Question 3 on midterm [ENDORSED]
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Question 3 on midterm
Why were we expected to know the makeup of nicotine? Every empirical and molecular formula problem we've seen in all the practice examples and everything we've covered in class included the elements Hydrogen and Oxygen in the makeup of the compound that underwent combustion. In this question you had to know that Oxygen wasn't apart of the compound in order to get the right answer. It was an assumption that I didn't even consider and I felt like it would've been much fairer to at least include which elements make up nicotine. Is there another way I should've looked at the problem? Did I miss something? Did anyone else struggle on this one?

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Re: Question 3 on midterm
So i assumed oxygen was present in the compound as well, so I started solving for oxygen. I took each of the masses of the products, divided by the molar mass of that product so I could get the moles, then I used another conversion factor (mole of one element/the mole of each product) to get the moles of N, H, and C. Then I used the molar mass of the element alone to get the grams of N, H, and C produced from the original reactant. I then took the mass given by the problem for nicotine and subtracted out each of the masses of N, H, and C. I was expecting to get a number of grams out that would have told me the grams of oxygen from the nicotine, but I got a number very close to 0, so I figured there was no oxygen in the nicotine. From there, I used the moles I had calculated previously to do the normal steps to find the empirical formula. Not sure if I got the correct answer but the numbers seemed to work out...

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Re: Question 3 on midterm
I just assumed that Nicotine would be made up of the elements Carbon, Hydrogen, and Nicotine. For this problem I bascially converted the mass that was given for carbon dioxide and water and solved for moles of carbon and moles of hydrogen, then converted mass of nitrogen to moles of nitrogen, then continued with the process of finding the empirical and molecular formula. I wrote the complete coompustion equation, then I balanced the equation. Did anyone else do this?

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Re: Question 3 on midterm
Honestly I took a bunch of random steps and hoped it worked out.
I started with the molar masses to figure out the moles being used during the combustion. Because we know that ~0.3 grams of nicotine were burned and what the products were, I calculated how many moles of nicotine was in the 0.3 sample and also how many moles were being produced of each substance (CO2, H2O, N2). Then I thought that if soandso fraction of a mole of nicotine could produce soandso amount of products, I multiplied the numbers to see how much product an actual mole of nicotine would create. I divided numbers here and there and came up with probably the worst answer of my life. But I managed to get an empirical and molecular formula.
TDLR: I used a bunch of roundabout ways to try to figure out this problem because there were a lot of assumptions to make.
I started with the molar masses to figure out the moles being used during the combustion. Because we know that ~0.3 grams of nicotine were burned and what the products were, I calculated how many moles of nicotine was in the 0.3 sample and also how many moles were being produced of each substance (CO2, H2O, N2). Then I thought that if soandso fraction of a mole of nicotine could produce soandso amount of products, I multiplied the numbers to see how much product an actual mole of nicotine would create. I divided numbers here and there and came up with probably the worst answer of my life. But I managed to get an empirical and molecular formula.
TDLR: I used a bunch of roundabout ways to try to figure out this problem because there were a lot of assumptions to make.

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Re: Question 3 on midterm
Namie Fotion1E wrote:So i assumed oxygen was present in the compound as well, so I started solving for oxygen. I took each of the masses of the products, divided by the molar mass of that product so I could get the moles, then I used another conversion factor (mole of one element/the mole of each product) to get the moles of N, H, and C. Then I used the molar mass of the element alone to get the grams of N, H, and C produced from the original reactant. I then took the mass given by the problem for nicotine and subtracted out each of the masses of N, H, and C. I was expecting to get a number of grams out that would have told me the grams of oxygen from the nicotine, but I got a number very close to 0, so I figured there was no oxygen in the nicotine. From there, I used the moles I had calculated previously to do the normal steps to find the empirical formula. Not sure if I got the correct answer but the numbers seemed to work out...
Thank you so much for your response! It is interesting to see how you thought about it. I worked on this problem for a very long time and I was trying lots of different methods to see if I could make it work. I eventually settled on finding the ratio of the elements using the moles of N, H, and C I calculated earlier. Then, I checked the gram/moles of the result and it was approximately 80 g/mol. Instead of realizing I should've multiplied the ratios by 2 to get the 160g/mol, I thought that Oxygen needed to be apart of the formula (because I hadn't seen any problem like this one before and didn't know nicotine isn't composed of O) so I added 5 Oxygens which unfortunately gave me the number of g/mol I was expecting (I wish 16*5 wasn't 80). When I saw this worked out I finished the rest of the problem. I realize now that those 5 oxygens should be gone and that my N,H, and C ratios should be 2 times what they are. Thank you again.

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Re: Question 3 on midterm
Noah Carey 1G wrote: Thank you so much for your response! It is interesting to see how you thought about it. I worked on this problem for a very long time and I was trying lots of different methods to see if I could make it work. I eventually settled on finding the ratio of the elements using the moles of N, H, and C I calculated earlier. Then, I checked the gram/moles of the result and it was approximately 80 g/mol. Instead of realizing I should've multiplied the ratios by 2 to get the 160g/mol, I thought that Oxygen needed to be apart of the formula (because I hadn't seen any problem like this one before and didn't know nicotine isn't composed of O) so I added 5 Oxygens which unfortunately gave me the number of g/mol I was expecting (I wish 16*5 wasn't 80). When I saw this worked out I finished the rest of the problem. I realize now that those 5 oxygens should be gone and that my N,H, and C ratios should be 2 times what they are. Thank you again.
Ya no problem! I'm so glad it helped. If any part of my explanation was unclear or vague to anyone, please don't hesitate to ask, but you seem to have understood it completely. Honestly I wouldn't have known how to approach it, but I went to a workshop and someone happened to ask a question about a problem in the book that was similar except in the problem, oxygen was included in the compound.

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Re: Question 3 on midterm [ENDORSED]
Hi, I also had a hard time figuring out this problem. When I checked my answer afterwards I got the correct molecular formula. I tried to explain it to a friend and came up with these pictures. I have absolutely no idea if I went about solving the problem the right way, but I thought I'd sort of show the way I went about it.
Re: Question 3 on midterm
Namie Fotion1E wrote:So i assumed oxygen was present in the compound as well, so I started solving for oxygen. I took each of the masses of the products, divided by the molar mass of that product so I could get the moles, then I used another conversion factor (mole of one element/the mole of each product) to get the moles of N, H, and C. Then I used the molar mass of the element alone to get the grams of N, H, and C produced from the original reactant. I then took the mass given by the problem for nicotine and subtracted out each of the masses of N, H, and C. I was expecting to get a number of grams out that would have told me the grams of oxygen from the nicotine, but I got a number very close to 0, so I figured there was no oxygen in the nicotine. From there, I used the moles I had calculated previously to do the normal steps to find the empirical formula. Not sure if I got the correct answer but the numbers seemed to work out...
Correct. No oxygen.
Re: Question 3 on midterm
Rebekah Kaufman 1L wrote:Hi, I also had a hard time figuring out this problem. When I checked my answer afterwards I got the correct molecular formula. I tried to explain it to a friend and came up with these pictures. I have absolutely no idea if I went about solving the problem the right way, but I thought I'd sort of show the way I went about it.
Correct answer. Well done.
Next week Wednesday the midterms will be returned in class and the detailed answers showing partial credit will be posted on my class website the same day.
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