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posted by  emc11 on 9/13/2008 6:14:33 PM  |  status: Live  

Elemental Analysis and Empirical Formula help

Course Textbook Chapter Problem
General Chemistry Chemistry (6th) by Zumdahl, Zumdahl N/A N/A
Question Details:
A particular carborane has the following mass percentages: 67.09%B and 11.62%H.
What is the empirical formula of this compound?
So far I got the C's %= 21.24%
 
If we say that we got a sample of 100g those percentages will be same in grams:
C=21.24g
B=67.09g
H=11.62g
 
Now If I divide them by their atomica mass we'll have :
 
C= 21.24g/12.011g/mol =1.7683 mol
B=67.09g/10.811g/mol = 6.205 mol
H=11.62g/1.00794g/mol = 11.578mol
 
Now If I divied them by the smallest of the above (1.7683) I should get the ratio right?
which is 1:4:7?
 
The computer says i'm wrong but i dont know what i'm doing wrong!
Can someone help me out?
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posted by leeo268 on 9/13/2008 6:47:25 PM  |  status: Live
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Response Details:
I see what is wrong.
If you do the ratio correct, you should get 1:3.5:6.5 and if time two, you should get whole number. Answer ----> 2:7:13,
Test that out on the computer
It is my pleasure to help you, please give me a rating. If not helpful, please explain why. I am only human, and I can make mistakes, but I need to know how I am wrong to improve and learn. . Thank You for your cooperation!
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posted by Jack Alope on 9/13/2008 7:19:11 PM  |  status: Live
Asker's Rating: Helpful   
emc11's comment:
"thanks soo much."
Response Details:
When you divide to get the ratio, you are not supposed to round unless it is very close, such as a .2 or .8. Since your numbers are closer to a .5 it is more accurate to double the entire ratio to achieve whole numbers, with final answer = 2:7:13
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