Hello, my name's Mrs. Nevin, and today we're going to be looking at how we can convert between mass and volume using moles, as part of our unit on calculations involving masses.

Now, you will have some experience of what we talk about in today's lesson from your previous learning, but today will help us to not only answer that big question of what are substances made of, but also help us to appreciate what we observe, and calculate, and experience in the laboratory, and how it applies to our understanding of chemical processes in the wider world, and how we can quantify those processes.

So by the end of today's lesson, you should feel more comfortable being able to combine those mathematical relationships used to calculate an unknown mass or an unknown volume.

Throughout the lesson, I will be referring to some keywords, and these include, moles, volume, molar gas volume, stoichiometry, and balance symbol equation.

Now, the definitions for these keywords are given in sentence form on the next slide, and you may wish to pause the video here to read through them or perhaps make a note of them, so you can refer back to it later in the lesson, or later on in your learning.

Today's lesson then is broken into two parts.

We'll first remind ourselves of those mathematical relationships between math, volume and moles before looking at how we can use those relationships to calculate an unknown gas volume or mass.

So let's get started by looking at mass, volume, and moles.

The first thing we need to do is to recall that all matter is composed of particles, and that mass is simply an indication of the amount of matter in a substance.

Now, the mass of a sample of that substance can be measured using a balance, and that the standard unit for mass then is grammes, which is indicated with a lowercase g.

Now, the number of particles or the number of moles in a sample affects the mass that will actually show on that balance.

Now, we don't individually count particles.

They're far too small, and they're far too difficult to count, but we can use that relationship of one mole equaling 6.

02 times 10 to the 23 particles to help us gain an understanding of the mass of a particular sample.

Let's take a closer look at this mathematical relationship.

The massing grammes is equal to the relative mass times moles.

Now, the massing grammes is simply a measurement of the amount of matter in a sample, whereas the relative mass is the mass of each particle, compared to a specific standard, and in this case it's an atom of carbon 12.

Moles, then, is the number of particles in a sample.

Now, if we work backwards on this particular mathematical relationship, essentially what this is saying is if I know the number of particles that are present and the relative mass of each, I can multiply those values together, and I will find the value of the mass that I should see if I was to put that number of particles on a balance.

Now, using this mathematical relationship, we could actually rearrange it and calculate, then, the number of particles or the number of moles that are contained in a particular sample.

For instance, if I know I have a sample of sodium chloride and I measure out two grammes of sodium chloride, I can get the value off the balance, find its relative mass using a periodic table, and dividing those values, I can then identify the number of particles that are in that sample.

So let's go through an example of how I can use that relationship.

I'd like to know how many moles are in a 2.

30 kilogramme sample of aluminium oxide, which has the formula AL203, and I've been instructed to give my answer to three significant figures.

Now, because I need to calculate the number of moles, I'm going to use that relationship of moles is equal to mass and grammes divided by relative mass.

Now, the mass then from my question is 2.

30 kilogrammes, but if I take a closer look at the equation I need to use, my mass needs to be in grammes.

So the first thing I'm going to do is convert the mass from kilogrammes into grammes by multiplying by 1,000.

The next thing I need to do, then, is to look at the relative mass, and that's related to the individual particles that make up that 2.

3 kilogramme sample.

So I need to look at the formula for that particle, and it's AL203.

So using a periodic table and adding up the relative mass for each of the atoms that make up that particle, I get a relative mass of 1020.

0.

Using now the mass in grammes and the relative mass.

I divide those values and get an answer of 22.

549.

However, I was instructed to give my answer to three significant figures.

So my final answer is 22.

5 moles of aluminium oxide that are in a sample that is 2.

3 kilogrammes large.

What I'd like you to do now then is to calculate how many moles of carbon dioxide are formed in a train journey from London to Edinburgh.

Now, I'd like you also to give your answer to three significant figures.

So use the calculation that's shown on the left as a bit of a guide to help you, but you will definitely need a periodic table, calculator, paper, and pen to get started here.

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

Okay, let's see how you got on.

So if you've done your calculations correctly, you should have got a final answer of 566 moles of carbon dioxide.

