hey everybody welcome to chapter eight uh we've gone over some of the basic skills of converting grams to moles and balancing reactions in previous chapters uh we're going to move on and work in quantities and chemical reactions or in other words we're going to tackle this difficult to spell or difficult to say word called stoichiometry so go ahead and you're going to feel silly if you're just sitting by yourself right now but pronounce it with me stoichiometry okay less scary if we can say it doesn't seem like quite a frightening topic so really just boils down to predictions it's predictions with ingredients lists predictions with recipes how much are we going to make how much are we going to need when you see these types of word problems you are dealing with stoichiometry stoichiometry deals in ratios just like recipes that we would use at the house so for example pancakes let's look at a sample recipe we've got one cup of flour we've got two cup of eggs and one half teaspoon of baking powder now this ratio of ingredients of one to two to one half is enough to make five pancakes now the assumption here is that the massive ingredients we start with are the mass of ingredients that we end up producing so just like in a chemical reaction and ratio is key so if we were looking at what we needed to make five pancakes we can't just up and increase one cup of flour to five cups of flour pancakes are going to be terrible we're not going to get the intended products so ratios are key they tell us how much we're going to need and how much we're going to make so let's focus in on two pieces here we want the ratio of eggs to pancakes so i need two eggs for every five pancakes so what we can start looking at this as is a conversion factor we're going to treat this just like the conversions that we did back in chapter two only we're going to be using coefficients from balanced reactions remember the coefficients are what we learned about when we started balancing they're the large numbers out front so this 2 would be the coefficient for eggs this 5 would be the coefficient for pancakes now these ratios will build the conversion factors that we use to move say between what we have and what we want so let's try one of these out this problem says how many pancakes so pancakes are what we want will eight eggs make so eight eggs is what i have if we have enough of everything else so it's key that it says if we have enough of everything else because let's say we didn't have any flour if we don't have any flour we can't follow this recipe and make the pancakes so we're assuming that we have enough of everything else like flour and baking powder lying around so let's start with what we're given we have eight eggs and our ratio from the coefficients this two to five ratio of eggs to pancakes i'm going to put eggs on the bottom here that way they'll cancel opposite sides top and bottom i'm going to put pancakes on the top and the recipe told us we'd make five pancakes for every two eggs that we used with the recipe so then we'll take eight times five divided by two that works out to a total of twenty pancakes so it's going to get a little more intimidating when we jump into stoichiometry because we're no longer dealing in eggs and pancakes we're dealing in chemical formulas but it's the exact same math so keep that in mind this may look like chemical formulas that we're jumping into but really it's just pancake science it's pancake mat okay there are two things that we have to have if we're going to jump into stoichiometry one is a balanced reaction the second moles so if you were going to get a tattoo from chapter eight this would be it so these are your two reminders every stoichiometry problem that you do you're going to be looking did they give me a balanced reaction and then did they give me moles so first off let's just kind of go back over the idea of a balanced reaction so here we've got a reaction it's called the haber process and in it we're taking hydrogen and nitrogen or forming ammonia and it tells me ammonia is an h3 so even if i forgot that typically in these space filling diagrams my little white spheres are hydrogen then because it's nh3 and i've got three of these little white spheres i know that i've got three h's 1n same thing here so based on the diagram on the product side of the reaction i have two ends total and i have one two three four five six h's so that gives me on the left side if i do the same thing start labeling i've got six h's total here and i've got two ends so the number and type of atom are balanced on both sides of the reaction so i'm not creating mass i'm not destroying mass so the law of conservation of mass follows that's why we're balancing these reactions but more importantly we have a chemical reaction happening now why do we have a chemical reaction happening because we have a rearrangement of the atoms and molecules so if we look at the reactant side i had nitrogen atoms bound to one another when i go to the product side i don't have nitrogen atoms bound to each other i have nitrogen atoms bound to hydrogen atoms so a rearrangement of atoms creation of a new substance here so i've got a chemical reaction occurring not just a phase change say if we were to melt ice or if we were to boil water so let's write a balanced chemical reaction i've got h2 molecules here remember h and n both show up as diatomic molecules i've got one two three of these molecules so three h2s i'm also going to have one of these n2 molecules and that's going to make one two of these nh3 molecules now we could