Sapling Hw Week 1 #10

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Raashi Chaudhari 3B
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Sapling Hw Week 1 #10

Postby Raashi Chaudhari 3B » Mon Oct 05, 2020 12:53 pm

A reaction was performed in which 0.20 mL of 2‑butanone was reacted with an excess of propyl magnesiumbromide to make 0.23 g of 3‑methyl‑3‑hexanol. Calculate the theoretical yield and percent yield for this reaction.

I don't know where to start with this problem. Am I supposed to find the empirical formula first? If so, how do I start?

Cassidy Cheng 1J
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Re: Sapling Hw Week 1 #10

Postby Cassidy Cheng 1J » Mon Oct 05, 2020 1:04 pm

I had different numbers, but the process is the same. I took .35 ml (amount of butanone I had) x the density they gave you for 2-butanone to find the grams of butanone. Then, I divided that by 2-butanone's molar mass (72.11 g/mol). This gives you the moles of 2-butanone. Next, I multiplied this by 1 mol 3 methyl-3-hexanol/1 mol 2-butanol, then multiplied by 116.2 g/1 mol 3-methyl-2-hexanol. This results in the theoretical yield for 3-methyl-3-hexanol.

Like this:

.35 ml x .81g/ml x (1 mol 2-butanone/72.11 g) x (1 mol 3 methyl-3-hexanol/1 mol 2-butanol) x (116.2g/1 mol 3 methyl-3-hexanol). I got .46 g, which is the theoretical yield. In the problem, my actual yield was .30 grams, so I then did .30g/.46g x100% = 65%.

Samantha Pedersen 2K
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Re: Sapling Hw Week 1 #10

Postby Samantha Pedersen 2K » Mon Oct 05, 2020 1:09 pm

I began solving this problem by converting the given mL of 2-butanone into grams of 2-butanone using the conversion factor d = 0.81 g/mL (given under the drawing of 2-butanone). Then, I converted the grams of 2-butanone into moles of 2-butanone using the molar mass of 2-butanone. To continue the problem, you'll need to use the mole-to-mole ratio to convert moles of 2-butanone into moles of 3-methyl-3-hexanol and finally use the molar mass of 3-methyl-3-hexanol to convert the moles of 3-methyl-3-hexanol into grams of 3-methyl-3-hexanol. This will be your theoretical yield. Then, you can plug these numbers into the formula for percent yield to calculate the percent yield. I hope this helps!

Juliet Cushing_2H
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Re: Sapling Hw Week 1 #10

Postby Juliet Cushing_2H » Mon Oct 05, 2020 1:16 pm

I also was confused at first because of the names of the compounds! Even though there are numbers in the names, the ratio in this chemical equation is 1 to 1 to 1. Therefore, if you find the mols of the starting compound you can use the 1 to 1 ratio to find the mols of the end compound. My numbers were different but I multiplied mL 2Butanol by the density to find grams 2But, and then divided the result by the molar mass to find starting mols. 1 to 1 ratio means that there are the same number of mols 3Met3Hex produced as mols 2But that started. Then multiply this number by the molar mass to find grams of 3Met3Hex produced.

Arezo Ahmadi 3J
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Re: Sapling Hw Week 1 #10

Postby Arezo Ahmadi 3J » Mon Oct 05, 2020 2:11 pm

Adding on to the previously great explanations given, when I approached this problem, I realized that the skeletal structures that were given also needed to be fully completed so that you could calculate the molar mass. To find the molar mass of each of the structures, take a look at the corners and the ends of the chain. These areas that do not show you the element that is being represented means that there is a carbon atom, and for these carbon atoms, you will need to attach hydrogen atoms to fill in the bonds to the carbon. This will help contribute to understanding how to calculate the molar mass that is needed to complete this problem, and it is the initial step you should take when starting to complete the problem.

After that, make sure that you convert the 0.20 mL of 2-butanone with the density of 2-butanone that they provide you so that you find its mass in grams, then take the molar mass that you calculated using the steps above to find the moles of 2-butanone by dividing the grams by the molar mass, and you will get the moles of 2-butanone. You can then use the ratio that you see from the problem in which it is 1 mole of 2-butanone to 1 mole of 3-methyl-3-hexanol, and then again make sure that you use the molar mass of 2-methyl-3-hexanol using the steps above so that you can multiply the moles of 3-methyl-3-hexanol by the molar mass to get its mass in grams.

This mass in grams is your theoretical yield, which is the amount that is calculated. You can then use the equation for percent yield and divide the actual yield, or the 0.23 g that they provided to you that they said was produced, by the theoretical yield which is the mass you calculated, and just multiply it by 100.

Thomas Vu 1A
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Re: Sapling Hw Week 1 #10

Postby Thomas Vu 1A » Tue Oct 06, 2020 9:32 pm

Wait so regarding how to find out what the molar masses are, is method of just assuming carbons and hydrogens as the parts of the chain a universal process? In other words, would we always assume this? thanks

Massimo_Capozza_1G
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Re: Sapling Hw Week 1 #10

Postby Massimo_Capozza_1G » Tue Oct 06, 2020 10:11 pm

Does anyone know what place we should round the final percentage? I'm confident I have the right answer, but every time I plug it in the system says I got it wrong.

