Lesson video

Hello, my name's Mrs. Navin and today we're going to be talking about how to calculate empirical formula as part of our unit on calculations involving masses.

Now you may have some experience of this from your previous learning, but what we do in today's lesson will help us to better understand the particles that we're talking about during a chemical reaction, and it also gives us another weapon in our arsenal of mathematical processing that helps us later on when we come to analyse different substances.

So by the end of today's lesson, you should be able to determine the empirical formulae of simple compounds using either reactant masses or their percentage composition.

Throughout the lesson, I'll be referring to some keywords and these include molecular formula, empirical formula, and relative formula mass.

Now the definitions for these keywords are given in sentence form on the next slide, and you may wish to pause the video here so you can jot down a quick definition for reference later on in the lesson or later on in your learning.

So today's lesson is broken into two parts.

Firstly, we'll look at empirical formula and then we'll move on to look at how we can determine the molecular formula.

So let's get started at defining what we mean by the term empirical formula.

All around us people have worked to simplify things and it could be something as simple as thinking about the way that different items are organised or even how information is presented.

For instance, here, by changing the fraction or by simplifying this equation.

Believe it or not, substances' formulas can be simplified as well.

Now we all know that substances have their own chemical formula.

If we look at a simple molecular substances' chemical formula, it's actually its molecular formula.

It's telling us the elements and the number of atoms of each element in it.

So if we look at this example of ethane, it's molecular formula shows us that we have two carbon atoms and six hydrogen atoms. So its molecular formula then is going to be C2H6.

The chemical formula for an ionic substance on the other hand, will indicate the ratio of the positive and negative ions that are present in its formula unit.

So if we look at this example of sodium chloride, its formula unit looks like this where for every sodium ion that is present, there is one chloride ion, and we could represent that with this NaCL, meaning that one positive sodium ion will bond with one chloride negative ion.

And then we have the chemical formula of NaCL giving us that ratio of positive to negative ions in that formula unit.

An empirical formula on the other hand is the simplest whole number ratio of atoms or ions in a substance.

And all substances have an empirical formula.

For instance, for water, it's its molecular formula.

We cannot reduce this formula any further.

So its empirical formula is also its molecular formula, H2O.

If we look at ethane, its molecular formula before we said with C2H6, but we had actually reduced that down to show the simplest whole number ratio of those atoms as CH3.

And if we look at something larger like silicon dioxide, we can actually show the ratio of the atoms in this macromolecular or giant covalent structure as SiO2.

So there are several ways in which we can determine a substance's empirical formula.

For instance, if we recall that the charges of the ions in an ionic substance must add up to equal zero so that the formula unit is uncharged overall, we could determine the empirical formula by looking purely at the charges of the ions in an ionic substance.

For instance, sodium chloride, it has a chloride ion with a one minus charge and a sodium ion with a one plus charge.

If we add those together in a one-to-one ratio, we'd have one minus one plus.

So we'd only need one of each ion for there to be an overall uncharged formula unit.

So NaCL is our chemical formula.

If we look at a different ionic substance, this time we have an oxide ion with a two minus charge and the sodium ion with a one plus charge.

In order for our formula unit to have an uncharged overall, we need to make sure that we have two positive charges to balance out that two minus charge of the oxide ion.

As a result, then the chemical formula for this substance of sodium oxide will be Na2O.

So two of those positive sodium one plus ions for every one of the two minus oxide ion.

What this means then is that the empirical formula for an ionic substance is gonna be the same as its chemical formula.

Let's stop here for a quick check.

What do you think is the empirical formula for the substance shown? Well done if you said D, this will be MgH2.

So we'd need two of the hydride ions to balance out the two plus charge of that magnesium ion so that we have an overall charge of zero for that formula unit.

Well done if you managed to get that correct.

Great start guys.

In some cases you may be provided with a model of the substance, in which case we could actually count up the number of atoms of each element and then determine a ratio from those values.

For instance, in glucose we can see that there are six carbon atoms, six oxygen atoms, and 12 hydrogen atoms. In order to find the empirical formula, all we'd need to do is be able to find the highest common factor of all these values.

So what's the largest number that we could divide all of these numbers by? We can do that by six.

So if we divide these numbers by six, carbon changes to one, oxygen changes to one and hydrogen changes to two, which means the empirical formula for glucose could be written as COH2.