have done this just from what they gave us by the description in saying this equation shows hydrogen which we would want to know as diatomic and nitrogen go to make ammonia remember it's not ammonium ammonium was our polyatomic ion nh4 instead this is ammonia which is an h3 we could put our blanks in and balance just as we did in chapter seven so if i have two ends here one times two gives me two n's i bounce to the other side rebalancing to either side of the reaction arrow i would put a 2 here 2 times 1 gives me 2 n's 2 times 3 gives me 6 h's bounce back to the other side 3 times 2 gives me 6 h's so we can balance based on the diagram we could balance based on the balancing skills that we've already picked up but what's most important here is that we know how to balance because these coefficients that i'm circling lay out the recipe for every three h2s i'll make two nh3s for every one eight one n2 that reacts i'll make two nh3s so what do these coefficients really mean they play into the ratio they play into the recipe for the math that we're going to be carrying out if we balance incorrectly and these coefficients are off and we're in trouble later in the problem so we have to be really careful balancing now this could represent moles it could represent molecules it just depends on what scale we're thinking of when we deal in the laboratory we're thinking in terms of moles so this would really be three moles of h2 molecules remember a mole is just 6.022 times 10 to the 23rd molecules we're just making it easier to discuss large numbers of molecules so this would be a mole ratio of 3 moles of h2 to 1 mole of n2 to 2 moles of nh3 produced so that's how we're interpreting these coefficients so let's actually do our first prediction so we have a problem that says we have three moles of n2 so i'm going to label that what we have and we've got it says more than enough h2 so i'm just going to assume i have what i need to get it done and move on how much nh3 can we make so i want nh3 so on my problem here they've given me a balanced reaction remember we said when we started we were going to check for two things it's a balanced reaction taken care of did they give me moles or grams the problem gives me moles good okay now if i have these two i can jump straight into the stoichiometry which means i'm going to be used using the coefficients of what i have so n2 so i'm going to focus on this one and what i want my nh3 so i'm also going to be using this coefficient of 2 and i'm going to take what i start with this 3 moles of n2 i'm going to use that recipe that ratio of coefficients that tell me for one mole of n2 i'm going to make two moles of product and h3 so my moles of n2 cancel 3 times 2 divided by 1 is going to give me 6 moles of nh3 and this step of going from one substance what we start with the n2 to the substance we're interested in nh3 and using this right here this was a ratio of coefficients this is stoichiometry balancing the reaction that was just getting ready for it making sure we're in in moles before we start again just getting ready for it but using the ratio coefficients to go from one compound to predicting what we want that's what stoichiometry is okay so remember you're going to hear this a lot during this lecture moles and balanced we need both if we're going to do stoichiometry so the previous problem that we did it was a mole to mole stoichiometry problem they gave us moles to start with and they asked us to predict how much nh3 we could make we predicted it in terms of moles this was a more straightforward problem if you think about what our checks look checklist is if i start in moles easier problem but let's add a twist the twist is we're now going to do mass to mass stoichiometry where i start in grams and i end in grams so i'm going to be using molar masses off the periodic table to convert between grams and moles so let's take a look at whoops let's see if i can hold it still let's take a look at where we start here we're going to start with what we're given we're given 51 grams of nh3 so i'm going to write have here and over here how much h2 is reacted in grams so this is what we want the reaction we're looking at is the same reaction that we've been dealing with so far and that was 1 n2 plus 3 h2s goes to 2 nh3 so it wants me to predict h2 and i have an h3 so the two coefficients that i'm most interested in are my two and my three it gives me well let's make our checklist moles and a balanced reaction since we're using the same reaction already taken care of already balanced but it doesn't give me moles it gives me grams so that's my good starting point here now if i look at nh3 nitrogen is roughly 14 grams per mole hydrogen is roughly one gram per mole so if i have nh3 i have one nitrogen and i have three hydrogens so when i add this up that adds to a total of 17 grams per mole for the molar mass of nh3 so if i'm going to convert this 51 grams to moles so i can do stoichiometry that's the best place to start by finding the molar mass i'm going to take that 51 grams of nh3 i'm going to divide by 17 grams for every one mole what we added above the periodic table grams cancel and i'm left with three moles of nh3 so that was my first step my first step was taking what they gave me and getting it in terms of moles so now i've got moles in a balanced reaction so now i can jump in and do the actual stoichiometry i've got three moles of nh3 that was what i started with and now i want moles of h2 so here's where i'll use my ratio of coefficients and up here it told me for every two nh3s