Lesson video

Hello, my name is Mrs. Collins and I'm going to be taking you through the learning today.

Today's lesson forms part of the unit industrial chemistry and is about percentage yield and atom economy.

During today's lesson, you're going to learn how to perform calculations relating to various aspects of industrial processes and explain decision-making about the chosen reaction pathways in those processes.

Here are the key words for today's lesson, theoretical yield, actual yield, percentage yield, atom economy, and useful product.

Pause the video here, read through those definitions, and write down any notes you feel you need to.

Today's lesson is divided into two parts.

We've got percentage yield and then atom economy.

So let's start with percentage yield.

Now in an ideal chemical reaction, all of the atoms in the reactants would end up in the useful products.

And the word useful there is important because we know all of the atoms in the reactants will end up in the products, but we want them to end up in the useful products.

And this reduces the amount of waste that's produced during a chemical reaction.

Now there's something called green chemistry.

So scientists are looking at how we can reduce the impact of chemical reactions on the environment, on health, and this is called green chemistry.

And there are lots of different ways that we can do that.

And we've got a list here of a few of them.

So we could try and prevent the production of waste.

We could make sure that some of the chemicals we're using in a chemical reaction or are produce during the chemical reaction less hazardous.

We could use safer solvents.

We could try and become more energy efficient.

So we want as little energy being used to produce these products as possible.

We might use catalyst to help speed up the rate of chemical reactions.

We might think about real-time pollution prevention.

So as soon as any pollution is produced, it's dealt with immediately.

And we obviously want to try and think about accident prevention as well.

So there's lots of different ways that we can improve the way that we produce chemicals.

Now it would be useful at this stage to pause the video and grab a periodic table if you don't have one available at the moment.

The theoretical yield of a chemical reaction is the maximum amount of product that could be produced if all of the reactants reacted completely.

So let's take a look an example.

So here we've got nitrogen reacting with hydrogen to form ammonia.

And normally this is a reversible reaction, but we're assuming that all of the reactants have reacted completely to form the products.

Now we can glean the amount of substance from that reaction.

So we've got one mole of nitrogen reacting with three moles of hydrogen to form two moles of ammonia.

And that's important.

We're using the information from the equation there.

We then need the relative formula mass of each of those.

Now this is where you'll need a periodic table.

So in the periodic table, the relative atomic mass of nitrogen is 14.

So the relative formula mass of a nitrogen molecule is 28.

And we can see each of those produced along there.

And then the mass in grammes, you can see is the amount of substance multiplied by the relative formula mass to form the mass in grammes.

So the theoretical yield can be calculated from a balanced reaction equation.

And we're going to have a look at how to do that in a moment.

So let's do a calculation together then.

And then I'm gonna ask you to do one by yourself.

So calculate the theoretical yield of water from four grammes of hydrogen with excess oxygen.

Now that word excess is important.

It means there's more oxygen than is needed for the reaction to take place.

So the hydrogen becomes the limiting factor in that reaction.

So we've got the equation for the reaction there.

Two moles of hydrogen react with one mole of oxygen to form two moles of water.

And a reminder of the equation that the amount of substance equals the mass divided by the relative formula mass.

And firstly, we're going to find the number of moles of hydrogen.

So for that, we need the mass of hydrogen from the question which is 4 divided by the relative formula mass, which we use the periodic table for.

And remember it's H2.

So we've got two hydrogen atoms. So therefore the value for the relative formula mass is 2.

So that gives us a value of two moles.

So there are two moles of hydrogen.

We then look at the ratio of hydrogen to water in the equation.

So if you go back up to the equation, we've got two moles of hydrogen forming two moles of water.

So the ratio is 2 to 2, and we can simplify that down to a 1 to 1 ratio.

That means if we've got two moles of hydrogen, we must have two moles of water being produced during that reaction.

And from that two moles of water, we can calculate the mass of water being produced.

So the mass of water is two moles multiplied by 18, which is the relative formula mass for water.

And that gives us a value of 36 grammes.

So what I'd like you to do is to mirror my calculation for this question here.

So pause the video here, answer that question, and I'll see you when you're finished.

Welcome back.

Let's go through the answer to that question then.

So very similar structure to the one that I've done already.

So firstly, we work out the number of moles.

So we've got 0.

5 moles there.

We then look at the ratio in the equation.

Now this time we've got a ratio of 1 to 3.

So go back up to that equation.

You can see a ratio of 1 to 3.

So that means we must have 1.

