Help with #9 of the homework - caproic acid

[August Blum Dis 3D]

How did you guys find the empirical formula of caproic acid with the weights of CO2 and H2O? I've been stuck for a little bit.

[Do Yeun Park]

Hi! So for that problem, find the molar mass of CO2 (44.01g/mol) and H2O (18.016) and divide the weight of each molecule with the molar mass: 2.275/ 44.01, 0.929/18.016.


Then, you get the moles. 0.0516 mol for CO2 and 0.0516 mol for H2O. Then, this tells you that there is a total of 0.0516 mol of C and 0.103 mol of H. Now, use the molar mass of C and H to get the grams used. Subtract the grams from 1.000g of the sample in order to find how many grams of O were used. After that, find the mass composition of each element, and you can do your normal empirical formula steps. Hope that helps!

[Natalie Keung 1D]

Hello, I am doing Q#9 and I got past the point where I found that there are .01720 mol O in the compound. However, when I continue with the empirical formula steps, I'm getting C3H3O and it seems to be incorrect when I check the answer. Can anyone help me with what I'm supposed to do after I found the moles of Oxygen please?

[Shria G 2D]

Hello, I am doing Q#9 and I got past the point where I found that there are .01720 mol O in the compound. However, when I continue with the empirical formula steps, I'm getting C3H3O and it seems to be incorrect when I check the answer. Can anyone help me with what I'm supposed to do after I found the moles of Oxygen please?

Hi, I got that there were 1.72 moles of oxygen, 5.17 moles of carbon and 10.297 moles of hydrogen so I just divided each by the smallest number. 5.17/1.72 = 3 and 10.297/1.72 = 6 so I got C3H6O as the empirical formula. I think the moles you got differ by a factor of 100 but the steps and the final answer should be the same.

[Lawrence Tran 2H]

Hello, I am doing Q#9 and I got past the point where I found that there are .01720 mol O in the compound. However, when I continue with the empirical formula steps, I'm getting C3H3O and it seems to be incorrect when I check the answer. Can anyone help me with what I'm supposed to do after I found the moles of Oxygen please?

I think a possible mistake you have is that you found the 0.0516 moles of H2O, but forgot to get the moles of H specifically. So you would multiply 0.0516 by 2 to get 0.103 moles of H. Afterward, finding the empirical formula should follow the regular steps as posted above. Hope that helps!

[Alexander Moroz 1B]

Using the values of CO2 and H2O given, convert it to grams of C and grams of H. Then subtract both of those values from the total amount of grams of the CHO compound (mine was 1 gram). This allows you to get the grams of O. Then follow the steps of finding the empirical formula by transforming grams of C, H, and O into moles of each. Then, divide by the smallest amount of moles to get your empirical formula of C3H6O.

[Bailey Wallace 3A]

Hi! So for that problem, find the molar mass of CO2 (44.01g/mol) and H2O (18.016) and divide the weight of each molecule with the molar mass: 2.275/ 44.01, 0.929/18.016.


Then, you get the moles. 0.0516 mol for CO2 and 0.0516 mol for H2O. Then, this tells you that there is a total of 0.0516 mol of C and 0.103 mol of H. Now, use the molar mass of C and H to get the grams used. Subtract the grams from 1.000g of the sample in order to find how many grams of O were used. After that, find the mass composition of each element, and you can do your normal empirical formula steps. Hope that helps!

Hi! I got to the part where you found the moles for CO2 and H20 but how did you find the moles for C and H individually?

[emmakvarnell]

Using the masses of 2.275g fo CO2 and 0.929g of H2O, and the law of conservation of mass, you can find the masses of C, H, and O in the reaction. Since C is only in carbon dioxide, use the molar mass of C and the molar mass of CO2 to find the percent mass composition of C, then that percent out of the original 2.275g. Use the same strategy to find the grams of H in H2O. Then since you have the grams of C and H, you can subtract that from the total mass of 1.000g to get the grams of O. From there, convert each element's grams to moles, divide by the smallest number, and you should get the number of each element in the empirical formula.

[Nico Towfighian 3L]

Hi! So for that problem, find the molar mass of CO2 (44.01g/mol) and H2O (18.016) and divide the weight of each molecule with the molar mass: 2.275/ 44.01, 0.929/18.016.


Then, you get the moles. 0.0516 mol for CO2 and 0.0516 mol for H2O. Then, this tells you that there is a total of 0.0516 mol of C and 0.103 mol of H. Now, use the molar mass of C and H to get the grams used. Subtract the grams from 1.000g of the sample in order to find how many grams of O were used. After that, find the mass composition of each element, and you can do your normal empirical formula steps. Hope that helps!

Can you further explain and elaborate after finding the total mols of CO2 and H2O, you kinda lost me and Im not sure what the step after that would be.

[Darren Apostol 2L]

Hi! So for that problem, find the molar mass of CO2 (44.01g/mol) and H2O (18.016) and divide the weight of each molecule with the molar mass: 2.275/ 44.01, 0.929/18.016.


Then, you get the moles. 0.0516 mol for CO2 and 0.0516 mol for H2O. Then, this tells you that there is a total of 0.0516 mol of C and 0.103 mol of H. Now, use the molar mass of C and H to get the grams used. Subtract the grams from 1.000g of the sample in order to find how many grams of O were used. After that, find the mass composition of each element, and you can do your normal empirical formula steps. Hope that helps!

Can you further explain and elaborate after finding the total mols of CO2 and H2O, you kinda lost me and Im not sure what the step after that would be.

For CO2, there is one mole of C per CO2, so 0.0516 mol of CO2 is just 0.0516 mol of C. For H2O, there is two mol of H per one mole of H2O as there are 2 hydrogens in H2O, so 0.0516 mol of H2O is doubled for 0.103 mol of H. After that, you take the moles of C and multiply that by the molar mass of carbon to get the total mass of carbon, and do the same for hydrogen. Once you have both of their masses, you can subtract their sum from the 1.000g of caproic acid to find the mass of oxygen used up. Then it's the regular mass composition and empirical formula steps.

[Natalie Keung 1D]

Hello, I am doing Q#9 and I got past the point where I found that there are .01720 mol O in the compound. However, when I continue with the empirical formula steps, I'm getting C3H3O and it seems to be incorrect when I check the answer. Can anyone help me with what I'm supposed to do after I found the moles of Oxygen please?

I think a possible mistake you have is that you found the 0.0516 moles of H2O, but forgot to get the moles of H specifically. So you would multiply 0.0516 by 2 to get 0.103 moles of H. Afterward, finding the empirical formula should follow the regular steps as posted above. Hope that helps!

Hi Lawrence,

Thank you for catching my mistake. That was the step I actually messed up on. Thank you for the help!

Natalie Keung