[Music] hello and welcome to this video on um amount of substance this is a revision video for AQA um specifically for the AQA exam board um and we're basically just going to go through an overview of all of the uh amount of substance um aspects that you need uh to know for the exam in particular um I should say that the Powerpoints that I'm using here these can be purchased um if you want them for revision and really good for that you can print them out you can go through them in your own time Etc um if you look at the um uh Link in the in the description box for this video you click on that link and that'll take you to the place where you can get them from okay so let's just um uh remind you that all these videos that basic they are tailored to the specification obviously these are the specification point points for AQA um and they just they're basically designed to cover um the content as an overview okay so let's start with um the mole this is something called avagadro number we're going to be looking at okay so um the mole is obviously very important in chemistry um it's used a lot uh and we need to know that uh basically it's just a way of measuring the amount of substance in chemistry and sometimes it shorten to the letters mole um and basically one mole of any substance contains 6.02 * by 10 23 atoms or molecules uh this number is known as avagadro's number um so for example we had one mole of water that would contain 6.02 * 10 23 molecules of water and one mole of ion would contain 6.02 6.02 * 10 23 atoms of ion so basically I think you get the idea um that one Mo of any substance whether it's an atom or molecule will contain that many of the atom or molecule okay so we can actually calculate the number of particles and some sometimes they get you to do this in a substance it's dead easy all we do is you just take avagadro's number and multiply it by the number of moles so if you had one mole of substance it's literally just avagadro's number so it's pretty straightforward so here's an example so if we're looking at how many particles uh make up uh 0.67 moles of ammonia this could be a question they could ask all we do is we use the formula number of particles is avagadro's number times the number of moles um so that's 6.02 * 10 23 multiply that by the number of moles 67 and we should get that many particles now it's a lot but obviously obviously particles are very small so you would have a lot of them um just be prepared to rearrange this formula as well um they could obviously ask you to work out um the number of moles of something and they'll give you the number of particles that are in there so just be kind of aware that they might get you to do do that okay so let's look at the mass and Mr or mass and Mis okay so this is um where we're looking at the number of moles using the um mass of something normally this is for solids um so um pretty straightforward really for this so the number of moles is just the mass over in grams over the M or ar so if you're using a a molecular formula or molecule or compound you'd use the Mr um or if you're using just an atom um then you or an element you would use the AR for that element so here's an example so if we calculate the number of moles once we calculate the moles of 23 g of gold um we're going to give our answer to two significant figures so the number of moles is mass over atomic mass which is a r so the number of moles is 23 g because that's what we're using divided by 197 that is the AR of gold um and basically we'd have a number moles of 0.12 moles and this is remember to two significant figures really look out for the accuracy there again be prepared to rearrange it um you will these will be integrated within questions so you've got to know how to rearrange this as well um basically a way in which you can kind of know if you're going to use this um look in your question um if you have any two of these uh in your question whether it's moles mass or a you obviously have to work that from the periodic table then you can use this equation um you might have worked one of these components out already previously in your questions so make sure you look back in your questions and make sure that you um see if you've got that information um I've got this saying really if in doubt work out the MS um because from the MS you can work out a lot of other things and you probably hear me say that again later on the video as well okay um so that's okay if it's for a mass of something like a solid but what about Solutions Solutions obviously are slightly different so the number of moles in a substance and a solution can be calculated from its concentration and volume so here's the formula that you need to know so the number of moles is concentration in moles per DM cubed times by volume now the really important thing is volume must be in decimeters cubed so um quite often they give you in centimet cubed and you've got to convert that all you do is you just divide by a th to get that into um decimet cubed so um that's pretty important you can also um just add * 10us three onto the end of your cm