Caproic acid has the odor of goats. (In fact, Capra is the genus of the domestic goat.) The compound contains only C, H, and O and was experimentally found to have a molar mass of 110±10 g/mol . When a 1.000 g sample of caproic acid is burned in excess oxygen, 2.275 g CO2 and 0.929 g H2O are collected. Determine the empirical formula and molecular formula of caproic acid. Insert subscripts as necessary
I am having trouble starting this problem. I know that the empirical formula of caproic acid is C3H6O because I used google. However, I don't know what the next steps are to find the molecular formula. If anyone could help explain step by step. It would be great!
Sapling Week 1 HW_problem #9
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 Posts: 65
 Joined: Wed Sep 30, 2020 9:55 pm
Re: Sapling Week 1 HW_problem #9
To determine the molecular formula, you find the molar mass of your empirical formula then divide that number by 110g/mol (molar mass that was given to you). Then, you multiply the quotient from that to your empirical formula, which would give you your molecular formula.
If you need help starting this problem, the answers provided under this post would help: viewtopic.php?f=11&t=63092
I hope that helps!
If you need help starting this problem, the answers provided under this post would help: viewtopic.php?f=11&t=63092
I hope that helps!

 Posts: 58
 Joined: Wed Sep 30, 2020 9:37 pm
Re: Sapling Week 1 HW_problem #9
Hi! I think I can help, I know that one was confusing for me as well. I'll break it down into the steps I used.
1) I divided the grams of CO2 by the molar mass of CO2 to find the mols of CO2. Because the Ratio of C to O in the CO2 formula is 1:2, I knew that the amount of C mols on the right side of the equation is the same number of mols I found for CO2 as a whole.
2) I did the same with the water. I divided the grams by the molar mass to find the amount of mols in H2O. Because the ratio of H to O is 2:1, I multiplied this number by 2 to find the mols of H on the ride side of the equation.
3) I multiplied these mol values by the molar masses of the individual atoms. I got .051 x 12.011 for carbon and .051x1.01 for hydrogen. My answers were .62 g of Carbon and .104 g of hydrogen.
4) I then took 1 gram and subtracted the H grams and C grams from it to find the mass of oxygen in CHO. I got .275 grams of O.
5) Then you take the masses of each atom, divide by their molar masses to find the moles of each atom, and divide by the smallest number of moles. It should work out to be C3H6O.
6) Then to find the molecular, you find the empirical mass by adding up the masses of the molecular and divide it by the given molecular mass. You should get a whole digit 2, knowing to multiply the subscripts by 2 :)
1) I divided the grams of CO2 by the molar mass of CO2 to find the mols of CO2. Because the Ratio of C to O in the CO2 formula is 1:2, I knew that the amount of C mols on the right side of the equation is the same number of mols I found for CO2 as a whole.
2) I did the same with the water. I divided the grams by the molar mass to find the amount of mols in H2O. Because the ratio of H to O is 2:1, I multiplied this number by 2 to find the mols of H on the ride side of the equation.
3) I multiplied these mol values by the molar masses of the individual atoms. I got .051 x 12.011 for carbon and .051x1.01 for hydrogen. My answers were .62 g of Carbon and .104 g of hydrogen.
4) I then took 1 gram and subtracted the H grams and C grams from it to find the mass of oxygen in CHO. I got .275 grams of O.
5) Then you take the masses of each atom, divide by their molar masses to find the moles of each atom, and divide by the smallest number of moles. It should work out to be C3H6O.
6) Then to find the molecular, you find the empirical mass by adding up the masses of the molecular and divide it by the given molecular mass. You should get a whole digit 2, knowing to multiply the subscripts by 2 :)

 Posts: 62
 Joined: Wed Sep 30, 2020 9:50 pm
Re: Sapling Week 1 HW_problem #9
Something that helps a lot when doing problems like these is to make sure to write down what you're given and what you're trying to find. For example, you're given the mass of the products and you need to find the mass of the reactant in order to determine empirical formula.

 Posts: 55
 Joined: Wed Sep 30, 2020 9:52 pm
Re: Sapling Week 1 HW_problem #9
to get to the molecular formula, you need the empirical formula, and molar mass of the actual compound. so if you already figured out that the empirical formula is C3H6O, then find its molar mass, which you can figure out with a periodic table. the molar mass is (12.01*3)+(1.008*6)+(16.00) = 58.078g. You must be given the molar mass of the actual compound to figure out the molecular formula, which is 110. and since the molecular formula's subscripts are always a multiple of the empirical formula's, you divide the molar mass of the molecular by the empirical. so you do 110/58.078 which is roughly 2. you multiply that 2 to all the subscripts of the empirical formula to get C6H12O2. hope that helps :)

 Posts: 55
 Joined: Wed Sep 30, 2020 9:31 pm
Re: Sapling Week 1 HW_problem #9
To find the molecular formula you want to use this equation: molecular formula mass/empirical formula mass. This fraction will equal a whole number that the empirical formula will be multiplied by. Since the problem states that the molecular formula mass is 110 +/ 10 just find the empirical formula mass (I got 58.08 g/mol). Since there is a margin of error, I did 2 fractions 100/58.08 and 120/58.08. I then rounded to the whole number that was closest to both numbers  which is 2. So just multiply the empirical formula by 2.
Hope this helps!
Hope this helps!

 Posts: 69
 Joined: Wed Sep 30, 2020 9:56 pm
Re: Sapling Week 1 HW_problem #9
I am also having trouble finding the empirical formula for this question. I wrote out the chemical reaction as C_H_O_ + O2 > CO2 + H2O because I believe "burning" is the same as combustion. From there I calculated the moles of CO2 and H2O, but I don't know what my next step should be. Any help would be greatly appreciated!

 Posts: 62
 Joined: Wed Sep 30, 2020 10:05 pm
Re: Sapling Week 1 HW_problem #9
I also had trouble with this homework problem. I feel like I understand how to do the material, but I can get confused in word problems on where to start. All of this helped a TON though thank you everyone!!

 Posts: 61
 Joined: Wed Sep 30, 2020 10:00 pm
Re: Sapling Week 1 HW_problem #9
I also had a tough time trying to work this problem out, as I had to do a lot of different conversions between elements in order to find my answer. These replies really helped me out and gave be a solid foundation as to how I had to start this problem and what I had to do in order to get the answer, so thank you!

 Posts: 55
 Joined: Wed Sep 30, 2020 9:38 pm
Re: Sapling Week 1 HW_problem #9
This problem is a bit wordy so I definitely understand why this problem would be difficult to piece together. If you've already found out the empirical formula, you just need to divide the given molar mass of the acid by the molar mass of the empirical formula you got.
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