Transcription
[Music] Hello and welcome to this video. This is a revision video for OCR A, and we're going to look through the amount of substance, compounds, formula, and equations. So, quite a bit in this video, and I can say this video is for OCR A. And my name's Chris Harris, and I'm from Albury Tutors dot com. And let me say, this video is going to look at an overview for revision purposes for these topics here.
Now, the PowerPoints that I'm using, they can all be purchased. If you just click on the link in the description box below the video, you'd be able to get ahold of them. They're probably great for revision. Print them off, and you can use them on your phone. You can supplement, obviously, the revision stuff that you've got already. So, it's just another way, another form of revising. And I can say these are very specific to OCR A. And here's the specification points for the syllabus for this topic. And all the content in here is kind of geared towards these these points here.
All right, okay. So let's look at the first bit, which is forming compounds. Right, so ions are created when electrons are transferred from one atom to another. Okay, so they attract each other to form compounds. So, here we've got two elements here. Okay, we've got chlorine and we've got sodium. Now, to gain a full shell of electrons, chlorine needs to gain one more electron to get a full shell. Sodium needs to get rid of this electron here to get a full shell, which will be underneath this outer shell. Okay. And then when they do that, when they give up an electron, they form negatively charged ions. While chlorine's negative charge, and sodium is a positive charge, and then they attract each other. So, let's have a look. So, it is that the electron goes up the chlorine, and we've formed two ions that are now attracted to each other. This is called an ionic bond. Okay, so opposite charges attract. Call this an electrostatic attraction between the positive and negative ion.
Okay, just looking at where these ions come from. And if you use the group number, it kind of helps. All group ones form one plus ions. Group two form two plus. Group three form three pluses. Fours, we can just ignore. Group fives, right, these find it easier to gain three electrons rather than lose five. So that's what they do. They form three minus ions. Group six form two minus, and group seven form one minus signs. Here's your molecular ions. So you'll need to know these molecular ions. You're going to see them quite a bit. Hydroxide ions, OH minus. Nitrate ions, NO3 minus. Ammonium is what the only positive one on here, which is NH4 plus. Sulfates, SO4 2 minus. And carbonates, CO3 2 minus. You've got to know these molecular ions. You're going to see them a lot when you're bonding them with things. So nitrates and sulfates, etc. Know the charges and make sure you know, obviously, the the basic structure of it, like CO3, for example.
Okay, so here's your compounds. Okay, this is how we form compounds are formed from these ions on here. Okay, so you also need to know about some of these ions as well. You might see compounds specifically to do these. So zinc ion is Zn2 plus, and silver is Ag plus. So you can see compounds specifically in this year one chemistry. You're going to see these ones as well.
All right, okay. So we can work out the formula of an ionic compound and by using what's called a swap and drop method. Okay, so the first thing we need to do is write the two ions. Let's look at calcium 2 plus and NO3 minus. It's just an example. Okay, we swap the charges. So instead, so we swap them over like that, and we have CA minus NO3 two plus. Then we drop the charges, basically. We take away the negative and the positive bit and we drop the numbers. So they're low. Obviously, there's a one in there, but we don't include it on there. And then we basically combine them and simplify if we need to. Now, because this is already in the simplified form, and we've already got the formula here, this is calcium nitrate. No, it's a little two outside the bracket, basically symbolizes we've got two lots of NO3. Look at an example, CA2 plus plus O2 minus. Right, two ions. Swap the charges as you can see. We swap them over. Drop the charges. So we have a two and a two. And then what we need to do is we need to write our combine the two and then simplify it. CA2OH2 can be simplified to CAO. So that's what you have to do. So always try and simplify it where you can. Obviously, the stuff's whole numbers, not half numbers.
Okay, and salts, an example of an ionic compound. Okay, so these are obviously salts. So these are and you'll see salts everywhere. So there are they are very important. So they can be hydrated with water or they can be anhydrous without water. Okay, so if we look here, we've got a sodium and this would have come from a salt. And you can see and that these are obviously interacting with the with the sodium and chloride ions. We'll do the same as well.