cubed volume um and that basically does the same thing it's just a little bit neater a bit quicker um if you're all right with mass you should be comfortable with that so here's an example calculate the number of moles of 200 cm cubed of 35 moles per DM cubed HCL so the number of moles is concentration times volume now notice the volume they've given us is in cenm cubed there it is look um so we got to convert that so the number of moles is 0.35 which is the concentration times by the volume and notice you see I've put the time 10us 3 on there that means the same thing is divide by th and we do that to convert that volume into decimeters cubed and obviously you put that in the calculator you get the number of moles to be 0.07 moles um of this substance okay right so um there you go that's just emphasizes your times SI 10 to Theus 3 um again just like the previous one be prepared to rearrange the formula and you've got be able to do that and again look in the question if you've got any of the two components that's mentioned in here you can use this um formula so look out for moles look out for concentration look out for volume again I'm going to say this again if in doubt work out the moles because if you know the moles and you know one of the other ones here you can work out a concentration or a volume so again the moles just like with the other equation you can work out a lot of things if you know the moles okay uh gases right gases are a bit strange because they occupy different volumes uh depending on the temperature so we have to take this into account um and presses obviously takes has an effect as well and to get round this we use something called the ideal gas equation so the number of moles in a specific volume of gas is basically what we're going to use for the ideal gas now the equation is this thing it looks a bit weird you've got five different factors here so it's PV equals n RT now really important thing you need to know what these mean so p is pressure and it's got to be measured in pascals if they give you in kilopascals or micr pascals or whatever I doubt be micro pascals but if they give you in kilop pascals convert to pascals these are standard units volume this is the tricky one volume's in meters cubed very often they will give you the volume in centimeters cubed make sure you can convert I'll go through conversion of units later on but it must be in meters cubed n is just the number of mes that's in mole that's pretty easy the gas constant which is R that's given 8.31 Jews per C per mole you'll be given this in the exam so don't worry about remembering that and temperature is in kelvin um that's important if they give you in degrees celsius you must convert to Kelvin um so um it's basically 273 plus whatever the degree Celsius is we'll get that into Kelvin for you okay so let's look at an example so calculate the volume in cm cubed of 0.36 moles of a gas at 100 kilopascals and 298 Kelvin so first thing write out your equation you get a mark for this and rearrange it so pval nrt and you rearrange this to give V equal nrt over p uh make sure you're familiar with rearranging um so volume uh is 0.36 this is the number of moles that we've been given Times by 8.31 that's the constant you'd be given that 298 Kelvin um this is obviously room temperature divide that by now the pressure look at the pressure they've given us 100 kilopascals our pressure must be in pascals that's why we've got 100,000 got to be in pascals put all that in calculator and we should get a value of 8.91 * 10 - 3 m cubed okay that's the standard units that's what we've got to use There It Is The Answer sorry the question said we need to convert it into centimet cubed now to convert from Meers cubed to centime cubed we have to multiply by a million again I'll show you the um a little bit later on how why we multiply by a million um but there's our volume in Meers cubed multipli by a million and we get um it in cm cubed 8 91 0 cm cubed okay so um yes so let's look at the we'll look at the converting units just on the next slide but the important things ideal gas make sure you use the units above they the standard units make sure you can rearrange it as well that's really important the gas constant will be given to you standard conditions are 298 Kelvin and 100 kilopascals that's pretty important okay let's convert these units because obviously this is pivotal to all this okay so I'm going to break it up um basically into meters decimeters and centimeters basically meter is the base unit that we're going to use um deci meter um is basically a tenth of a meter so this is 10 cm effectively so like decade Dei centimeter is like Cy is 100 so it's a 100th of a meter so you might have obviously seen these before obviously meters decimeters centimeters now the little squared bit on the top tells you the dimensions that we've got so for an area you would use met squared decimet squar centime squar and volumes are three dimensions so it's three so we've got the little three there so we've got three different dimensions and we have to take into account all of