If you didn't get that answer, please do pause the video here so you can compare your calculations to what's shown to see if you can identify where you may have gone wrong so you can make adjustments going forward, but incredibly well done if you managed to get that correct, guys.

Great start.

Now, you may also recall that how much 3D space matter occupies is known as its volume, and volume is usually measured in the units of centimetres cubed or decimeters cubed.

And we can find the volume of a sample in many different ways.

We could, for instance, with a cube, measure the sides and calculate it out.

We could potentially for a liquid, use a apparatus, and simply read the scale on that apparatus.

Or for instance, with a gas, we might actually use a mathematical formula, and calculate out the volume for that sample.

Now, we need to recall that when temperature and pressure are kept constant, an equal number of moles of all gases will occupy the same volume.

Also at room temperature and pressure, so that's 20 degrees Celsius and one atmosphere, one mole of any gas will occupy 24 decimeters cubed or equivalent to 24,000 centimetres cubed.

Now, this is known then as the molar gas volume for room temperature and pressure, RTP.

This gives us then, a mathematical relationship where we can calculate the volume of any gas if we know the molar gas volume and the number of moles of gas present.

So the volume then is equal to the moles times that molar gas volume.

Chemists then can rearrange that mathematical relationship to calculate the number of moles, the number of particles that are present in a specific volume, as long as we also know that molar gas volume.

Let's look at how I can use that mathematical relationship.

I found out that a mature maple tree can absorb 5,440 decimeters cubed of carbon dioxide every year, but at RTP, how many moles of carbon dioxide is that maple tree absorbing in a year to two significant figures? Well, because I'm being asked to find the number of moles of a gas, I'm going to use that relationship of moles is equal to volume divided by the molar gas volume.

And if I look at the question, I've actually been given the volume 5,440 deters cubed.

I've not been provided that molar gas volume, but there are some hints to help me decide what value I need to put in.

It's at room temperature and pressure, and it's decimeters cubed.

Now, the molar gas volume at RTP in decimeter cubed is 24.

So moles then is equal to 5,440 divided by 24 to give me a value.

However, I have to give my final answer to two significant figures.

So using that calculation, a mature maple tree can absorb 230 moles of carbon dioxide in one year.

What I'd like you to do now, then, is to calculate how many moles of oxygen a mature oak tree can produce daily to two significant figures.

Use the calculations that are shown on the left to help guide you in your own calculations.

I would recommend that you pause the video now, and come back when you're ready to check your answer.

Okay, let's see how you got on.

So if you've done this correctly, you should have got a final answer to two significant figures of 11 moles of oxygen.

So incredibly well done if you managed to get that correct, and if not, do pause the video and go back to check your calculations.

But great, great job guys.

Now, mathematical relationships allow chemists to interconvert between the mass and volume of a gas, and if we look at our two equations that we've been using, and take a closer look at them, we can see both contain that variable of moles and that's quite useful.

If we go back to our example before of that mature maple tree, we found that it could absorb a volume of 5,440 decimeters cubed of carbon dioxide, and calculated that that volume was equivalent to 230 moles of carbon dioxide per year.

Now, if somebody is calculating their carbon footprint or how to offset their carbon footprint, that is usually calculated in grammes or kilogrammes.

So what I'd like to know is what mass of carbon dioxide does a mature maple tree absorb each year? Well, I already have the number of moles from my previous volume calculation.

What I need now is the relative mass for that gas, which is carbon dioxide.

For this, all I need is a periodic table.

I'll add up the relative mass for each of the atoms, and that gives me a value of 44.

0.

Putting, then, the number of moles and the relative mass into my mass in grammes equation, I get a final value of 10,120 grammes of carbon dioxide is absorbed per year by one mature maple tree.

Let's look at how I can bring those two calculations together in an example.

I found that a flight between London and Edinburgh releases a whopping 105,000 decimeters cubed of carbon dioxide, but I'd like to know what mass that is in kilogrammes.

So the first thing I need to do is find the number of moles of carbon dioxide that is present in that volume that's released.

Now, I've been given that volume, but I don't have the molar gas volume that I need for my equation, but there's a clue in the volume.

