HW Week 1 Question 9

[Claire Guazelli]

For question 9 does anybody know how to find the mass percentage of each element in the compound since we are only given the mass of the 2 compounds as a whole?

[Jessica Cornelia Hongarta 1G]

First, find the mass of the element in the compound. Multiply the atomic mass of the element and the number of atoms of the element in 1 mol of the compound. Then, divide that by the molar mass of the compound. After that, you need to multiply that with the mass of the compound to get the mass of the element in the compound. To find the percentage mass, divide the mass of the element by the mass of the compound and multiply by 100%.

[Ivy Nguyen 3I]

Hi Claire,

Because H2O and CO2 each have components of each element that you want, I used the values given for each compound and used stoichiometry to find individual element masses.
I used the given value 2.275 g CO2 to find the total amount of C: 2.275 g CO2 * 1 mol CO2/44.01 g CO2 * 1 mol C/1 mol CO2 * 12.01 g C/1 mol C = 0.621 g C
I did the same for H with the given H2O value: 0.929 g H2O * 1 mol H2O/18.02 g H2O * 2 mol H/1 mol H2O * 1.01 g H/1 mol H = 0.104 g H
From there, I took the total amount and subtracted the H and C masses to find the mass of O. 1.000 g sample - (0.621 g C + 0.104 g H) = 0.275 g O
(I recommend finding C and H through the given CO2 and H2O values as finding O would result in overlap because it appears in both CO2 and H2O which makes solving a bit more time consuming)

You now have all of the masses of each element needed and you continue the question by solving to find the empirical formula from there. Hope that helped!

[905744270]

Hey guys, after completing the steps listed above, don't forget to calculate the moles from the found masses of each element to determine the empirical formula! I forgot to and it would have saved me a lot of time if I remembered that step lol.

Good luck! :)

[Kaira Shibata 1E]

For question 9 does anybody know how to find the mass percentage of each element in the compound since we are only given the mass of the 2 compounds as a whole?

Hi Claire!
You would need to convert the mass of each compound into mass (g) of the element you are trying to find. You can do that by dividing by the molar mass of the compound and then multiplying by the mole ratio and the molar mass of the element. You will want to do carbon and hydrogen first and then use simple addition to find that of oxygen. Once you have the molar masses of each element, you can divide the calculated mass of each element by the molar mass and then divide by the smallest number. From here, you probably know how to find the empirical formula. Hope this helps!

[Kayla Arellano 1K]

I was able to solve for the moles of C, H, and O. I divided each mole by the smallest value, however, I was still struggling to get whole numbers. I remember in one of the lecture examples, Dr. Lavelle multiplied each mole by a value after dividing it by the smallest value. I'm not sure if this question makes sense, but how do you find the number you have to multiply it by to get the whole number?

[Kayla Arellano 1K]

I was able to solve for the moles of C, H, and O. I divided each mole by the smallest value, however, I was still struggling to get whole numbers. I remember in one of the lecture examples, Dr. Lavelle multiplied each mole by a value after dividing it by the smallest value. I'm not sure if this question makes sense, but how do you find the number you have to multiply it by to get the whole number?

Nevermind, I figured out what I was doing wrong haha.

[Meagan Kimbrell 1I]

For question 9 does anybody know how to find the mass percentage of each element in the compound since we are only given the mass of the 2 compounds as a whole?

Hi Claire!
You would need to convert the mass of each compound into mass (g) of the element you are trying to find. You can do that by dividing by the molar mass of the compound and then multiplying by the mole ratio and the molar mass of the element. You will want to do carbon and hydrogen first and then use simple addition to find that of oxygen. Once you have the molar masses of each element, you can divide the calculated mass of each element by the molar mass and then divide by the smallest number. From here, you probably know how to find the empirical formula. Hope this helps!

Hi!! I need a little clarification for the solving of this problem.. Is there a reason that we use the given mass of CO2 and H2O to find the mass of C and H produced first, and not the mass of O produced too?

[Chem_Mod]

Hi Claire,

We don't use the provided masses for CO2 and H2O to determine the mass of O produced because oxygen is present in both the carbon dioxide (CO2) product and the water (H2O) product. This complicates the calculation for the mass of O produced if we were to use the masses of CO2 and H2O. It is simpler to determine the mass of oxygen produced by simply subtracting the combined masses of C and H produced from the total caprioic acid sample (1g) as Ivy described in her answer.

-Gabby (TA for 1E, 1K, and 1L)

[Shria G 2D]

Hi!! I need a little clarification for the solving of this problem.. Is there a reason that we use the given mass of CO2 and H2O to find the mass of C and H produced first, and not the mass of O produced too?

I think it is because all the hydrogen in the caproic acid is converted to H2O and all the carbon from the caproic acid becomes CO2 so it is easy to find the mass of C and H from that. But since the caproic acid is burned in oxygen, the oxygen in H2O and C2O is from both the caproic acid and the O2, which makes it harder to find the mass of O in only caproic acid.

[Jenna 1G]

Hi Claire,

Because H2O and CO2 each have components of each element that you want, I used the values given for each compound and used stoichiometry to find individual element masses.
I used the given value 2.275 g CO2 to find the total amount of C: 2.275 g CO2 * 1 mol CO2/44.01 g CO2 * 1 mol C/1 mol CO2 * 12.01 g C/1 mol C = 0.621 g C
I did the same for H with the given H2O value: 0.929 g H2O * 1 mol H2O/18.02 g H2O * 2 mol H/1 mol H2O * 1.01 g H/1 mol H = 0.104 g H
From there, I took the total amount and subtracted the H and C masses to find the mass of O. 1.000 g sample - (0.621 g C + 0.104 g H) = 0.275 g O
(I recommend finding C and H through the given CO2 and H2O values as finding O would result in overlap because it appears in both CO2 and H2O which makes solving a bit more time consuming)

You now have all of the masses of each element needed and you continue the question by solving to find the empirical formula from there. Hope that helped!

