Moles and Molarity | Nick Maths | Skillshare
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8 Lessons (30m)
    • 1. Why are moles important?

      3:54
    • 2. What is a mole?

      1:13
    • 3. Calculating grams to moles

      2:21
    • 4. Calculating moles to grams

      3:06
    • 5. What is Molarity?

      1:27
    • 6. Prefixes of Molarity

      4:13
    • 7. Introduction to Molarity worked examples

      1:31
    • 8. Example Molarity Calculations

      12:40

About This Class

This course looks at moles and molarity, and why an understanding of moles and molarity is important for chemistry, biology, and pre-med students.

This course will look at moles and molarity, and Avogadro's constant, and then later move on to look at Amounts, Volumes and Concentrations, and it contains a number of worked examples of how to calculate the number of moles present, how many grams of something you will need to get a certain number of moles, and how to work out the molarity of a solution.

Transcripts

1. Why are moles important?: in this video. We're going to look at the ports of moles on my clarity on why you need to understand them . Let's say we've got two compounds or atoms A and B. Let's say the aiways one and be ways to now. This could be one gram per mole or it could be two grams per mole. Doesn't really matter, but be is twice as heavy as a If we wait out six, let's say grams of a We would get six atoms or molecules of A. Whereas if we wait out six grams of B, we would only get three atoms or molecules of B because each one weighs two. So we've weighed out six grams, six atoms of A and we've weighed out six grams, which gives us three atoms or molecules of B and let's say thes react together. Oneto one. So one a will react with one be So is the reaction runs and a binds with a B, another A. Then bind with a B. And then finally, 1/3 a will bind with a B. At the end of the reaction, we've used up all be, but we've still got three a left so we haven't had an equal reaction. Yes, we've got three molecules off a be formed, but we have three of a leftover. What we need to do is no measure out the number of grams off the atoms or molecules am be We need to measure out the same number of each. Now A has a molecular weight of one. Where's bee has a molecular weight off to so one mole of a would weigh one on one mole of B would wait to in our reaction, because it's 1 to 1, we will be reacting one mole of a with one mole off B. Therefore, we need the same number of a and the same number of being not the same mass, but the same number. And this is the critical thing. So if we way out four moles for a that were way four. In the case off B, we need to match A and B. So we need four moles or four molecules or atoms, whatever they may be off be. So we've got four A and four b, however, be weighs twice as much. Therefore, to way out four months or molecules or atoms. Whatever you want to consider it? As of be, we will need to way out eight grams or eight kilograms, whatever the units. Maybe so. We would need to weigh out twice. A much off be as we would A because B weighs twice the amount off A by using molds. We've now got a balance here between A and B. We've got four A and four B. So we ran this reaction we would get for a B molecules being formed. And that is why the understanding of moles is important. Malaria T is just the number of moles per given volume of solution. It's a concentration. 2. What is a mole?: So what is a mold? The number of moles present is equal to the number of grams of the compound divided by the molecular weight. So we can write this out as moles is equal to number of grams divided by the molecular weight. Now we can ride this out as em is equal to G. The number of grams divided by the molecular weight began big W This equation we can rearrange as M multiplied by molecular weight is equal to the number of grams. The two important equations that you have to remember here will. In fact, it's only really one is grams divided by molecular weight is the order number of moles from that, you can rearrange it to give you the second equation. So you don't really need to remember the second equation in which grams is equal to most times molecular weight 3. Calculating grams to moles: in this video, we're going to look at how to calculate the number of moles of a substance for a given number of grams to convert grams to moles. It's just a case of dividing the number of grams off the compound by the molecular way to the compound. We can express this as an equation as follows in the equation. N is equal to G, divided by M. W. Where M is equal to the number of moles. G is the weight of the compound ing grams and M W is the molecular weight of the compound in grams per mole. So let's have a look at work calculation. Let's say you have three grams of compound eggs and Compound X has a molecular