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Offline Aud

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Need help ASAP empirical formula
« on: June 12, 2014, 06:56:23 AM »
hey guys i need help this is the problem:


A 1.254g sample of a compound containing C, H, and O + a jet of CL2 reacted and made 4.730g of HCL and 9.977 g of CCl4.

What is the empirical formula of the compound?

I dont understand how theres no Oxygen in the product


This is what ive done so far but I feel like im doing it wrong :


CHO+cl2 -----> HCL+CCl

CHO=1.254g
HCL=4.730g
CCl4=9.9977g

CHO= 29g/Moles:

C- number of moles: nC= 12/29x1.254=0.52moles
H: ------nH= 1/29x1.254=0.043moles
O------nO=16/29x1.254=0.69 moles

HCL= 36g/Moles:

H= 1/36x4.730g= 0.131 moles
CL=35/36x4.730g=4.73moles

CCl4= 152g/moles:

C 12/152gx9,977=0.79 moles
Cl4 140/152gx 9.977= 9,78 moles

Im feeling very dumb atm this is Grade 11 Chem and ive been on this for over 3 hours.  I really need some help as I value to get better in chemistry


If you have any tip on how to do this id be happy

Offline DrCMS

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Re: Need help ASAP empirical formula
« Reply #1 on: June 12, 2014, 07:26:58 AM »
Your approach is wrong and you are using the wrong molecular weight for chlorine.

In 9.977g of CCl4 there are X grams of Carbon and A grams of Chlorine  (X+A = 9.977)
in 4.730g of HCl there are Y grams of Hydrogen and B grams of Chlorine  (Y+ B = 4.73)

X + Y + Z g of O in your starting sample (X+Y+Z = 1.254)

Solve for X, Y and Z.
Turn those grams of C, H, and O in moles of C, H and O
Divide the rest by the smallest number to end up with an empirical formula with whole numbers

Offline Aud

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Re: Need help ASAP empirical formula
« Reply #2 on: June 12, 2014, 07:35:10 AM »
Wow I appreciate the response, will try aagain and will post update.

Thanks Very much DrCMS

Offline Aud

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Re: Need help ASAP empirical formula
« Reply #3 on: June 12, 2014, 08:44:37 AM »
im still stuck  :-\  I have:


 Cx+Hy+Oz=1.254g
H+CL=4.730g
C+Cl4=9.977g



C+CL4 n=9.977/152 = 0.06moles

nM=m so 0.6x12=0.72

so C= 0.72g???

am I on the right path ?
         

Offline DrCMS

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Re: Need help ASAP empirical formula
« Reply #4 on: June 12, 2014, 09:55:32 AM »
As I said earlier you are using the wrong molecular weight for chlorine it is 35.5 not 35.

So CCl4 has a molecular weight of 154 of which 142 is from the Cl and 12 is from C. 
So in 9.997g there are ((12/154)x9.977)g of C.

Now try it yourself.

Offline Aud

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Re: Need help ASAP empirical formula
« Reply #5 on: June 12, 2014, 09:59:22 AM »
You're right, I corrected it... It's 0.78g

Offline Aud

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Re: Need help ASAP empirical formula
« Reply #6 on: June 12, 2014, 10:05:13 AM »
So now for H+Cl

4.730g/36.5 g/mol

C=0.133


Is the answer H6C3O or not enough precised?
« Last Edit: June 12, 2014, 10:28:12 AM by Aud »

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