5 moles of carbon dioxide because there's three times as much.

So we multiplied it by 3.

And then we can calculate the mass of carbon dioxide.

So we have the amount of substance multiplied by the relative formula mass of 44, and that gives us a value of 66 grammes.

So well done if got that correct.

If you've got the answer wrong, you might want to pause the video here and just go through that working and see where you made your mistake.

So the percentage yield shows how the actual yield compares with the theoretical yield, and we can actually carry out a calculation to show this.

So percentage yield equals actual yield divided by theoretical yield times by 100.

And the actual yield of a chemical reaction will depend on the type of reaction and the conditions.

The higher the percentage yield, the more efficient the reaction.

So we're going to carry out a percentage yield calculation here.

So in the reaction, 30 grammes of water was produced, but the theoretical yield was calculated to be 36 grammes.

So we thought we were going to produce 36 grammes of water in the reaction, but we only produce 30 grammes of water.

So step one is to write out the equation, percentage yield equals actual yield divided by theoretical yield times by 100.

Then we substitute in the values making sure we get them the right way round.

So 30 divided by 36, and that gives us a percentage yield of 83.

3.

So what I'd like you to do now is go through this calculation and calculate the percentage yield.

So pause the video here and I'll see you when you're finished.

Welcome back.

So let's go through that calculation then.

So firstly, we write out the equation, then we substitute in the value.

So we've got the value for the actual yield there of 44.

6 and the theoretical yield of 66 multiplied by 100.

And that should give you a value of 67.

6.

So well done if you've got that answer correct.

If you've got it incorrect, just pause the video here and check through your answer carefully.

Now the percentage yield of a reaction is unlikely to ever be 100%.

And this is because the reaction might not go to completion because it's a reversible reaction for example.

So the earlier example we used of hydrogen reacting with nitrogen to form ammonia is a reversible reaction.

The product may be lost when it's transferred between reaction vessels or when separated from the reaction mixture.

So we might lose some of the product.

And reactants might react in unexpected ways, forming different products to those expected.

And those are side reactions.

So those are different ways that the percentage yield of a reaction is unlikely to be 100%.

Sometimes percentage yields may appear to be over 100%.

Now you can't get a percentage yield of over 100%, so there must be something happening here.

So this is because the measured weight of the product will include the weight of any impurities that might be present.

And one of the most common impurities that might be present is water.

Because if the product is not dried enough, then that means that will contribute to the mass of the product.

And you can see there, there's some wet crystals drying on filter paper.

So if there's still water present in the product, that will appear to increase the percentage yield over 100%.

So here's a question based on that learning, which of the following statements are correct? Pause the video here and I'll see you when you're finished.

Welcome back.

So hopefully you've recognised that the answer to this question is most reactions have a percentage yield lower than 100%.

We've now got a second question this time true or false.

So a sample of product may appear to have a percentage yield greater than 100% due to impurities.

Is this true or false? And then choose one of those statements to justify your answer.

So pause the video here and I'll see you when you're finished.

Welcome back.

So hopefully you've recognised that the answer to this question is true, and that's because if the product has not been dried correctly or dried enough, the crystals will contain water.

So well done if you've got that correct.

We are now gonna have a go at task A.

So we've got a series of questions there with calculations.

What I'd like you to do is pause the video here and spend some time answering those questions and I'll see you when you're finished.

Welcome back.

So let's go through those questions one at a time.

So firstly, we need to calculate the mass of ammonia that can be formed from 24 grammes of hydrogen.

Now initially this might look a little bit confusing, but I'm gonna go through it with you step by step.

So if you find hydrogen in that equation there, the numbers relating to hydrogen are below it.

And then if you find the ammonia, the numbers relating to ammonia are below that.

And the arrows indicate the direction in which we're going to carry out the calculation.

So firstly, at the top under hydrogen, so we've got 24 grammes of hydrogen and we're going to divide that by the relative formula mass.

And the relative formula mass for hydrogen is 2.

And then underneath that, we've got the amount of substance which is formed from that calculation.

So you should end up with 12 moles.

So 24 divided by 2 is 12.

Then at the very bottom, we've got the mole ratio and we got this from the equation.

So three moles of hydrogen forms two moles of ammonia, so it's a 3 to 2 ratio.

So from that, we can work out that the amount of substance for ammonia is 8.

Next up, we've got the relative formula mass of ammonia you should have worked out as 17.

And 8 multiplied by 17 is 136.

So you should have worked out that 136 grammes of ammonia will be produced from 24 grammes of hydrogen.

