Self Test 1 #5

Moderators: Chem_Mod, Chem_Admin

alexis castro 1B
Posts: 30
Joined: Wed Nov 16, 2016 3:02 am

Self Test 1 #5

Postby alexis castro 1B » Sun Jul 02, 2017 1:17 pm

6.40g of a compound was burned in air and produced 8.80g CO2 and 7.2g of H2O. find the empirical formula of the compound.

Can someone please help. I am so lost on this question.

Ben Rolnik 1D
Posts: 33
Joined: Fri Jun 23, 2017 11:39 am

Re: Self Test 1 #5

Postby Ben Rolnik 1D » Sun Jul 02, 2017 3:55 pm

Yeah... this is pretty tricky... I got C1H4.

I'm not sure if my method was correct but here's how I thought about it. Would be great if one of the TA's checked this....

1. 6.4g X + O2 --> 8.8g CO2 + 7.2g H2O

2. With this in mind, I'm assuming we're trying to finding the empirical for a CxHx compound (because O is a constant).

3. Since there's a conservation of mass, I would expect whatever the molar ratio of C::H in the product to be the empirical formula. I'm not sure we can find the molecular formula or anything more with the information given [that is, I'm not sure we can calculate the stoichiometric coefficients for each reactant and product].

4. Find the moles of C and H in the product:

8.8g CO2 = .2 mol CO2 = .2 mol C
7.2g H2O = .4mol H2O = .8 mol H
(approximately)

5. Divide by the lowest # of moles (.2)

1 mol C
4 mol H

6. That's the empirical formula: C1H4

.... like I said, I'm not sure this is correct, but I wasn't sure how else we could solve it.... and since i'm not utilizing the 6.4g number I may be way off in terms of how I thought about this!

Ben Rolnik 1D
Posts: 33
Joined: Fri Jun 23, 2017 11:39 am

Re: Self Test 1 #5

Postby Ben Rolnik 1D » Sun Jul 02, 2017 4:17 pm

Okay... I think C1H4 has to be the answer because I actually solved this a completely different way. Still, it would be great to get a TA to explain this more succinctly.

Here's how to solve the problem using the 6.4 g CxHx information...

1. 6.4g X + O2 --> 8.8g CO2 + 7.2g H2O

2. factor out the weight of O from the products:

8.8g CO2 = .4 mol CO2 = .8 mol O = 72.7272% O = 6.39999g O = 2.4g C
7.2g H2O = .4 mol H2O = .4 mol O = 88.8888% O = 6.4g O = .8g H

3. 6.4g CxHx = 2 ( 2.4g C + .8g H)

4. With this in mind, we can take the relative percentages of both C and H and solve this as we would any other empirical formula problem (assuming 100g).

2.4gc / 3.2g CH = 75%
.8g H / 3.2g CH = 25%

5. Assuming 100g .... (skipping some steps) you get 6.25 mol C and 25 mol H.

6. Divide by the smallest number (6.25) = empirical formula = C1H4

alexis castro 1B
Posts: 30
Joined: Wed Nov 16, 2016 3:02 am

Re: Self Test 1 #5

Postby alexis castro 1B » Sun Jul 02, 2017 6:33 pm

I got 1 for C and 4 for H also but what I was confused about is the oxygen. I got 4 for Oxygen but it doesn't seem right.

alexis castro 1B
Posts: 30
Joined: Wed Nov 16, 2016 3:02 am

Re: Self Test 1 #5

Postby alexis castro 1B » Sun Jul 02, 2017 6:34 pm

Or is oxygen not in the compound?


Return to “Empirical & Molecular Formulas”

Who is online

Users browsing this forum: No registered users and 1 guest