Lesson video

Hello, my name's Mrs. Nivin, and today we're going to be talking about masses in a chemical reaction using moles.

Now you will have some experience of aspects of what we talk about in today's lesson from your previous learning, but what we do in today's lesson will help us to not only answer that big question of what are substances made of? But also to better appreciate some of the calculations that take place in industry, chemical engineering, or something as simple as deciding how many trees to plant to offset one's carbon footprint.

By the end of today's lesson, you should feel more comfortable interpreting a balanced symbol equation and calculating, therefore predicting, the mass of a reactant or product for a specific reaction.

Throughout the lesson, I'll be referring to some keywords, and these include: balanced symbol equation, coefficient, mole, stoichiometry, and relative formula mass.

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 make a quick note of what each means, so you can refer to it later on in the lesson or later on in your learning.

So today's lesson is broken into two halves.

We'll first look at the ratios in a balanced symbol equation and then look at how we can use those ratios to calculate an unknown mass.

So let's get started by looking at where the ratios are in an assembly equation and how they can be interpreted.

Now those of you who are familiar with cooking or baking will be also familiar with a recipe.

Now, recipes indicate a lot of information, including what ingredients are needed, how much of each is required of those ingredients, but they also tell us what's going to be made and how much you can expect to produce using that recipe.

Now in chemistry, it's these chemical equations that act as the recipe because they indicate what we need for this reaction and also what will be made in the reaction.

So we have then the reactants on the left hand side of our arrow and the products on the right hand side of our arrow.

When we're talking about a chemical reaction, it's important to remember that reactant atoms simply reorganise themselves to form all of the products, and because of this, the atoms and therefore the mass in a chemical reaction are conserved.

Now, a balanced symbol equation shows this conservation of both atoms and mass using coefficients, and you can see that here in how they are circled.

So what that shows me with this coefficients, if I wanted to look at the conservation of atoms, for instance, I could draw out a diagram to represent two magnesiums, one oxygen molecule, and the formula units for my magnesium oxide.

And I can see I have two magnesium symbols on the left, two on the right, two oxygen symbols on the left and two on the right.

So it's balanced and it's showing me that conservation of atoms. But coefficients tell me so much more about a reaction than simply this conservation of atoms and mass.

In fact, coefficients can be interpreted in two different ways.

We can look at that in the small scale by looking at the individual particles that it's representing.

So here we have two atoms of magnesium, one molecule of oxygen, that they react together to form then two formula units that is composed of one magnesium ion and one oxide ion.

Another way of interpreting the coefficients in this reaction is looking at it in terms of a larger scale where we're looking at packages of moles, so that's 6.

02 times 10 to the 23rd particles representing a mole.

So we could actually reread this particular equation as two moles of magnesium atoms reacting with one mole of oxygen molecules to form two moles of those formula units made up of a magnesium and oxide ion.

Let's stop here for a quick check.

Which statement correctly describes what this chemical equation is representing? Now, you may wish to pause the video here so you can really read these statements carefully, maybe discuss your ideas with the people nearest you and then come back when you're ready to check your answer.

Well done if you chose C and D.

Both of these statements correctly describe this chemical equation.

A and B are incorrect because they're referring to sodium chlorine as a molecule, and it's actually a formula unit because it is a ionic substance and we know it's ionic because it's made of a metal and a non-metal.

So very well done if you've got either C or D.

And incredibly well done if you managed to choose both of those statements as being correct.

What a great start, guys.

Keep it up.

Let's return to this idea of a cooking or a baking recipe.

Not only will it tell you what you need and how much you need of it, but it also indicates the ratio of those ingredients that are needed.

For instance, if I wanted to make three pancakes, these are the ingredients that I would need, but if I wanted to make 12 pancakes, I can use that recipe for three pancakes and just decide how to increase those ingredients because I know the difference from three to 12 is a four times difference.

What I would need to do then is multiply all the ingredients for my three pancake recipe times four, to indicate how much of each ingredient I need in order to make 12 pancakes, because of that ratio of the ingredients within the recipe.

Now in chemistry then, the ratio of my ingredients, or the reactants, and the products is shown by the coefficients in the reaction's balanced symbol equation.

