Empirical Formula of Hydrocarbon from Reacting 10mLCxHy with O2 and NaOH

  • Thread starter Thread starter Grhymn
  • Start date Start date
  • Tags Tags
    Analysis
Click For Summary

Discussion Overview

The discussion revolves around determining the empirical formula of a hydrocarbon (CxHy) based on its reaction with oxygen and sodium hydroxide, involving calculations related to the volumes of gases produced and absorbed during the reaction. The scope includes homework-related problem-solving and chemical reasoning.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Initial conditions involve 10 mL of CxHy reacting with excess O2, producing a gas volume of 55 mL after cooling.
  • Participants question the composition of the 55 mL gas, with some suggesting it consists solely of CO2, while others propose that it includes unreacted O2.
  • There is a discussion about the role of NaOH in absorbing CO2, with some participants asserting that 20 mL of CO2 was absorbed, leaving 35 mL of gas remaining.
  • Confusion arises regarding the presence of gases after the reaction, with some participants questioning whether all CO2 was absorbed or if excess O2 remained.
  • Participants explore the implications of gas volume ratios and their relation to moles, with references to ideal gas behavior.
  • Some participants express uncertainty about the information provided in the problem and its sufficiency for deriving the empirical formula.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the exact composition of the gases involved or the sufficiency of the information provided to determine the empirical formula. Multiple competing views remain regarding the interpretation of the gas volumes and the reactions taking place.

Contextual Notes

There are limitations in the information provided, particularly regarding the completeness of the question and the assumptions about gas behavior and reactions. The discussion reflects uncertainty about the reactions and the quantities involved.

Grhymn
Messages
16
Reaction score
0

Homework Statement


10mLCxHy reacts with excess O2
After reacting and cooling until 298K the volume GAS = 55mL. Then NaOH is added and then the final volume of GAS = 35mL. Give the emperical formula of the hydrocarbon used in this reaction.



Homework Equations



-CxHy + (X+Y/4)O2 -> X CO2 + Y/2 H2O

-CO2+ 2NaOH -> Na2CO3 + H20


The Attempt at a Solution


10ml CxHY in the beginning and 55ml CO2 at the end => 1/X=10/55
55ml/X=2*VolumeH20/Y
 
Physics news on Phys.org
Why do you think there was 55 mL of CO2?
 
Last edited by a moderator:
Because at the end of the reaction after cooling there is 55mL gas, H20 is liquid then and I think you shouldn't take acount of the O2 in the 55mL
 
So what was the gas adsorbed by the NaOH?
 
Last edited by a moderator:
You are right, I meant the amount of CO2 after the first reaction.
 
I think you are still misunderstanding the question. What gases were present in 55 mL, what gas is present in the remaining 35 mL?
 
Last edited by a moderator:
I think the answer is CO2 for both questions
 
So why part of the CO2 was adsorbed and part not?
 
Last edited by a moderator:
20 mL absorbed of the CO2 created by the first reacting absorbed and 35 mL remaining after the second reaction
 
  • #10
What second reaction? There was just a one reaction producing carbon dioxide, and that was combustion.
 
Last edited by a moderator:
  • #11
NaOH reacts with CO2 (I guess)

-CO2+ 2NaOH -> Na2CO3 + H20
 
  • #12
So how come some of CO2 was left?
 
Last edited by a moderator:
  • #13
Because there wasn't enough NaOH, or because there is a dynamic equilibrium. I am not sure what kind of reaction this is (<=> or ->)
 
  • #14
No. All CO2 was absorbed. What other gas was present in the mixture? Reread the question.
 
Last edited by a moderator:
  • #15
I see, so the gass at the end is the excess O2, this would mean that the volume CO2 after the first reaction = 20mL

Molfraction=volumefraction for ideal gasses so if I am not mistaken :

20mL/X=2*VolumeH20/Y=10mL/1=VolumeO2/(X+Y/4)

But I still don't get the formula out of this.
 
  • #16
From ideal gas ratio of volumes is identical to ratio of moles, isnt't it? 10 mL of gas gave 20 mL of carbon dixode, how many carbon atoms per mole?

Edit: not sure what to do with hydrogen, as far as I can tell there is no enough information. Or I have a senior moment
 
Last edited by a moderator:
  • #17
20mL/X=10mL so 2
 
  • #18
Correct. And as far as I can tell that's all that can be said.

Unless I am missing something.

Have you posted whole question?
 
Last edited by a moderator:
  • #19
I also think there is missing something, but that's the whole question.

Thank you for the help anyway
 

Similar threads

Replies
1
Views
4K
  • · Replies 3 ·
Replies
3
Views
19K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 2 ·
Replies
2
Views
9K
Replies
2
Views
9K
  • · Replies 1 ·
Replies
1
Views
8K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 4 ·
Replies
4
Views
7K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 7 ·
Replies
7
Views
3K