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Empirical Formula of a Compound with only percentages given

  • Thread starter dacoolguy
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Homework Statement


Determine the Empirical Formula of a compound which, when analyzed, is shown to consist of 62.07% carbon, 27.59% oxygen and 10.34% hydrogen.
The answer is supposedly C3OH6

Homework Equations





The Attempt at a Solution


First,I started by writing the ratio: C : O : H
Then, under each, I wrote the respective percentages.
I divided everything by the lowest percentage (10.34), giving me a ratio of 6:2.668:1, (C6O3H).
I have no real idea of what to do or where I went wrong. Thanks so much for your help!
 

Answers and Replies

  • #2
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I'm really sorry- worked it out from another thread:)

Percent to mass,
Mass to mole,
Divide by small,
Multiply 'til whole.

Assume 100g of compound. Change % to g. Divide these masses by their molar masses to get moles. Divide each mole amount by the smallest mole quantity of the group. You'll get numbers that are either close to a whole number or close to half (such as 1.5, 2.5, etc.) multiply all of the numbers by a common factor until they are all whole numbers, and this will be the empirical formula.
Thanks to Gannon on thread: https://www.physicsforums.com/showthread.php?t=224348 :)

PS: how do I delete a thread once I found the solution (before any posts)?
 

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