them so let's look at the first one so to go from meters to decimeters we multiply by 100 to go from decimeters cenm we multiply by 100 as well um so because we got 10 decim in a meter 10 cm in a decimet and 100 cm in a meter so if we want to go um if we want to go the full hog going from meters to centimet we do 10 times by 10 and that effectively means we're times it by 100 so that's how we get there obviously going backwards it's just the opposite okay so the general rule is basically for every Dimension we add a Zero from the previous Dimension so if you look at this one this is one dimension so if we're going to go from me squ to decimet squar we need to multiply by 100 we add an extra zero onto there okay from the previous Dimension so me Square to decimet squ Time by 100 times by 100 for this one we just added there and obviously go in the um the full hog um we just times 100 by 100 and we get 10,000 so to go from me squ to CM squ we multiply by 10,000 obviously in chemistry we're looking at volumes so again we add a zero for the extra Dimension so it's Times by a th000 to go from me Cub to decimet cubed decimet Cub to centimet cubed we multiply by a th000 as well and obviously to go the whole hog we um from me cubed to centimet cubed like in the previous example we do a th times a th which will means Times by a million obviously going back the other way you just divide by the same numbers that's shown here so really important you need to know how to convert your units okay so there we go that just confirms it right so ionic equations now these are pretty important um basically these show um the just the ions uh which are formed in solution and they show which particles are reacting um now you've got to be able to do these pretty straightforward but normally the things which are in solution the ions in solution are acids bases and salts so we'll look at this equation here this is a standard balanced equation this is for an acid reacting with a base so this is acid there it is there sulfuric acid plus base gives salt plus water now what we're going to do is we going to write the full ionic equation for this reaction now all we do is we take the bits which are uh basically um ionic uh in solution and we split them into the ions so what we mean is the positive ions and the negative ions sulfit is negative H is positive so we split them up and we get this potassium hydroxide is aquous again we split them up we've got two k pluses and two o minuses potassium sulfate again there's your acid there's your base these can form ions in the solution and your sulfate is a salt that forms ions in solution so you get 2 k+ 2 S so4 2 minus the water is neither an acid base or salt so we just leave that as it is and we form H2O now we need to form an ionic equation we're looking for ones where the important ions effectively so what we need to do is cancel any ions that appear on both sides of the equation to form the simplest ionic equation so let's cancel them out there you go so you see the sulfate and the potassium they appear on both sides so we cancel them out and we rewrite the equation to form whatever is left so 2 H+ plus 2 oh minus will form 2 H2O that is effectively a neutralization reaction so these sulfates and potassiums are called spectator ions they don't actually get involved with the reaction okay so this is the full ionic equation and the charges must balance uh on the left and the right and they should do it already if you've done it correctly so there's an ionic equation pretty simple actually when you look at it um right so we're going to use these equations to work out masses okay now you need to know state symbols um state symbols are s for solid L for liquid G for gas it's pretty straightforward and AQ is aquous basically this just means anything that's dissolved in water is aous acids for example are aquous okay so balanced equations we can use these to work out theoretical masses this is ideal when you're using it to work out um the percentage yield of something you need to work out the theoretical Mass that's produced um to be able to work out percentage yield so this is where it could fit in okay so let's look at this one calculating the theoretical Mass from an equation so how much calcium oxide coo can be made when 34 gram of calcium is burned completely in oxygen okay so the first thing we need to do is write out our equation you can see two lots of calcium plus oxygen will form two lots of calcium o oide make sure it's balanced as well the next thing we have to do is basically work out the Mr or the AR of the species involved and we going to write these masses in grams so you can see here um the Mr is uh 40 for calcium because we've got two of them it's going to be 80 g um and the two lots of calcium oxide calcium oxide is obviously 56 times it by two we got two of them is2 so basically what we're saying is 80 g equals 112 G okay then what we have to do is divide the calcium Side by 80 to find 1 G then multiply by 34 to get 34 G so because we want to work out we don't want to know what 80 gram of calcium would