Help with #10 of the homework
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Help with #10 of the homework
How did you guys approach this problem? I'm not sure what information is relevant in the calculations. In case it's unclear, I'm working on the problem where there's a reaction between .50mL of 2-butanone and propyl magnesiumbromide to make 3-methyl-3-hexanol.
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Re: Help with #10 of the homework
So first, multiply the density of 2-butanone with 0.55mL to find grams used. Then, find the molar mass of 2-butanone and multiply that with the grams you found earlier.
We know that 2-butatone and 3-methyl-3-hexanol is 1:1 ratio, so the mole used for 2-butatone and 3-methyl-3-hexanol should be the same. Using that mole, find the molar mass of 3-methyl-3-hexanol to get the theoretical yield. Now you can find the percent yield too.
We know that 2-butatone and 3-methyl-3-hexanol is 1:1 ratio, so the mole used for 2-butatone and 3-methyl-3-hexanol should be the same. Using that mole, find the molar mass of 3-methyl-3-hexanol to get the theoretical yield. Now you can find the percent yield too.
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Re: Help with #10 of the homework
And then for the percent yield, you would divide the actual yield (given in the problem) by the theoretical yield (whatever answer you came up with) x 100%
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Re: Help with #10 of the homework
Hi August,
I found it easiest to start by figuring out the molar masses of 2-butanone and 3-methyl-3-hexanol, because we'll need those calculations later. We'll need the formulas of these to calculate molar mass. To figure out the formulas of these compounds, we'll need to use the diagrams they gave us. 2-butanone is C4H8O and 3-methyl-3-hexanol is C7H16O.
Once we've calculated molar mass, take 0.40 mL of 2-butanone and use the given value 0.81 g/mL to convert to g. Once we have the g of 2-butanone, we can convert to moles using the molar mass we calculated earlier from the formula. From there, convert to moles of 3-methyl-3-hexanol. Take the moles of 3-methyl-3-hexanol and convert to g using the molar mass we calculated earlier from its formula. The answer, in g of 3-methyl-3-hexanol, is our theoretical yield.
Now to calculate percent yield, take 0.449 g (actual yield)/theoretical yield x 100%.
Hope this helps!
I found it easiest to start by figuring out the molar masses of 2-butanone and 3-methyl-3-hexanol, because we'll need those calculations later. We'll need the formulas of these to calculate molar mass. To figure out the formulas of these compounds, we'll need to use the diagrams they gave us. 2-butanone is C4H8O and 3-methyl-3-hexanol is C7H16O.
Once we've calculated molar mass, take 0.40 mL of 2-butanone and use the given value 0.81 g/mL to convert to g. Once we have the g of 2-butanone, we can convert to moles using the molar mass we calculated earlier from the formula. From there, convert to moles of 3-methyl-3-hexanol. Take the moles of 3-methyl-3-hexanol and convert to g using the molar mass we calculated earlier from its formula. The answer, in g of 3-methyl-3-hexanol, is our theoretical yield.
Now to calculate percent yield, take 0.449 g (actual yield)/theoretical yield x 100%.
Hope this helps!
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Re: Help with #10 of the homework
Just an FYI for anyone trying to compare solutions across different steps; the numbers might not be the same since I've seen people talk about starting with 0.30, 0.40, and 0.50 mL of 2-butanone.
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Re: Help with #10 of the homework
I also had a question about #10 in the homework. I found it hard to do because I felt like I needed to balance the equation before I found the theoretical yield, but no empirical formulas to do so. Are we supposed to have a knowledge of most empirical formulas or should we assume the stoichiometric coefficients are all 1?
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Re: Help with #10 of the homework
I also was confused on this question—I wasn't sure how to find the molecular formulas based on the diagrams given to us. That information is necessary in order to find the molar mass of the compound and therefore convert to moles, but I wasn't sure how to go about finding that.
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