Combustion analysis is performed on 0.50 g of a hydrocar-bon and ...

Question

Combustion analysis is performed on 0.50 g of a hydrocar-bon and 1.55 g of CO2, and 0.697 g of H2O are produced. The mass spectrum for the hydrocarbon is provided below. What is the molecular formula? (LO 3.12 and 3.13)

(a) C5H11 (b) C8H18 (c) C11H10 (d) C10H22

Accepted Answer

Hi everyone. This problem reads 0.75 g of a hydrocarbon is subjected to combustion analysis 2.30 g of carbon dioxide and 1.10 g of water are produced if the hydrocarbon has the following mass spectrum, what is its molecular formula? Okay, so we want to write out its molecular formula Before we can write out the molecular formula. We first need to write out the empirical formula. And the first step that we're going to do to write the empirical formula is figure out the moles of both carbon and hydrogen. Starting with what we're given. All right, so let's go ahead and write out. Step one is the empirical formula. Okay. And so let's go ahead and start off by figuring out our moles of carbon. So, the way that we're going to figure out our moles of carbon, we know that we're starting off with 2.30 g of carbon dioxide. So, our goal here is we want to go from g of carbon dioxide, two moles of carbon. So this is where we're starting. And this is where we want to go. So let's go ahead and figure this out. Alright, so we have grams of carbon dioxide and we want to go to moles of carbon dioxide. So we need the molar mass. So in one mole of carbon dioxide, The Molar mass is 44. g of carbon dioxide. So looking at our units are grams of carbon dioxide cancel and we're left with moles of carbon dioxide. So now we want to go from moles of carbon dioxide to moles of carbon. Alright. And we need to look at our multiple ratio here. Alright, so in one mole of carbon dioxide we have one mall of carbon. Alright, so Making sure our units cancel properly our moles of carbon dioxide cancel. And we're left with moles of carbon. So once we do this calculation we get 0. 5 2-6 moles of carbon. So now that we figure figure it out our moles of carbon, we need to figure out moles of hydrogen. All right. So We'll start off with what's given and we know we have 1.10 g of water. Alright, so our roadmap here is we want to go from g of water. two moles of hydrogen. Alright, so We're starting with 1. g of water. So now we want to go from grams of water, two moles of water. And we need the molar mass of water. So in one mole of water, The Molar mass is 18.016 g of water. So our grams of water cancel. And we're left with moles of water. So now we want to go from moles of water to moles of hydrogen. Alright, so in one mole of water There are two moles of hydrogen. Okay, so moles of water cancel and we're left with moles of hydrogen will do this calculation and we get 0.1221 moles of hydrogen. Now to write out our empirical formula, we need to calculate the mole ratio. We're going to calculate the mole ratio by dividing both of our both are moles of carbon and moles of hydrogen by the lowest number of moles. Okay, so that is what we mean by mole ratio. Alright, so our mole ratio, So our lowest moles is our moles of carbon. Alright, so for our moles of carbon We're Gonna Divide zero. We're going to take both of them and divide by the lowest molds. Okay, so we have 0.52 to six moles of carbon. And we're going to divide that by 0.52 to six. Alright, so this gives us a mole ratio of one carbon. Alright, so we're going to do the same thing for hydrogen. We have 0.1221 moles of hydrogen. And we're going to divide it by the lowest number of moles out of the two and that is 0.52 to six. And so we get 2. hydrogen. Okay, so we have one carbon and 2.3364 hydrogen. So what are empirical formula is going to look like right now is c one which we don't need to put the one there, but we'll put it there for right now and we have a church, 2.33. Okay, so we can't have a decimal. So we're going to need to multiply by the lowest number. That will make the 2.33 whole number. So we're going to multiply by three. Okay, so I'll put it to the side, we're going to multiply by three. So that means our empirical formula is now going to become C three H seven. Alright, so there's our empirical formula. So we're gonna box this in. Oops. I'll drive, I'll draw a box around it. This is our empirical formula. Now that we know what our empirical formula is. We can calculate the empirical mass because the empirical mass is what will need to be able to calculate the molecular formula. But we wouldn't know what the empirical masses without first knowing what our without first knowing our empirical formula. So now our empirical mass we're going to take so let's right here, our empirical mass. We're gonna take we know we have three carbons and seven hydrogen. We're gonna multiply that by their molecular masses. Okay, so we have three carbons times its molecular masses. 12.1 based off the periodic table. One carbon is 12.1 g per mole. So we're going to take the three that's in our empirical formula and multiply it by 12.1 plus we have seven hydrogen. And we're going to multiply those by its um Molecular weight which is 1.008. So we get an empirical mass equal to 43.086. Alright, so now that we know the empirical mass, our molecular weight is the most intense peak with the highest mass charge. So let's go back to our mass spectrum. So our most intense peak with the highest mass charge ratio here Is going to be 86. All right, so here we go. This is our most intense peak with the highest mass charge ratio. So that's the molecular weight. Alright. So with that we know that mole. We need to now take the molar mass and divide it by the Empirical mass or molecular weight and divided by our empirical mass. So we're going to take our molecular weight which is 86, which is what we saw up above. And we're going to divide it by our let me write that out instead. So we're gonna take molecular wait and divided by let me write molecular mass instead. Okay, so we're gonna take molecular mass and divided by the empirical mass. Okay, so based off our mass spectrum are molecular masses 86. And the empirical mass that we just calculated is 43.086. So we get a ratio of two. So now what we're gonna do is take this ratio of two and multiply our empirical mass or empirical formula by that. Okay, so we're gonna go back up here where we wrote empirical formula. Okay and now we know that we need to multiply our sub scripts By two. Alright, so right underneath, so we have c. three times two. Okay, And then we have H seven times two. We needed that ratio. Okay, So now what we get is C. Six H. 14. Now this is our molecular formula, This is going to be our final answer. Alright, so right over to the side, this is the molecular formula. Alright, so that's it for this problem. I hope this was helpful.

Combustion analysis is performed on 0.50 g of a hydrocar-bon and 1.55 g of CO2, and 0.697 g of H2O are produced. The mass spectrum for the hydrocarbon is provided below. What is the molecular formula? (LO 3.12 and 3.13) (a) C5H11 (b) C8H18 (c) C11H10 (d) C10H22

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Combustion analysis is performed on 0.50 g of a hydrocar-bon and 1.55 g of CO2, and 0.697 g of H2O are produced. The mass spectrum for the hydrocarbon is provided below. What is the molecular formula? (LO 3.12 and 3.13)
(a) C5H11 (b) C8H18 (c) C11H10 (d) C10H22