So well done if you've got that correct.

You might want to pause the video at this stage and just check through your answer.

So now on to B.

And for this one, you've got the percentage yield calculation of actual yield divided by theoretical yield multiplied by 100.

So 40.

5 divided by 136 times by 100, and that should give you a value of 29.

8.

So well done if you've got that correct.

Going on to C, calculate the mass of iron that can be formed from 112 grammes of iron oxide.

So again, look at the equation.

We've got the iron oxide there, Fe2O3, and we've got the iron, which is 2Fe.

So underneath the iron oxide, you should have 112 grammes of oxide divided by the relative formula mass of iron oxide which is 159.

6.

And that gives you 0.

70175 moles.

And then if you go to the equation, the mole ratio is 1 to 2.

So that means we've got 1.

4035 moles of iron multiplied by the relative formula mass of iron, which is 55.

8.

And that then gives us a value of 78.

3 grammes of iron.

So well done if you've got that correct.

And then for B, the percentage yield, actual yield divided by theoretical yield times by 100, 60.

5 divided by 78.

3 multiplied by 100, giving us a value of 77.

3.

So well done if you've got that correct.

And then for C, give three potential reasons why the percentage yield was less than 100%.

The reaction might not go to completion, some of the product might be lost, and some of the reactants might react in unexpected ways.

So well done if you've got that correct.

And particularly if you've got those calculations correct.

If you were struggling with the calculations, just pause the video, go back to the answer pages, and have a look through those again.

So let's move on to atom economy now.

So atom economy is a measure of the efficiency of how a reaction uses its reactants.

Here's the equation for that.

So you've got the relative formula mass of useful product on the top.

So we need to look at the products, decide which products are the useful products, and work out the relative formula mass of those.

And then underneath, you've got a sum of the relative formula masses of all the reactants.

So we're trying to work out how much of the reactants have ended up in the useful product.

So when doing this calculation, we assume that the reaction goes to completion.

So it's not reversible reaction and the reaction does complete.

And that there are no side reactions happening, so no unexpected reactants.

So we are making a couple of assumptions there.

So in this reaction, hydrogen is the useful product.

If you look, we've got a methane there, CH4, reacting with water to form carbon monoxide and hydrogen.

And hydrogen is the useful product.

So that means the carbon monoxide is not useful.

So before calculating the atom economy, we already expect it to be low as the atom economy is based on the relative masses of the chemical substances.

Let's carry out a calculation for atom economy.

So we're gonna calculate the atom economy of this reaction when hydrogen is the useful product.

So the hydrogen useful product is gonna go on the top and the sum of the relative formula masses of all the reactants is gonna go on the bottom.

So we need to know the relative formula mass for hydrogen which is 2, 'cause it's hydrogen molecule so H2, and then the relative formula masses of methane and water there.

And then we're going to substitute in those values into the equation.

But note at the top, we've done 3 times by 2.

0.

Now 3 we've got from the equation, there are three moles of hydrogen in the equation, and the 2 is the relative formula mass of the hydrogen molecule.

And then underneath are the relative formula masses of the reactants, methane and water.

We've carried out the calculation and got an answer of 17.

6.

Now what I'd like you to do is in a moment pause the video and carry out the calculation for this atom economy.

So I'll join you when you finished.

Welcome back.

So here's the answer to that.

So first of all, we should calculate the relative formula masses of the three different substances we're interested in, it's the useful product and the two reactants.

And then substitute in those values into the equation.

This time we haven't got anything in the equation that's gonna complicate the calculation.

And we end up with a value of 76.

7.

So well done if you got that correct.

If you've got it incorrect, just pause the video for a moment and check through your answer carefully.

So in reactions where there's only one product, the atom economy will be 100%.

And we've got an example here.

So we've got nitrogen and hydrogen forming ammonia.

Remember we make assumptions here, so we're assuming it's gone to completion even though it's a reversible reaction.

And this is because all the reactant atoms end up in the useful product.

So remember, for any reaction that forms one product, the atom economy is going to be 100%.

So we're going to show that using this equation here again and this equation for atom economy.

And we're gonna carry out that calculations.

The first thing is we're interested in the relative formula masses of the various substances.

So we've got nitrogen, hydrogen, and ammonia there.

And remember, you need your periodic table to calculate those.

Substitute those values into the atom economy equation using information from the equation.

And that gives us a value of 100%.

So now we have a true or false question.