produce we want to know what 34 G is so we divide by one to find 1 G multiply to find the 34 so here we go there's divide by 80 to get 1 G multiply to find out what 34 G is but the great thing about this method is um the little units here you're divid and multiplies we do exactly the same on this side so we divide that so 1 G of calcium will form 1.4 G of calcium oxide but 34 G of calcium will form 47.6 G so you see we're using the same division and multipliers on this side as we are on this side so this is our leading side um and this basically we're using this to work out what we divide and multiply by and we get the same on this side so it's pretty straightforward so this is the theoretical mass that we've worked out um and in theory if we use 34 G of calcium we should um in theory in a very perfect world we should get 47.6 G of calcium oxide we can't get any more that is the maximum we can potentially get obviously we lose products when we do reactions Etc so the amount you get in a practical might be lower okay using equations to work out volumes of gases okay so we can use balanced equations to use to work out the volume of a gas so here's an example uh what volume of hydrogen H2 is produced when 12 G of potassium reacts with water at 100 kilopascals of pressure and 298 Kelvin um the gas constant is 8.31 okay so the first thing we need to do is write out equation just like the previous step so write it out make sure it's balanced okay then what we have to do is then have to work out the number of moles of potassium so the number of moles of potassium um again like I said if in doubt work out the moles so so we can work out the moles because we've got a mass and we know the AR of pottassium because we work that out from the periodic table so by doing that we've got 12 G ided 39 gets you 0.31 moles of potassium remember we're want to work out the volume of H2 so if I know the moles of potassium I can work out the number of moles of hydrogen it's a 2:1 ratio so two moles of pottassium weact to produce one mole of hydrogen the moles of hydrogen in this case is 0.31 because that was the moles of pottassium divide by two and that gets us 0.155 moles of hydrogen we've worked out the moles of hydrogen because from that we can now work out a volume so use your pval nrt because we're dealing with gases here rearrange it to get V equal nrt over P so volume is number of moles which we've just worked out Times by the ideal gas constant times by the temperature we're told the temperature is 2 2 98 divided by the pressure is 100 kilopascals remember to convert to pascals 100,000 so the volume is 3.84 * 10us 3 m cubed okay so it's pretty much as simple as that again we've used the moles look once you work once you know the moles of one of these substances you can work out the moles for any one of them just make sure you use your mol ratio okay titration absolutely crucial in chemistry so these can be used to work out the concentration of an acid or an alkal so you could for example have an acid or an Alkali either or in a buet with a known concentration so we know the concentration of it but in your conical flask we have an unknown concentration but we have a known volume of a substance in there and again it could be an acid Aly it doesn't really matter which way around it is we're going to add an indicator in there so we can detect the end point of our titration so we add the chemical into the buet uh into the uh conical flask uh until the indicator changes color this is known as the end point we've got to add it drop by drop near the end point we're do want to miss the end point it's very very subtle only one drop can make a massive difference uh and at the buet point if you look on this bit here there's the buet we've got a read on the miniscus okay so the miniscus is this bubble that we see on the on the water on the liquid that we're reading we always read at the bottom and you always read at I level so you can see this one um is reading that's 21 so this one's reading at 20 it's not reading at 19.8 which is this line we're reading at the bottom of the miniscus very very important that you do that okay there it is okay so and then all you have to do is you record your results to two decimal places um and basically you keep repeating to you get two concordant results that basically means they are within .1 cm cubed of each other um as soon as you got two concordant results that's it you take the average and that is your your average tighter okay indicators you could use um you can use phenoline in acid it's colorless and it turns pink and an alkaline or you could use methyl Orange it's yellow in acid and red in alkaline uh now obviously titration you got to be able to calculate these things so here's an example of a a a calculation so basically titrations can be used to work out the concentration of an acid or an alkaline so here's an example we've got 18.3 cm Cub of .25 moles DM Cub of HCL this is required to neutralize 25 cm cubed of potassium hydroxide what going to do is calculate the concentration of potassium hydroxide so uh let's have