And this is known as the stoichiometry of the reaction.

But what do we mean by this word stoichiometry? Well, if we break it apart, it comes in two halves and both halves come from the Greek: "stoichio" comes from stoichein, which means element, and the "metry" comes from metron, which means to measure.

So we're essentially measuring elements.

But a better definition of stoichiometry is that it refers to the quantitative relationships, those molar ratios between all the substances in my chemical reaction.

So when we're talking about the stoichiometry of a reaction, what we're talking about is that molar ratio between my reactants and products that is indicated by the coefficients in that reaction's balanced symbol equation.

So there are a few ways that we could interpret the coefficients within this particular symbol equation.

So for this one, I could say that in this reaction, twice as much magnesium is going to be required to react with all of the oxygen.

But that's quite vague actually, because we know that these are molar ratios and a mole represents an actual number of particles.

Remember, one mole is equal to 6.

02 times 10 to the 23 particles.

So a better description of this particular reaction might be that in this reaction, two moles of magnesium atoms react with one mole of oxygen molecules to produce two moles of the formula unit magnesium oxide.

So we get a lot of information from this stoichiometry or the molar ratios represented by those coefficients in a balanced symbol equation.

Now, the best thing about stoichiometry or that molar ratio of a reaction is that it remains unchanged.

Just like in my cooking recipe, that ratio doesn't change even if the quantitative amount might change, that I'm using.

So the stoichiometry of this particular reaction means that I've got those two atoms of magnesium, one of the oxygen molecule, and then two formula units of the magnesium oxide.

But what if this time I need four magnesium atoms? Well, I've got twice as much magnesium, which means in order to figure out the molar ratio for the rest of this reaction, I'm simply going to multiply the coefficients for the rest of this reaction by two as well.

So oxygen becomes a two and the magnesium oxide becomes a four.

Likewise, if I have only one mole of magnesium rather than two, how would the molar ratios change for the rest of the reactants in products? Well, I need to decide how I got from two to one, and for here I divided by two.

That means then for the rest of the coefficients in this particular reaction, I'm going to also need to divide those by two.

So one becomes a 0.

5 or one half and the two becomes a one.

So I've multiplied to get higher numbers and divided to get those lower numbers, but what I've done is ensure that the molar ratio across the entire reaction remains unchanged.

Let's stop here for another quick check.

Which chemical equation or equations show or shows a similar molar ratio to the reaction of 2Al plus 3I2 makes 2AlI3? You may wish to pause the video here so you can do a bit of comparison, maybe discuss your ideas with the people nearest you and then come back when you're ready to check your answer.

Well done if you chose B and C.

To get to B, all of the coefficients have been multiplied by three.

And for C, all of the coefficients have been divided by two.

So the molar ratios have stayed unchanged.

A and D are incorrect because not all of those coefficients have changed.

So very well done if you've got at least one of those correct and incredibly well done if you managed to choose both of the correct equations.

Great job, guys! Right.

Let's move on to the first task of today's lesson.

What we have are some students that are describing what this chemical equation actually means, and I'd like you to decide who you agree with and why? So Jacob reckons that "one carbon atom reacts with one oxygen molecule making one carbon dioxide molecule." Izzy thinks that "each carbon atom reacts with one oxygen molecule making a carbon dioxide molecule." Laura thinks that "carbon and oxygen react to make carbon dioxide," And Andeep says that "1 billion carbon atoms react with 1 billion oxygen molecules making 1 billion carbon dioxide molecules." Now, there are very subtle differences between each of these statements, so you may wish to discuss your ideas with the people nearest you and then come back when you're ready to check your answer.

Okay, let's see how you got on.

What I'm gonna do is to go through each of these statements separately and then we'll see who you agree with and maybe who you don't.

So I'm gonna start with Laura.

Now, Laura says, "Carbon and oxygen react with carbon dioxide." Now, she's not wrong.

It's entirely correct.

All she's done though is changed her symbol equation into a word equation and she hasn't recognised that actually that symbol equation is providing us some quantitative information.

So it's far too vague.

Jacob said that "one carbon atom reacts with one oxygen molecule making one carbon dioxide molecule." And at first glance, that does appear to be correct, but what he hasn't recognised is that that carbon actually represents a really large number of carbon atoms that are reacting, not just one carbon atom.

