Lesson video

Hello, my name's Mrs. Niven, and today we'll be looking at how we can determine a chemical equation from given values as part of our unit on calculations involving masses.

What we go through in today's lesson, you may have some experience of from your previous learning, but what today's lesson is all about is being able to link up the values that we are able to collect from a practical experiment and how we can use those to communicate information about the reaction that's taken place to other people and also give us some insight about the particles that are involved in that reaction.

So by the end of today's lesson, you should hopefully feel more comfortable being able to deduce a balanced symbol equation from the masses that you have about the reactants or products for a given reaction.

Now, I'll be referring to some keywords throughout this lesson, and they include ratio, stoichiometry, balanced symbol equation, and mole.

Now, the definitions for these keywords are provided in sentence form on the next slide, and you may wish to pause the video here so you can have a quick read through or maybe jot down a note of what each means so that you can refer back to it later on in this lesson or later on in your learning.

So today's lesson is broken into three sections.

First we'll look at general chemical ratios before moving on to looking at how we can calculate the reacting molar ratios for a reaction and finally how we can use those ratios to then balance a chemical equation.

So let's get started by looking at what we mean by chemical ratios.

Now, whenever we use that term ratio, what we're talking about is a relationship between two or more quantities.

So it doesn't matter if you're talking about a ratio in chemistry, in building or in maths.

What a ratio does is it allows us to determine the size of one quantity in relation to another, and I'll give you an example.

Whenever somebody makes concrete, standard concrete has a ratio of one part cement, two parts sand, four part aggregate, which is like pebbles or stones, and then all of those different parts are mixed together with some water simply to bind them, but it's a standard ratio, and that ratio could change for concrete depending on its use later on, but it's a ratio.

It's telling us the quantity of one material compared to another for that final product.

Now, ratios exist in chemistry in a variety of ways.

For instance, there are ratios of elements within the substances that we're using for different reactions, and those ratios show up as the subscript numbers in the chemical formula.

For example, here in H2O, those subscript numbers would be a 2 and a 1, but what do those values actually mean? Well, it means I have two atoms of hydrogen for every one atom of oxygen, and if I wanted to represent one molecule of water then, I would need to show two hydrogens, shown here as the white balls, and one oxygen, shown here with one red.

And the thing is, if I have multiple molecules of a substance present, the ratio of the atoms within that substance doesn't change.

So if I have multiple molecules of water, the ratio of the atoms within each molecule of water stays the same, two hydrogens for every one oxygen.

And the same is true if we look at larger substances.

So if we're looking at a giant covalent substance or even an ionic substance, the chemical formula for these substances indicate those ratios of atoms or ions in a formula unit rather than the molecule.

So no matter the size of the structure, the ratio of the atoms or the ions in those substances stays constant.

So if I look at this example of silicon and oxygen that have bonded together in this giant covalent structure and I take a closer look at a part of it, I can see that for every two silicon atoms, there are four oxygen atoms bonded around it.

So I could say that its formula was Si2O4, but we simplify it down for its chemical formula, and so this is SiO2, or silicon dioxide.

If I look at an ionic substance, here I've got sodium ions bonded to chloride ions, and I click a really close look at the number of ions in these, I can see that for each formula unit, I have one sodium ion for every one chloride ion.

So its ratio is one to one, and so its chemical formula will be NaCl for sodium chloride.

Let's stop here for a quick check.

True or false? The ratio of atoms or ions in a substance changes as the number of particles increases.

Well done if you said false, but which of these statements best justifies your answer? Well done if you said a, the ratio of atoms and ions in a substance is indicated by its chemical formula.

It does not change if the size of the structure increases.

Well done if you got those correct, guys.

Great start.

So while ratios exist between the elements within a substance, ratios also exist between substances, and those ratios are indicated by the stoichiometry or that molar ratio that we can see in a chemical equation, and what that indicates is that regardless of the amount that you're actually using of each substance, that molar ratio for a chemical reaction doesn't change.

So let's look at an example.

I have a reaction here, and the stoichiometry, or that molar ratio between substances, has been circled.

So these are the coefficients for my balanced symbol equation, those large numbers before each chemical formula, and it tells me, regardless of how much I'm starting from with each of my substances, the molar ratio or the relationship between those substances in this chemical reaction.

