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Topic: empirical fomula  (Read 4491 times)

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Offline mol

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empirical fomula
« on: February 09, 2011, 08:29:11 AM »
If for example, C=1 and H=1.45 for a hydrocarbon and it contains only c and h,
 how should I  caluclate the empirical formula if it is not a whole no for H?
also , from here how do i go for molecular formula? is there any formula involved for both of these formulas?
 pls help.
 thanks
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Offline sjb

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Re: empirical fomula
« Reply #1 on: February 09, 2011, 10:02:49 AM »
Quote from: mol on February 09, 2011, 08:29:11 AM
If for example, C=1 and H=1.45 for a hydrocarbon and it contains only c and h,
 how should I  caluclate the empirical formula if it is not a whole no for H?
also , from here how do i go for molecular formula? is there any formula involved for both of these formulas?
 pls help.
 thanks

What is the highest common factor of 1 and 1.45?
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Offline mol

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Re: empirical fomula
« Reply #2 on: February 09, 2011, 10:50:02 AM »
iam sorry i dont get it. is it 1.45?
 so should i divide them both?
pls help
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Offline sjb

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Re: empirical fomula
« Reply #3 on: February 09, 2011, 10:55:50 AM »
Quote from: mol on February 09, 2011, 10:50:02 AM
iam sorry i dont get it. is it 1.45?
 so should i divide them both?
pls help

Can you simplify the ratio 1 : 1.45 so that both are whole numbers?
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Offline AWK

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Re: empirical fomula
« Reply #4 on: February 10, 2011, 03:26:09 AM »
Quote from: mol on February 09, 2011, 08:29:11 AM
If for example, C=1 and H=1.45 for a hydrocarbon and it contains only c and h,
 how should I  caluclate the empirical formula if it is not a whole no for H?
also , from here how do i go for molecular formula? is there any formula involved for both of these formulas?
 pls help.
 thanks
You should find a factor that gives you approximately the whole and simultaneously even number of hydrogen atoms with the error not exceeding 0.4 % for H (and eventually 0.3 % for C). The lower factor fulfilling this requirement in your case is 18. (C18H26 with the error of 0.38% for H)
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Offline DrCMS

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Re: empirical fomula
« Reply #5 on: February 10, 2011, 05:54:15 AM »
Quote from: AWK on February 10, 2011, 03:26:09 AM
Quote from: mol on February 09, 2011, 08:29:11 AM
If for example, C=1 and H=1.45 for a hydrocarbon and it contains only c and h,
 how should I  caluclate the empirical formula if it is not a whole no for H?
also , from here how do i go for molecular formula? is there any formula involved for both of these formulas?
 pls help.
 thanks
You should find a factor that gives you approximately the whole and simultaneously even number of hydrogen atoms with the error not exceeding 0.4 % for H (and eventually 0.3 % for C). The lower factor fulfilling this requirement in your case is 18. (C18H26 with the error of 0.38% for H)


Is the answer not 20 with an error of 0%
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Offline AWK

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Re: empirical fomula
« Reply #6 on: February 10, 2011, 07:02:29 AM »
Quote
Is the answer not 20 with an error of 0%
29 hydrogen atoms in hydrocarbon?
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Offline DrCMS

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Re: empirical fomula
« Reply #7 on: February 10, 2011, 08:59:56 AM »
Yes C20H29 as an empirical formula.

Which could then be C40H58, C80H116 etc etc as the molecular formula.
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Offline mol

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Re: empirical fomula
« Reply #8 on: February 10, 2011, 10:55:23 AM »
thank you all for the infos. i get it now.
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