He's misinterpreted the idea that there are coefficients here, they're not actually written because if it's at least there, all the coefficients are one.

But his answer's just far too specific 'cause he's talking about one atom and we can never just get one atom to react at one time.

Not in the science laboratory anyway, it's far too difficult.

So again, this one is too specific while Laura was too vague.

Now if we look at Andeep's suggestion, he's actually got the right idea that the chemical equation is actually representing a large number of particles, but he's given a specific number of atoms, okay? So his description, again, very similar to Jacob's, it's just too specific.

He's not indicating a ratio.

He's giving us a specific number.

So the best answer, I reckon is gonna be Izzy.

She said that "each carbon atom reacts with one oxygen molecule to make a carbon dioxide molecule." And she's correctly interpreted what this actually means, so what's actually happening.

She's kind of taken into account, subtly, that stoichiometry, those molar ratios.

Each atom reacts with one molecule and makes a carbon dioxide molecule.

So as I said, they were really subtle differences, but the best of the four descriptions here was going to be Izzy's.

Very well done if you managed to get that correct.

Okay, for the next few questions of this task, I'd like you to use the stoichiometry, remember, that's the molar ratio, to determine the new molar quantities that are suggested for A, B, and C.

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

Okay, let's see how you got on.

Now the first thing to remember is that if there isn't a coefficient actually written in front of a particular substance, that coefficient is a 1.

So you may wish to write those in to start with when you're thinking about new molar quantities.

But when I do think about that and I want to see how it's changed.

So for a, I can see that the tin oxide has changed from a one coefficient to a two, and the only way that can happen is by multiplying by two.

So I need to multiply the coefficients for the rest of the equation to get the new molar quantities.

So I then have four moles, two moles, and four moles.

For B, I needed to multiply by six.

So my new quantities then for the rest of the materials is 12, 6 and 12.

And for C, I needed to divide by two to go from one to one half, and therefore the rest of my quantities will be one, 0.

5 or one half, and one mole.

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

Good job, guys! For this next question, I'd like you to use the stoichiometry similar to what you did in the last question to determine the new molar quantities.

But what we've done is given you that molar quantity under a different reactant or product.

So be careful about how you consider what those new molar quantities are.

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 for A, the coefficient changed by two, a multiplication of two, and therefore multiplying the rest of those coefficients gives me the answers of 2, 4, 2, and 2.

For B, we multiplied by six.

So our new coefficients reading across from left to right will be 12, 12, 6, and 6.

And for C, we divided by four going from a coefficient of 1 to 0.

25, which is equivalent to one fourth.

So our new coefficients now are 0.

5, 0.

25, 0.

5 and 0.

25.

Don't worry if you put these as fractions as well, it is preferred to use decimals in chemistry, but if you kept them as fractions, that would be acceptable as well.

Very well done if you got those correct.

Good job, guys! And for this last question of task A then, we're gonna do the same thing.

I'd like you to use that stoichiometry to determine the new molar quantities.

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

Okay, let's see how you got on.

So for A, we needed to multiply by four.

So your new molar quantities are these.

For B, we needed to multiply by three, and so our new molar quantities are as that.

And for four, then, to get from 3 to 0.

5, this was a tricky one, but you needed to divide by six.

And these new molar quantities may look a little bit odd, but are correct, 0.

17, 0.

33 and 0.

5 moles if we've divided all of those coefficients by six.

Very well done if you managed to get that last one, it was a tricky one.

So good job, guys! I'm really impressed with your start to this lesson.

Fantastic job.

Let's keep it up.

Now that we're feeling a little more comfortable interpreting and determining ratios in a balanced equation, let's look at how we can use them to calculate an unknown mass.

Now the thing about having a balanced symbol equation is that stoichiometry, those molar ratios that they provide chemists, because what we can do with that then is exploit that stoichiometry in order to calculate any unknown mass that we need from that balanced symbol equation.

Now, these calculations then that are carried out, assume that every single atom has reacted, that nothing's been lost to the environment, that none of our reactant maybe was stuck to the measuring cylinder as we measured out a particular volume, anything like that.

