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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: jonnyset on July 03, 2007, 06:22:33 PM

Title: empirical formulas and stoichiometry
Post by: jonnyset 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 =)

Title: Re: empirical formulas and stoichiometry
Post by: Borek 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?

Title: Re: empirical formulas and stoichiometry
Post by: jonnyset 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 =)

Title: Re: empirical formulas and stoichiometry
Post by: Borek 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.

Title: Re: empirical formulas and stoichiometry
Post by: jonnyset 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?

Title: Re: empirical formulas and stoichiometry
Post by: AWK 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 Cl_xO_y, where x and y are real numbers.
Finally find a scale factor that cause both numbers to be  integers (with a good approximation).

Title: Re: empirical formulas and stoichiometry
Post by: Morgrothiel 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)

Cl_xO_y + 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

Title: Re: empirical formulas and stoichiometry
Post by: enahs 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.

Title: Re: empirical formulas and stoichiometry
Post by: Morgrothiel on July 05, 2007, 12:32:36 PM

Ah, I forgot a 6 in there. It´s supposed to be 0,00638..