i will get three moles of h2 and again that came from this 2 and this 3. my moles of nh3 cancel 3 times 3 divided by 2 that's going to give me 4.5 moles of h2 so just to recap this first step was converting grams to moles and that was for nh3 this second step was the actual stoichiometry and for our third step remember the problem asked me not to predict in moles for nh2 it asked me to predict in grams so my last step is going to convert moles of h2 to grams of h2 so for h2 if i looked on the table h weighs roughly one so if i'm starting with h2 that means that i would have two times this value for h which gives me roughly two grams for every mole so then i can take that four and a half moles that i just found of h2 and we would say that there are for every one mole of h2 two grams that was our molar mass moles of h2 cancel four and a half times two divided by one that gives me nine grams of h2 the answer that i was looking for and then this last step just as a reminder to ourselves the whole purpose was to convert moles to grams and that was for h2 so don't uh don't just focus on the answer focus on how we got to the answer so if you want to try and pause this later try and remind yourself why you're doing each step this whole problem banked on us being able to balance the reaction before we started it banked on us being able to move comfortably between grams and moles so if you're getting stuck there head back to our previous chapter rework the skills on converting between grams and moles and balancing i'm here to help you shore up those foundations because we've got to have those strong if we're going to jump into stoichiometry okay so let's try this one again we're going mass to mass so we're going to look for two things did they give me moles and did they give me a balanced reaction so let's look through the problem what mass of carbon dioxide is emitted by an automobile per 5.0 times 10 to the 2 grams of pure octane used so used that tells me it must be a reactant we know that this is just oxygen this is diatomic oxygen so it makes sense if this was octane and also think oct8 that's that eight carbons so i have 5.0 times 10 to the 2 well that's 500 grams of octane so did they give me moles no did they give me a balanced reaction no they just gave me these blanks so let's start there i don't have moles and i don't have a balanced reaction yet but we will get there so let's clear this off and we'll start off by balancing the reaction okay good place to start this is a combustion reaction so we're going to be balancing in the order cho i'm going to put a 1 here to begin with 1 times 8 gives me 8 carbons i'm going to bounce to the other side to get eight carbons here i need to put an eight in front eight times one gives me eight c's both sides have eight carbons carbons are balanced how about h's i've got 1 times 18 i've got 18 h's on this side so over here if i put a 9 9 times 2 gives me my 18 h's h's are balanced o's are the tricky one because they show up in three places eight times two gives me 16 o's and then nine times one gives me 9 more o's that's going to equal 25 so if i want to balance this i need 25 total o's on this side what times 2 would give me 25 12.5 now here's our issue we have to have the lowest set of whole numbers can't reduce them any further once we have the best set and right now we don't have that we've got a decimal we've got a 0.5 so the easiest way to get rid of a half is to make it a whole we're going to double it but we can't just double the 12 and a half we have to double everything so i'm going to double this to i'm going to double this to 12 and a half whoops i'm going to double that to sorry 25 i'm going to double this to 16 and 18. so those are our coefficients remember we've got to take our time get a good set of coefficients or it all goes downhill from there so balanced check and let's see let's go through the problem a little more it said we had 500 grams of pure octane used so let's put that note back down here 500 grams of c8h18 this is what i have so this 2 is going to be a coefficient that i'm going to use later what i want is mass of carbon dioxide so i'm going to be using this 16 coefficient because that's the coefficient for what i want so step one was to balance we're done next we're going to move on to step two so step two is going to take this 500 grams and get it to moles so if i look at c and h off the periodic table carbon weighs roughly 12. hydrogen weighs roughly 1. if i have c8h18 then i'm going to have a total molar mass of about 140 grams per mole so i'm going to use this 114 grams per mole to take me from grams of c8 h18 moles so let's start with that 500 grams and i've got 114 grams for every one mole put the grams on the bottom they cancel with the grams on the top and 500 divided by 114 gives me 4.386 and this is moles of c8 h18 okay we got a balanced reaction we got moles to start now so now we can jump into the stoichiometry so step three stoichiometry i'm just going to abbreviate it here here's where we're going to go between moles of c8h18 which is what we have here to moles of what we want moles of co2 so let's start with our 4.386 moles of c8h18 i'm going to use that ratio of coefficients so there are two moles of c8h18 for every one let's see 16 moles of co2 so i set it up so that moles of what i have will cancel i'm left with moles of co2 i'm going to take 4.386 times 16 divided by 2 and that gives me 35.09 moles make that point a little clearer moles of co2 we're not done yet though because the problem specified it didn't just want