weight of 30 grams per mole. How many moles do you have for this? We need the equation, which we introduced earlier. Em is equal to G, divided by M. W. Where M is equal to the number of moles. G is the weight of the compound used in grams on M. W is the molecular weight of the compound in grams per mole. If we substitute in the values from the question, we get G is equal to three grams on M W is equal to 30 grams. If we then perform the calculation we get N is equal to G, divided by M. W. We have m equals three divided by 30. Therefore, em is equal to no 300.1 moles. Therefore, if we take three grams of the compound and it's got a molecular weight of 30 grams per mole , that would actually contain 0.1 mould of compound eggs, so, in summary to convert grams to moles. It's just a case of dividing the number of grams of the compound by the molecular weight. This could be easily expressed as an equation, as we have seen where M is equal to G, divided by M. W. Where M is equal to the number of moles. G is the weight of the compound in grams on M. W. Is the molecular weight of the compound 4. Calculating moles to grams: in this video, we're going to look at how to calculate the number of grams of the substance for a given number of moles. In the previous lecture, we looked at converting grams to moles on. What we found was it was just a catered dividing the number of grams president by the molecular weight of a compound and that gave us the number of moles. We could express this with a fairly simple equation in which M equals g divided by M. W. Where M is the number of moles, G is the weight of the compound and M w is the molecular weight of the compound expressed as grams per mole. What we need to do now is rearrange this equation and isolate for the term G. That is the weight Now the equation we start with is em is equal to G divided by M w. We're rearranging this toe. Isolate Fergie. So, looking at our equation, what we need to do is move m w over from the right hand side to the left hand side, so isolating g. So if we take our equation removed, m w over and as we've moved it from the right hand side to the left hand side. We also have to inverse it so the divide becomes multiplied. So what we end up with is M more to plant by M. W is equal to G, so just rewriting that out G is equal to M multiplied by him. W where G is the weight of the compound. M is the number of moles present, and M W is the molecular way to the compound in grams per mole. So let's have a look. A worked example say you have no 0.1. Moles of Compound X and Compound X has a molecular weight of 30 grams per mole. How many grams do you have for this? We take the equation. G is equal to M multiplied by MW, where G equals the weight of the compound. Ingram's M is the number of moles, and M W is the molecular weight of the compound. From the question, we know that em is equal to North 0.1 moles and that the molecular weight it's 30 grams per mole. So taking the equation weaken substitute in these values, so G equals 0.1 multiplied by 30. The final answer there is, of course, three grams. Therefore no 0.1. Moles of a compound with a molecular weight off 30 grams per mole would be three grams of material. So in summary, to convert moles to grams is just a case of multiplying the number of moles of the compound by the molecular weight. As we've seen, we can express this with this fairly simple equation in which G is equal to M multiplied by M. W. Where G is the weight of the compound in grams and is the number of moles present and M w is the molecular weight in grams per mole. 5. What is Molarity?: What is my clarity? Minority is just a measure of concentration. It is the number of moles per liter of solution. So one Moeller solution is one more of a compound made up toe one leader. We can express this as an equation as follows where big M for Mullah is equal to little M for moles divided by V volume. Now, as you remember from a previous lecture, little M moles is equal to grams off compound divided by the molecular weight. Therefore, we can substitute into the first equation for the little em to give us Big em is equal to open brackets G divided by molecular weight. Closed bracket divided by V We can expand this equation and it gives us G divided by molecular weight times. Volume is equal to Moller. If we move over the molecular weight and volume part to the moulder side, we get em multiplied by molecular weight multiplied by V is equal to G. Just rearranging that G is equal to the McGarity, multiplied by the molecular weight multiplied by the value, and that gives us a very easy way to derive the equations that we need and to work out the number of grams of a compound we require to make up a solution 6. Prefixes of Molarity: 7. Introduction to Molarity worked examples: an understanding of mill arat e is important. In the next few lectures, we