Now, water is a polar molecule. Okay, it has a delta positive and a delta negative bit. Okay, what these do, these but this means that different parts of the water molecule are attracted to certain parts of the ionic compound. For example, the delta negative oxygen is obviously attracted to the Na+ ion, and the delta positive hydrogens are attracted to the Cl- ion. So what this is doing is it's effectively trying to interact with the salt ion, to the ions that make up the salt. And this is obviously how water works.
Now, water molecules, these can exist and as within crystal structures. And we can call this water of crystallization. You'll see this quite a bit. So for every mole of salt, you have a certain number of moles of water of crystallization. And you'll see formulas written like this, dot X H2O. This basically tells you how many molecules of water do we have for every mole of substance or salt that we that we have as well. For example, copper sulfate, CuSO4 dot 5H2O. And basically, this means that this salt has five moles of water for every one mole of copper sulfate. And this is called a water of crystallization. This is basically water interacting with your copper sulfate. Silly.
Okay, and obviously, we can calculate the water of crystallization as well. I mean, carry out an experiment to work out the value of water crystallization. Basically, it's a really straightforward experiment. All we do is we heat the salt. Removes the water from the moon's the water from the crystals. And what you get left with is an anhydrous salt, which means salt without water. Salt on hydrous means.
Okay, so let's look at this example. One point eight grams of hydrated calcium sulfate X H2O. We don't know the value of the X, just as a number. What we don't know it. Well, as heated until there is no more water of crystallization left in the sample. The mass of this anhydrous compound was one point one three grams. Work out the value of X, which is the amount of water of crystallization. Okay, so the first thing we need to do is to write out the two molecules involved. So calcium sulfate bit, and we've got a water bit. Guys, we've got the two halves. Then we have to write the masses of each of the molecules. So the mass of calcium sulfate on its own with no water was one point one three. We've worked it out. The mass of the water was literally, if this was the mass with water in it, that was the mass without, then if we subtract these two numbers away from each other, we can find out the mass of water. In this case, naught point seven five grams of water is being lost. Then what we need to do is divide these by the relative molecular mass. And what we're doing is effectively calculating the number of moles. So the relative molecular mass of calcium sulfate is 136, and the relative molecular mass of water is 18. We divide these two numbers together. We should get the number of moles to be naught point naught naught 83 moles of calcium sulfate, naught point naught 416 moles of water. Then what we do is we divide all the numbers by the smallest number of moles. So the smallest number of moles in this case is this naught point naught naught 83. So we divide both sides by that number. So it divides naught point naught naught 83 by naught point naught naught 83, that gets us one. Could divide this by the same number, we get five. And and then all we do is we say, right, this is basically saying for every one mole of calcium sulfate, we've got five moles of water. So the water of crystallization is five. So the value of X here is going to be five. Calcium sulfate, five H2O. Pretty straightforward.
Okay, right, let's look at some ionic equations. Okay, so ionic equations show the ions that are formed in solution. And they show which particles are reacting as well. Okay, so normally acids, bases, salts, these form ions in solution. Okay, so here's a standard balanced equation of an acid reacting with a base. So this is sulfuric acid reacting with potassium hydroxide. And what we're going to do is you're going to write a full ionic equation underneath this one and basically work out what we've got. So here it is. Book. So we split the sulfuric acid into two H+ and the sulfate ion. Potassium hydroxide is split into two lots of K+ and two lots of OH-. Potassium sulfate, two lots of K+ and a sulfate. And then the water isn't ionic, so we just leave it as it is. So we're only looking at ionic compounds here. Okay, and obviously, these ions are formed when them in solution. So water is just left alone. We don't split that up. Okay, so we've got all our ions here. So when we're looking at our ionic compound, and what we need to do is we need to cancel any ions that appear on both sides of the equation. And then what we get as our simplest ionic equation. So you can see here we've got sulfates and sulfate. K+, K+. These are all the same. But you can see what we're left behind is H+ and then OH-. So let's get cancelling. There you go. So we get rid of them. And we should be left with whatever's left is our simplest ionic equation. So in a neutralization reaction, H+ ions come from the acid, the OH- ions come from the alkali or the base, and then what we form is water. The sulfates and the potassium ions are called spectator ions. So these ones here, they don't actually get involved with the reaction. But obviously, they are there. But we don't need to worry about them because nothing has happened to them on the left compared to the right. Okay, so in this equation here, in the final ionic equation, the charges must balance on the left and the right. And make sure they balance. Obviously, we can have one. We can just have H+. So it's minus one. H minus one. Two. We can have that as well. And it's it doesn't really matter.
