Sapling Week One Homework Question
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 Joined: Wed Sep 30, 2020 9:31 pm
Sapling Week One Homework Question
Hi! So I'm in the process of completing the Sapling assignment for week one. However, I am stuck on the question 10 and can't seem to do it right. Did any of you run into the same problem as me? If yes, then what did you do to overcome it? If not, then what did you do to begin with?

 Posts: 100
 Joined: Wed Sep 30, 2020 9:31 pm
Re: Sapling Week One Homework Question
I think my issue has to do the values I'm using for the molar masses of 2butanone and 3methyl3hexanol. When I googled the molar mass of 2butanone it said that it was 72.11 g/mol, however when I calculated it myself it came out to be 72.12 g/mol. With 3methyl3hexanol, it was 130.23 g/mol on google and 130.26 g/mol when I calculated it myself. Nonetheless, I computed the theoretical and percent yield using a combination of these values with no luck. I'm on my fifth attempt for this problem and I really just want to know what I'm doing wrong. Please let me know!

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Re: Sapling Week One Homework Question
3methyl3hexanol has the molecular formula C_{7}H_{16}O. Its molar mass is therefore (7*12.011)+(16*1.008)+15.999=116.204 g/mol instead of 130.26 g/mol.

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Re: Sapling Week One Homework Question
For me, I attempted the problem this way: I got the mass of 2butanone by multiplying the .25 mL given by the g.mol that is also given. That gives us the mass of 2butanone which I then divided by the molar mass of 2butanone which I think you can derive from the visual but I just googled it. That gives us the mol. Then when you compare 2butanone to 3methyl 3hexanol, you can see it's a 1:1 ratio since the pic is showing one 2butanone and one 3methyl 3hexanol. So knowing that information, you multiply the mol you found earlier with the molar mass of 3methyl 3hexanol (which I also just googled) and you'll get a mass that is the theoretical yield. To get percent yield, you take the given mass since it's the actual yield and divide it by the theoretical yield and multiply by 100%.

 Posts: 100
 Joined: Wed Sep 30, 2020 9:31 pm
Re: Sapling Week One Homework Question
Hi! So after I read your responses I went back and googled the molar masses one more time to see how I butchered them so horribly. It has come to my attention that I was looking at the molar mass of 3ETHYL3haxanol instead of 3METHYL3hexanol. Totally my fault! Thank you for your responses though! I will reevaluate the homework problem, and my life, and update you all!

 Posts: 100
 Joined: Wed Sep 30, 2020 9:31 pm
Re: Sapling Week One Homework Question
So final update on this, I did it! Thank you all for your help!

 Posts: 84
 Joined: Wed Sep 30, 2020 10:09 pm
Re: Sapling Week One Homework Question
Hi,
I have also been trying to solve the last question on the homework. I believe I am following the right procedure, but it seems like I can't find the right answer. I start by multiplying 0.30 ml by 0.81 g/ml in order to find the mass of 2butanone. Then, I multiply the mass I found by the molar mass of 2butanone (72.11 g/mol), which gives me the amount (in moles) of 2 butanone in 0.30 ml. Since there is a 1:1 ratio, I would expect obtaining the same amount of 3methyl3hexanol as the product. Multiplying the answer I found by the molar mass of 3methyl3hexanol (116.2 g/mol) would give me the theoretical yield. Lastly, I divide the actual yield (0.30 g) by the theoretical yield and multiply it by 100 to obtain the percent yield. I checked my calculations a few times and they seem to be correct. Could you tell me if there is anything wrong with my method of solving this question? Do you know any other way of solving this question?
Thanks!
I have also been trying to solve the last question on the homework. I believe I am following the right procedure, but it seems like I can't find the right answer. I start by multiplying 0.30 ml by 0.81 g/ml in order to find the mass of 2butanone. Then, I multiply the mass I found by the molar mass of 2butanone (72.11 g/mol), which gives me the amount (in moles) of 2 butanone in 0.30 ml. Since there is a 1:1 ratio, I would expect obtaining the same amount of 3methyl3hexanol as the product. Multiplying the answer I found by the molar mass of 3methyl3hexanol (116.2 g/mol) would give me the theoretical yield. Lastly, I divide the actual yield (0.30 g) by the theoretical yield and multiply it by 100 to obtain the percent yield. I checked my calculations a few times and they seem to be correct. Could you tell me if there is anything wrong with my method of solving this question? Do you know any other way of solving this question?
Thanks!

 Posts: 84
 Joined: Wed Sep 30, 2020 10:09 pm
Re: Sapling Week One Homework Question
Hi again,
I was trying to be very careful about the significant figures throughout my calculations, but I realized which step I was wrong. I would suggest everyone to think about the significant figures at the very end of the calculations.
I was trying to be very careful about the significant figures throughout my calculations, but I realized which step I was wrong. I would suggest everyone to think about the significant figures at the very end of the calculations.

 Posts: 85
 Joined: Wed Sep 30, 2020 9:45 pm
Re: Sapling Week One Homework Question
Are we able to redo the homework problems in sapling? Like are we given multiple tries?

 Posts: 84
 Joined: Wed Sep 30, 2020 10:09 pm
Re: Sapling Week One Homework Question
Yes, you have multiple tries. As long as you solve all the homework questions you will get the credit for it. It doesn't matter how many times you try to solve a question. The only thing that is important is that you eventually solve all the questions.
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