are going to look at some worked examples. Malarkey is just an expression of concentration. It is the number of moles of a substance per liter, so we can write this out. Asmal Ara T is equal to concentration, which is equal to the number of moles divided by later. The volume we can express mill ality is began moles, a small M and volume as V. This gives us m equals m divided by V. The molecular weight of a compound is equal to one mole. Therefore, if we have one mole in one litre, we will have a one molar solution. So essentially the molecular weight of a compound dissolved in a litre will give you a one molar solution. If you can remember this relationship between moles on Moeller and molecular weight, then you can answer any of the types of questions you may encounter, in which you have to calculate the concentration of a sample or you have to calculate the number of grams of a compound you require to make a solution of a given mill. Arat e 8. Example Molarity Calculations: you have two moles of a compound dissolved in one liter. What is the mill aridity of the solution Mill? Arat E is just concentration. It is the number of moles per liter. We can express this as an equation where big em is equal to little m divided by V where big em is equal to mill Ara t small em is equal to the number of mom was present in this case too. And V is equal to the volume in liters which in this case is one therefore substituting in these values we get em is equal to two divided by one. Therefore em is equal to two Moeller. So the concentration of the solution is to Moller We have two moles of a compound dissolved in 20.5 liters of water. We know that big em is equal to small m divided by V where big em is equal to mill arat e which is what we want to calculate. Small M is equal to the number of moles present, which in this case is too. And V is the volume in leaders which in this case is 0.5 Therefore, mill arat e is equal to two divided by north 20.5. Therefore, hilarity is equal to four Moola. We have two moles of a compound dissolved in 250 mils of water on, we wish to calculate collaborative solution. We know that big em is equal to small m divided by V where big em is equal to my clarity, which is what we wish to calculate Em is equal to the number of moles which in this case is equal to two on V is the volume in leaders. Now here we have 250 mils of water. Therefore, we need to remember to convert this to leaders, which is in fact 0.25 liters. Therefore we get big em is equal to two divided by 8.25 Therefore, M mill Arat E is equal to eight Moola. In this example, we have two grams of a compound with a molecular weight of two grams per mole dissolved in one litre of water. And we wish to calculate the mill aridity of the solution. Now, two important points to keep in mind is that the molecular weight of a compound dissolved in one litre is a one molar solution. We can express this as big em is equal to G divided by M W multiplied by V where big em is equal to Molnar G is the number of grams. MW is the molecular weight and V is volume in liters. Now one approach we can take toe entre nous question is to work out the number of mold present. So the number of moulds small m would be equal to the number of grams present divided by the molecular weight. We know the number of grams is too. We know the molecular weight is to therefore em is equal to two divided by two. Therefore, em is equal to one mole. We know that this one mole is present in one litre. We know that mola began is equal to the number of moles divided by the volume. We know that small M molds is equal to one. We know that volume is equal to one liter. Therefore, big em is equal to one divided by one which is one. So the answer is one Moeller. Alternatively, we could use the equation that I showed in two. In this case, we've got big em is equal to small G, Divided by MW multiplied by V. We know that big em is equal to Moola and that is what we wish to calculate. G is the number of grams present, which is to m w is the molecular weight which is two grams per mole on V. The volume in leaders which we know is one so big em is equal to two divided by two times one, so that effectively gives us two divided by two. So the answer is one Moeller. In this example, we have one gram of a compound with a molecular weight off two grams per mole dissolved in one litre of water. And we wish to work out the mill aridity of the solution. Now there are two ways we can go about this. We can work through it in a syriza logical steps starting by working out the number of moles present on, then saying those molds are in the litre, all we can use an equation that was introduced in an earlier example. So first, let's have a look at working out the number of moles. We know that the number of moulds small em is equal to the number of grams small g present , divided by the molecular