Now because these calculations make those assumptions, they are referred to as a theoretical mass.

They are our starting point for determining how much of a reactant might be needed or how much of a product could possibly be made using these stoichiometric relationships.

Now, in order to carry out these calculations, you're going to need a few things.

You need a calculator, definitely, you need a periodic table, you'll need a balanced symbol equation.

And then we'll need a step-by-step strategy to look at how we can process then the information that we're provided in order to find that unknown mass.

Now over these next few slides, I'm going to outline a strategy that you can use in order to process the information to get that final answer of a mass that's unknown for a particular reaction.

Questions like these tend to give you the balanced symbol equation from which you can work, but the processing can take a little bit of time to get your head wrapped around.

So I will outline these strategies step-by-step, and this works consistently correctly.

You may find another method that works well for you, but the method I'm showing you tends to match most mark schemes.

If you try any exam style questions, you'll be able to hopefully follow through to find whether or not you'd achieve a mark for the working out that you've done.

Now, previous students have found, when I've outlined strategies like this, to use some colour coding.

So they might write out the step instructions in one colour and then follow with that colour showing the working out elsewhere so they can work through it.

And it's one of these processes that becomes easier the more and more you work with it.

So they might stop using the colour coding eventually and then being able to skip some steps because it just becomes second nature to it.

So this is definitely gonna be one of those practise makes perfects and persevering with these types of mathematical processings, okay? So, if you don't already have a calculator and a periodic table to hand, please pause the video and collect those and come back when you're ready to proceed.

If you'd like to do some colour coding, you may wish to grab some of those colours now, and again, pause the video and come back when you're ready to proceed.

So let's get started.

The very first step that you need to do when you are being asked a question where you need to calculate an unknown mass is to identify the known and unknown substances in your balanced symbol equation.

And that is gonna come from the question itself.

In this particular question, we've been asked what mass of magnesium oxide is produced when 48 grammes of magnesium burns in oxygen? So the known substance is the substance for which you've been given a mass.

So I've identified it as magnesium.

And so above it on my balanced symbol equation, I'm going to put a little tick because that's the one I have information for.

The unknown substance then, is that mass that you are trying to calculate.

And for this, it's the mass of the magnesium oxide.

So above this formula, I'm going to put a question mark because that was what the question was about.

Now the other substances within this balanced symbol equation I don't need any information about.

There's no processing involved with it, so I'm going to cross it out.

So the following steps after this then do not involve any of the reactants or products that are not a known or unknown substance, okay? So now that you have those, the next thing we need to do is we are going to create a calculation grid below the known and unknown substances.

And that's gonna include four essential pieces of information.

That's gonna be the mass in grammes, their relative mass.

So that could be the relative atomic mass or the relative formula mass, depending on the substance, the number of moles, and then the ratio.

So the grid for this particular balanced equation is gonna look something a little bit like this: Now once you have that calculation grid completed, we are going to follow through with our processing in a U formation.

You are always going to divide down the known column of information, and then we're going to multiply up the unknown column of information.

Now, it doesn't always happen that your known is a reactant and your unknown is the product.

So sometimes those arrows might be reversed, but it's always down the known and up the unknown.

This type of processing though, this kind of shortcut to remember the mass if you can't remember, only works if you create your calculation grid in the following order: So the mass in grammes, then your relative masses, then the moles, and finally the ratio.

And hopefully that will all become clear as we go through these next steps.

But this is a quick and easy shortcut that you can use to remember the mass if you're not quite sure on the day of how to carry this out.

Now that we've identified our known and unknown substances and created our calculation grid below them, we need to populate that calculation grid with as much information as we can before we start the calculations.

So the first thing we can do is the easiest.

We're simply gonna copy the mass of our known substance from the question into our grid.

The next thing you need to do then is to find the relative masses for both the known and unknown.

And for this you're going to use the periodic table.

So in my particular question here, the magnesium then is going to be 24.

3, its relative atomic mass 'cause it's just the element.

For my magnesium oxide though, because it's a formula unit, I need to add the relative masses for the magnesium and oxygen together.

And that gives me a final answer then for its relative mass as 40.

3.

Finally, then I'm going to include the stoichiometry, that molar ratio from my balanced symbol equation into my grid.

