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Topic: Finding empirical formula for unknown hydrocarbon  (Read 2259 times)

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Offline Luddus

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Finding empirical formula for unknown hydrocarbon
« on: January 19, 2024, 02:31:54 AM »
Hi. I have a bit of a conondrum. Its probably easy to find out, but i cant for the life of me understand how. Here Goes:

I have an unknown hydrocarbon (CxHx) being incinerated with a surpluss of oxygen. From the incineration i am left with 206,4mg of CO2 and 56,2mg of H2O.

What is the empirical formula for the hydrocarbon and how can i calculate this?

Offline Hunter2

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Re: Finding empirical formula for unknown hydrocarbon
« Reply #1 on: January 19, 2024, 03:22:22 AM »
Get the moles of water and carbondioxide, these correspond to the moles of C and H.
Then compare both and get ratio with both.

Offline Luddus

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Re: Finding empirical formula for unknown hydrocarbon
« Reply #2 on: January 19, 2024, 04:34:46 AM »
Thank you very much!

It should be something like this then?

206,4g*10-3/44,01g/mol=4,689*10-3mol
56,2g*10-3/18,01g/mol=3,12*10-3
CxHy +O2 -> CO2 + H2O
Cx/Hy= 4,689*10-3mol/3,12*10-3mol=1,50
CH1,5
C2H3

Offline Borek

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Re: Finding empirical formula for unknown hydrocarbon
« Reply #3 on: January 19, 2024, 05:11:32 AM »
56,2g*10-3/18,01g/mol=3,12*10-3

3.12×10-3 of what?
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Offline Luddus

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Re: Finding empirical formula for unknown hydrocarbon
« Reply #4 on: January 19, 2024, 05:20:02 AM »
my bad, it should say 3,12*10^-3mol

Offline Hunter2

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Re: Finding empirical formula for unknown hydrocarbon
« Reply #5 on: January 19, 2024, 06:34:46 AM »
But that is not the question. Recheck moles of H. How many moles H you have, if you have 1 mol H2O.

Offline Luddus

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Re: Finding empirical formula for unknown hydrocarbon
« Reply #6 on: January 19, 2024, 10:23:50 AM »
i would still have two moles of H, right?

Offline Hunter2

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Re: Finding empirical formula for unknown hydrocarbon
« Reply #7 on: January 19, 2024, 10:25:15 AM »
 Yes, so what does it mean for your exercise value.
« Last Edit: January 19, 2024, 10:35:51 AM by Hunter2 »

Offline Luddus

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Re: Finding empirical formula for unknown hydrocarbon
« Reply #8 on: January 19, 2024, 10:39:17 AM »
i would have to double my H value? thus getting C2H6?

Offline Hunter2

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Re: Finding empirical formula for unknown hydrocarbon
« Reply #9 on: January 19, 2024, 10:43:01 AM »
No

You have 4,68  x10^-3 C and 6,24 × 10^-3 H
This number you have to convert to unbroken number.
Start convert C to 1. What is the result for H then.
If you have that continue to multiply until both unbroken.

Offline Luddus

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Re: Finding empirical formula for unknown hydrocarbon
« Reply #10 on: January 19, 2024, 12:26:34 PM »
i am not sure, but taking the %weight of each atom and converting it to moles gave me 0,006 moles of H, 0,01 moles of O and 0,0046 moles of C. I am not sure how to proceed from this though

Offline Hunter2

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Re: Finding empirical formula for unknown hydrocarbon
« Reply #11 on: January 19, 2024, 12:32:49 PM »
No


4,68  x10^-3 C and 6,24 × 10^-3 H
1 C to  6,24×10^-3/4,68x10^-3 H means
1 C to 1,33333 H
Now calculate further that you dont have broken number of H

Offline Luddus

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Re: Finding empirical formula for unknown hydrocarbon
« Reply #12 on: January 19, 2024, 12:51:26 PM »
did you switch up your numbers? You have different values on line 1 and 2

Offline Hunter2

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Re: Finding empirical formula for unknown hydrocarbon
« Reply #13 on: January 19, 2024, 12:58:47 PM »
What do you mean.
You have xC and yH

C = y/x* H

Offline Luddus

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Re: Finding empirical formula for unknown hydrocarbon
« Reply #14 on: January 19, 2024, 01:59:33 PM »
No


4,68  x10^-3 C and 6,24 × 10^-3 H
1 C to  6,24×10^-3/4,68x10^-3 H means
1 C to 1,33333 H
Now calculate further that you dont have broken number of H

you have 4,68 x10^-3 C on the first line, and on the second you have 6,24x10^-3 on the second. I'm just a little confused, but thank you so very musch for your help, i really appreaciate it!

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