Example

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Chem_Mod
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Example

Postby Chem_Mod » Sun Dec 04, 2011 2:13 am

6.40g of a compound was burned in air and produced 8.80g of CO2 and 7.20g of H2O. Find the empirical formula.

Chem_Mod
Posts: 18913
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 737 times

Re: Example

Postby Chem_Mod » Sun Dec 04, 2011 2:13 am

The first thing is to realize that when burning you are just reacting the unknown compound with O2. First, you have to use the mass of CO2 produced to find the number of mols of CO2 and therefore the mols of C present. Then you have to do the same with H2O to find the mols of H present. Understanding that the compound is made of C, H and O in some ratio, you can convert mols C and mols H to grams of each, add those two up and subtract their combined weight from the weight of the compound to get the mass of O in the compound. You can then convert grams of O into mols. Now you know the mols of C, H, and O present in your compound. Divide all of these by the lowest of the 3 and you will be able to determine the ratio of the elements. For example if you have 6 mol C, 2 mol H and 8 mol O: you divide all of those numbers by 2 and find that the ratio is 3:1:4 C:H:O so your formula is C3HO4.

lnlyzelda_3J
Posts: 7
Joined: Tue Nov 15, 2016 3:00 am

Re: Example

Postby lnlyzelda_3J » Tue Jul 11, 2017 9:04 pm

At first, I was confused with the answer for this problem. From looking at other posts I assumed the answer was C3HO4 just like some peers did. No matter how many times I tried I got a different answer. I finally realized that the answer was not provided only and example of how to reduce the moles in order to get an accurate ratio.
It would have been too easy if the answer had been posted.


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