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 formula 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, relative formula mass and the mole.

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 that you can jot down a quick definition of each for reference later on in the lesson or later on in your learning.

So today's lesson is broken into two parts, firstly empirical formula, and then we'll move on to look at how we can determine the molecular formula.

So let's get started by looking at how we can define and determine the empirical formula of different substances.

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.

Its 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, 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 macro molecular 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 going to 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 in which to determine empirical formula is to calculate it either from the percentage composition of the elements it's composed of or the reactant masses of the substance.

Now the mathematical processing involved, the steps that you need to carry out are exactly the same regardless of the data that you're given, whether it's percentage composition or reactant masses.

And the first step in this process then is to divide each percentage composition or reactant mass by the relative atomic mass of that element.

What that does then is calculate the number of moles that are present of each element in that substance.

You'll then divide each answer from step one by the smallest answer from step one.

And what that does is it creates a unitary molar ratio, and one answer will always be a one.

Some of these answers then may be very close to a whole number or very easily rounded to it.

And that's very good because if you remember, an empirical formula then is the simplest whole number ratio of atoms of each element in a substance.

Okay, let's look at how we can take some of that processing and put it into action.

I'd like to determine the empirical formula for a substance that contains 71.

4% calcium and 28.

6% oxygen.

Now I'm going to try to keep track the maths within a grid structure.

So I've set out this table like setup where I'm going to keep all the information about calcium in one column and all the information about oxygen in another.

And the far left hand column then is outlining either the information that I need or the steps that I need to take place throughout this calculation.

So the first thing I'm going to do is simply copy down the percentages for each of my elements.

So 71.

4 for the calcium and 28.

6 for the oxygen.

The next thing I'm going to do is grab my periodic table and find the relative atomic mass for each of these elements and add them to it.

So calcium is 40 and oxygen is 16.

Now I've put these values in my set like this on purpose where the mass comes first and then the relative atomic mass comes second because what I'm doing next then to find the moles is I need to divide these values.

So 71.

4 divided by 40 gives me a value of 1.

785.

28.

6 divided by 16 gives me a value of 1.

7875.

So this is my moles, but I need to turn this into a unitary molar ratio.

And in order to do that, I need to divide by that lowest number.

For here, that answer is from calcium.

So when I divide that calcium changes to one and oxygen changes to 1.

0014.

I need to double check though that I have whole numbers for my empirical formula and the calcium is, but the oxygen isn't, but it is very close to one and I could therefore round it.

So my molar ratio now here then is a one calcium for every one oxygen.

Now I could write that formula is Ca1O1 and it would be correct, but we don't really use ones when we're writing these formula, so it's better if the final answer is CaO.

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

2% aluminium and 79.

8% chlorine.

You may wish to use the kind of layout that I've shown here on the left as a guide or go back to some previous slides or previous notes to help just remind you of the processing involved here.

But whatever you do, grab a calculator 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.

Well, if you've done all that processing correctly, you should have had a final answer of AlCl3.

Now if you didn't get that answer, I have shown my processing here, so you may wish to pause the video and just double check my values against yours and see if you can identify any of the errors so that we can correct them going forward.

If you've got the correct answer though, very, very well done.

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 going to 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 not 0.

5, that's equivalent to roundabout one half.

So to remove the decimal, both of the values in my unitary ratio are going to 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 is 1.

52 to one.

In order to remove that decimal, I'm going to have to multiply by two for both of those.

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

04 to two, but 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 want to 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 not 0.

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

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

Great job.

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

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

70 grammes of iron and 0.

30 grammes of oxygen.

So I'm going to set up a calculation grid very similar to what I used before with my percentage composition.

And like before, I'm simply going to copy down the masses underneath the appropriate elements.

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

When I then divide those two values, iron comes out for the moles of 0.

0125, and for oxygen 0.

01875.

I need to then take these mole answers and create a unitary molar ratio.

So a divide by the lowest value.

And for this, it's iron.

When I do that, iron comes out as one and oxygen at 1.

5.

Now it's clear that the iron is a whole number, but oxygen isn't.

And in order to remove that 0.

5 decimal, I'm going to need to multiply my unitary ratio values by two.

And when I do that, iron comes out as two and oxygen comes out as three.

So the empirical formula for this substance is Fe2O3.

What I'd like you to do now then is to determine the empirical formula for a substance that was formed from 0.

529 grammes of aluminium and 0.

471 grammes of oxygen.

Now, you may wish to create a similar calculation grid to what I show the left or perhaps work with the people nearest to you, but whatever you do, pause the video and then come back to check your answer when you're ready.

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.

So if you want to double check your working out, if you didn't manage to get that correct, and then we can identify hopefully any errors that you've made and avoid them going forward, but incredibly well done if you manage to get that correct.

Great job guys.

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 elements 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 manage 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 going to do is I'll provide you with all the working out and then the final answer I was looking for.

Now, the reason I'm doing it that way is that you're looking, first of all, if you've got the correct answer that we were looking for.

If it doesn't match, that's when you're going to go back to that working out to see if there are any errors that you could have found in your own working to see how you can improve going forward.

For part A then of this question you should have had as a final answer, BaCl3.

And when we were finding our whole number ratio and our molar ratio, we simply needed to round those numbers.

So well done if you got that correct.

Now, B was a little bit more complicated because we had three elements rather than two that we've been using in our examples.

But if you've managed to do all the calculations correctly, you might have got an answer of O4H2S, or you may have rearranged it a little bit so that it reads H2SO4.

And again, to find those whole number molar ratios, we simply needed to round those lowest answers.

So well done.

And then for part C, we needed to find the empirical formula of a substance that was composed of phosphorus and oxygen.

And the working out is shown here, and you'll see that rather than rounding like we have done for the previous two for this one, we needed to multiply in order to achieve that whole number molar ratio.

And when you've done it that way, you should have a final answer that is P2O5.

So very well done if you managed to get that.

I'm really impressed with your perseverance on this.

There's a lot of processing involved.

It's very easy to go wrong or get a little confused, but it's really systematic approach and you stick with that and we will be able to go very quickly afterwards.

I'm so impressed with your start to this lesson, guys, keep it up.

Now that we're feeling a little bit more comfortable talking about empirical formula, let's look at how we can use it to determine a 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 it's ratio of atoms is C2H5, but it's molecular formula than 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 their 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 managed 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 going to call that the empirical formula mass.

And you're going to 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 take the relative molecular mass to find the multiplier.

And 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 is the simplest whole number ratio of atoms of each element in that substance.

And that if we use the masses or the reactant masses that came together to form that substance or their percentage compositions, then we could actually calculate the empirical formula for that substance.

And we do that by determining the moles of each element and the molar ratio between them in order to find that empirical formula.

We also discovered that if we have the relative formula mass for a substance and its empirical formula, we could actually calculate out and determine the molecular formula for that substance.

So a lot of stuff that we can do if we just know how to manipulate those numbers usually by ratios in order to come to a final answer.

I had a great time learning with you today.

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

Bye for now.