And for this, you simply copy those coefficients into the ratio boxes for each substance.

Now once your grid has been populated with as much information as it can be, if you look at the known and unknown columns, you'll see that the known column has more information in it.

So that's a really quick and easy check to make sure that you have filled your column in correctly.

Okay, once you've done that, you're going to start with the known column.

You have more information there, so it's gonna be easier to calculate.

The first thing we need to do then is to calculate the moles of your known substance.

And we're gonna use that equation of moles is equal to the mass in grammes divided by its relative mass.

And you'll notice in the calculation grid that it works very easily.

Because we've put it in this order, the mass in grammes is above its relative mass, so it's already in place almost like a fraction ready for dividing.

And when we do that, we get an answer then of 1.

975 as a number of moles of magnesium atoms in a 48 gramme sample.

The next thing you're going to do then is to use that stoichiometry to calculate the moles that you would have in your unknown sample.

Now this maths looks a little complicated at first glance.

You're gonna take the moles of known, divide it by the ratio of your known, and then multiply that answer by the ratio of your unknown.

And when we do that, we get a value of 1.

975.

Now that can be a little tricky to remember, but if we use that guidance of the U formation of our calculation grid, we're doing the same thing.

We're still dividing down the known column.

And then as we move across to the unknown column, we start to multiply, so dividing and then multiplying to give us that molar ratio that if I have 1.

975 moles of magnesium in this reaction, I will also need 1.

975 moles of my magnesium oxide.

We are now on the home straight.

We can see in our calculation grid there is only one box that has yet to be completed, and that is to find your final answer to calculate that mass of your unknown.

And we're going to use the equation then of mass in grammes is equal to the relative mass times the moles.

And all you need to do then is exactly what we have been, going up that unknown column, and that is to multiply those values together.

And when we do that, we get an answer of 79.

59.

And if we round that then to our standard three significant figures, we get a final answer of 79.

6 grammes of magnesium oxide is produced from 48 grammes of magnesium.

Now, I said before that a strategy like this is going to take a bit of practise.

Practise makes perfect.

So let's go through another example.

But this time, rather than talking through each individual step, I'm going to just outline it to you using my calculation grid.

So for this question, I need to know the mass of hydrogen that's needed to produce 100 grammes of ammonia.

I'm going to give my answer to three significant figures and I'm going to be using the balanced symbol equation that's been provided.

The first thing I need to do then is to identify my known and unknown substances from the question in my balanced symbol equation.

And because I've been asked about the hydrogen, I'm going to put a question mark above that symbol, and I've been given information about ammonia, so I'm going to put a tick above that one.

The next thing I'm gonna do is draw my calculation grid below those two substances.

Then I'm going to populate that grid with as much information as I can.

And if I pause here, I can double check that I have more information under my ticked column than I do under my question mark column.

And if I have, I've put things in the right place and I can start to move forward.

So the first thing I'm going to do is to divide down my known column.

And by doing so, I find the number of moles that I have in my known sample.

So for this example, it's 5.

8823 moles of ammonia particles that are found in a 100 gramme sample.

The next thing I'm going to do then is to continue dividing down that column until I get to the bottom.

And then as I move across to the unknown column, I multiply.

And by doing this, I will be able to find the number of moles of hydrogen that's needed for this question.

And when I do that, I have 8.

8235 moles of hydrogen.

The last thing I need to do then is to multiply, continuing up that unknown column.

And when I multiply those values, I get 17.

647 as an answer.

But I've been instructed to give my final answer to three significant figures and therefore my final answer will be 17.

6 grammes of hydrogen that's needed to produce 100 grammes of ammonia.

Now what I'd like you to do is to use this processing that's been outlined on the left to help you answer that question on the right.

I'd like to know what mass of aluminium can be produced from a thousand grammes of aluminium oxide? I'd also like you to give your answer to three significant figures, and to help you, I've provided that balanced symbol equation and I've shown a little bit more clearly the processing that I did, using my calculation grid on the left hand side.

Now this is gonna take a little bit of time, so definitely pause the video, maybe double check your answers with the people nearest you as you go through and come back when you're ready to check your answer.

Okay, let's see how you got on.