Okay, so let's look at the mole. This is really important. Everything in chemistry and to do with your calculations. And what nearly everything in chemistry has got to do with the mole. Okay, so we need to know what this is. And we need to know about this person called Avogadro, because he played quite an important role in this. Okay, so the mole is just something in these and chemistry to measure the amount of substance, hence the name of the topic. Okay, we can shorten it to mole, MOL. And basically, one mole of any substance, doesn't matter what it is, whether it's in a smaller molecule, will contain 6.02 times by 10 to the 23 atoms or molecules. This is known as Avogadro's number. And we normally see it written as N with a little A as a subscript. So one mole of water contains 6.02 times 10 to the 23 molecules of water. And one mole of iron also contains the same number, but it's atoms of iron. So it's pretty straightforward. And obviously, from this, we can work out the number of particles in the substance. We can use Avogadro's number multiplied by the number of moles. So, for example, if we had one mole of a substance, it would just be Avogadro's number. So that basically tells us the number of particles in there.
So let's have an example. How many particles make up 0.67 moles of ammonia, which is NH3? So the number of particles is Avogadro's number times by the number of moles. 6.02 times 10 to the 23 times by 0.67 moles. That's how many particles we've got. 4.03 times 5 to the 23. Also, be prepared to rearrange this formula as well. Okay, obviously, we've written it down as number of particles. That could get you to work out any number of these.
Okay, so let's look at the mole in terms of mass and MR, okay, which is a molecular mass. So mole and moles can be calculated for mass and MR, or it could be AR, is the equation. Now, you've got to know this one, really. So the number of moles is mass in grams divided by the MR or the AR, atomic mass or molecular mass. Okay, so this is ideal for things like solids. So calculate the number of moles, 23 grams of gold. Very nice. Okay, so we give you altitude and figures. So the number of moles is mass over AR. Mass is 23, exceeding 23 grams of gold. The AR of gold is 197. Okay, so the relative atomic mass. Number of moles is 0.12 moles. And this is given to two significant figures. Okay, so make sure you give your answers to the appropriate degree of precision. Basically, look in your question. You have any two of these components, any of these, then you can use it. So look in your question. Look at the numbers you've worked out. Do you have any of these? Any two of these, and you can use it. So you might have an atomic mass, obviously worked out from the periodic table, and the number of moles. You can work out the mass of the substance. And if you have the mass and then a revolving, work out the AR. So anything like that. And also, like with any equation that you'll see on here, be prepared to rearrange it.
Okay, so let's look at the mole in solution. So then ones were mainly for solids when we're using a mass and or anything as measuring mass in grams. And so this one is for solutions. So the number of moles of a substance in the solution can be calculated from concentration and volume. Here's the equation here. So number of moles, concentration in moles per DM cubed times by the volume. The volume has got to be in decimeters cubed as well. Quite often they give you in centimeters cubed, so you'll have to convert. So like I say, so how you would convert is you divide by a thousand to get it into decimeters cubed. So they give you in centimeters cubed, divide by a thousand. Or what you can do, and this is what you're going to see a lot in this in this PowerPoint, it all you have to do is take your number that you've got in centimeters cubed and just add times 10 to the minus three to the end of it. This means exactly the same as divided by a thousand. Okay, so it's a good example. Calculate the number of moles of 200 centimeters cubed, 0.35 moles per DM cubed, to HCl. So the number of moles is concentration times by the volume. So concentration is 0.35 times by 200 times by 10 to the minus 3. The number of moles is 0.07 moles. So all don't take the concentration, take the volume. There's my end up times 10 to the minus 3. The reason why I've put that on the end is because our volume is in centimeters cubed. So again, into decimeters cubed. Up to times by 10 to the minus 3. That means the same thing as divided by a thousand. See, see where I've got that from. Okay, and obviously again, look in your question. If you have any two of these components here, any of them, you can use this equation. So look in your answers. If you've got a concentration, a volume of something, you can work out the number of moles. And then you can use that moles to work out the mass in the previous equation that you've seen in the previous slide. So you can interchange these equations. And you will be expected to be able to do that as well. And also, like with any of them, rearrange the formula. You've got to know how to be able to do that as well. To make sure you can, you're prepared to rearrange formulas if they need to.
