Lesson video

Hello, my name's Mrs. Nevin 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 a 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.

First, we'll look at the ratios in a balanced equation and then we'll look at how we can use those ratios to calculate a mass.

So let's get started by looking at where the ratios are in a balanced symbol equation and how we can interpret them.

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're 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 an 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 stoicheion which means element and then 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 coefficient 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 and 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 nought.

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 one billion carbon atoms react with one billion oxygen molecules making one 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, 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 one.

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, six 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, nought.

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 twp, four, two and two.

For B, we multiplied by six so our new coefficients reading across from left to right will be 12, 12, six and six.

And for C, we divided by four going from a coefficient of one to nought.

25 which is equivalent to 1/4.

So our new coefficients now are nought.

5, nought.

25, nought.

5 and nought.

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 is that.

And for four then, to get from three to nought.

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, nought.

17, nought.

33 and nought.

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 perfect 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, 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're 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're 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.

So we've now identified the known and unknown substances from our question and we've created our calculation grid.

So the next thing we need to do is populate that grid with as much information as we possibly can before we start doing any calculations.

So the easiest thing to fill in first of all is the mass of our known substance because that was given in the question, the mass in grammes.

The next thing you'll need to do then is to find the relative masses for both the known and unknown substances and for this, you're going to use the periodic table.

So in my grid, I'm going to put 24 for magnesium, that's straight off the periodic table, the relative atomic mass and then 24 plus 16 gives me the relative formula mass then for my magnesium oxide, finally then, I can input the stoichiometry, those molar ratios, and all I'm doing then is literally copying that coefficient from the balanced symbol equation into the relevant ratio boxes below the known and unknown substances.

Now, once we've populated our calculation grid with as much information as we can, we can see then that we have more information under the known column and that's the way it should be.

So it's a quick and easy check to make sure everything's in the right place before you get started.

You're also then going to start your calculations on the known side, the one where you have more information.

And the first thing you need to do then is to calculate the number of moles that are present in your known sample and you're going to do that by dividing the mass in grammes by the relative atomic mass and you can see in our calculation grid, it looks a little bit like a fraction.

So for this particular one, we're going to do 48 divided by 24 and that gives us a value of two which means that in 48 grammes of magnesium, we have two moles of magnesium atoms. The next thing we need to do then is we need to use the stoichiometry or those molar ratios of our balanced symbol equation to calculate the number of moles of our unknown and to do that, we're going to take the moles of known divided by the ratio and then multiply that answer by the ratio of our unknown substance.

And the calculation looks a little bit like what's shown but if we use our grid, if we use our calculation grid, remember we're dividing down the known, so if we continue to divide down, two divided by two equals one, and then as we move over to the unknown column, we're multiplying and that gives us then that final answer which is the number of moles of our unknown substance.

So if you can't remember that method of how to use that stoichiometry to calculate the moles of unknown, you can use your shortcut of divide down the known and multiply up the unknown.

We can see now that our calculation grid is very near complete.

The very last thing we need to do now is to calculate the mass of our unknown and we're gonna go back to that hold all equation of the mass in grammes is equal to the relative mass times the moles.

And for this then is simply two times 40 and that gives us an answer of 80.

So the answer to my question of what mass magnesium oxide is produced when 48 grammes of magnesium burns in oxygen is 80 grammes.

Now, you may remember I said that this kind of processing is a practise makes perfect technique.

So let's go through another example but this time, I'm not going to outline each individual step.

I'm just gonna talk you through my calculation grid.

So I need to know the mass of hydrogen that's needed to produce 100 grammes of ammonia and to give my final answer to three significant figures.

And as before, I've been given my balanced symbol equation.

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

So I've been asked about the hydrogen so I'm going to put a question mark above that and I have information about the ammonia so I'm going to put a tick above that.

I then form my calculation grid and I always have to do it mass, relative mass, moles and then the ratio.

So MRMR if you want to come up with some kind of mnemonic to remember the order.

