Friday Oct 6 Test

Moderators: Chem_Mod, Chem_Admin

sandros
Posts: 19
Joined: Fri Sep 29, 2017 7:07 am

Friday Oct 6 Test

Postby sandros » Fri Oct 06, 2017 3:14 pm

Concerning the test we've had today, we were asked in question 6 to find the molecular formula of a compound X that has a MM= 46g.mol-1. They also gave us the mass of the compound and the mass of the products. They mentioned that the reaction was a combustion and that X was only made of the elements C,H and O. My final result was a little bit strange so can anyone clarify the method we had to use please?
Thanks

Yifei Wang 3G
Posts: 19
Joined: Fri Sep 29, 2017 7:07 am

Re: Friday Oct 6 Test

Postby Yifei Wang 3G » Fri Oct 06, 2017 3:28 pm

Here's how I worked out this problem:

1. apply the Law of Conservation of Mass to find out the mass of O2 in reactant
2. calculate the moles of X, O2, CO2, and H2O
3. find out the net molar ratio of C, H, and O (subtract the moles of O from the reactant O2) of the products, which in turn should be the molar ratio of C, H, O in X
4. multiply the ratio and try to make every number a integer, the result should be the empirical formula of X
use the molar mass of X to find out how much you should multiply to the numbers in the empirical formula to get the chemical formula of X

BTW I got a weird answer too... And I didn't write down my answer eventually...

Taylor 1F
Posts: 17
Joined: Thu Jul 27, 2017 3:00 am

Re: Friday Oct 6 Test

Postby Taylor 1F » Fri Oct 06, 2017 4:56 pm

Also, are we supposed to do the pre-modules for week 2 before next week?

Mariane Sanchez 1E
Posts: 57
Joined: Fri Sep 29, 2017 7:07 am

Re: Friday Oct 6 Test

Postby Mariane Sanchez 1E » Fri Oct 06, 2017 7:07 pm

Taylor 1F wrote:Also, are we supposed to do the pre-modules for week 2 before next week?


I think the pre-modules are just optional stuff and not graded. :)

Curtis Wong 2D
Posts: 62
Joined: Sat Jul 22, 2017 3:00 am

Re: Friday Oct 6 Test

Postby Curtis Wong 2D » Fri Oct 06, 2017 10:24 pm

For the problem, here's what I did. I converted all grams of the products into moles, and then used the moles of those to figure out the moles of the individual elements of Carbon and Hydrogen, because we can't figure out Oxygen due to it coming from the two sources of compound X and the Oxygen used for combustion. So then we convert the moles of Carbon and Hydrogen into grams, subtract the grams from Compound X, and that's the grams of Oxygen. We convert the grams of that oxygen into moles of oxygen.

And then, we basically have the moles of the individual element of Carbon, Hydrogen, and Oxygen of Compound X. Since we know that, we can now treat it as an empirical formula question and just divide by the smallest number of moles of the element to figure out the ratio. As soon as we figure out the empirical formula, we just find the molar mass of the empirical formula to figure out the ratio of it to the molecular formula.

And now that we know the molecular formula, we balance the equation.

Mirrat
Posts: 20
Joined: Fri Sep 29, 2017 7:04 am
Been upvoted: 1 time

Re: Friday Oct 6 Test

Postby Mirrat » Sat Oct 07, 2017 6:48 pm

What was the correct answer for the identity of Compound X?

Suchita 2I
Posts: 59
Joined: Fri Sep 29, 2017 7:04 am

Re: Friday Oct 6 Test

Postby Suchita 2I » Sun Oct 08, 2017 4:34 pm

I got C2H6O as the empirical formula and the molecular formula. Did anyone else get this?

Kaylin Krahn 1I
Posts: 57
Joined: Fri Sep 29, 2017 7:04 am
Been upvoted: 1 time

Re: Friday Oct 6 Test

Postby Kaylin Krahn 1I » Sun Oct 08, 2017 4:48 pm

^I believe that was the molecular and empirical formula since we were also given the molar mass


Return to “Empirical & Molecular Formulas”

Who is online

Users browsing this forum: No registered users and 1 guest