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

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The discussion revolves around determining the empirical formula of a hydrocarbon (CxHy) based on its reaction with excess O2 and NaOH. Initially, 10 mL of CxHy produces 55 mL of CO2 after combustion, indicating that 20 mL of CO2 is absorbed by NaOH, leaving 35 mL of excess O2. Participants clarify that the final gas volume consists of unreacted O2, while all CO2 was absorbed by NaOH. The calculation of moles and gas volumes leads to confusion regarding the hydrogen content, with participants noting insufficient information to derive the complete empirical formula. The conversation highlights the importance of understanding gas reactions and stoichiometry in determining hydrocarbon formulas.
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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
 
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Why do you think there was 55 mL of CO2?
 
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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?
 
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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?
 
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I think the answer is CO2 for both questions
 
So why part of the CO2 was adsorbed and part not?
 
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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.
 
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  • #11
NaOH reacts with CO2 (I guess)

-CO2+ 2NaOH -> Na2CO3 + H20
 
  • #12
So how come some of CO2 was left?
 
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  • #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.
 
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  • #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
 
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  • #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?
 
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  • #19
I also think there is missing something, but that's the whole question.

Thank you for the help anyway
 
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