carbon dioxide it wanted mass so i need my final answer in grams so i need to take these moles and convert to grams so step four is going to be take grams of co2 or excuse me take moles of co2 which is what we just solved for and convert it to grams of co2 for our answer to do that we're going to use the molar mass of co2 for co2 carbon roughly 12 oxygen off the table roughly 16. so i've got 1c i've got 2 o's when i take that and add it up i've got 44 grams per mole so what i can do is take that 35 .09 moles of co2 that i started with i want to cancel moles for every one mole i have 44 grams of co2 so i get 35.0 times 44 and that gives me 1544 approximately grams of co2 and we solve for mass of carbon dioxide my final unit that didn't cancel was grams so let's just go back and do a 30 000 foot view of what just happened we started out knowing that we had to have moles in a balanced reaction the problem didn't give the reaction to us balance so our first step was to balance it we then had to go from grams of pure octane which is what we started with this is our second step we converted grams of what we started with to moles now we have moles in a balanced reaction so we could do step three this is the actual stoichiometry we used a ratio of the coefficients or the big numbers out front that we used to balance the reaction we used that as a conversion factor to go from moles of what we have to moles of what we want step four we needed a mass for our answer so we took moles of what we wanted and converted to grams of what we wanted we had to use the molar mass of co2 to get that done so the molar mass was our conversion factor to move from moles of co2 to grams of co2 and then final answer so again the focus here understand why you're doing each of these steps not just copying because problems can be very different when you jump from word problem to word problem but if you understand why you're doing each step that's how you're going to really ace the next exam that's how you're going to really understand this topic so we're going to go more real world scenario now we're going to jump into limiting reactants and we're going to look at a recipe that we used at the start of the lecture we're just going to take a slightly different spin on it now if you look at the problem we just did they gave us one value to start with they said that we had 500 grams of octane and then said how much product do we make an even earlier problem they said we had 51 grams of ammonia how much h2 do we make or we have three moles of n2 how much nh3 can we make what do all three of those have in common they only gave me one starting value if i look at the problem that we're about to tackle do they give me a balanced reaction yeah they give me the balanced reaction it has coefficients in it it's a recipe but the different thing is they don't give me one value to start with they tell me that i have three cups of flour or 10 eggs or four teaspoons of baking powder i'm going to abbreviate that bp and then they ask me how many pancakes i can make if you look in your cupboard at home and you pulled out a random recipe chances are you don't have the exact amounts of each ingredient that you would need sitting in your cupboard you're going to have extra of some you're going to have not enough of others so that's where limiting reactants come in limiting reactants are where they give you information on more than one of your ingredients or reactants and we have to figure out what runs out first what we have extra of there are extra ways we can look at these problems so let's do this we're going to focus on one reactant at a time and predict the number of pancakes that we could make so let's start out doing flour so in starting with flour i've got three cups of flour and if i use a ratio of these coefficients for every one cup of flour i'm going to produce five pancakes cups of flour cancel so three times five based on my flour i have enough to make 15 pancakes so now we looked at it based on the flour what we could make now let's look at the eggs for the eggs i've got ten and the recipe says that for every two eggs i could make five pancakes eggs cancel 10 times 5 divided by 2 so based on the eggs that i have i could make 25 pancakes now let's look at the baking powder it says i have four table or excuse me teaspoons of baking powder and it says one half of a teaspoon of baking powder for every five pancakes so i'm going to put i could write a half but makes a little easier for me to think of it as 0.5 teaspoons of bp for every five pancakes so 4 times 5 divided by 0.5 is going to give me 40 pancakes so now the question is do we add these up do we subtract them where do we go from here well take a look at this from the top with me so we made three predictions one with the flower on how many pancakes we make one with the eggs on how many pancakes we'd make and one with the baking powder on how many pancakes we would make each one predicted a different number of pancakes what that translates to is i only had enough flour to make 15 pancakes so that means that my flour is going to run out first which means that my flour is going to limit how many pancakes i'm going to make so i would call my flour my limiting reactant because it predicts the least amount of product that i can make so while the flour runs out making only 15 pancakes theoretically this is the most pancakes we can make with the ingredients that we have so we would call this our theoretical yield theoretical yield so 25 pancakes we can disregard because we have enough eggs to make 25 