Then I'm going to populate this calculation grid with as much information as I can and when I do that, I'm just gonna do my double check that my tick column has more information which it does and that's also the column I'm going to start with.

I'm going to divide down that known column and that tells me then the number of moles that I have in a 100 gramme sample of ammonia.

That is 5.

8823.

I'm going to continue dividing down that column and then as I get to the bottom, I need to move across to the unknown column and when I move across, I'm going to multiply.

What this is doing then is helping me to use the stoichiometry of my balanced symbol equation to find the number of moles of my unknown substance which here is 8.

8235.

I'm going to continue then multiplying up the unknown column and I get a mass then of my unknown of 17.

647.

However, I've been asked to give my final answer to three significant figures so my final answer will be 17.

6 grammes of hydrogen is needed to make 100 grammes of ammonia.

What I'd like you to do now then is to use this as a guide to answer this question on the right.

What mass of aluminium can be produced from 1000 grammes of aluminium oxide? And I'd like you to give your answer to three significant figures.

I've given you the balanced symbol equation and if you look back on the calculation grid on the left hand side as the example, I've shown my processing a little bit more clearly this time.

So use this as a guide to help you in your calculations.

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

Okay, let's see how you got on.

So first things first, we needed to find our known substance and our unknown substance in our balanced symbol equation.

And then if you have completed your calculation grid correctly, you should have got a final answer of 529 grammes of aluminium is produced.

Now, if you didn't get that answer, definitely pause the video and double check your processing which is shown in the calculation grid above but very, very well done if you managed to get that correct, guys.

What a fantastic effort.

Great job.

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 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 for 2A, the first thing you needed to do was identify your known and unknown.

So the iron was known and the copper was your unknown.

Using our calculation grid then, if you've done it correctly, you should have got a final answer of 56.

7 grammes of copper that can be extracted using 50 grammes of iron.

Now, for each of these feedback slides, what I'm gonna recommend you do is if you didn't get the correct answer, definitely pause the video and review your calculations to those in the calculation grid to double check and identify where you may have gone wrong so we can try to avoid those in the future but very well done if you've got this first one correct.

Good job, guys.

And for part B then, we should have seen that the sodium was your unknown substance whilst the titanium was your known substance and if we follow through with our calculation grid, you should have calculated that we needed 288 grammes of sodium in order to extract 150 grammes of titanium from that titanium chloride.

And for this final one, we're trying to find out the amount of oxygen that reacts that produces 35 grammes of carbon dioxide.

So our known substance was the carbon dioxide and our unknown substance was the oxygen.

Following through with our calculation grid, we should have got a mass of 19.

1 grammes of oxygen reacted to produce those 35 grammes of carbon dioxide.

So very, very well done if you managed to get these correct, guys.

I'm so impressed with your perseverance.

Well done.

Wow, we have done a lot of work in today's lesson and I'm sure that for some of you, it might've felt more like a maths lesson than a chemistry lesson but your perseverance will really pay off in the long run and I'm so impressed with the work you've done but let's take a moment now to summarise what we've done in today's lesson.

Well, we reminded ourselves that the atoms in all of the reactants of any chemical reaction are rearranged and form then all of the products in that chemical reaction.

So the same atoms we start with are the ones that we end with.

And we also realise that in the balanced symbol equation, we are given coefficients and they're most correctly interpreted, not in terms of the individual particles but as in terms of molar ratios.

Saying something like one mole of substance A will react with two moles of substance B, something like that.

So to interpret those coefficients as a molar ratio.

And that those molar ratios are actually known as the stoichiometry of that particular chemical reaction and the stoichiometry of a reaction is represented by those coefficients.

These molar ratios don't change.

The stoichiometry of a chemical reaction do not change.

And that the mass of each substance in a reaction can actually be determined using that relationship of mass in grammes is equal to the relative mass times moles and the stoichiometry of a reaction as long as you've got that really systematic strategy that you can employ.

You guys have done such a fantastic job in today's lesson.

I'm really proud of you.

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

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

Bye for now.