So let's say, for instance, that rather than having two moles of my aluminium, I only have 0.

5.

I can use this molar ratio to find out what that change in molar ratio will be using my balanced symbol equation and my understanding of ratios.

To get from 2 to 0.

5, I need to divide by 4, which means I'd need to divide all the other coefficients by 4 as well to get that new molar ratio.

Similarly, let's say now rather than 2 aluminium, I have 10 moles of aluminium.

Well, to get from 2 to 10, I need to multiply by 5, and therefore I need to multiply all the other coefficients by 5 as well to get that new molar ratio, and it's all because of that relationship between the substances in this chemical equation.

Let's stop here for another quick check.

Which chemical equation or equations shows a similar molar ratio to the reaction shown? Now, you may wish to pause the video here so you can take a closer look at each of these equations before making your final choice.

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

Okay, let's see how you got on.

Well done if you chose b and d as showing a similar molar ratio to the reaction shown.

B, all you needed to do is multiply every one of those coefficients by 2, and for d, all of those coefficients were divided by 2.

A and c do not work because not all of those coefficients were changed.

Only one or two of them have been.

So you've gotta be very careful in identifying those molar ratio reactions, but very well done if you managed to get at least one of those correct, and incredibly well done if you managed to find both of those correct answers.

Great job, guys.

Okay, time for our first task.

So what I'd like you to do to start with is to match each of the chemical formulas that are shown to the correct diagram.

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

Okay, let's see how you got on.

So C6H12O6, I'm looking for something that has at least three different atoms in it, and that's gonna be the third diagram.

Na2O is going to be the last diagram or the one on the far right.

KI, for me, I have two different atoms in there, but it's a metal and a non-metal, which means I should be looking for an ionic structure, and that's going to be the first diagram.

So C4H10 then is the second diagram from the left.

Well done if you managed to get those correct.

Great job, guys.

For the next part of this task, I'd like you to use the stoichiometry or those molar ratios to determine the new molar quantities for each of the parts a through d shown below the chemical equation.

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

Okay, let's see how you got on.

So the first thing you needed to do was identify what the stoichiometry was of the original equation, and then for a, you needed to multiply by 2 and do that for the rest of those values, so 4, 2 and 12.

For b, you were dividing by 2 and giving you values of 1.

5, 0.

5 and 3.

For c, you were dividing by 6 and getting values of 0.

5, 0.

33 and 0.

17, sorry, and for d, which was particularly tricky, you needed to multiply by 3 and divide by 4.

So that's essentially multiplying by 3/4, and that gives you then final values of 2.

25, 1.

5 and 4.

5.

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

Great job.

For the next part of this task then, what I'd like you to do is to consider the reaction of methane burning in oxygen, and when it does so, it produces, when it completely combusts, carbon dioxide and water.

So I'd like you to decide which of the four chemical equations that are shown is correct and also identify what is incorrect about the other chemical equations.

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

Okay, let's see how you got on.

Hopefully you have chosen b as being the correct chemical equation.

Very well done if you managed to get that.

The other three, then you needed to identify what was incorrect about them, and for a, carbon dioxide had the incorrect ratio.

It should be CO2.

For c, it also had an incorrect ratio, this time for water, which should be H2O, and for d then, it had the incorrect stoichiometry.

So the molar ratios were incorrect.

It should have been 2O2.

Well done if you managed to get those correct, guys.

What a fantastic start to today's lesson.

Keep it up.

Now that we're feeling a little more comfortable talking about chemical ratios in general, let's look at how we can calculate then reacting molar ratios.

Now, we said earlier that the stoichiometry of a reaction is shown by a balanced symbol equation and that the coefficients in that symbol equation represents the molar ratio of the substances involved in it.

Now, we need to remember that one mole is equal to or represents 6.

02 times 10 to the 23 particles of any substance.

So if we look at this particular reaction of creating ammonia, we could read it as one mole of nitrogen molecules reacts with three moles of hydrogen molecules to form two moles of ammonia molecules.

Now, the thing is sometimes we don't actually know what the reacting ratio of some chemicals are at the start, but we can at least use the mass of the reactants that were used to calculate that reacting molar ratio, and these calculations then are going to use both that mathematical relationship we've been using throughout the entire topic of mass in grammes is equal to the relative mass times moles and also our understanding of ratios.