Hi!
I just wanted some clarification on the molar ratio you used to get to grams of H. You said there were 2 moles of H, is that because of the subscript 2 in H2?

[Riya Sawhney 1C]

I was able to solve for the moles of C, H, and O. I divided each mole by the smallest value, however, I was still struggling to get whole numbers. I remember in one of the lecture examples, Dr. Lavelle multiplied each mole by a value after dividing it by the smallest value. I'm not sure if this question makes sense, but how do you find the number you have to multiply it by to get the whole number?

Hi! Once I have a ratio of elements in a compound (make sure that it is moles), this is what I do.
First, I divide each value by the lowest one. This means that one of the numbers will be 1.00, and the rest will be numbers above it. Using #9 as an example:

C:H:0
0.0517 : 0.1040. : 0.0172
(divide each value by 0.0172

3: 6: 1

In this case, I ended up with nice whole numbers. If I didn't, I would try and think of a number to multiply all the terms by that would lead to whole numbers. For example, if you have terms that end in .5, multiply by 2. If you have terms that end in .333 or .666 then multiply by 3. Remember, it doesn't have to be exact because it is representing experimental data.

[Claire Kim 1F]

I was able to solve for the moles of C, H, and O. I divided each mole by the smallest value, however, I was still struggling to get whole numbers. I remember in one of the lecture examples, Dr. Lavelle multiplied each mole by a value after dividing it by the smallest value. I'm not sure if this question makes sense, but how do you find the number you have to multiply it by to get the whole number?

I'm pretty sure that you convert the grams to moles and then divide each value by the smallest mole value. My values weren't 100% whole numbers, but were very close in value. I may be misunderstanding the question, but I don't think you need to multiply, just divide.

[Minoo Bastani 2J]

I was able to solve for the moles of C, H, and O. I divided each mole by the smallest value, however, I was still struggling to get whole numbers. I remember in one of the lecture examples, Dr. Lavelle multiplied each mole by a value after dividing it by the smallest value. I'm not sure if this question makes sense, but how do you find the number you have to multiply it by to get the whole number?

I'm pretty sure that you convert the grams to moles and then divide each value by the smallest mole value. My values weren't 100% whole numbers, but were very close in value. I may be misunderstanding the question, but I don't think you need to multiply, just divide.

While you're right that for this particular problem you should get whole numbers right off the bat, it can be necessary in some instances to multiply in order to get a whole number after the initial division where you divide the number of moles of each element by the smallest number of moles. But that is really only if it is too far away to round up to a whole number, like if you have "2.5" for example. In that case you would multiply the subscripts for all the elements by 2 in order to maintain the ratio while still getting whole numbers for all the subscripts. Hope that helps!

[Jessica Li 1G]

For this question, you can first use the given grams of CO2 and H2O to find the moles of each. Using the moles of CO2 and H2O, you can directly find the moles and then grams of C and H, respectively. To find the moles of O, use the given 1.000g sample of caproic acid and subtract the grams of C and H from that to get the grams of O. Use this to convert from grams of O to moles of O. Now, you have the moles of all three elements separately, so you can divide the elements with the larger moles (C and H) by the one with the smallest # moles (O) to find the empirical formula.

[Rio Gagnon 1G]

First, you are going to want to find the mols of C (bc it is only present in one of the reactants) by converting the grams of CO2 to moles and using the molar ratio (1:1) to convert it to moles of C. Then use that number to find the grams of C.
Repeat the same process to find moles and then grams of H from the given amount of H2O.
We know that the sample of caproic acid is 1.000g, so subtract the mass of C and H from that, and the result is the grams of O. Convert grams of O to moles.
The result is .05169 mol C : .1031 mol H : .01721 mol O
Divide by the smallest number (.01721) and you are given the ratio 3 C: 6 H: 1 O, so the empirical formula is C3H6O
Find the molar mass of that and divide the given mass (110 g/mol) by the calculated molar mass of the compound (58.078), which is about 2, so you multiple the empirical formula by two to get the molecular formula of C6H12O2

[Rebekah Han 2K]

I was wondering what the +/- sign meant for the 110 and 10 g/mol. Does it just mean that there is a margin of error of about 10 grams? Do we ignore it and just use 110 g when finding the molecular formula?

[Rachel Martinez 1A]

I was wondering what the +/- sign meant for the 110 and 10 g/mol. Does it just mean that there is a margin of error of about 10 grams? Do we ignore it and just use 110 g when finding the molecular formula?

I believe if it falls in the range of 110 +/- 10 g/mol. so I am guessing 100 g/mol through 120 g/mol would work to find the molecular formula.

[Shay Crookall 2L]

I was wondering what the +/- sign meant for the 110 and 10 g/mol. Does it just mean that there is a margin of error of about 10 grams? Do we ignore it and just use 110 g when finding the molecular formula?

Hey Rebekah! This just means that the molar mass of the compound is within 100 and 120 g/mol. You shouldn't ignore it, I used it kind of like a hint that I had gotten the correct molecular formula.