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Topic: empirical formulas and stoichiometry  (Read 6383 times)

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Offline jonnyset

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empirical formulas and stoichiometry
« on: July 03, 2007, 06:22:33 PM »
Hi, I have a question about this problem that just really got to me.
It's a level 3 (hardest) problem in my text and I was just stumped.

A sample of a compound of chlorine and oxygen reacts with an excess of H2 to give 0.233 g of HCl and 0.403 g. of H2O. Determine the empirical formula of the compound.

answer is Cl2O7.

Can someone help me?
Thanks =)

Offline Borek

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Re: empirical formulas and stoichiometry
« Reply #1 on: July 03, 2007, 06:28:46 PM »
How many moles of oxygee in given mass of water? What about moles of chlorine in hydrochloride? What is ratio of these?
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Offline jonnyset

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Re: empirical formulas and stoichiometry
« Reply #2 on: July 03, 2007, 07:02:25 PM »
how do you find how many moles of O in given mass of H2O?
once I know, I can find it for Cl.
Sorry, thx =)

Offline Borek

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Re: empirical formulas and stoichiometry
« Reply #3 on: July 04, 2007, 02:33:28 AM »
How many moles of oxygen per one mole of water? (hint: look at formula).

How many moles of water in given mass of water? (hint: what is a mass of one mole of water).

These are very basic mole/mass calculations, they are for sure covered in your textbook - and in many places over th enet. Just google them.
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Offline jonnyset

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Re: empirical formulas and stoichiometry
« Reply #4 on: July 04, 2007, 08:39:24 PM »
ok - would this be it?

16/18 =  .8889  X .403 =  .3582

35.45/36.45 =  .9726  X .233 = .2266

Then what?

Offline AWK

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Re: empirical formulas and stoichiometry
« Reply #5 on: July 05, 2007, 06:11:42 AM »
These are masses of an oxygen and a chlorine in your sample.
Now find an empirical formula of your compound  dividing these masses by appropriate atomic masses (O and Cl, respectively). You will get the formula ClxOy, where x and y are real numbers.
Finally find a scale factor that cause both numbers to be  integers (with a good approximation).
AWK

Offline Morgrothiel

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Re: empirical formulas and stoichiometry
« Reply #6 on: July 05, 2007, 11:13:53 AM »
k, I tried to solve it. I think it's right

You have: (For each Cl you get one HCl, and for each O you get one H2O)

ClxOy + zH2 ==> xHCl + yH2O

You see now that y = z

x = 0.233/36.5  =  0.0038356
y = 0.403/18     =  0.02238889

So we have the ratio, but we want whole numbers so we do this:

x = (0.0038356/0.0038356)*2 = 2
y =(0.02238889/0.0038356)*2 = 7

Offline enahs

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Re: empirical formulas and stoichiometry
« Reply #7 on: July 05, 2007, 12:21:55 PM »
Your method is fine Morgrothiel, but the numbers you use are wrong.
0.233/36.5 does not equal 0.0038356.

Offline Morgrothiel

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Re: empirical formulas and stoichiometry
« Reply #8 on: July 05, 2007, 12:32:36 PM »
Ah, I forgot a 6 in there. It´s supposed to be 0,00638..

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