Hello, my name's Mrs. Niven, 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 balanced 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 the relationships regarding mass, volume and moles, and then we'll look at how we can use those relationships to calculate an unknown gas volume or mass.

So let's get started by reminding ourselves about 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 mass in grammes is equal to the relative mass times moles.

Now the mass in 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 of 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 take a closer look at how we can actually use that mathematical relationship.

I've been given the question here.

I want to know how many moles there are in a 2.

30 kilogramme sample of aluminium oxide, which has the formula Al2O3, and I've been instructed to give my answer to three significant figures.

Now, I'm going to use that relationship of moles is equal to mass in grammes divided by the relative mass.

So the first thing I need to do is double check the mass.

Now when I look here, I can see though that the mass I've been provided is in kilogrammes, but the mass that I need for my mathematical relationship is in grammes.

So I need to convert 2.

30 kilogrammes into grammes, and I do that by multiplying by a thousand.

So my mass in grammes is sorted.

The relative mass then is related to the individual particle that I am measuring out.

So I need to look at the formula for that substance.

And here then what I'm going to do is take the relative mass of each individual atom, multiply it by the number of those atoms and add those values together.

And for aluminium oxide, we get a value of 102 for the relative mass.

Taking now my relative mass and the mass in grammes dividing those values, I get a number of moles of 22.

549.

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

So rounding it, my final answer is 22.

5 moles of aluminium oxide.

What I'd like you to do now then is to calculate the number of moles of carbon dioxide that are released during a train journey from London to Edinburgh.

You can use the calculations that are shown on the left as a bit of a guide, and I'd like you to give your answer to three significant figures.

Now, you're going to need to calculator, periodic table, pencil and paper in order to be able to calculate this out.

So 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 carried out your calculations correctly, you should have got a final answer of 566 moles of carbon dioxide that are formed in a train journey from London to Edinburgh.

Now, if you didn't get that answer, please do pause the video and compare your calculations to what are shown to see where you may have gone wrong so you can make your adjustments going forward, but incredibly well done if you've got that correct, guys, great, great job.

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 centimetres 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 decimeters 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 decimeters 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 manage 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.

So we have our two equations that we've been using, but if we take a closer look at them, we can see that both contain that variable of moles.

Now, if we go back to that example we looked at earlier, a mature maple tree can absorb 5,440 decimeters cubed of carbon dioxide each year.

That was a volume.

And we calculated that that was equivalent to 230 moles of carbon dioxide per year.

But trying to calculate one's carbon footprint or how you can offset one's carbon footprint is usually given in calculations of grammes or kilogrammes of carbon dioxide.

So I want to know what mass of carbon dioxide then does a mature maple tree absorb each year? Well, I already have the moles of carbon dioxide that are absorbed from my previous calculation.

What I need now then is simply the relative mass of the carbon dioxide.

And for this, I need my periodic table.

I simply adding up the relative mass for each of the atoms, and that gives me a value of 44.

If I take, then the mass, grammes, is equal to 230 times 44, I get a final value of 10,120 grammes of carbon dioxide is absorbed every year by a mature maple tree.

Let's look at how I can bring those two equations together.

I found that a flight between London and Edinburgh releases a whopping 105,000 decimeters cubed of carbon dioxide, but I want to know what mass of carbon dioxide is released during that flight.

Now, the first thing I need to do then is to find the number of moles that is present in that volume of gas.

So I've been given the volume of 105,000, but the molar gas volume has not been provided.

But the clue, if we remember, is in the units for our volume, because it's decimeters cubed, I'm going to use a value of 24 for my molar gas volume.

Using those values then of the volume and molar gas volume, I can calculate that there are 4,375 moles of carbon dioxide released during this flight.

I need to now convert that moles into a mass.

So I'm going to move to the mass in grammes is equal to the moles times relative mass equation.

Now I have the moles from my previous calculation, but now I need the relative mass.

So I need to look at the gas that I am concerned about.

And for here it's carbon dioxide.

Using my periodic table, I find that a relative mass is 44.

So I now have the moles and the relative mass, so I pop those in my calculator and I get 1,902, sorry, 192,500 grammes of carbon dioxide is produced.

But I've been instructed to give my answer in kilogrammes.

So I need to take that value and divide it by a thousand to get a final answer of about 190 kilogrammes of carbon dioxide is released during a flight between London and Edinburgh.

What I'd like you to do now then is to calculate for me 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 a calculator, pen, pencil and periodic table, 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 got a final answer of 35 kilogrammes of carbon dioxide are produced during that train journey.

If you didn't get that answer, make sure you pause the video so you can go back through your calculations to see where you may have gone wrong and hopefully make some adjustments going forward.