a look here's an example HCL is in our bu this time and our pottassium hydroxides in our chicle flask uh the reaction that's going to happen when we add the HCL to the pottassium hydroxide is this reaction here it's already balanced so we don't need to add any numbers to it then again just like the the first step we calculate the number of moles we can work out the number of moles this time um of our HL because we have a volume and we have a concentration so we can use these two numbers to work out the moles so that's what we're going to do concentration times volume at .25 is the concentration volume's 8 18.3 this is in cenm cubed remember we need to convert to decimet cubed times by 10us 3 that means divide by a th000 effectly turns it into decimeters cubed that should give us a volume uh a mole sorry of 4.58 time 10us 3 again you can work it out and try it okay so then we look at our equation 1: one ratio between HCL and pottassium hydroxide so the number of moles of potassium hydroxide is the same as the number of moles of HCL so that is 4.58 * 10us 3 again very useful um for effectively we need to work out the um concentration of potassium hydroxide so if we know the moles of it then we can use that okay and this is how we work it out so concentration using that equation shown you before rearranged concentration is moles divided by volume we know the moles just worked it out volume we've been told is 25 cm cubed 4.58 * 10us 3 divid 25 again Times by 10us 3 that will give you a concentration 0.18 moles per DM cubed so that's basically how we work out the concentration here's another example so we can work out the the volume of something as well so is the 115.7 cm cubed of 45 moles DM Cub of h204 was required to neutralize 0.120 moles per DM cubed of sodium hydroxide calculate the volume of sodium hydroxide being neutralized in cenet cubed so again here's the same thing different different acids obviously write the equation out again balance it same thing then we need to work out the moles um the moles of the stuff that we've got sulfuric acid this time got a volume got a concentration so this is going to be concentration 0.45 Times by the volume again look at the time 10 Theus 3 bit means divide by thousands we got to convert to decimet cubed gives us the number of moles 7.07 * 10us 3 then we use the ratio to work out the number of moles of sodium hydroxide this time it's a 1 to2 ratio so we're basically saying the moles of sodium hydroxide is 2 times by the number of moles of sulfuric acid because it's a 1 to2 r IO so the number of moles of sodium hydroxide is 2 * by 7.07 * 10us 3 and that will give 0.041 moles and then we need to work out the volume so um the volume is basically the number of moles divided by the concentration um so the volume is basically 0.041 that was the number of moles that we've worked out before divided by the concentration which we already have there it is there's the concentration 0.120 stick it in and we get 0.118 decim cubed now we need to convert this into cm cubed remember from the conversions 0.118 Times by 1,000 we'll get it in cm cubed so that's 118 cm cubed right empirical formula this is the simplest whole number ratio of elements in a compound so here's an example a compound contains 23.3% of magnesium 30.7% sulfur and 46% oxygen what is the empirical formula of this compound right first thing write your elements we've got magnesium we've got an oxygen we've got a sulfur then we need to write the percentages that we've been given as masses so 23.3 G of magnesium 30.7 G of sulfur 46 G of oxygen once we've done that we need to divide these by the relative atomic mass to get the uh number of moles so R Top Max of magnesium is 24.3 sulfur is 32.1 Oxygen 16 divide them and we get the number of moles for each there we go okay then we need to divide by the smallest number of moles now here the smallest number of moles is 0.96 so divide all of them by 0.96 that is going to give us our whole number ratio if it doesn't give you a whole number ratio round it to get the whole number ratio if it's very close to a whole number if not if say if that was 2 and a half you would have to double all of them to get your whole number in this case it's 1: 1:3 so you got one magnesium one sulfur one oxygen there's your empirical formula don't worry if that looks a bit odd um because with empirical formula it doesn't have to be um it doesn't have to be obviously the same as the molecular formula it's the simplest whole number ratio but to work out the molecular formula all we do um is we just work out the Mr of this this is empirical formula um divide it by the Mr of the molecular formula that they will give you and we use this to multiply all the atoms in the empirical formula so we can use that to work out that okay so um let's look at another one so this is basically where um we're using a hydrocarbon and it's combust completely to make 0.845 G of carbon dioxide 0.173 G of water what is the empirical form with the hydrocarbon this is very common in terms of burning of fuel so you all you're given now is a mass of