Now, in order to do these calculations, you will need a calculator, a periodic table, you'll need the masses for the reactants in those reactions, and you'll also need a step-by-step strategy.

Now, you'll be provided the masses and the strategy in a moment, but if you don't have a calculator or periodic table to hand, I recommend you should pause the video and come back when you have those before we continue.

So you have your periodic table.

You have your calculator.

You'll be provided the masses of your reactants, and what I'm going to do now then is use an example to talk you through a strategy that you could use in order to calculate those reacting molar ratios.

So I want to know what the reacting ratio is when 0.

892 grammes of sulphur dioxide reacts with 0.

224 grammes of oxygen to form this sulphur trioxide.

So the first thing I'm going to do is I want to write down the chemical formula for each reactant.

So I find my reactants, write down their names.

Below them, I'm going to put the formula.

So sulphur dioxide is SO2 and oxygen is O2.

It's the formula for the element, not its symbols.

So it's important that oxygen is O2 and not just O because it is a diatomic element.

Then below those reactants, I'm going to create a calculation grid that includes some important information that I'm going to need for these calculations, and that's going to be the mass in grammes, their relative masses, the number of moles within that mass in grammes, the unitary ratio, and then finally its molar ratio.

So you should have something that looks a little bit like this once your calculation grid has been created.

The next thing you'll need to do then is to populate that grid with as much information as you can.

So the first thing is going to be the mass in grammes that was given to you in the question.

So simply copy those into your calculation grid.

Then you also need to be able to find the relative mass.

Now, you're just going to use a periodic table for this.

So once you calculate those, pop those into your grid as well.

The next thing you're going to do then is you're going to calculate the number of moles, so the number of particles of each substance in that mass sample, and you're going to use that equation, the relationship, anyway, of moles is equal to the mass in grammes divided by that relative mass.

So there's a reason why your grid was organised in this way.

The mass being above the relative mass means you're literally showing the numbers as you'd put them in your calculator to divide, and then the moles row is for you to put your answers in.

So once you've found the number of moles or number of particles that are present in each sample of your reactants, you need to find the unitary ratio, and to do that, you're going to look at the values in your moles row and decide which is the smallest value.

Once you've done that, you're going to divide all of the values in that row by that smallest value.

So for this example, my lowest moles value is 0.

007, and when I divide all of the values in that row by 0.

007, I get these values.

Now the key here is, in this step, you should have at least one value for your unitary ratio that comes out as 1.

If you don't, you've gone wrong somewhere, okay? So that's a quick and easy check to make sure you've created that unitary ratio correctly.

Finally then, you're going to determine that molar ratio.

Now, molar ratios should be whole numbers, and some of the unitary ratio values that you calculated could be rounded in order to achieve this.

So we look at our examples here.

We've got 1.

991, which is not a whole number, but it could be easily rounded to one, and the other value for oxygen is already a whole number.

So using that rounding ability, I could say the reacting ratio for this example of sulphur dioxide to oxygen is going to be a 2:1 ratio.

Let's go through another example, but this time I'm not gonna show you the different steps.

I'm just gonna talk you through them.

So first things first, I'm going to write the formula for my reactants and create my calculation grid below them.

I'm going to then populate that grid with as much information as I can, starting from the top down.

So the masses are coming straight from my question, and the relative masses then are coming from calculating using my periodic table.

Dividing those values, then I get the number of moles in each of the mass samples, and to find that unitary ratio, I need to identify the lowest moles value and divide all of them by that smallest number.

So here, my smallest number is for the magnesium chloride of 0.

01653, and that gives me a unitary ratio of 1 to 1.

996.

To find my molar ratio, remember, I'm looking for whole numbers, and some of those unitary ratio values can be rounded, and that's exactly what happens here for my sodium hydroxide.

For this then, I have a reacting ratio of magnesium chloride to sodium hydroxide of 1:2.

What I'd like you to do now then is to have a go at finding the reacting ratios for this reaction of propane reacting with oxygen, and as a little reminder, I have put up the calculation processing that I used in my example back up on the left-hand side to use as a guide.