But incredibly well done if you have managed to calculate that correctly, guys, great job.

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.

Well, we can see from part A that we have a sample that is in grammes.

That means I need to use my equation of moles is equal to the mass in grammes divided by the relative mass.

And for glucose that relative mass is calculated as 180.

So if I take the mass 655 and divide it by 180, I get a value to three significant figures of 3.

64 moles of glucose in that sample.

For B, following that same process, I should have a value of 0.

175 moles of cuprite Cu2O in that 25.

0 gramme sample.

For C, I can see that I still have a mass, however my mass has been given in kilogrammes.

So the first thing I need to do is convert that mass from kilogrammes into grammes and find then the relative for my sodium chloride.

And when I do that, I get a value of 5.

98 moles of sodium chloride is in that sample.

For D, I'm gonna follow the same processing.

I need to change my kilogrammes into grammes, find the relative formula mass.

And the answer then to three significant figures is 54.

4 moles of hematite in that sample.

Well done if you got those correct.

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 find the mass that is in each description.

And because I have a gas that has been described in each one, I need two equations.

Both the moles equals volume divided by molar gas volume, and the mass in grammes is equal to moles times relative mass.

So the first step is to find the number of moles that are present in that volume.

And I've used 24 for this calculation because the units for my volume were in decimeters cubed.

So I now know the number of moles that were present in my gas.

Now, my gas was described as steam, but steam is really simply water in the gas state.

So finding the relative mass for water, I find that it is 18.

Using now the number of moles and the relative formula mass, I find to two significant figures that 2.

8 grammes of steam are present in a 3.

68 decimeters cubed sample.

Now, if we apply that same process to the gas described in letter B, we should have a final answer of 1.

1 grammes of hydrogen in a 12.

7 decimeters cubed sample.

And for letter C, if you've calculated that correctly, you should have got a final answer that was 1.

9 times 10 to the three grammes of butane.

Now, there are two places where people may have got caught out on this particular question.

Firstly, the answer was given in standard form to ensure that the answer was given to two significant figures.

And if you didn't get an answer that was similar to that, it's possible that you used the incorrect molar gas volume.

We used 24,000 in this particular calculation because the units for the volume were given was centimetres cubed and therefore our molar gas volume choice of value needed to match those units.

So if you didn't get that answer, please do pause the video and go back through to check your calculations.

But if you did, very, very well done.

And finally then for D, we should have had a final answer that was 2.

3 times 10 to the three grammes of hydrogen sulphide.

So incredibly well done if you manage to get these correct guys, great job.

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

So the first step I need to do is identify my known and unknown substance in my balanced 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 all set up.

I know I'm gonna have to do some maths in it, but I need some values to populate it with before I can do those calculations.

Those values are gonna come from a variety of places, and we're gonna start with information from the question.

I have 20 grammes of hydrogen peroxide, so I'm gonna pop that in the mass in grammes row.

My reaction is taking place at RTP, which is room temperature and pressure.

So I'm gonna get the molar gas volume from that.

Now, because I've not been given a unit for the volume, I'm gonna keep the easiest value I can use, and that's simply 24, which tells me my volume is going to be in decimeters cubed at the end.

I'm also gonna use a periodic table.

I'm going to use that to fill in the row for my relative mass.

I'm also going to use then that balanced symbol equation.

Remember, the coefficients in a balanced symbol equation are telling me the molar ratio of my substances from my known to unknown.

So I'm going to take those coefficients and simply copy them into the relevant rows for my ratios, from my known and unknown.

That leads me then with three boxes that need filling, and that's gonna happen using those calculations where we say we are going to divide down the known substance and multiply up the unknown.

Now setting up your calculation grid like this and then dividing down the known substance, what you're doing in the first step is actually calculating the number of moles of that known substance using that relationship of the moles is equal to the mass in grammes divided by that relative mass.

And that tells you then the number of moles or the number of particles that are present in that 20 gramme sample of hydrogen peroxide.

The next step then is to calculate the number of moles of your unknown substance.

And we do this by continuing to divide as you go down the known column and then multiply as you move across to the unknown column.

And your answer then is the number of moles that we have for our unknown substance.

So at this point, you have just one box left that needs filling in through your calculations, and that is essentially to answer the question.

And here we were asked what volume of oxygen forms in this reaction? And because I used that relationship of volume equals moles times the molar gas volume to fill in the labels for my calculation grid.

I have volume at the top and I know that all I need to do now then, is multiply up the rest of that unknown column and I get a value then of 7.

059 for my volume.

But units, if you remember, I used that molar gas volume of 24 to keep the maths simple.

24 is the molar gas volume in decimeters cubed.

So my final answer to two significant figures is 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'll simply talk you through it.