pancakes but we don't have enough flour the 40 pancakes same thing that number really doesn't mean much other than 40 pancakes is how much baking powder we have to make 40 pancakes so we would say that our eggs and our baking powder they aren't limiting instead they're called excess they're excess or i have extra laying around when i'm finished because they're predicting i could make more product than i can actually make so let's turn this into a yield question so when i looked at this at first we calculated how many pancakes we could make this is theoretical we're just putting pen to paper and saying in a perfect world if nothing goes wrong i could make 15 pancakes for anybody who has ever cooked pancakes you know things never go 100 perfectly so let's say that we dropped two of our pancakes we burned two of our pancakes so theoretically we could have made 15 pancakes but really we only made 11 that survived to the breakfast table so if we were to calculate a percent yield our percent yield would be 100 oops would be 100 which makes it a percentage times the actual yield what we actually made that survived cooking divided by the theoretical or how much we could have made so let's label this did make and could have made and when we work that out in the calculator 100 times 11 divided by 15 that works out to about 73 which makes sense if we had had 16 pancakes 4 out of 16 would have been about 25 so i would expect a percent yield or how much we actually collected 11 out of 15 is pretty close to that so we made about 73 percent of the pancakes that we could have made so not a bad yield all right so we've gone over some new language let's just catch up on that the limiting reactant or limiting reagent it's completely consumed so we could also see this as used up first is another way that you can see the limiting reactant listed the theoretical yield is what we could make it's not what we actually make that's what we could make theoretically based on what we start with the actual yield is what's actually produced so if we carried out this reaction we spilled some we lost some when we transferred it we didn't let it completely react it's not necessarily all we could make but it is what we did make and then the percent yield compares what we did make to what we could have made as a percentage we multiply by a hundred percent so you will see percent yield calculations following up these stoichiometry problems all right so let's go back and see if we can approach a few different types of problems first we're going to look at limiting reactant problems and we're going to start in moles we're going to end in moles so let's see why this is a limiting reactant problem how we could recognize it well it says what's the limiting reactant first it doesn't always say that sometimes it says things like what's the reactant that runs out first but just mechanistically what tells me this is a limiting reactant problem is that i have two values to start it tells me i have 1.8 moles of titanium and 3.2 moles of chlorine so it gives me two halves and it has a single want it wants the theoretical yield or what we could make of ticl4 so i'm going to use this reaction that they've given me up top i'm going to do my normal checks did they give me a balanced reaction yes did they give me moles yes so that means i can jump straight into the stoichiometry now this problem tells me i have 1.8 moles of my first reactant and i have 3.2 moles of my second reactant and i'm going to use each one of these independently to predict how much of my product i can make so first we're going to start with our 1.8 moles of titanium i want to cancel titanium so that's going to go on the bottom and for every one mole of titanium i'm going to make one mole of titanium chloride so one mole of ti for every one mole of ticl4 my moles cancel 1.8 times 1 over 1 that's my kind of math i'm going to get 1.8 moles of my product so again stoichiometry we're just pulling directly from the coefficients to get a conversion factor so we did the first reactant now we're going to do the second reactant and figure out how much product we'll make so i'm going to use 3.2 moles of cl2 i want moles of cl2 to cancel my coefficient for cl2 is two so two moles of cl2 coefficient for my product is one for every 1 mole of ticl4 moles of cl2 cancel 3.2 times 1 over 2 gives me 1.6 moles of product now remember what we did when we were looking at pancakes when we were looking at pancakes we did three calculations each one predicted that i would make a different number of pancakes in the end i went with the smallest most conservative projection of how much we would make so here 1.8 versus 1.6 this will be my theoretical yield the smaller amount so the problem asks for the theoretical yield theoretically we can make 1.6 moles of ticl4 if this asks for the theoretical yield in grams i would have had to go an additional step and use the molar mass of ticl4 to take it to grams of ticl4 because it didn't ask that then i'm good just to stop here where i've predicted the smaller of the two amounts to be my theoretical yield now if we backtrack a little bit what did we use to predict this smaller amount we used cl2 so cl2 must run out first so that's going to be my limiting reactant that means that my titanium because it predicted a larger amount would be my excess reactant so just a word of caution here if i ask you to identify a limiting reactant well it's a reactant so it's going to have to be in this case either titanium or chlorine it's not going to be the product because remember it's