The units are decimeters cubed, and therefore for the molar gas volume, I'm going to be using 24 decimeters cubed.

And when I put that value into my equation, I find that there are 4,375 moles of carbon dioxide released during this flight.

I then need to move to my mass in grammes equals moles times relative mass equation.

Now, I have the moles from my previous calculation, but now I need the relative mass for the gas that I'm looking at, and here it's carbon dioxide.

Using that formula in a periodic table, then, I calculate that the relative mass of carbon dioxide is 44.

Using now the moles and the relative mass, I can calculate that there are 192,500 grammes in that volume of carbon dioxide that's released.

But I was instructed to give my answer in kilogrammes.

So I'm going to take that answer that I had, divide by 1,000, and I get a final answer of 190 kilogrammes of carbon dioxide are released in one flight between London and Edinburgh.

What I'd like you to do now, then, is to calculate what mass in kilogrammes of carbon dioxide is released in a train journey between Inverness and Dover.

So use the calculations on the left as a guide, grab yourself a periodic table, calculator, paper, and pen, pause the video, and come back when you're ready to check your answer.

Okay, let's see how you got on.

If you've done it correctly, you should have had a final answer of 35 kilogrammes of carbon dioxide that are produced in that train journey.

If you didn't get that answer, please do pause the video, so you can check your calculations and see where you may have made an error, so you can adjust as you go forward.

But a fantastic job if you've got that correct guys.

You are doing so, so well.

Keep it up.

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

What I'd like you to do is to calculate the number of moles that are present in each sample that is described below, and to give your answer to three significant figures.

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

Okay, let's see how you got on.

Now, I can see from the description in part A that my sample is made of grammes, and because of that I'm going to use the relationship of the moles is equal to the massing grammes divided by the relative mass.

So if I calculate the relative mass for glucose, it comes out as 180.

0, and if I pop that and the mass of my sample into my equation, I get a value of 3.

64 moles of glucose that is present.

For B, following that same process, I get an answer of 0.

175 moles of cuprite or CU2O in a 25.

0 gramme sample.

For part C, I can see that I still have a unit for mass.

However, it's been given to me in kilogrammes, instead of grammes.

So I need to convert my sample mass into grammes before continuing with my calculation.

And when I do so, I get a final answer now of 5.

98 moles of sodium chloride.

Now, for D, I need to do the same thing, and I get a final answer then a 545 moles of hematite, or Fe2O3, in an 8.

70 kilogramme sample.

So well done if you managed to get those correct.

Very, very impressive.

For the second part of this task, then, I'd like you to calculate the mass of gas that is present in each of the descriptions below.

Now, each gas is at room temperature and pressure, and I'd like you to give your final answer this time to two significant figures.

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

Okay, let's see how you got on.

Now, we've been asked to calculate the mass of differently described gas samples, and because of that, we need two different equations.

We need the moles as equal to the volume divided by the molar gas volume, and we'll need the massing grammes is equal to the moles times the relative mass.

Now, the first thing I need to do is calculate the number of moles, and I've used the value of 24 for the molar gas volume because the units of my volume were given as decimeters cubed.

So I now have the value of 0.

153 moles of steam that are present in that volume.

Now, steam is simply water in the gas state.

So if I calculate the relative mass of water, I get a value of 18.

0.

Using that relative mass and the number of moles that were calculated earlier, I get a final answer, then, to two significant figures of 2.

8 grammes of gaseous water is present in a 3.

68 decimeters cubed sample.

Using that same process, then, for letter B, I get a final answer that was 1.

1 grammes of hydrogen in a 12.

7 decimeter cubed sample.

Now, if you've calculated letter C correctly, you should have had a final answer of 1.

9 times 10 to three grammes of butane that were in your sample.

Now, there are two places where you may have got caught out in your calculations, and the first is that the answer was given in standard form to ensure we had that two significant figure final answer.

Secondly, we needed to ensure we had the correct value for the molar gas volume that matches the units for the volume we were provided.

Because it was sent to me just cubed, I needed to use 24,000 for my molar gas volume.

So if you did get that answer, incredibly well done.

If not, make sure that you're pausing the video to double check your calculations going forward.

