Week 1 Sapling HW Chem 14A Problem 10

Moderators: Chem_Mod, Chem_Admin

Vivian Leung 1C
Posts: 221
Joined: Wed Sep 30, 2020 10:09 pm

Week 1 Sapling HW Chem 14A Problem 10

Postby Vivian Leung 1C » Sun Oct 04, 2020 9:44 pm

Consider the nucleophilic addition reaction of 2‑butanone with excess propyl magnesiumbromide, made in situ by reacting 1‑bromopropane with metallic magnesium, to make 3‑methyl‑3‑hexanol.

Reaction scheme where 2-butanone (d = 0.81 g/mL), 1-bromopropane (d=1.35 g/mL), and magnesium react to form 3-methyl-3-hexanol (d = 0.82 g/mL)

A reaction was performed in which 0.55 mL of 2‑butanone was reacted with an excess of propyl magnesiumbromide to make 0.63 g of 3‑methyl‑3‑hexanol. Calculate the theoretical yield and percent yield for this reaction.


Hello, I am having trouble with how to do this problem. I reasoned that since propyl magnesiumbromide is in excess, 2-butanone would be the limiting reactant. Is this reasoning correct? Also, how do I do the calculations for the theoretical yield and percent yield given just the above information?

Any help would be greatly appreciated!

Carolina 3E
Posts: 92
Joined: Wed Sep 30, 2020 9:42 pm
Been upvoted: 2 times

Re: Week 1 Sapling HW Chem 14A Problem 10

Postby Carolina 3E » Sun Oct 04, 2020 9:54 pm

You are correct. 2-butanone is the limiting reactant.

To find the theoretical yield, you would use the given mass of 2-butanone and it's density (0.81g/ml) to determine the moles of 2-butanone. Using the mole to mole ratio of 2-butanone and 3-methyl-3-hexanol, you can determine the mole and the mass of 3-methyl-3-hexanol, which is the theoretical yield.

And, percent yield = (actual yield/theoretical yield) x 100%

Selena Quispe 2I
Posts: 127
Joined: Wed Sep 30, 2020 10:01 pm
Been upvoted: 3 times

Re: Week 1 Sapling HW Chem 14A Problem 10

Postby Selena Quispe 2I » Sun Oct 04, 2020 9:58 pm

Hi I thought the same way as you did! The first step would be to convert the given density for 2 butanone into grams so you would multiply the density (0.81 g/ml) by the given volume (0.55mL) to get 0.445g of 2 butanone. Then you would find the moles so divide 72.1g (the molar mass of butanone) from 0.405g and get 6.18*10^-3 mol 2 butanone. From this we know that the total product for 3-methyl-3-hexanol will be 6.18*10^-3. To find the theoretical yield you would multiply 6.18*10^-3 by 116.2g (the molar mass of 3-methyl-3-hexanol). Then to calculate the percent yield you would divide the 0.63g that was produced (given in the problem) from your theoretical yield and multiply by 100 percent! I hope this helps you!!
Last edited by Selena Quispe 2I on Sun Oct 04, 2020 10:24 pm, edited 3 times in total.

Emmeline Phu 1G
Posts: 116
Joined: Wed Sep 30, 2020 9:52 pm
Been upvoted: 2 times

Re: Week 1 Sapling HW Chem 14A Problem 10

Postby Emmeline Phu 1G » Sun Oct 04, 2020 10:04 pm

The 2-butanone would be assumed as the limiting reactant since its the only information provided by the question regarding the reactants. For percentage yield, use the density of 2-butanone to convert the 0.55 mL to grams. After finding the grams, convert the grams to moles by dividing by 2-butanone's molar mass. Multiply the moles of 2-butanone by the ratio of moles of 3-methyl-3-hexanol to moles of 2-butanone (1:1) to find the number of moles of the 3 methyl-3-hexanol. Multiply the moles by its molar mass to find the grams of 3-methyl-3-hexanol which is the theoretical yield. Now that you've found the grams of the product use the equation percent yield= actual yield/theoretical yield x 100% to find the the percent yield. The actual yield is already provided in the problem (0.63 grams). Hope this helps! :)

Vivian Leung 1C
Posts: 221
Joined: Wed Sep 30, 2020 10:09 pm

Re: Week 1 Sapling HW Chem 14A Problem 10

Postby Vivian Leung 1C » Sun Oct 04, 2020 10:31 pm

Thank you Carolina, Selena and Emmeline! How did you know the molar mass of butanone and 3-methyl-3-hexanol? Was that something that you calculated by knowing the chemical formula of butanone and 3-methlyl-3-hexanol? Also, how did you know that the ratio of 2-butanone to 3-methyl-3-hexanol is 1:1 in this problem? Is that illustrated by the picture of the reaction equation?

