Loading...

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 keywords 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 a 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 catalysts 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.

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

So let's look at a reaction, a theoretical reaction between nitrogen and hydrogen forming ammonia.

And we can see all of the reactants in this example would react completely to form the ammonia.

It's not a reversible reaction in this example.

Firstly, if we look at the amount of substance, the moles, so we can see we've got one mole of nitrogen reacting with three moles of hydrogen to form two moles of ammonia.

We also know the relative formula mass of each of those and this is where a periodic table comes in useful.

So the relative atomic mass of nitrogen is 14 and we can get that from the periodic table.

So N2 would therefore be 28 because there are two nitrogen atoms in a nitrogen molecule.

So we can now calculate the mass in grammes for each of those.

So we've got the amount of substance multiplied by the relative formula mass to give us the mass.

And the theoretical yield can be calculated from a balanced chemical reaction in this way.

So if you look at that equation and look at those calculations, we can see that's happening.

Now, we're going to calculate the theoretical yield of water from four grammes of hydrogen in excess oxygen.

Now that word excess is quite important.

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

So the oxygen is not going to limit the reaction.

The availability of hydrogen will limit the reaction.

So the hydrogen is the limiting reactant.

We are going to need the equation moles = mass divided by the relative formula mass.

And the first thing we are going to do, is we're going to calculate the number of moles of hydrogen present so we know the mass and we'll know the relative formula mass.

And this is where your periodic table comes in useful.

So we've got a mass of four grammes, divided by two, and the two is the relative formula mass of H2.

So that means we've got two moles of hydrogen.

We then look at the equation and we look at the ratio between hydrogen and water in the equation.

So if you go back up to the equation, you can see two moles of hydrogen reacts with one mole of oxygen to form two moles of water.

So the ratio of hydrogen to water is 2 : 2 or 1 : 1.

So we know there's a 1 : 1 ratio between hydrogen and water.

So if there are two moles of hydrogen, that means two moles of water must be produced in that reaction.

And then we take that two moles and we work out, well what mass must that be? So the mass of water would be two moles multiplied by 18, which is the relative formula mass of water.

And that gives us a value of 36 grammes.

So pause the video here, read through my calculation again and have a go at answering that second question.

Welcome back.

So hopefully you've used my model there to answer the question.

So in this case, the moles of the C3H8 should be 0.

5 moles.

Next step is the ratio.

So the ratio to carbon dioxide is 1 : 3 in the equation.

So that means there must be 1.

5 moles of carbon dioxide produced.

And that works out at 66 grammes.

So again, pause the video here, check through your answer.

If you've made a mistake, go through the example there and see if you can work out where the error is.

But well done if you've got that correct.

So the percentage yield shows how the actual yield compares with the theoretical yield.

So here's the equation for that.

So it's the actual yield divided by the theoretical yield multiplied by 100.

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

The higher the percentage yield, the more efficient the reaction, and that's important to remember.

So we want as higher percentage yield as possible.

So let's have a go at calculating percentage yield then.

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

So we thought we produced 36 grammes, but we only produced 30 grammes of water.

And we can use that information to calculate the percentage yield.

So the first step is to write out the equation.

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

We then substitute in the values.

So we've got 30 grammes and 36 grammes, and notice we've put the units there and that's quite important because it helps you ensure that you don't end up with numbers where you don't know what they mean.

So including the units is helpful.

And then we carry out the calculation and come up with the answer 83.

3%.

So here's one for you to have a go at.

So pause the video here, follow my example to answer that question, and I'll see you when you finished.

Welcome back.

So step one is to write out the equation.

Step two, substitute in the values from the question.

So that's 44.

6 divided by 66.

And then step three, carry out the calculation and you should have come up with 67.

6%.

So well done if you got that correct.

So 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're 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.

Now you may have found some of that challenging, but we're just going to go through the answers to those questions.

So for One A, calculate the mass of ammonia that can be formed from 24 grammes of hydrogen.

So you can see there we've gone through the whole process and we've ended up with 136 grammes of ammonia.

So we've worked out, first of all, what's the mass of hydrogen? So underneath the 3H2 in the equation, you can see we carried out the hydrogen calculations there.

So we had 24 grammes of hydrogen.

We divided that by the relative formula mass of hydrogen, which was two, and that gave us a moles of 12.

And then we looked at the ratio in the equation and it was a 3 : 2 ratio.

And from that we worked out that the number of moles of ammonia was eight.

We knew the relative formula mass of ammonia is 17.

So we multiply together eight by 17 to give 136.