Okay, the gas one. Okay, so this is for gas. It's a bit strange because they occupy different volumes and different amounts of space depending on the obviously depending on the temperature and the pressure of it. So the number of moles in a specific volume of gas. So we use this equation PV equals NRT. And we call it the ideal gas equation. You've got to know the standard units for this. You've got to be really vigilant in these questions. Pressure is P, always measured in Pascals. They've given you in kiloPascals, you must convert to Pascals. Volume is in meters cubed. Okay, make sure you're able to convert to meters cubed. I'll show you how to convert units as well. N is just a number of moles. R is what we call a gas constant. We'll give you this in the exam. It's 8.31 joules per Kelvin per mole. And the temperature is measured in Kelvin. Again, if they've given it in degrees Celsius, you must convert to Kelvin.
Okay, so let's look at an example. Calculate the volume in centimeters cubed of 0.36 moles of a gas at 100 kilopascals and 298 Kelvin. So we've got PV equals NRT. Rearrange this and we get V equals NRT over P. Okay, as the volume is not 0.36 and of the volume, sorry, we use the volume is not 0.36. It which is the number of moles times by 8.31, which is the ideal gas constant. 298 is in Kelvin at the temperature divided by another pressure. Look, 100,000. Don't put 100 because we've got to convert to Pascals. So it's 100,000 Pascals. Volume is 8.91 times by 10 to the minus 3 meters cubed. But look, the answer, the question said we need to work out in centimeters cubed. So what we've done is you're taking that number and multiply it by a million. And that gets you the volume in centimeters cubed. Now, if you're not sure why you're doing it in a million, I'll show you on the next slide. But and just bear with me for two moments. Okay, so um, important things though, just want to show you this and is when we use the ideal gas, you should use the units above. Make sure these units are standard units. And make sure you can rearrange it as well. You've got to be prepared to rearrange it. The gas constant, you'll be given. So don't worry too much about like 8.31. Standard conditions for this reaction is 298 Kelvin and 100 kilopascals. So make sure you are able to recall these standard conditions.
Okay, so let's look at these units because units are quite important here. And it's just a little way of trying to explain how these units work. Now, you see what here I've got three different types of dimension here. One dimensional units, two dimension, three dimension. So one dimension is just a line. Okay, now if we want to convert to meters to decimeters and decimeters to centimeters, we use this unit quite a lot. Decimeters and decimeters cubed, or to say in chemistry. Now, we really use it in everyday life. So it is important to know where it fits with the rest of them. So basically, and a decimeter means decade, like 1/10 of a meter. So a decimeter is just 10 centimeters. That's all it is. And we've got 10 of these in a meter. So that's why it's go from a meter to a decimeter when multiplied by 10 because we have 10 of them in a meter. Going from a decimeter to a centimeter, sense meaning century, make 100. So one hundredths of a meter is basically and there's 10 centimeters in a decimeter. So it's multiplied by 10. You know, obviously, if we're going the full way, like I say, what 10 decimeters in a meter, 10 centimeters in a decimeter, and 100 centimeters in a meter. Factory. So we should know that bit. Now, if we are going for the whole way, you can see if we're gone from meters to centimeters, what you have to do is multiply by hundreds because you've got 100 centimeters in a meter. Okay, now this is fine for one dimension, but obviously in chemistry, don't use one dimension. So we're going to show you how we work out to get to volumes. So big dimension, we add a zero from the previous dimension. Okay, so you can see here, this little number here tells you we've got two dimensions because this is an area. So let's have a look. So you can see here, to convert from meters squared to decimeter squared, times by 100. All we've done is added a zero onto the end there. It is. We do exactly the same here. Now, that's the extra dimension. There's the extra number. They're going from meters squared to centimeter squared. We just do 100 times 100, which is times by 10,000. So if I want to convert meters squared to centimeter squared, times by 10,000. Obviously, if we go backwards, you're just dividing by the same number. Okay, obviously, the important thing is volume in chemistry. Adds an extra zero on there because we've got next to dimension. Look, so now we're multiplying by a thousand to go from meters cubed to definite this cube. Multiply by a thousand. The Gulf and decimeters cubed to centimeters cubed. Okay, look, there's the extra dimension. There's the extra zero. And basically, it's go from meters cubed to centimeters cubed. We do a thousand times by a thousand because we've got two steps here. That's times by a million. And that is effectively where we get the million prompt. Planned by a million. So obviously, just divided by the same number if you're going backwards.