Selena Quispe 2I
Posts: 127
Joined: Wed Sep 30, 2020 10:01 pm
Been upvoted: 3 times

Re: Week 1 Sapling HW Chem 14A Problem 10

Postby Selena Quispe 2I » Sun Oct 04, 2020 10:43 pm

I don't think we're expected to know the molar mass of 2-butanone or 3-methyl-3-hexanol so I simply found the molar mass from the internet. As for molar ratios a good way to think of it is the product is determined from the limiting reactant so in this case because 6.18*10^-3 mol of 2-butanone was produced then only 6.18*10^-3 3-methyl-3-hexanol will be produced. I hope this makes sense :)

Emmeline Phu 1G
Posts: 116
Joined: Wed Sep 30, 2020 9:52 pm
Been upvoted: 2 times

Re: Week 1 Sapling HW Chem 14A Problem 10

Postby Emmeline Phu 1G » Sun Oct 04, 2020 10:45 pm

Yes, you're correct! In the reaction equation, only one of each is illustrated which indicates a 1:1 ratio of 3-methyl-3-hexanol to 2-butonane.

Jeffrey Hablewitz 2I
Posts: 71
Joined: Wed Sep 30, 2020 9:33 pm

Re: Week 1 Sapling HW Chem 14A Problem 10

Postby Jeffrey Hablewitz 2I » Sun Oct 04, 2020 11:03 pm

I believe you can also determine the chemical formula, and therefore molar mass, of the reactants based on their given structures. The reactants and products are shown in bond line notation, which is similar to a Lewis structure. This website has a really brief explanation of bond line structures and how to interpret them: https://orgomadesimple.com/how-to-draw- ... 0atoms.%20

Vivian Leung 1C
Posts: 221
Joined: Wed Sep 30, 2020 10:09 pm

Re: Week 1 Sapling HW Chem 14A Problem 10

Postby Vivian Leung 1C » Mon Oct 05, 2020 10:13 am

Oh I see. Thank you so much everyone!

isha dis3d
Posts: 101
Joined: Wed Sep 30, 2020 9:47 pm

Re: Week 1 Sapling HW Chem 14A Problem 10

Postby isha dis3d » Mon Oct 05, 2020 10:44 am

Hi! I know this problem was already solved but in the case that a problem does not state which reactant is in excess, an easy to I have found to solve for limiting reactant is to first balance the chemical equation if not already given. Then I convert both reactants from grams to moles and use stoichiometric ratios to convert the reactants to grams of one of the products given. I do this for both reactants and see which reactant produced less grams of that said product. This reactant is the limiting reactant. If the question asked for how much reactant that was in excess is leftover, use stoichiometric ratios to convert grams of the limiting reactant to grams of the other (excess) reactant. This is how many grams of excess reactant is used in the reactions. Subtract this mass from the total mass of the excess reactant given in the problem, and this is how many grams of excess reactant is left over.

Andrew Yoon 3L
Posts: 93
Joined: Wed Sep 30, 2020 9:36 pm

Re: Week 1 Sapling HW Chem 14A Problem 10

Postby Andrew Yoon 3L » Mon Oct 05, 2020 12:22 pm

Hello!

I was also struggling with this problem. As I was balancing out the equation, I realized that on the product side, there was no magnesium or bromine. So I was wondering how to balance out the equation and figure out that 2-butanone would be the limiting reactant. I mean I understand that since magnesiumbromide is reacted in excess, it makes sense that 2-butanone is the limiting reagent. However, how would you balance out the equation without magnesium or bromine on the product side? Thank you in advance for the help!


Return to “Limiting Reactant Calculations”

Who is online

Users browsing this forum: No registered users and 1 guest