So atom economy and percentage yield can be both 100% for a reaction, is that true or false? Pause the video here and I'll see you when you're ready.

Welcome back.

So hopefully you've recognised that that's false.

So when there is only one product in a reaction, the atom economy can be a 100%.

So there's often more than one route to synthesise a chemical.

When deciding which method to use, industry will weigh up various factors.

So as well as considering atom economy and percentage yield, they'll also consider the reaction rates, so how quickly the reaction happens, how easy it is to isolate the product that's wanted, the equilibrium position if it's a reversible reaction, how much we want the byproducts 'cause those byproducts might actually be quite useful, the cost of the reactants and how much energy is required for that reaction to happen, and the environmental impact of the particular chemical reaction.

So you might choose one particular route because it's quite a fast reaction.

And you quite like the byproduct.

You want the byproducts as well.

And it has less environmental impact for example.

Now ethanol is a useful product.

It's got many uses.

It's used as a solvent, it's used as feedstock for other reactions, a fuel disinfectant, and obviously it's used in alcoholic beverages as well.

Now we can make ethanol by fermentation using yeast.

And this takes glucose producing ethanol and carbon dioxide.

Or we can produce it by hydration of ethene.

And this is where we react ethene with steam and that produces ethanol.

So look at those two different equations there.

Look at the products of those equations and think about how easy they might be to carry out, how much the reactants might cost for example, what's the environmental impact of those? And you can see choosing which reaction pathway to use is quite complicated.

You need to consider lots of different factors.

So here we have a comparison of the two different reaction pathways.

So we've got the atom economy first of all.

So fermentation has a much lower atom economy than the hydration of ethene.

The yield in fermentation is much lower than with ethene.

And the reaction rate is very slow as well.

So you'd think that wouldn't be a potential choice.

However the raw materials for fermentation are renewable.

So we're using glucose, normally sugar.

We're obviously using yeast for that process.

Whereas hydration of ethene uses crude oil, which is finite.

The product needs to be isolated with fermentation, whereas it's pure with hydration of ethene.

Equilibrium position is not a reversible reaction for fermentation, but it is for ethene.

And byproducts.

Now carbon dioxide we know is a greenhouse gas, but it is used for a number of different purposes, so it could be useful and that would increase the atom economy for that reaction.

So here's a question based on that learning so far.

So which of the following statements are correct? Pause the video here, answer the question, and I'll see you when you're finished.

Welcome back.

So hopefully you've recognised that the answer to that question is a reaction pathway describes a sequence of reactions needed to produce a useful product.

So well done if you got that correct.

We are now going to move on to task B.

So first of all, what I want you to do is pause the video here, answer those questions, and then we'll go through them together.

Welcome back.

So let's go through question one first of all.

So we need to calculate the atom economy for each reaction and use that information to decide which is most efficient.

Now this layout may be a little bit confusing to start off with but I'll just explain it to you.

So firstly there at the top, we've got the relative formula mass of aluminium chloride as 133.

5.

So check that you've got that correct first of all.

And then on the left-hand side, we've got reaction A.

And on the right-hand side, we got reaction B.

So check that you've got the relative formula masses correct first of all for the four different chemicals for the two reactions.

And then underneath, we've got the two calculations for atom economy.

So check that you've ended up with 71.

2% for reaction A and 58.

9% for reaction B.

Which shows that reaction A is more efficient, it produces the least amount of waste.

Now if you've got the answer incorrect, just pause the video and check through your answers.

But the most common mistake people make is to forget to include the multipliers from the equation.

So if we look at A at the bottom, we can see we've got 3 lots of 36.

5.

And that's because if you go back up to the equation, you can see the number 3 there from the equation.

So we need to be take care with the calculation, but just check through and make sure you got that correct.

For two, other factors that must be considered when determining which synthetic route to choose.

We've got percentage yield, reaction rate, equilibrium position if it's a reversible reaction, how easy it is to isolate the product, how desirable are the byproducts, the cost of the reactants, and the energy needed for the reaction to take place, and the environmental impact.

So hopefully you've got at least three of those.

Here's a summary of today's lesson.

In an ideal chemical reaction, all atoms in the reactants would end up in useful products with no waste.

The theoretical yield of a chemical reaction can be calculated from the reaction equation alone.

The percentage yield shows how actual yield compares with theoretical yield.

Atom economy is a measure of the efficiency with which a reaction uses its reactants.

And where more than one route to produce a product is available, many factors will be considered when choosing a reaction pathway.

So thank you very much for joining me for today's lesson.