the products of combustion so again what we're going to do this time instead of writing the elements at the top we're going to put carbon dioxide and water at the top now if we write down the masses of these just like we did last time uh we've got 845 and. 173 G again divide each of these by the relative molecular mass this will get us the number of moles relative molecular carbon dioxide is 44 and of water is 18 so we're going to get .019 .0 96 of water okay now this is where it gets a little bit different one mole of carbon dioxide has one mole of carbon atoms so the original hydrocarbon must have 0.019 moles of carbon atoms okay so that's really important and if we look on um and this well the main reason is because the carbon and the carbon dioxide and the hydrogen in water they only come from the hydrocarbon so if we look at the hydrogen side this time so we've got one mole obviously the this tells us the number of carbons in the hydrocarbon the water will tell us the number of hydrogen in there one mole of H2O has 2 moles of hydrogen because it's H2O so the original hydrocarbon must have 0.096 Times by 2 number of moles of hydrogen so it has .01 192 moles of hydrogen atoms it's really important okay the carbon and the hydrogen from here have only come from the hydrocarbon so that's what we're using here then now we've got the number of moles we basically just do the same setup as we did last time number of moles of carbon uh divid it by the smallest number so we've got a one to one ratio in this case so dead straightforward parle formula in this case is just ch make sure you understand this little bit here okay and you can use it okay percentage yield this is what I was talking about before percentage yields actual yield over theoretical yield Times by 100 so the theoretical yield is the amount of a product produce assuming no reactants uh and no products actually are um are lost and all reactants react fully so we've got a complete reaction so that was the method that I shown you before um when you worked out the theoretical yield so um let's use this one so in a reaction involving the complete combustion of calcium 32.6 G of calcium oxide was produced the theoretical mass is 47.6 G now you would have to work out the theoretical Mass but save time I've included it here now the percentage yield is dead easy okay so like I said you got to calculate that by yourself percentage yield is actual yield over theoretical 32.6 over 47.6 Times by 100 this case the percentage yield is 68.5% okay remember yield is never 100% you always when you do these reactions you always lose some reactants um or some of your products sorry may be lost if it's a gas for example it might Escape um not all of your reactants might have react Ed so or if you transferred it from be to Bea you might have lost some so yields very it's never in fact 100% so but we need to know what we're going to produce theoretically okay atom economy how efficient a reaction is atom economy is the molecular mass of the desired product divided by the sum of the molecular masses of all reactants Times by 100 different to yield um so you got to watch out for that here's an example so ion oxide Fe 203 can be reduced using carbon to make pure ion and carbon dioxide calculate the atom economy for the extraction of iron so here's our reaction look Fe2O3 plus half Carbon 2 Fe one and half CO2 so it's just balanced so the atomic mass of the desired product we're making iron is two lots of 55.8 so remember we' going to multiply it by two the uh mass of the desired product is 111.6 the atomic mass okay so the sum of the molecular masses of all reactants this is these here so we've got to add them all up so we've got two lots of 55.8 that's two lots of ION three lots of 16 that's the oxygen bit plus 1 and 1/2 * by 12 because we got 1 and A2 carbons we add all that up and we get 177.1 so we've got the mass molecular mass of the desired product and some of the molecular masses of all reactant stick that into your equation and we should get um 62.8% as an Asom economy different because we're not using masses actual masses we're using molecular masses we're only looking at the desired product and finally importance of atom economy why do we need it well to chemists um it basically tells us obviously how efficient reaction is as we know but companies will try and use reactions that tend towards 100% atom economy um basically this means that the raw materials used efficiently um it's more sustainable um we're using all of the raw materials to make our product uh this means that we get less waste we're not binning as much um byproducts um and if we do produce a byproduct companies try and make use of it they try and make something with it uh higher atony means less byproducts less time less money spent separating them so that's quite useful um so you just want your desirable product and that is it that is the amount of substance um again then this PowerPoint um you can purchase it if you just click on the link in the description box um you can um purchase it from there um but um that's it bye-bye