So well done, if you got that correct.

For B, this is the percentage yield.

So you should have written out the equation before answering it, remember, but the actual yield divided by the theoretical yield multiplied by 100.

So you should have had 40.

5 divided by 136 giving a value of 29.

8%, so that's quite a low percentage yield there.

Question two, again, calculate the mass of iron that can be formed from 112 grammes of iron oxide.

So the calculations under the iron oxide there FE2O3.

So we had a mass of 112 grammes divided by the relative formula mass of 160, giving us an amount of substance in moles of 0.

7.

From the equation we can see the ratio is a 1 : 2 ratio.

So that means we've got 1.

4 moles of iron.

The relative formula mass of iron is 56 and 1.

4 multiplied by 56 gives us 78.

4.

So well done, if you got that correct, that's 78.

4 grammes.

And for Two B, calculating the percentage yield, we've got 60.

5 divided by 78.

4, giving us a percentage yield of 77.

2%.

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 a 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.

So here's the calculation, we've got the equation across the top there.

So relative formula mass of the useful product divided by the sum of the relative formula masses of all the reactants.

So here's my example, calculate the atom economy of this reaction, when hydrogen is the useful product, give your answer to two significant figures.

So we've got methane reacting with water to form hydrogen and carbon monoxide.

So firstly, we need to work out the relative formula masses of the different substances and this is where you need your periodic table.

So for hydrogen we've got two, for methane 16, and for water 18.

And then for atom economy, we want the relative formula masses of the useful products.

So if we have a look, the useful products is the hydrogen.

So then the hydrogen molecules.

So you can see we've got three multiplied by two.

Now the three has come from the equation and the two is the relative formula mass of hydrogen.

And underneath we want the sum of the relative formula masses of all the reactants.

So if we look at the equation, we've got methane and water, so that's 16 plus 18.

So we carry out that calculation and we've got a percentage of 18%, which is quite a low percentage.

So what I'd like you to do now, is mirror my calculation by calculating the atom economy in this example.

So you'll need your periodic table 'cause you'll need to work out the relative formula masses of the different substances in that chemical equation.

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

So let's go through the answer then.

So firstly, we need to find the relative formula mass of all of the different substances.

So that's the useful product and the reactants there.

So hopefully you got that correct first of all.

And then we place all of that information into the equation there.

And you should get 77% as your answer.

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

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

So we've got an example here, although this is a reversible reaction, so it won't go to completion.

All of the reactant atoms will end up in the useful products.

The atom economy is 100% 'cause we are making that assumption that it's gone to completion.

For any reaction that forms one product, the atom economy will be 100%.

So again, back to that example there, if we put the information into the equation, so the relative formula mass of nitrogen is 28, the relative formula mass of hydrogen is two, and the relative formula mass of ammonia is 17.

We put that information into the equation and we can see that we get an answer of 100%.

So now we have a true or false question.

So atom economy and percentage yield can both be a 100% for a reaction.

Is that true or false? And then justify your answer using the statements below.

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

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, it's 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're 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 those answers then.

So first of all, calculating the atom economy of those two reactions and deciding which one is the most efficient.

Now, we've got all of the calculations on one page here, but we are gonna work down it from the top.

So we've just got the two equations there at the top, and then you'll notice we've calculated the relative formula mass of aluminium chloride.

At the top there is 133.

5.

So check you've got that correct.

First of all, we've written out the equation for atom economy, and then underneath, on the left-hand side, we've got the reaction calculation for A and on the right-hand side for B.

So just check you've got the relative formula masses calculated correctly, first of all.

So those four values, check those first.

And then, check the atom economy percentages.

Have you got those correct? So 71.

2%, 58.

9%.

So reaction A is more efficient as it produces least waste.

Now, if you've got those answers incorrect, then just pause the video here and check through the calculations.

The most common mistake to make is to forget to use the multipliers.

So if you look on the left hand side, you've got 133.

5 divided by 78 plus three times by 36.

5.

So you have to remember to use the multiplier from the equation at the top.

So that's where you may have gone wrong.

So just pause the video here, check through your answer, make sure you got that correct, but well done if you did get it correct.

For two, what are the factors must be considered when determining which synthetic route to choose? So we've got a whole series here, so hopefully you've got at least two or three of those.

So we need to consider the percentage yield, and the reaction rate, the equilibrium position, how easy it is to isolate the product? How desirable the byproducts are? The cost of the reactants, and how much energy is required? And the environmental impact.

So again, well done if you got that correct.

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 reactant.

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.