And finally then for D, I'm going to use the same process that I used for letter C, and I get a final answer then of 2.

3 times 10 to the three grammes of hydrogen sulphide that are present in my sample.

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

You guys are machines.

I'm so impressed.

Keep it up.

Now, that we're feeling a little more comfortable talking about and calculating those relationships between the mass, volume, and moles of a sample, let's look at how we can use those relationships to calculate an unknown gas volume or mass.

Now, you may recall that chemists can calculate unknown values using the stoichiometry or those molar ratios that are found in a balanced symbol equation, and they're able to do this because we can calculate the number of particles or the moles of each substance.

And this idea of moles is found across chemistry in our balanced symbol equations, and also in our mathematical relationships for mass and volume.

So what I'd like to do now is take a moment to look at how chemists are able to combine these ideas, those balanced symbol equations, the molar ratios, and, then, those mathematical relationships in order to calculate an unknown.

I'd like to know what volume of oxygen forms when 20 grammes of hydrogen peroxide decomposes and the reaction takes place at RTP, so room temperature and pressure.

To help me with this, I've been given the balance symbol equation.

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

Because I've been given the mass of hydrogen peroxide, that is my known substance, and so I'm going to put a little tick over the formula for hydrogen peroxide in my balanced symbol equation to remind me that's my known substance.

It also gives me an indication of where I'm gonna start my calculations.

My unknown substance then is what I've been asked about, and in this question it's the oxygen.

So I'm going to put above my oxygen formula a little question mark.

For the water that's left, I'm gonna cross it out because any calculation going forward doesn't involve it, and I don't wanna get confused by seeing that and thinking I have to do some extra calculations with it.

The next step then is to create a calculation grid below the known and unknown samples, and that needs to have at least four rows.

And you'll see here it also consists of four columns.

Now, those outside columns are actually going to be used for labels for what we're going to be putting in the columns directly below my known and unknown substances.

So I can keep track of what each value represents.

So the third row is going to be labelled moles, and the fourth row is going to be labelled ratio.

To label the other rows.

I need to use the question to help guide me.

If I look at the hydrogen peroxide, I have 20 grammes, and because of that, I'm going to use that mathematical relationship of the mass in grammes is equal to the relative mass times the number of moles.

Well, moles has already been labelled in that third row underneath my known substance in that grid.

So I need to put the mass in grammes and the relative mass in those last two spots.

For my unknown sample, I've been asked to calculate the volume of oxygen, and for this then I'm going to need to use that calculation of volume is equal to moles times the molar gas volume.

And as with my known substance, moles has already been labelled.

So I need to put in those other two variables of volume and the molar gas volume into those last two rows.

Okay, now the calculation grid is going to be completed in a U formation, and what we're going to do is we're going to divide all of the values down that known column, and then we're going to multiply the values that we put into our calculation grid as we move up the unknown column.

So essentially, if you're not sure what to put on that top row for your unknown substance, it's always gonna be what you've been asked to find, either the mass that you've been asked to find or the volume that you've been asked to find.

So that's always that top row.

Now, this U formation calculation, where you divide down your known and multiply up the unknown, only works if the grid is constructed with those labels in the order that has been shown.

Okay? So be careful that you are ensuring the volume column is always under your gas substance.

Okay, so we have our calculation grid ready to go.

We know how we're going to do those calculations, but I need values in those boxes for both my known and unknown substances in order to carry out those calculations.

And I'm going to get those values from a variety of places.

And we're gonna start with information from the question.

I know that I'm starting with 20 grammes of hydrogen peroxide, so I'm putting 20 down in the mass in grammes row.

I also know that the reaction is taking place at RTP, room temperature, and pressure, and that is gonna impact the value I put down for my molar gas volume.

Now, I haven't been given a unit to calculate the volume in, so I'm going to use the easiest value to remember it, which is 24.

And remember, that's a decimeters cubed.

So you may wish to put a little note next to the volume, what units you expect to finish in, if you use that particular value for the molar gas volume.

I'm then gonna move on to use a periodic table to calculate the relative mass for the substance that requires a relative mass in its calculations.

And then I need the balanced symbol equation to tell me those molar ratios, how the ratio of my known substance compares to my unknown substance.