So that's showing the simplest ratio of atoms of each element in this particular substance.

Another way to determine the empirical formula of a substance is by calculating it either from the percentage composition of each element that it's composed of, or using the reactant masses of the elements it's composed of.

Now the mathematical processing, so the steps involved in those calculations is exactly the same regardless of whether or not you have been given percentage composition or reactant masses.

Now the first step in this calculation is to divide each of the percentage compositions or the reactant mass by the relative atomic mass of that particular element, and you'll end up with some answers.

Step two then is to divide each of those answers by the smallest answer you got from step one, and the reason is you will form then a unitary ratio in that process, okay? So you'll have at least one answer that has a value of one.

Some of the other values may be very close to a whole number or very easily rounded to it, and that's good because remember, an empirical formula is the simplest ratio of atoms of each element in our substance.

Let's look at an example.

I'd like to know what the empirical formula is for a substance that contains 71.

4% calcium and 28.

6% oxygen.

So the first thing I'm going to do is set up a little table in order to keep track of my calculations and we'll see that I've got a vertical column set up for calcium and a vertical column set up for oxygen.

And the far left column then is outlining is the information that I need for these calculations or the step that I'm going to be carrying out.

So the first thing I'm going to do is simply copy down those percentage compositions for each element under the appropriate element.

Then I'm going to use my periodic table and copy over the relative atomic mass for each of these elements.

So for calcium it's 40, and for oxygen it's 16.

What I'm going to do then is to divide those and I've put them in this order mass first and then the relative atomic mass second because that's how I'm dividing them.

71.

4 divided by 40.

That gives me an answer then of 1.

785 for the oxygen then.

So 28.

6 divided by 16 will give me an answer of 1.

7875.

Between these two then, I need to determine which is the lower answer, okay? And for this one, it's calcium.

What I'm going to do then is to take that value and divide both of my answers by the value for calcium.

And when I do so, I get a one for calcium and a 1.

0014 for oxygen.

I need to decide then do I have whole numbers? Well, for calcium I do, but oxygen isn't a whole number, okay? But it's very close to one and can therefore be rounded.

So that tells me then I have a ratio of one calcium to one oxygen atom in the empirical formula for this substance.

Now I could write that empirical formula as Ca1O1, but we don't really write ones in a formula.

So the final answer for this will be CaO.

What I'd like you to do now then is to determine the empirical formula of a substance that contains 20.

2% aluminium and 79.

8% chlorine.

What you are going to need then is a calculator and a periodic table.

I might recommend you setting out a table very similar to what I have shown here on the left, just as a guide, but pause the video here and come back when you're ready to check the empirical formula for this substance.

Okay, let's see how you got on.

Well, if you've done all your calculations correctly, you should have got a final empirical formula of AlCl3.

If you're not sure how to do that, I have the processing laid out here.

You may wish to pause the video and just check your answers quickly against that to see whether or not you agree or identify any errors in your working out.

But incredibly well done if you managed to get that correct.

Great job, guys.

Now, sometimes in our processing we get values for that unitary ratio that aren't whole numbers or aren't close to whole numbers.

And in those cases, we're gonna need to multiply accordingly to achieve a whole number ratio to get rid of that decimal.

So for instance, if we have a unitary ratio and it ends in roundabout, 0.

5, that's equivalent to roundabout one half.

So to remove the decimal, both of the values in my unitary ratio are gonna need to be multiplied by two, essentially doubling those values and achieving whole numbers.

So I've got an example here, if my unitary ratio was 1.

52 to one, in order to remove that decimal, I'm gonna have to multiply by two for both of those.

And when I do that, I get a value now of 3.

04 to two.

Well, 3.

04 could easily be rounded at this point.

So my whole number ratio now will be three to two.

Let's stop here for a quick check.

Which ratios listed below do you think need to be multiplied accordingly to achieve a whole number ratio? If you wanna challenge yourself a little bit, what number would you multiply these values by to achieve that whole number ratio? Well done if you chose A and D, they definitely do need to be multiplied to get that whole number ratio.

B already is, you don't need to do anything with that and C could be easily rounded, but A and D will need to be multiplied by two because both of those decimal places are round about 0.

5, and therefore would need to be multiplied by two to get that whole number ratio.

So well done if you managed to suggest at least one of the ratios that needed to be multiplied and very well done if you managed to get both of them.

Great job.