So this is gonna take your periodic table, your calculator, a little bit of working out.

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

Okay, let's see how you got on.

So if you've done everything correctly, you should have a ratio of a 1:5, so 1 propane to 5 oxygen.

So if you didn't get that answer, I would recommend that you pause the video and compare your calculations to what has been shown for the working out, but if you did get that, fantastic job.

Really impressive, guys.

Just, yeah, keep it up.

Now, it's important to understand that sometimes that unitary ratio value is not close enough to whole numbers that it could actually be easily rounded, and in those situations, what you need to do is multiply accordingly in order to achieve a whole number in order to remove the decimal, and I always found it useful to have a little bit of a guide, and that's what this table represents.

If you have a decimal that ends in around about 0.

5, well, if we represented that decimal as a fraction, it would come up as about 1/2, and I know that if I wanna get a whole number from 1/2, I need to multiply by 2.

The same goes for the other decimals then.

Round about 0.

33 is about 1/3.

So I'd have to multiply by 3.

So if you can change your decimal into a fraction, the denominator of your fraction tells you what you need to multiply by in order to get a whole number ratio.

So let's look at an example.

If I have a unitary ratio of 1.

52 to 1, I can see that that 0.

52 is very close to around about 1/2.

So I need to multiply by 2, but the key here is that when you're multiplying, you're not just multiplying that one value.

You need to multiply all of those unitary ratio values by 2, and when we do that, we get values that are closer to whole numbers.

So this time, 3.

04 can very easily be rounded now to 3, and because I had to multiply all of my values by that value of 2 to remove the decimal, my overall final molar ratio would be 3:2.

Let's stop here for a quick check.

Which of these ratios do you think need to be multiplied accordingly in order to achieve a whole number ratio? And if you wanna challenge yourself a bit, what number would you multiply those by in order to achieve that whole number ratio? Well done if you chose a and d.

Both of those have decimals that cannot be easily rounded to whole numbers, so they'll need to be multiplied accordingly.

For a, you would've needed to multiply by 2 because 0.

45 is very close to 0.

5, and d, you would've needed to multiply by 4 because 0.

77 is very close to 0.

75.

Very, very well done If you managed to identify the two ratios that needed multiplying, and incredibly well done if you managed to choose the correct value to multiply those by.

Great job, guys.

Okay, let's go through another example.

So again, I need to identify my reactants and write their formula.

I then need to create my calculation grid and populate it with the mass from the question, and then I'm going to use my periodic table to write in the relative masses.

I divide those values now and I get the moles that are found in each sample of my reactants, and dividing by the lowest value, I get my unitary ratio.

For this unitary ratio though, it cannot be easily rounded, and therefore I need to multiply in order to remove that decimal, and that gives me then values that can more easily be rounded to whole numbers, giving me a final molar ratio of aluminium to iodide is a ratio of 2:3.

What I'd like you to do now then is to calculate the reacting ratio when 8.

7 grammes of gallium reacts with 3.

0 grammes of oxygen, and to remind you of the processing I used, I've put that up on the left-hand example that I've just shown.

So you'll need your calculator, periodic table, bit of time.

So 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've done your calculations correctly, you should've had a ratio of gallium to oxygen as being 4:3.

If you didn't get that value, make sure that you're pausing the video so you can double check your calculations, find out where you've gone wrong, and so going forward, you can correct any of your mistakes, but very well done if you got the correct reacting ratio here.

Great job, guys.

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

What I'd like you to do is to use the data that's been provided to calculate the reacting molar ratio in each instance.

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, you should've had a ratio of barium to chlorine as 1:2.

Now, as always, if you didn't get that answer, please do pause the video and double check your calculations to see if you could identify where you've gone wrong, but very well done if you got that first one correct.

For b then, you should've had a reacting ratio of potassium hydroxide to sulfuric acid as being a 2:1 ratio.

Very well done if you managed to get that correct.

And for c then, you should've had a final reacting ratio value of magnesium oxide to phosphoric acid as being a 3:2 ratio.

Great job, guys.

I'm really impressed with your perseverance here.

Keep it up.

Now we're feeling more comfortable being able to calculate the reacting molar ratios, let's look at how we can use those to balance a chemical equation.