a limiting reactant okay so let's jump a step further we're going to look at gram-to-gram stoichiometry we're going to still jump into limiting reactants and wrap up by calculating percent yield so looking at this problem here we have two values given to start with we have sodium we also have chlorine and it says how much product is made our product is nacl it's table salt so this is what we want and it says what is the percent yield so we're going to have to figure out how much we could theoretically make and then we'll be able to calculate our percent yield now let's do our checks we've got to look for a balanced reaction and we've got to look for moles i can tell you right off the bat they gave us grams so that's a problem we're going to have to get over to moles and for a balanced reaction this was intended to be given to you balanced but small typo here i'll go ahead and correct it i've got two cls i've got two cls i've got two sodiums there should be a two out front to give us our balance of sodium so now it's balanced but we do have to get moles to start off the problem so first let's make our first goal to take grams of n a to moles of n a and from there we can just carry out our calculation to calculate our theoretical yield really in one step so if i look up sodium on the periodic table n a is about 22.99 grams per mole it says that i start out with 53.2 grams of sodium so i can use this molar mass off the table to go between grams and moles it's 23.8 grams of sodium for every one mole of sodium my grams of sodium cancel and that pauses me there with moles of sodium now let's keep going since i now have moles and the reactions balanced let's just go ahead and calculate how much product we would actually make so i want to cancel moles of sodium and my coefficient for sodium is two so i could say for every two moles of sodium i get two moles of product my moles of n a cancel i got this second conversion factor from my coefficients because this is where my stoichiometry came in i'm left with moles of an acl when i'm done so i would take 53.2 multiply by what's in the top here and divide by what's in the bottom so that equals 2.2175 that is moles of an acl so we looked at the amount of product that we could make based on the sodium so now let's look at the chlorine so for my second step here i have grams of chlorine so i'm going to start by going grams of cl2 to moles of cl2 so i've got 65.8 grams of cl2 if i look up chlorine on the periodic table chlorine is 35.45 grams per mole and this is cl2 so i'm going to double that value which is going to give me 70.9 grams of cl2 for every one mole of cl2 so again this first step was just using our molar mass and that took us from grams to moles of cl2 our second step we can also build in just like we did in the first calculation and we could say for every one mole of cl2 we're going to get two moles of our product so let's look at how everything canceled out grams of cl2 canceled moles of cl2 canceled and left with us with moles of product which is what we were looking for so 65.8 times 2 divided by the molar mass of chlorine and that gives me 1.8561 moles of nacl now just like when we did the pancakes we looked at each of our ingredients we looked at sodium we looked at chlorine we predicted how much product we would make the sodium said that we could make about 2.2 moles of nacl the chlorine said we could make about 1.85 moles of nacl so here we're predicting less so this is going to be our answer this is going to be our theoretical yield in other words this is what we could make now we didn't have to go all the way to grams to make this prediction we could just say we're making fewer moles of product so this is going to be my theoretical yield now we know that when we actually carried out the reaction that 86.4 grams was actually collected so that's what we did collect so to be able to wrap up this calculation and do a percent yield i need to have these both in terms of grams or both in terms of moles so i'm going to convert the amount of nacl that i could make into grams so we can go ahead and calculate our percent yield so let's take that 1.8561 moles of nacl and i'm going to take the mass of a single n a the mass of a single cl to calculate the molar mass of nacl that works out to 58.44 grams per mole now for every one mole of nacl i'm going to have 58.44 grams remember that comes from a single n a a single cl and we know how much each of these weighs so this 58.44 is really just the sum of 35.45 and 22.99 straight off the table so my moles cancel i'm left with grams 1.8561 times 58.44 that's going to give me about 108.46 and this is grams of nacl now again this is my could have made this is my theoretical it's just now i'm expressing that theoretical yield in terms of grams because my actual yield was in grams they've gotta agree for percent yield calculations so step four just wrapping this up calculating percent yield i put what i did collect the problem told me that was 86.4 grams i'm going to divide that by what i did or what i could have collected which was 108.46 grams so actual divided by theoretical and then i'm going to multiply that by one hundred percent and that gives me about a seventy nine point seven percent yield all right so buckle up this is our very last problem of this lecture uh we're going to work in the idea of limiting reactants we are given two values so we have 11.5 grams of copper one oxide we have 114.5 grams of carbon and it says how much copper will be made so what we want is copper and it says what reactant runs out