Okay, so let's use equations to work out the masses of things. Now, also, you need to know state symbols. You're going to get these quite a lot in your equations. So S stands for solid, L is a liquid, G is a gas, AQ is aqueous, and which just means that anything is dissolved in water. So really, kind of look out for these. These are going to be pretty important.
Okay, balanced equations. We can use these to work out the theoretical mass of something. So let's have a look at an example. Calculating the theoretical mass from an equation. So how much calcium oxide can be made when 34 grams of calcium is burnt completely in oxygen? So the first thing we need to do is write down our equation and balance it out. So that calcium reacts with oxygen forming calcium oxide. Make sure it's balanced as well. There's our state symbols, solid, gas, solid. Then what we have to do to work out the the theoretical mass, then we basically work out the MR or the AR of the species involved. So because this is only talking about calcium oxide, calcium, we can ignore the oxygen bit. We write these masses in grams. So the relative atomic mass of calcium is 40 times 2 because we've got two of them is 80. And then the relative molecular mass of calcium oxide is 56 times 2, or formula mass 56 times 2 is 112. So basically, what we're saying is that 80 grams of calcium will give 112 grams of calcium oxide. Now, this is theoretical. Okay, in a perfect world where you're not losing any product, you know, 100% reaction, this is how much you should get. But the problem is, we want to know, we don't want to know what 80 grams and of calcium, and we don't want to know that. We want to know how much of this is made when we burn 34 grams of calcium. So all you have to do is change this number to 34, and that will tell us how much calcium oxide would produce. So the first thing we're going to do is we're going to divide the calcium side by 80 to find one gram. Then we're going to multiply by 34 to find 34 grams. And we're going to do exactly the same on the calcium oxide side. So let's go for it. So 80 divided by 8 is 1 gram. So that tells us what 1 gram is. And then multiplied by 34, nervously, that tells about 34 grams is. I'm going to use these numbers here and use them to do exactly the same on this side to work out how much the 34 grams of calcium give us for calcium oxide. So for 1 gram, it's 1.4 grams. A 1 gram of calcium will produce 1.4 grams of calcium oxide. Again, divided by 80, multiplied by 34. The nut lapel is our final answer. So 34 grams of calcium will produce 47.6 grams of calcium oxide. And it's as simple as that. Now, this is the theoretical mass. It's really useful. This this particular method for working out the air yield, percentage yield, because you're going to need to calculate a theoretical mass in order to work out a percentage yield. Okay, so this is in theory, how much calcium oxide we can produce.
Okay, so we're going to use equations to work out the volumes of gases. Okay, so balanced equations can be used to work out the volume of a gas. So we're going to calculate a volume of a gas from an equation. So for example, what volume of hydrogen is produced when 112 grams of potassium reacts with water at 100 kilopascals of pressure and 298 Kelvin? The gas constant, remember, is 8.31 joules per Kelvin per mole. So first of all, we write out our equation and we balance it. So there it is. S is two lots of task you map for two molecules and water, forming two multitasking hydroxide and a molecule of H2, hydrogen gas. Notes, all state symbols, solid, liquid, aqueous, gas, that's there. This is the amount that we're going to work out. Then we have to do is work out the number of moles of potassium. And so we work out the number of moles. Moles is mass over AR of potassium. So this is going to be M 12 divided by 39. Bassitt aciem. And that's going to give us 0.31 moles of potassium. Okay, so what we could do, we can work out the number of moles of this substance here. Once we know the moles of one of these, we can use the molar ratios here, the twos and ones in front of them, and we can use that to work out the moles of any one of these substances. And once we know the moles of that substance, we can work out quite a few things. So that's exactly what we're going to do here. We're going to use the equation to find out the molar ratio. Now, this two moles of potassium react to produce one mole of hydrogen. It's a two to one ratio. So the moles of hydrogen is 0.31 divided by 2, 0.155 moles of hydrogen is produced. So because we know the number of moles, and if we go back here, we need to work out the volume of hydrogen. So let's go for it. PV equals NRT because it's a gas, we're using. Rearrange it to get V equals NRT over P. So number of moles, ideal gas constant, temperature and pressure. So we know the number of moles is 0.155. We've just worked out here. Ideal gas is 8.31. We know the temperature is 298 because we've been told there. And the pressure is 100 kilopascals. Convert that into Pascals, 100,000. So the volume is 3.84 times of 10 to the minus 3 meters cubed. And that's the volume of hydrogen gas produced. So it's as simple as that. So we're into it. We're basically into using the mass equation for moles and using ideal gas equations. We're using both of them in the same question. But you can see how they work together.