Now, we said earlier that the reacting ratios that we did previously indicate then the relative amounts of each reactant that's required for a reaction to occur, and for this one then, if we looked at our example from earlier, sulphur dioxide to oxygen, we calculated it as being a 2:1 ratio.

Now, that essentially means that I'm going to need twice as much sulphur dioxide to react with oxygen in order to make my product.

Now, there are two ways that I can use this information in order to write a balanced equation for that reaction.

One is to simply use these as my coefficients and then write a balanced equation.

The other is a little bit more complicated but still quite easy.

If I know the mass of my products as well as the mass for my reactants, I could actually calculate the stoichiometry or those molar ratios for my products as well as my reactants in the same way that I calculated those reacting ratios in the previous part of this lesson, and that then allows a chemist to calculate the stoichiometry for the entire reaction rather than just the reactants, and then I can use the values from those calculations to write a balanced symbol equation from those known masses.

So overall, once I know the reacting ratios, I can write a balanced symbol equation.

So again, like we said, I'm gonna need twice as much sulphur dioxide to react with oxygen to make a product, and if you recall, that product was sulphur trioxide.

Now, I told you earlier that there were two ways that we could use this information to write a balanced symbol equation, and the first one was simply to balance that equation.

So I'm gonna go through that process first.

The first thing you want to do is write the chemical formula, including any of the reacting ratios they have for your reactants and your products.

Now, the reactants for this one is particularly easy 'cause we already had our sulphur dioxide and oxygen, and we already knew that that reacting ratio was 2:1.

Sulphur trioxide as a formula you could figure out is SO3, but we don't know the reacting ratio here, and we're gonna need to figure that out, and we do that simply by balancing the equation the way we do simply understanding the idea of conservation of atoms. So I'm going to separate my reactants from my products.

I'm going draw a little diagram to represent each of my substances.

So I've got my molecules of sulphur dioxide, and I've drawn two because I have a 2 for my coefficient, one of my oxygen molecules, and then sulphur trioxide as a molecule as well, and then keeping track of the amount of sulphur and oxygen on my reactants and products, so for my reactants, I have 2 sulphur and 6 oxygen atoms, and on the products side, I have one sulphur and three oxygen atoms. Now, these are clearly unbalanced, and in order to balance these numbers, I need to have another molecule of sulphur trioxide, and when I add that diagram to my picture here and then recalculate the number of sulphur and oxygen atoms, I can see it's now balanced.

So I count up the number of diagrams I have for my sulphur trioxide, which is two, and I have now my final balanced symbol equation of 2SO2 plus O2 makes 2SO3.

Now, the thing about that process where we find our reacting masses and then balance the equation using our conservation of atoms, we're using two different processes, and you kinda have to swap your brain a bit to think about how you're going to do this.

If you're given the mass for the products, you can actually use the one process to find those molar ratios for the entire reaction rather than just the reactants and then swapping processes to find that balanced symbol equation, and I'll show you what I mean.

In this example here, we have the same masses for our sulphur dioxide and oxygen as before, but this time I also have the mass for the sulphur trioxide.

So I'm gonna do the same as before, identify the substances in my reaction and write the chemical formula below them.

Below that, I'm going to create a calculation grid, and I'm going to populate it with the information I can, so the mass in grammes and then grab my periodic table and fill in the relative atomic masses.

I'm going to divide those values then to find the number of moles or the number of particles that are present in each one of my mass samples.

Find the smallest value of moles and divide all of them by that value.

So in this case, it's 0.

007 to give me my unitary ratio for all the substances.

I then can easily round my answer for the sulphur dioxide to 2, and I now have molar ratios of a 2:1:2.

Now, that's all well and good, but how do I then turn this into a balanced symbol equation? Well, all you need to do is identify your reactants from the question and write down those chemical formula.

Then you're going to do the same for your products and just draw that on the other side of your arrow.

Then you're going to go to your molar ratio row in your calculation grid and simply copy the values there in front of the correct substance that it represents, and that is your final balanced symbol equation.

So let's go through another example of how we can use molar ratios to determine the balanced equation for a reaction.

So same as before, I identify my substances, and I'm going to write down the chemical formula, being careful that chlorine is diatomic.