Ryan Laureano 3I
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Re: Sapling Hw Week 1 #10

Postby Ryan Laureano 3I » Tue Oct 06, 2020 11:56 pm

Yeah I've been having the same problem too. My theoretical yield is correct but for some reason I can't seem to get the percentage correct. I think its the sig figs but I put multiple answers and all seem to be wrong.

nayha a 1E
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Re: Sapling Hw Week 1 #10

Postby nayha a 1E » Wed Oct 07, 2020 7:12 pm

Can someone explain to me how we convert the known masses of Br and Mg to moles? I'm very confused and stuck on this step. I looked up the molar masses but don't know what to do from here. Thanks!

Javier Perez M 1H
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Re: Sapling Hw Week 1 #10

Postby Javier Perez M 1H » Wed Oct 07, 2020 8:00 pm

Was this chemical reaction balanced already or did we have to balance it? The one provided in the question

Javier Perez M 1H
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Re: Sapling Hw Week 1 #10

Postby Javier Perez M 1H » Wed Oct 07, 2020 8:01 pm

Also where can we learn to read such symbols or way of displaying a chemical reaction without the lettering we usually use?

Nick Saeedi 1I
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Re: Sapling Hw Week 1 #10

Postby Nick Saeedi 1I » Thu Oct 08, 2020 12:01 am

Javier Perez M 2J wrote:Also where can we learn to read such symbols or way of displaying a chemical reaction without the lettering we usually use?


For 2-butanone and 3-methyl 3-hexyl, there are Carbon molecules in between each bond represented by the lines in the picture. Each Carbon molecules needs to have 4 bonds so add hydrogens to the carbons that do not have 4 bonds.

Hope this helps

StephanieGrigorian2J
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Re: Sapling Hw Week 1 #10

Postby StephanieGrigorian2J » Thu Oct 08, 2020 12:11 am

How do we know that the ratio for the problem is 1 to 1 to 1? Are we just assuming this or do we have to figure that out? If so, how do we find the ratio?

Lizbeth Garcia 1F
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Re: Sapling Hw Week 1 #10

Postby Lizbeth Garcia 1F » Thu Oct 08, 2020 12:55 am

Stephanie Grigorian 2C wrote:How do we know that the ratio for the problem is 1 to 1 to 1? Are we just assuming this or do we have to figure that out? If so, how do we find the ratio?


I believe the ratio is 1 to 1 between the reactant 2-butanone and the product. You can get the ratio if you do the same steps as the reactant to the product. You should arrive to the same amount of moles.

SavannahScriven_1F
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Re: Sapling Hw Week 1 #10

Postby SavannahScriven_1F » Thu Oct 08, 2020 3:05 pm

Massimo_Capozza_1H wrote:Does anyone know what place we should round the final percentage? I'm confident I have the right answer, but every time I plug it in the system says I got it wrong.


I had the same issue. I finally got the right answer when I wrote 90%. The solution showed the answer as 9.0 x 10^1% even though their work showed 0.24/0.26. So they maintain the 2 significant figures by writing 9.0, but for some reason round down. 0.24/0.26 should be 0.92. Not sure why they do this.

Gian Boco 2G
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Re: Sapling Hw Week 1 #10

Postby Gian Boco 2G » Thu Oct 08, 2020 8:06 pm

Stephanie Grigorian 2C wrote:How do we know that the ratio for the problem is 1 to 1 to 1? Are we just assuming this or do we have to figure that out? If so, how do we find the ratio?


I assumed that it would be 1:1 because there's no way would be expected to balance it even if it wasn't. Also if you work out the problem using the 1:1 ratio you should get the right answer. Hope you get it(:

rhettfarmer-3H
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Re: Sapling Hw Week 1 #10

Postby rhettfarmer-3H » Fri Oct 09, 2020 12:09 pm

Hey, so to start this problem there is some information you have to find: the molar mass of both 2-butanone and 3-methyl-3hexanol which ill give as 72.11 and 116.2. Granted they also give a density for me it was .81g/ml might be different. But I attach my work. To explain, convert the ML of butanone to g by multiplying the density and given ML. Then just molar conversions. Get to moles and then molar ratio is 1:1. Then convert back to grams of 3-meth-3hexanol and then actual/theo X100 gives the percent yield follow my work attached.
Attachments
IMG_2756.pdf
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emmaferry2D
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Re: Sapling Hw Week 1 #10

Postby emmaferry2D » Fri Oct 09, 2020 1:26 pm

Javier Perez M 2J wrote:Was this chemical reaction balanced already or did we have to balance it? The one provided in the question


The chemical equation is balanced already giving a ratio of 1:1:1:1

Mari Williams 1K
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Re: Sapling Hw Week 1 #10

Postby Mari Williams 1K » Fri Oct 09, 2020 3:18 pm

I haven't completed all the modules on the website yet, is there a module that helps teach how to work through this type of problem?

keely_bales_1f
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Re: Sapling Hw Week 1 #10

Postby keely_bales_1f » Sat Oct 10, 2020 5:17 pm

When I was doing this problem there was this equation that was shown in the feedback system.

Mass of Reactant (g) = density of reactant (g/mL) X volume reactant (mL)

Where and when did we learn this equation?

Aimee Alvarado 3J
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Re: Sapling Hw Week 1 #10

Postby Aimee Alvarado 3J » Sat Oct 10, 2020 5:21 pm

Javier Perez M 2J wrote:Was this chemical reaction balanced already or did we have to balance it? The one provided in the question


The chemical reaction that was provided was balanced already, so you did not have to balance it.


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