In this example, I want to know what mass of calcium carbonate has 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.

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

Now, I've been asked to find the mass of the calcium carbonate, so that's my unknown substance, and I'm going to put a question mark above the formula for that.

I've been given information about carbon dioxide, so that's my known substance, and I put a tick above that.

The next thing I'm going to do is to create my calculation grid.

And the third and fourth rows will be labelled as moles and ratio respectively.

I then need to fill in the rest of my labels.

So because I'm asked the mass of calcium carbonate, I'm going to need to know the mass and the relative mass for that substance.

And because I have a volume that's been given to me for the carbon dioxide, I'm going to need to use the volume and molar gas volume in that calculation grid label.

I now need to populate that grid.

So the information I've been given is 65,000 centimetres cubed, so the volume then is 65,000.

And because the units are centimetres cubed at RTP, the molar gas volume value I need to use is 24,000.

Again, because those units are centimetres cubed.

I then move to using the periodic table and the relative mass for my calcium carbonate then is 100.

And then I need to use my balanced symbol equation to fill in the ratios, which for here is one each.

Now I'm going to do my calculations, which starts by dividing down the known column.

So I go back to the substance that has a tick above it, and I divide down that column to find the number of moles that were present in 65,000 centimetres cubed of carbon dioxide.

Continuing to divide down that column and then multiplying across, I find that the number of moles of calcium carbonate is 2.

708.

To then finally answer to that question, I'm going to continue multiplying up that unknown column, and I now have a mass of 270.

8.

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

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

And I've outlined a little bit more detail, the calculations that I followed on the left hand side that helped guide you in that calculation.

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

Okay, let's see how you got on.

If you've carried out these calculations correctly, you should have a final answer of 51 grammes of potassium chloride decomposed.

Now, if you didn't get that answer, do pause the video and check the values that are in our calculation grid to see if you have carried them out correctly.

And if not, maybe talk with someone nearest to you to see if you can identify where you've gone wrong so you can avoid it in future, but incredibly well done if you got that correct, 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 your 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.

If you remember, by conservation of mass, the mass of my reactants has to equal the mass of my products.

So by subtracting the values I've been provided in A, I find that 2.

2 grammes of carbon dioxide was produced in this reaction.

For part B, I was asked to calculate the number of moles of carbon dioxide which was produced, and I'm going to need to use that mathematical relationship of moles is equal to the mass in grammes divided by the relative formula mass.

Well, the mass in grammes was actually calculated in part A, so I can use that value along with the relative formula mass, which is calculated using a periodic table, pop those values into my mathematical relationship, and I find that 0.

050 moles of carbon dioxide are produced.

For part C, I've been asked to find the volume in decimeters cubed of carbon dioxide.

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

Well, I calculated the number of moles in part B, so I can use that value going forward.

And the volume that I'm asked is decimeters cubed, so my molar gas volume value that I use is 24, giving me a final answer of 1.

2 decimeters cubed of carbon dioxide produced.

So very, very well done if you got these correct.

One thing I want to point out here is that whenever you get a question where there's parts A, B, and C, don't be afraid to go back and use any of those values going forward in your subsequent questions.

More often than not, it's expected for you to be able to do that.

So do feel free to go back in previous parts of the question and see if there's anything that you can use to answer 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 azide 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 for the first part, you were asked to calculate the volume of nitrogen in centimetres cubed, and if you carried out your calculation grid correctly, you should have a final answer to two significant figures of 8.

0 times 10 to the four centimetres cubed of nitrogen produced.

If you didn't get that answer, please do pause the video and check your calculation grid to see if you can identify where you may have gone wrong.

But very well done, if you got that correct.

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

Now, if you recall in part A, we found that 8.

0 times 10 to the four centimetres cubed of nitrogen was produced in that decomposition reaction.

So using that mathematical relationship of moles equals volume divided by the molar gas volume, we find that the number of moles that that represents is 3.

33 moles of nitrogen.

To calculate the mass then, we need to move to that mass in grammes equals moles times relative mass relationship, finding the relative formula mass for nitrogen and then we get a final answer of 93 grammes of nitrogen that was produced in this particular reaction.

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

Fantastic work, guys.

Great job.

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

So let's just take a moment to summarise what we've managed to go through.

Well, 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 or one atmosphere, is 24 decimeters cube or 24,000 centimetres cubed.

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

But the choice of equation for calculating those moles depends on the context of the problem or the calculation you've been asked to carry out.

But if we use a combination of those mass to volume calculations and a balanced symbol equation, chemists can calculate at unknown value.

And that's really useful for things like calculating one's carbon footprint, how to offset that carbon footprint, or how we might be able to ensure that reactions take place safely or even on an industrial scale.

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

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

Bye for now.