I'm gonna create my calculation grid and populate it with the mass from the question.

I'm going to then use my periodic table to fill in the relative mass and then divide those values in order to find the number of moles present in each sample.

I divide by the smallest value then to find my unitary ratio, and in this case, I can see that the chlorine value is not going to be easily rounded.

So it's very close to 0.

5 and therefore nearly 1/2.

So I know I need to multiply that and the rest of my unitary ratio values by 2, and when I do that, I get values that are far easier to round to a final answer, and that gives me then my final balanced symbol equation of 2Fe plus 3Cl2 makes 2FeCl3.

What I'd like you to do now then is to have a go as well.

I'd like you to use the molar ratios to determine the balanced symbol equation for when hydrogen peroxide decomposes to form water and oxygen.

So 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've done all your calculations correctly, you should have got a final answer of 2H2O2 makes 2H2O plus O2.

Very well done if you got that correct.

If not, definitely pause the video and double check your calculations to make sure that you have carried it out correctly, but very well done if you got that correct, guys.

Well done.

Time for the last task of today's lesson.

Now, you may or may not know, but a lot of the metals that we use in today's society are actually needing to be isolated from a compound using some chemical processing.

So what I'd like you to do is use molar ratios to determine the balanced symbol equation for each of the reactions that are described to isolate a metal.

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

Okay, let's see how you got on.

For letter a, trying to see about aluminium oxide decomposing, you should've had a final answer of 2Al2O3 makes 4Al plus 3O2.

So if you didn't get that answer, definitely pause the video and check your calculations.

Now, part b was a little bit trickier, however, if you managed to carry out your calculations correctly as well as identify the reactants and products correctly, you should have ended up with a final answer that was Fe2O3 plus 3CO makes 2Fe plus 3CO2.

So if you didn't get that answer, please do make sure you pause the video and double check your calculations versus what has been shown here, but very, very well done if you've managed to correctly determine that balanced symbol equation.

Great job, guys.

Now, for the next part of this task, I'd like you to help Alex.

He's used masses to calculate the molar ratios of a reaction that uses magnesium to extract titanium from titanium chloride, but when he's double checked his final chemical equation, he found out that the number of atoms was not conserved.

He's made an error somewhere.

Now, the calculations that he did are shown here, and what I'd like you to do is to identify and correct Alex's error within his calculation grid, and then using that, I'd like you to then write the correct balanced equation for this reaction.

Now, you may wish to discuss your ideas with the people nearest to you.

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 the first thing I'd do is go to Alex's final answer and try and work back from there.

How did he realise that there was a problem with the conservation of atoms? And what I can see is that the number of magnesium and the number of chlorine do not match up, and so I'm going to go to the only chemical that has both of those atoms in it, and that's the magnesium chloride that he's written, and what I can see immediately is that he's actually written the incorrect formula for magnesium chloride.

It should be MgCl2, and there's a knock-on effect for having the incorrect formula here.

His relative mass for that substance is going to be incorrect and that has a knock-on effect for his calculations.

So that should have been 95, not 59.

5, if the formula for magnesium chloride had been written correctly.

So now that we've fixed Alex's errors from the formula going downwards, we have a final fixed balanced symbol equation that shows conservation of atoms of TiCl4 plus 2Mg makes Ti plus 2MgCl2.

So well done if you managed to identify his errors and correct them to get the correct balanced equation.

Fantastic work today.

Wow, we have done a lot in today's lesson, and you guys have been absolute superstars as we go through this, but let's take a moment now and just summarise what we've learned.

Well, we learned that there are ratios throughout chemistry and that the subscript values that we find in chemical formula indicate the ratio of the atoms or ions in a molecule or a formula unit, for instance, in our individual molecules or in a larger structure.

We also found that there are ratios in the stoichiometry of a chemical reaction, and that molar ratio is shown in the balanced symbol equation using those coefficients, so the larger numbers that come in front of the chemical formula, and that we can find those stoichiometric values, those coefficients using known masses or ones that are obtained by an experiment and using our understanding of ratios and that relationship of the mass in grammes is equal to the relative mass times the number of moles.

So I'm just really impressed with the work you guys have done today.

Fantastic job.

I hope you had a good time learning with me.

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

Bye for now.