And remember, those are simply the coefficients that are found at the front of each formula.

So I'm copying down those coefficients into the ratio rows.

At this point, then, we are left with simply three boxes to fill, and I'm going to fill them using those calculations that we talked about earlier, using that U formation.

So the first step then is to divide down your known column.

And by doing so, what you're doing is calculating the number of moles.

Using that relationship of moles equals the mass in grammes divided by the relative mass, and that gives you your value for the number of particles that are in that 20 gramme sample of hydrogen peroxide.

We then need to calculate the number of moles in my unknown substance.

So I do this by continuing to divide down that known column and then multiply as I move across to the unknown column.

And my answer then goes into the unknown moles box.

At this point, then we have just one box left in our calculation grid to fill, and that will be done by simply answering the question we were asked.

And here it was to find the volume of oxygen that forms in this reaction.

Now, because I used that mathematical relationship of volume equals moles times the molar gas volume earlier on in my creation of that calculation grid, everything is already set up.

So all I need to do is continuing to multiply up that unknown column, and I get a value of 7.

059 for the volume.

But what units do I need to include? Well, if you recall, I used 24 for my molar gas volume because it was a small number.

It's easier maths, but that is specific to decimeters cubed.

So my final answer then, to two significant figures, 7.

1 decimeters cubed of oxygen is produced when 20 grammes of hydrogen peroxide decomposes.

Okay, let's go through another example, only this time, rather than writing out each step, I'm going to simply talk you through them.

For this example, I'd like to know what mass of calcium carbonate decomposed to form 65,000 centimetres cubed of carbon dioxide when the reaction took place at room temperature and pressure.

And I have my balanced symbol equation here.

On it, I need to identify my known and unknown substances.

Now, I'm asked about the calcium carbonate.

So I'm going to put a question mark above it.

And because I have information about the carbon dioxide, that's my known substance, and I'll put a tick above the formula for carbon dioxide.

Next, I need to create my calculation grid that has at least four rows and space for labels on those rows on the outsides.

And the third and fourth row are going to be labelled moles and ratio, respectively.

I need to fill in the rest of those labels now.

And because I've been asked about the mass for calcium carbonate, my top row underneath the unknown substance here, then will have mass.

And the other relationship that I need for that calculation is the relative mass of the calcium carbonate.

For the carbon dioxide calculation grid, I've been given the volume, and for the relationship between volume and moles, I need the molar gas volume as well, that Vm.

The next thing I'm going to do, then, is start to populate that grid, and I start with information from the question.

So I have 65,000 is the volume, and because the reaction is taking place at RTP, and the volume units are centimetres cubed, the value for the molar gas volume I'm using is 24,000.

I then need to use my periodic table to find the relative mass for my calcium carbonate, which is 100.

1, and then I'm going to use my balanced symbol equation to fill in the ratios.

So simply copying those coefficients at the front of each formula down into my grid, I now start my calculations.

The first step is to divide down that known column to find the number of moles that were present in that volume of carbon dioxide.

Continuing to divide down that column and then multiply across, I find the number of moles of calcium carbonate.

Continuing to multiply up that column, I now have a mass of calcium carbonate that decomposed, which is 271.

1.

So to two significant figures, 270 grammes of calcium carbonate decomposed to form 65,000 centimetres cubed of carbon dioxide.

What I'd like you to do now, then, is to calculate the mass of potassium chlorate that decomposed to form 15 decimeters cubed of oxygen when the reaction took place at RTP.

And I've outlined some of the calculations that I used on the left hand side to help guide you in those calculations.

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

Okay, let's see how you got on.

So if you carried out your calculations correctly, you should have a final answer of 51 grammes of potassium chloride decomposed to form those 15 decimeters cubed of oxygen.

Well done, if you managed to get that correct.

If you didn't, double check the value that you used for your molar gas volume, and pause the video if you want to check any of the other values in that grid.

But incredibly well done if you managed to get 51 grammes.

Guys, great job.

Okay, onto the last task for today's lesson, and we're going to start by looking at the decomposition of copper carbonate.

