Ok...I cannot figure out this chemestry problem in my book...can you help me?
What is its empirical formula for 18.6g C, 3.1g H, and 55.02g Cl?
I am litterally freaking out because I have a B in Chem...and well, let's just say it will not be good if I get anything less than an A. My parents need me to be perfect...Can you please explain how you got your answer so that I can learn how to do it? THANK YOU SO MUCH!!!! I would give you a huge hug if I could, but...well ur on the other end of a computer screen...Thank you!
-Someone desperate for answers
+394 What is its empirical formula for 18.6g C, 3.1g H, and 55.02g Cl?
0.24% C 0.04% H 72% Cl of 76.72g total
Convert to moles and find molar ratios
18.6/12.01 =1.55 mol C
3.1/1.00 = 3.10 mol H
55.02/35.45 = 1.55 mol Chlorine
The carbon and chlorine have the same number atoms and hydrogen will have very near twice as many
Playing with it gives These
CClH2
These are closer to the ratios
C3Cl3H6
C31Cl31H61
C67Cl67H132
_7UP_
+394 What is its empirical formula for 18.6g C, 3.1g H, and 55.02g Cl?
0.24% C 0.04% H 72% Cl of 76.72g total
Convert to moles and find molar ratios
18.6/12.01 =1.55 mol C
3.1/1.00 = 3.10 mol H
55.02/35.45 = 1.55 mol Chlorine
The carbon and chlorine have the same number atoms and hydrogen will have very near twice as many
Playing with it gives These
CClH2
These are closer to the ratios
C3Cl3H6
C31Cl31H61
C67Cl67H132
_7UP_