Okay, let's do another example, but this time we're going to use reactant masses.

I'd like to know the empirical formula for a substance that is composed from 0.

70 grammes of iron and 0.

30 grammes of oxygen.

So I'm going to set up my calculation grid similar as before and like before, I'm simply going to copy over then the reactant masses from the question.

Then I'm going to grab my periodic table and copy over the relative atomic masses for each of those elements.

When I divide those values then, I get 0.

0125 for iron and 0.

01875 for oxygen.

Then I need to make my unitary ratio.

So I divide by the lowest value and I get iron comes out at one and oxygen is at 1.

5.

Now, whilst iron is a whole number, oxygen is not and I need to be able to remove that decimal.

In order to do that, I need to div, sorry, multiply both of those values by two.

And when I do that, iron now is two and oxygen is three, which gives me an empirical formula for this substance of FE2O3.

What I'd like you to do now then is to find the empirical formula for a substance that's composed from 0.

529 grammes of aluminium and 0.

471 grammes of oxygen.

You can use the guide calculation grid that I've shown you here on the left, maybe work with the people nearest to you.

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

Okay, let's see how you got on.

If you've done all your calculations correctly, you should have got a final answer of Al2O3.

If you didn't get that, you may wish to pause the video here and go back through to check your working out similar to what I've done to see if we can find out where you've gone wrong and fix that as we go forward.

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

Time for the first task in today's lesson.

For this first part, what I'd like you to do is to match each term to the correct description.

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

Okay, let's see how you got on.

So molecular formula best fits the bottom description, which is a chemical formula for a simple molecular substance, and molecular being that keyword that links the two.

RAM or AR is an element's relative atomic mass.

So the A being so significant here for atomic and it's all found on the periodic table.

The empirical formula then is the simplest whole number ratio of elements in a substance.

So well done if you managed to match those up correctly.

Great start guys.

For this next part, I'd like you to determine the empirical formula for each substance.

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

Okay, let's see how you got on.

So for substance A, what you could have done is counted up each of the atoms for each element.

And when you do that, you get a molecular formula of C4H8O2, but the empirical formula is the simplest whole number ratio.

So the final answer should be C2H4O.

For substance B, you needed to identify the ion charges for the lithium ion and the oxide ion, first of all.

So the lithium is a one plus and the oxygen is a two minus.

So we needed to have these ion charges add up to have an overall charge of zero for the formula unit.

And when you do that, you get an empirical formula of Li2O.

So well done if you managed to get those correct.

For this next part, what I'd like you to do is use the data that's been provided to calculate the empirical formula for each substance.

Now people work at different rates, so don't compare yourself to anybody near you.

Just go slow and steady at your own pace to try to answer these questions.

And don't worry if you don't get to all of them.

If you want to go from A to C and then back to B, that's absolutely fine, but whatever you do, grab your calculator, your periodic table, pause the video and come back when you're ready to check your work.

Okay, let's see how you got on.

Now what I'm gonna do in this feedback session then is I'm going to provide you then with the working out and then the final answer.

Now, the reason I'm doing it this way is that if you don't get the correct final answer, you can then go back to that working out and double check if yours matches and identify where you've gone wrong.

So for A, where we had barium and chlorine then, your final answer should have been BaCl2, and there's the working out to help you.

Now, B was a little bit more complicated 'cause we've got three elements here rather than just two.

So there was quite a bit more working out, but when you're finished, you should have had an empirical formula of O4H2S, or you may have rearranged it so that it looks like H2SO4.

Okay, and then for part C then, we needed to find the empirical formula that had phosphorus and oxygen in it, and the processing then is as shown.

And for this one, rather than rounding like we have done in the other two, this one we needed to multiply in order to get that whole number ratio.

And when you do so, you get a final answer of P2O5.

So very well done if you've managed to get that.

I'm so impressed with the start that you guys have made to this lesson.

It's not an easy thing to keep clear in your head and you are doing superbly well.

What a fantastic start.

Now that we're feeling a little bit more confident talking about empirical formula, let's look at how we can determine molecular formula.

Now when we're talking about molecular formula, what we're really talking about is a simple covalent substance.

Those substances that have a really definite number of atoms that have been covalently bonded together.

And an example of that would be this one, butane.

Now if we look at butane, it's empirical formula, so its ratio of atoms is C2H5, but it's molecular formula then the actual number of atoms in this substance would be C4H10.