I'd like you to calculate the mass of carbon dioxide that's produced, the number of moles of carbon dioxide produced, and then the volume of carbon dioxide that was produced.

So you're going to need to use a variety of mathematical relationships.

You may wish to work with somebody near you, but definitely show you working out.

So when we go through this, you can check to see if you did the working out correctly or not.

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

Okay, let's see how you got on.

Using conservation of mass, we can remember that the mass of my reactants has to equal the mass of my products.

So using that idea, I calculate that 2.

2 grammes of carbon dioxide was produced in this reaction.

For part B, I need to know the number of moles of carbon dioxide that were produced.

So I need that mathematical relationship of moles is equal to the mass in grammes divided by the relative mass.

Well, I calculated the massing grammes in part A, so I'm gonna use that value for part B.

Once I find, using my periodic table, the relative mass of carbon dioxide is 44.

0, I calculate that 0.

050 moles of carbon dioxide were produced in this reaction.

For Part C, then, I need to find the volume in decimeters cubed of carbon dioxide gas produced.

And for this, I need that equation of volume is equal to the moles times the molar gas volume.

Well, I calculated moles in part B, so I'm going to use that going forward.

And my molar gas volume has to match the units for the volume that I've been asked to find.

And for this, that's 24.

So when I calculate everything out, I get a final answer of 1.

2 decimeter cubed of carbon dioxide gas was produced.

Very well done, if you managed to get that final answer correct.

What I'd like to point out here is that don't forget if you have a question that comes in parts A, B, and C, don't be afraid to go back and see if there's any information from those previous questions that you can use going forward.

More often than not, you can.

So definitely go back and see if you can use anything in subsequent questions.

For the last part of today's task, I'd like to look at vehicle airbags.

Now, these are designed to inflate instantly, and that occurs because a large volume of gas is produced very quickly.

Now, one reaction used for that purpose is the decomposition of sodium azide, and you have the balanced symbol equation provided.

What I'd like you to do is to calculate both the volume in centimetres cubed and the mass of nitrogen that is produced when 145 grammes of sodium aside decomposes.

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

Okay, let's see how you got on.

So firstly, you were asked to calculate the volume of nitrogen in centimetres cubed that was produced when 145 grammes of sodium azide decomposed.

And if you carried out your calculation grid correctly, you should have got a final answer to two significant figures of 8.

0 times 10 to the four centimetres cubed.

Very well done if you managed to get that correct.

If not, please do take a closer look at the calculation grid to see if you can identify where you may have gone wrong and see if we can take that information forward with our next calculations.

But very, very well done, if you got that correct.

You were then asked to calculate the mass of the nitrogen that was produced in that decomposition reaction, and to give your answer to two significant figures.

Now, we can use the volume of nitrogen that was produced, that was calculated in part A, to help us because of that mathematical relationship of moles equals volume divided by the molar gas volume.

Now, the value we use for that molar gas volume has to match the units for our volume.

And because that was centimetres cubed, we used the value of 24,000.

So we now know that the number of moles of nitrogen in that volume was 3.

33.

I then need to find the mass, and for that, I need to know the relative formula mass of nitrogen using my periodic table, taking the moles and that relative mass, multiplying them together and to two significant figures, my final answer is 93 grammes of nitrogen was produced in this reaction.

Very, very well done if you managed to get that correct.

Superb work, guys.

I'm so impressed with it.

Wow, we have gone through a lot in today's lesson.

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

We started by reminding ourselves that the volume of one mole of any gas at room temperature and pressure, so that's 20 degrees Celsius and one atmosphere, is 24 decimeters cubed, or 24,000 centimetres cubed.

And that the amount of substance, the number of moles, can be calculated using multiple equations.

But the choice of equation for calculating moles depends on the context of the question or the calculation you've been asked to complete.

But chemists can use a combination of these mass and volume calculations and a balanced symbol equation to calculate an unknown by that link, that understanding of the number of particles being such an important part of understanding how chemical reactions take place in quantifying those chemical reactions.

And that's very useful for things like calculating one's carbon footprint, how we might offset that carbon footprint, and for other things as well.

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

I certainly had a good time learning with you, And I hope to see you again soon.

Bye for now.