Empirical formula. So this is the simplest whole number ratio of elements in a compound. So this is pretty important. So let's look at an example. A compound contains 23.3% magnesium, 30.7% sulfur, and 46.0% oxygen. What is the empirical formula of this compound? Okay, let's work through a worked example. First thing you have to do is write out all the elements involved. We've got magnesium, got sulfur, and we've got oxygen. Then we write out the percentages as masses. So they've given us the percentages of each of the elements. Obtain their percentages and I've just put grams on the end of them. Then we divide these by the relative atomic mass. And what we're working out is the number of moles. So 23.3, the relative atomic mass of magnesium is 24.3. Well, for a massive sulfate, 32.1, and oxygen is 16. So it divided these two numbers and we get BCA. This is just the number of moles. Then we have to do is divide the numbers by the smallest number of moles. So if you look here, the smallest number is 0.96. So we divide all of them by 0.96. And here are our answers. We get one magnesium, one sulfur, three oxygen. And then we write our formula. Empirical formula is MgSO3. Now, you might look at that and think, oh, that's a bit strange. It doesn't look and might not look complete. And sometimes these empirical formulas don't look complete. Okay, so you've got to kind of have faith in your and your calculations about because you might get so much look a bit strange. But we can use this empirical formula and we can use it to work out the molecular formula as well. So all we have to do is work out the MR of the molecular of the empirical formula that I've just worked out, divided by the MR of the molecular formula that they would have given you. Okay, and then we can use this number to multiply all the atoms in the empirical formula. And so in half, Mg2S2O6, for example. So basically, we'll work out what the multiplier is. And we'll multiply everything in the empirical formula by whatever that number is because they could get you to do that in the exam.
Okay, so let's look at more empirical formulas. Simple whole-number ratios in an element. Okay, now this one's slightly different because actually, we've now got a combustion reaction. So a hydrocarbon combusts completely to make 0.845 grams of carbon dioxide and 0.173 grams of water. What is the empirical formula of the hydrocarbon? So all we've got here are just combustion products. So it's similar, but you've just got to think there's one little extra step that we need to think about here. So instead of elements at the top, we've now got our combustion products, carbon dioxide and water. So write these as the headings at the top. Then we write the masses of each molecule. We've been told about how much carbon dioxide and how much water we've got there. And there's the masses. Okay, then we're going to divide each of them by the relative molecular mass of each. So the molecular mass of relative molecular mass of carbon dioxide is 44. Relative molecular mass of water is 18. Now, if we put them in our calculator, we should get 0.019, 0.096. Now, this is the number of moles of carbon dioxide and water. This is where it gets a little bit tricky. Okay, one mole of carbon dioxide has one mole of carbon atoms in it. Okay, so the original hydrocarbon that we burnt here must have 0.019 moles of carbon atoms. The only place where the carbons have come from to make carbon dioxide is within the actual hydrocarbon. So because we know the number of moles of carbon dioxide, this is the same number of moles of just carbon in the carbon dioxide. So now we know the number of moles of carbon. Now, if we come on this side, one mole of water has two moles of hydrogen. Okay, so because we've got one mole of this has two atoms of hydrogen per molar per molecule of water. So the original hydrocarbon must have had that number of moles times by two times. So this is 0.0192 moles of hydrogen atoms in the original hydrocarbon. Okay, this is the tricky bit. You've got to make sure you kind of be able to link the number of moles of this to the number of moles of atoms in the original hydrocarbon. Okay, so the carbon and carbon dioxide and hydrogen, the water can only come from the hydrocarbon. Okay, so what we do is we then use them numbers and we divide them by the number of moles of C and H atoms by the smallest number of moles. So the smallest of number of moles was 0.019. Okay, so and just about. So and we divide them both by the small amount and we obviously get a one-to-one ratio. So the empirical formula of this hydrocarbon is CH. Now, this is what I mean. It looks a little bit odd. We just see it written like that. It looks as though it doesn't look like a formula at all. But that is the empirical formula. Remember, it's the simplest whole number ratio.