weight. In this case, we know that the number of grams present is one. And we know that the molecular weight is to therefore the number of moles present small em is equal to one divided by two and that gives us no 0.5 moles. So we have thes nor 0.5 moles. Now we know that big M hilarity is equal to the number of moles present divided by the volume in leader So em is equal to the mill Arat e small em is equal to the number of moles present, which is 0.5 in this case and V is the volume in liters which is one liter therefore big em is equal to nor 0.5 divided by one. Therefore, big em is equal to north 10.5 Moola. So the concentration of the solution is 0.5 Moola. Now the alternative approach we can take is where big em is equal to G divided by MW multiplied by V where big m equals mola which is the mill arat e that we wish to calculate . G is equal to grams which in this case is one M. W. Is equal to the molecular weight which is equal to two grams per mole and V is equal to the volume in leaders which in this case is one litre. Therefore, substituting these values into the equation we get Biggin is equal to one divided by two, multiplied by one therefore began is equal to nor 0.5 Moola. So again, the answer is no 0.5 Moeller Okay, In this example, we have one gram of compound with a molecular weight of two dissolved in half a liter of water and we need to work out the mill Ara, tee off the solution now. One approach we can take is to work out the number of moles off the compound present. We know that M equals G divided by the molecular weight where small m equals moulds on G equals grams which in this case is one gram. The molecular weight we know is two grams per mole. Therefore, the number of moles president is one divided by two, which is north 20.5 moles. Now this nor 0.5 moles is present in half a liter of water. We know that big M equals small m divided by V, where big em is equal to mola small em is equal to Moulds, which in this case is equal to 1.5 on V equals volume in liters which in this example is 0.5. Therefore, substituting these values into the equation, we get 0.5 divided by 0.5, which is one Moeller. Now, alternatively, we can work this out using an equation. Big em is equal to small G, divided by the molecular weight multiplied by the volume. We introduced this equation in an earlier lesson. Big em is equal to Moola. G is equal to grams which in this case is one m. W. Is equal to the molecular weight which in this case is two grams per mole and V is equal to the volume in liters which is nor 20.5 liters. Therefore, substituting these values into the equation we get big em is equal to one divided by to multiplied by 0.5 as two times 20.5 is equal to one. This means we have one divided by one. Therefore, the concentration off the solution is one Moeller In this example, we have one gram of a compound with a molecular wage of two grams per mole dissolved in 250 mils of water. Now there are two approaches we can take to working out the mill aridity of this solution. We can work out the number of moles present in the volume or we can use an equation. So to work out the number of moles present we use small em is equal to G divided by M. W. Where M is equal to moles, G is equal to grams. In this case, one gram and M w is equal to molecular weight which is two grams per mole. Therefore, substituting these values into the equation, we get one divided by two. Therefore the number of moles president equals 20.5 moles. Now this number of moulds is present in 250 mils of water. So using the equation, Big M equals small m divided by V where big m is the mill arat E that we're trying to work out. Small M is the number of moles present, which in this case is 0.5 and V is equal to the volume. Now remember, the volume is in liters on the question gave it in Mills. So the volume is 250 mils. But we need to convert this to leaders, which is 2500.25 liters. Therefore, substituting in the values we get big em is equal nor 0.5 divided by north 0.25 Therefore em is equal to two molder. Therefore, the concentration off the solution is to mola now using the equation that was introduced in an earlier lesson. We have big em is equal to G, divided by molecular weight MW multiplied by V. In this equation, em is equal to mola, which is the concentration of the solution. G is equal to grams, which is one gram m. W is equal to molecular weight which in this case is two grams per mole and V is equal to volume. Now remember again, volume is in leaders and this question has given in mills, so volume is equal to 250 mils but we need to convert that to leaders and 250 milliliters is 2500.25 liters. Therefore, we just substitute the values into the equation. We get big em is equal to one divided by two multiplied by 0.25 Therefore, em is equal to one divided by north 10.5. Therefore, the concentration of solution is to moulder.