first so this would be where when we look back at our terminology the limiting reactant or limiting reagent this is what is used up first it's what runs out first what is fully consumed so different ways of wording it where really it's just looking for the limiting reactant so be really familiar with these terms okay so i'm going to make a note to myself i want oops the limiting reactant okay and it does give me an actual yield now let's back up for a minute when you read through this problem it says how much copper will be made and what reactant runs out first really what it's having us calculate this problem wants us to find a theoretical yield it doesn't say anything about calculating a percent yield so really this right here is just extra information and you'll find that in some problems so don't get tricked by it if it's not asking me for a percent yield i really didn't need the actual yield to begin with so let's do our our limiting reactive problem where we look at one reactant at a time we're going to look at the copper one oxide and predict how much copper will make look the carbon and predict how much copper will make then we select the smaller of the two predictions that will be our theoretical yield or how much copper will be made and whatever predicts the smaller amount that'll be our limiting reactant so step one i'm going to take that 11.5 grams of copper one oxide and i'm going to need to take this from grams to moles because look at me i forgot to uh write this down moles in a balanced reaction i think i just do that part automatically in my head which really that's where you want to be you want to go through that moles and balanced reaction check in your head did they give me a balanced reaction yes i've got two coppers two coppers one oxygen one oxygen one carbon one carbon so it's balanced but they give me grams instead of moles so that's why we're starting here so here i'm going to need the molar mass of copper one oxide if i look this compound up i can look up copper on the table and copper is 63.55 grams per mole and i have two of those in the compound oxygen is 15.99 grams per mole and i've got one of those so 63.55 times 2 plus 15.99 times 1. is going to give me the value 143.09 and that's the number of grams of copper one oxide that show up in one mole so grams cancel and that's just taken me from grams to moles so now if i have and let me make a little extra room for us here if i have moles of copper oxide then i can go straight to what i'm looking for which is how much copper i'm going to make so our next step is going to be the stoichiometry where we look at the coefficients now copper one oxide there's a coefficient of one so for every one mole of copper one oxide i'm going to make two moles of copper moles of copper one oxide cancel i'm left with moles of copper so now i have 11.5 times 2 divided by 143.09 that's going to give us our moles of copper so that gives me about whoops zero to get my pen back here that gives me about 0.161 moles of copper so based on the copper one oxide that we had this is how much copper that we could make so now let's repeat this calculation only now we're going to focus on the carbon so i'm going to take that 114.5 grams of carbon and i'm going to repeat the same calculation i'm going to look at the molar mass and that's going to be for carbon and that's going to take me from grams of carbon to moles of carbon because i need moles to do my stoichiometry for carbon when i look it up on the table it's 12.011 uh so we'll go with 12.01 here i've got 12.01 grams for every one mole grams cancel and that's going to take me to moles of carbon so that's our first stepping stone just like in the top here now i can do my stoichiometry because i've got moles i've got a balanced reaction so now i can go to the coefficients for every one mole of carbon i'm going to make two moles of copper moles of carbon are going to go on the bottom moles of copper are what we're trying to finish with they're going to go in the top moles of carbon cancel and i'm going to end up with 114.5 times 2 divided by 12.01 and that is going to give me 19.07 moles of copper so let's digest this for a minute based on the amount of copper one oxide that i have i could make .161 moles of copper based on the amount of carbon that i have i could make way more copper but when i actually carry this out in the lab one of these is going to run out the one that predicts the smaller amount this is going to be my theoretical yield so when it asks how much will we make how much could we make this is how much we could make when i backtrack and see what helped me calculate that that means that this copper one oxide is going to be limiting or in other words like the problem asked it's going to run out first that means that my carbon which predicted the way larger amount of copper that could be made is not limiting but it's excess so uh just public service announcement this was a shorter lecture than we normally have but that's because lecture isn't as important in this chapter practices i've posted a lot of extra tutorial videos a lot of web resources you want to pause these videos and be able to work them yourself because stoichiometry problems are coming they're a culmination of our skills they test do we know whether we can balance do we understand moving between grams and moles because once we understand those we can start to get into stoichiometry but make sure please practice this week because that's going to pay off big when it comes to exam time i'm here if you have any questions and good luck