Okay, so look at percentage yield. Percentage yield is the actual yield divided by the theoretical yield times by a hundred. Okay, remember we looked at the theoretical yield last time and we said, right, we needed to know how to calculate the theoretical yield. Well, theoretical yield is the amount of a product produced assuming no products are lost and all the reactions react fully. So we showed you that bit already with the three step divides from 1 gram, multiply final that step. You would use to work out the u.s. kahlil's. In this example, I'm just going to give you the theoretical yield and just just to show you how to use this equation, really. So any reaction involving the complete combustion of calcium, 32.6 grams of calcium oxide was produced. Theoretical mass is 47.6 grams. So this is just an example showing how to use the equation only. Don't worry too much about the numbers here. So calculate the percentage yield of this reaction. Okay, so obviously, have to calculate the theoretical mass itself. But again, I've just picked an unregistered kind of show you how it works. But the percentage yield is actual yield divided by theoretical yield times by a hundred. So all we do is you take the actual yield that we've made, 32.6, divided by the theoretical yield, you will have calculated this on your own. This is just an example number, times by a hundred. Percentage yield, 68.5% in this case. But the theoretical, the actual yield, you'll always get in an exam because obviously, you can only get this from actually doing the practical. So pretty obvious, really, as opposed.
Right, okay, so yeah, the other thing is, remember that yield is never 100%. It is virtually impossible to get a reaction to have 100% yield. You lose some of your products when you're transferring it from beaker to beaker. Not all of your reactants would have reacted. You may have lost some product if as a gas, that might have escaped, and loads of things like that. And you've just got to, you may have own impurities in there, which is obviously reacted. So you've got a few reasons why you can't get 100%. So as long as you know that, obviously, we try to get as close to the percent as possible. But that's about it.
Okay, and atom economy. My atom economy is how efficient a reaction is. Okay, so atom economy is the molecular mass of desired product, what we're wanting, divided by the sum of the molecular masses of all products times by a hundred. Okay, so let's look at an example. Iron oxide, Fe2O3, can be reduced using carbon or coke to make pure iron and carbon dioxide. Calculate the atom economy of the extraction of iron. The obviously iron is our n is our desired product. So let's bring up our equation here. There's iron oxide reacting it with coke, that forms iron and carbon dioxide. So that's our equation. Make sure it's balanced as well, because it's really, really important. So the atomic mass of the desired product, okay, iron is 55.8 because we've got two of them. Okay, we have to multiply it by 2. So it's 2 times 55.8, and the desired product mass, tonk master design post is 111.6. So that's this bit here, molecular mass or atomic mass. We've done the top bit. So now we have to do is the sum of the metric masses of all the products. So all the products, two times 55.8 plus one and a half times 12 plus 16 plus 16, that's carbon dioxide. That adds up to 100 and 1777, sorry, point six. No, we do stick it into equation. 111.6 divided by 177.6 times by 100 gives us an ass of economy, 62.8%.
Okay, so we need to know, just finally, the importance of atom economy. Why do we bother? Well, quite a few things. It tells you how efficient a reaction is. So it's actually really important for chemists. And so companies will try and get reactions that tend towards 100% atom economy. And high, high atom economies means that the raw materials are used efficiently. So this is a lot more sustained, and we're not wasting raw materials. High atom economies produce a lot less waste. And so they benefit the environment. So we're not polluting the environment with harmful pollutants. And a higher atom economy means you get less byproducts. So you don't need to spend loads of money separating the products from your desirable ones that you want, which could be energy intensive, which a drug should cost money and could be bad for the environment as well. So that's pretty much it. And that's just an overview of this part of OCR A specification. I like to say, you can see the little tiles in there. If you just click on the little circle in the middle and subscribe to my channel, and I'll be really great. Also, just remind you, you can buy these PowerPoints as well. Every just click on the link in the description box for this video and you'll be able to find them there. That's it. Bye bye.