So the first thing you needed to do was to identify your known and unknown substances.

So I had information about the aluminium oxide, so that gets a tick, and I need to know about the mass of aluminium, so that's my question mark.

Once you've created your grid then and done all your processing, you should have got a final answer of 529 grammes of aluminium that is produced to three significant figures.

Now, if you didn't get that as an answer, I highly recommend you pause the video here and double check your processing against the working out that's been shown.

And the idea here then is you can identify any errors that you may have made so that we can try to avoid them as we go forward.

But if you did get the correct answer, fantastic work, guys.

Brilliant job.

So proud of you.

Okay, guys, let's move on now to the last task of today's lesson.

For this first question, what I'd like you to do is use the numbers in the box to complete the statements below about the stoichiometry that's involved when we calculate the mass of carbon dioxide that might be produced when six grammes of carbon reacts with oxygen.

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 if you've completed these statements correctly, they should read like this: The mass of one mole of carbon atoms is 12 grammes.

The mass of one mole of carbon dioxide molecules is 44 grammes.

Now if you're unsure how those numbers were calculated, you need to remember that equation of mass in grammes is equal to the relative mass times the number of moles.

The next statement then is for each mole of carbon, the number of moles of carbon dioxide produced is one.

From above, 12 grammes of carbon produces 44 grammes of carbon dioxide.

This means that six grammes of carbon produces 22 grammes of carbon dioxide.

Well done if you got those correct.

So the final part of today's lesson's tasks is for you to use your understanding of stoichiometry to answer the following questions and to give your answers to three significant figures.

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

Okay, let's see how you got on.

So the first thing we needed to do with each of these questions was to identify our known and unknown substances.

And for part A, iron was our known and copper was our unknown.

Now, if you've carried through your calculation grid correctly, you should have got a final answer of 56.

9 grammes of copper that's extracted using 50 grammes of scrap iron.

Now, what I would recommend is if you didn't get that correct answer is that you pause the video then so you can compare your working out to those in the calculation grid that's shown to see where you may have gone wrong.

Perhaps you've calculated a relative mass incorrectly, maybe you multiplied when you should have divided or vice versa, but double check so that we can identify where your errors are so we can work on improving them as we go forward.

Well done if you've got this first one correct.

For part B then, what you should have noticed is that the sodium then was your unknown whilst the titanium was your known substance.

And if we follow through with our calculation grid, then we would've calculated that 288 grammes of sodium was needed in order to extract 150 grammes of titanium from that titanium chloride.

And for our final question, then, you should have identified the carbon dioxide as our known substance and the oxygen as our unknown.

And again, following through with that calculation grid gives us a final answer of 19.

1 grammes of oxygen that's reacted to produce 35 grammes of carbon dioxide when burning butane.

So very, very well done with this, these calculations, guys.

So impressed with your perseverance.

Just really, really well done.

Wow, you guys have done just amazingly well today.

I'm so proud of you, and I wouldn't be surprised if many of you thought that today's lesson was more a maths lesson than a chemistry lesson, but your perseverance has really impressed me.

I'm so proud of you.

Let's take a moment now then to summarise what we've done in today's lesson, while we reminded ourselves that the atoms in the reactants are simply rearranged and reorganised to form all of our products, and therefore the atoms that we start with are the same as the atoms that we end with, and that these reactions can be represented using a balanced symbol equation, which is most correctly interpreted in terms of moles and molar ratios rather than individual particles.

So a better way to read a balanced symbol equation might be one mole of substance A reacts with two moles of substance B and so forth.

These molar ratios then, that a balanced symbol equation represents, is actually known as the stoichiometry of that chemical reaction.

And that the stoichiometry then is represented by those coefficients in the balanced symbol equation, those molar ratios, and these relationships do not change.

And finally, if we want to be able to calculate an unknown mass or the mass really, of any substance in a particular chemical reaction, if we have the stoichiometry of that reaction, we can use, then that calculation of mass and grammes is equal to the relative mass times moles and be able to calculate any of the masses for those substances.

I am so.

I still am flabbergasted by how you guys have persevered with today's lesson.

It's not the easiest to get through, but it is a practise makes perfect, and you've really impressed me.

Just fantastic job.

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

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

Bye for now.