to see you again here soon.

Bye for now.