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Current time:0:00Total duration:12:06

Worked example: Calculating amounts of reactants and products

AP.Chem:
SPQ‑4 (EU)
,
SPQ‑4.A (LO)
,
SPQ‑4.A.1 (EK)
,
SPQ‑4.A.2 (EK)

Video transcript

we're told that glucose c6h12o6 reacts with oxygen to give carbon dioxide and water what mass of oxygen in grams is required for complete reaction of 25.0 grams of glucose what masses of carbon dioxide and water in grams are formed so pause this video and see if you can have a go at this and then we'll work through this together all right now first let's just set up the reaction so this is going to be we have glucose so that is c6h12o6 it's going to react with oxygen now oxygen in its molecular form is going to be o2 and what it gives is carbon dioxide and water plus water now the next question is are we balanced do we have a conservation of mass here and let's just go element by element so first let's focus on the carbons so we have six carbons on the left hand side of this reaction how many carbons do we have on the right hand side well right now we only have one carbon in this carbon dioxide molecule so if we want the carbons to be conserved we need to have six carbons on the right hand side as well so let me see if what will happen when i throw a 6 there so now my carbons are balanced 6 on the left 6 on the right now let's look at the other elements so on the left hand side i have 12 hydrogens in total now let's see on the right hand side i only have 2 hydrogens so if i want to balance that i could multiply the water molecule instead of instead of just one here i could have six and now this would be 12 hydrogen atoms so both the carbons and the hydrogens are now balanced by putting that six there i haven't messed with the carbons and now last but not least let's think about the oxygens here so on the left we have six oxygens there and then another two oxygens there so that's a total of eight oxygens and on the right i have 6 times 2 i have 12 plus another 6 oxygens so i have 18 oxygens on the right hand side and i only have 8 oxygens on the left hand side so i have to increase the number of oxygens on the left left-hand side so let's see this six you can see you can imagine matches up with this six so if i could somehow make this into 12 oxygens i'll be good so the best way to make this into 12 oxygens is to multiply this by 6. so let me do that so if i put a 6 right over here i think i'm all balanced i have 6 carbons on both sides i have 12 hydrogens on both sides and it looks like i have 18 oxygens on both sides so i'm a fully balanced equation here so let me get a periodic table of elements and i only need to think about carbon hydrogen and oxygen so for the sake of space let me scroll down a little bit i just need to look at this stuff over here and so let's first let's see hydrogen's right up here and we can see in this periodic table of elements it gives the average atomic mass but you can also view that number 1.0 as its molar mass so it's 1.008 grams per mole of hydrogen now we can move on to carbon carbon is 12.01 12.01 grams per mole of carbon and then last but not least oxygen over here that is 16. grams per mole of oxygen and now we can use that information i can now remove our periodic table of elements we can use this information to figure out the molar masses of each of these molecules so for example glucose right over here if we're talking about c6h12o6 how many grams per mole is that going to be gram how many grams is a mole of glucose going to be well it's six carbons 12 hydrogens and six oxygens so one way to think about it's going to be six times this so it's going to be six times twelve 12.01 plus we have 12 times 1.008 plus 6 times 16.00 and then this is going to be grams per mole and this is going to be equal to let's see 72.06 plus 12.096 plus 6 times 16 is 96.00 and of course all of this is in grams per mole and so this is going to be equal to let's see 72 plus 12 is 84 plus 96 is 180 180 point and let's see we have 60 thousands plus 96 thousands which would be 156 thousandths so and 156 thousands grams per mole and let me make sure i got the significant figures right the 6 the 12 and 6 are pure numbers so i can still i'm still good with this and then when i add these numbers together i need to round to as much precision as i have in the least one so here i go up to the hundredths up to the hundreds up to the thousands so i need to round this actually to the hundredths place so this is going to be 180.16 grams per mole of glucose and now let's think about the other molecules if we think about oxygen i'll do that over here to save space oxygen that's pretty straightforward a molecular oxygen molecule just has two oxygen atoms so it's going to be two times this it's going to be 32.00 grams per mole and then carbon dioxide carbon dioxide is going to be it's two oxygens plus one carbon so it's going to be this plus one carbon so that's this plus 12.01 so that's 44.01 grams per mole and then last but not least we have the water and so h2o this is going to be two hydrogens so that's two times 1.008 so that's 2.016 plus an oxygen so that's going to be 2.016 plus 16 is going to be 18 point once again we only go to the hundreds place here so i'm going to round to the hundredths place here so 18.02 grams per mole now the next step is to think about all right we're reacting with 25 grams of glucose how many moles of glucose is that so we have 25.0 25 grams of glucose and we want to turn this into moles of glucose and we know the molar mass is 180.16 grams per mole or we can multiply and say that this is for every one mole per mole it is 180.16 grams all i did is take the reciprocal of this over here and notice the grams will cancel with the grams and i'll be left with mole of glucose or moles of glucose is going to be equal to 25.0 divided by 180.16 moles of glucose 25 divided by 1 180.16 is equal to this and let's see i have three significant figures divided by five significant figures so i'm going to round to three significant figures so 0.139 so this is approximately equal to 0.139 mole of glucose now they say the first question is what mass of oxygen is required for a complete reaction of this well for every mole of glucose we need six moles of molecular oxygen and so we're going to need let me just multiply that times six so times six is equal to that and remember i had three significant figures so 0.833 so we're going to need 0 three 0.833 moles of molecular oxygen and then i just multiply that times the molar mass of molecular oxygen so times 32.00 grams per mole of molecular oxygen 0.833 times 32 is equal to that if we go to three significant figures it's 26 26.7 grams of oxygen of molecular oxygen and so we've answered the first part of the question what mass of oxygen is required for complete reaction so that's this right over here and then the next question is what masses of carbon dioxide and water in grams are formed remember for every one mole of glucose we needed six moles of molecular oxygen and we produce six moles of carbon dioxide and we produce six moles of water so this is the number of moles of glucose we input into the reaction six times that was this so this is actually also the number of moles of carbon dioxide or water that we're going to produce so if we take our original 0.833 mole of carbon dioxide and i multiply that times carbon dioxide's molar mass 44.01 grams per mole this is going to be approximately equal to 0.833 times 44.01 is equal to three significant figures 36.7 grams 36.7 grams of carbon dioxide once again you can see that the moles cancel with the moles just as they did before to give us grams of carbon dioxide and then last but not least we can do that with the water 0.833 mole of h2o times its molar mass times 18.02 grams per mole of water is going to give us approximately 0.833 times 18.02 gives us three significant figures is going to be 15.0 15.0 grams of water and once again those moles canceled out to give us the right units and we're done which is pretty cool this is really useful in chemistry to be able to understand based on a balanced chemical equation to be able to understand hey if i have a certain mass of one of the inputs one of the things that are one of the reactants how much do i need of the other and then how much mass of the of the products am i actually going to produce all of which we just figured out