Now the molecular formula of a substance then could actually be deduced using its empirical formula and its relative formula mass.

Let's stop here for a quick check.

True or false, the molecular formula of a molecule indicates the simplest ratio of atoms in it.

Well done if you said false, but which of these statements best justifies that answer? Well done if you said B.

The molecular formula indicates the total number of atoms in a molecule, not just the elements.

So very well done if you manage to choose false and fantastic work if you managed to choose the correct justifying statement.

Great job, guys.

Now, if we're going to try to determine the molecular formula of a substance from its empirical formula, there are a few mathematical processing steps that we need to follow.

The first thing that we need to do is find the formula mass for the empirical formula.

So I'm gonna call that the empirical formula mass, okay? And you're gonna find that the same way you would find the relative formula mass.

Then what we're going to do is take the relative formula mass of the substance and divide it by the empirical formula mass.

And when we do that, what we get is what's known as a multiplier for the empirical formula.

So what we'll do is take that multiplier and we'll multiply the subscript numbers of our empirical formula by that answer, and that should then give us the molecular formula.

Let's look at an example of how we can use that processing then.

I'd like to know the molecular formula for a compound that has a relative molecular mass of 92 and an empirical formula of NO2.

So the first step I'm going to do is to calculate the empirical formula mass.

So the empirical formula is NO2, and if I find the formula mass for that, it should be then 46.

The next thing I'm going to do then is take the molecular mass, the relative molecular mass of my substance and divide it by the empirical formula mass.

So the molecular mass is 92 divided by the empirical formula mass of 46, and my multiplier then is going to be two.

So the next step then is to multiply the subscript numbers of my empirical formula NO2 by the answer to step two, my multiplier two.

And when I do that, I get a molecular formula for this compound of N2O4.

Let's stop here for a quick check.

I have a molecule with an empirical formula of C3H6O and it has a relative molecular mass of 116.

What I'd like you to do is determine its molecular formula.

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

Okay, let's see how you got on.

So the first step you needed to do was to calculate the empirical formula mass, and for this one, it was 58.

We're then going to the relative molecular mass to find the multiplier.

And for this instance, again, it was two.

We're going to take our multiplier and multiply the subscript values in our empirical formula to give our final answer, which was C6H12O2.

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

If you went wrong, some of the most common errors in this particular calculation is you maybe have calculated the empirical formula mass incorrectly.

So double check you've done that right.

Or sometimes the values tend to get swapped around in step two when trying to calculate that multiplier.

So those are two really quick and easy ways that you might be able to correct your work if you didn't get the correct answer, but very, very well done guys.

You're doing brilliantly.

Okay, let's move on to the second task of today's lesson.

For this first part, Alex has been tasked with calculating the relative formula mass for a molecule, and his answer doesn't seem to match his peers.

So what I'd like you to do is identify and fix Alex's errors.

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

Okay, let's see how you got on.

Well, if you looked closely at his calculations, you'll have noticed that Alex accidentally added these values when he should have multiplied them.

So if we correctly multiply then these values out, the carbon changes to 24, the nitrogen changes to 28, and that then gives us the relative formula mass for this molecule as being 88.

Well done if you managed to find that error and even better if you managed to correct it as well.

Great job, guys.

Okay, for this next task then I'd like you to use the information that's been provided to determine the molecular formula for each of these substances.

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

Okay, let's see how you got on.

So similar to before, I will show you the processing and the final answer.

And what you want to do is just check your final answer, and if that's correct, well done.

And if it's not, you've got the working out so you can go back and identify where you've gone wrong.

So for part A then you should have had a final molecular formula of I2Cl6, because our multiplier was a two after that processing.

For part B then, you should have had a molecular formula of C8H14O2, because again, our multiplier was two after that processing.

So well done if you got that answer.

Now for C, you should have had a molecular formula of C6H6, because this time our multiplier was six.

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

Great job, guys.

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

So let's take a moment to summarise what we've learned.

Well, we've learned that the empirical formula of a substance represents the simplest whole number ratio of atoms of each element in a compound.

And that using the masses of the elements that reacted together or the percentage composition of the elements in that substance, we could actually calculate then a substance's empirical formula by using ratio relationships.

And finally, we discovered that if you have the relative formula mass for a substance and its empirical formula, you can actually calculate then the molecular formula for that substance.

Wow, I hope you 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.