[ahawk1]

The ternary compounds called chlorofluorocarbons, or Freons, have proved to be very valuable as refrigerants and as cleaning agents for circuit boards. Unfortunately, in the atmosphere these compounds produce chlorine atoms that catalyze the decomposition of the ozone that protects the earth from ultraviolet radiation. One of these compounds has the following mass percentages: 54.6%F and 11.5%C.

What is the empirical formula of this compound?

i have no idea at all how to start this problem. thanks!

[macman104]

Are those the only mass percents that are given?  I'm not sure if they want you to assume that the rest of the mass belongs to Chlorine or whether there is hydrogen in there.

[ahawk1]

Are those the only mass percents that are given?  I'm not sure if they want you to assume that the rest of the mass belongs to Chlorine or whether there is hydrogen in there.

thats all that is given but im pretty sure they want us to assume the rest of mass belongs to chlorine

[macman104]

Ok, then you have all of the percents, and it is just a typical mass % problem.

http://www.chem.tamu.edu/class/majors/tutorialnotefiles/empirical.htm

[ahawk1]

Ok, then you have all of the percents, and it is just a typical mass % problem.

http://www.chem.tamu.edu/class/majors/tutorialnotefiles/empirical.htm

alright so i went there found mass of each element in moles. Times the mass in moles by percentage. divided by smallest number and came out with C=1,Cl=8.5,F=7.5, so i times everything by two and got C2Cl17F15 but that answer was wrong. Can u tell me what i did wrong here?

[macman104]

Can you please show the work, it's hard to help without me knowing what you did.

[ahawk1]

Can you please show the work, it's hard to help without me knowing what you did.

i did it two ways:
1)1mol c=12g @ 11.5% =1.38          1.38/1.38   = 1x2        =2
   1mol cl=35g@ 33.9% =11.865  11.865/1.38     = 8.5x2     =17
   1mol f=19g@54.6%   =10.374  10.374/1.38      =7.5x2      =15
 
2)1mol c=12g @ 11.5% =7.59          7.59/7.59       = 1       
   1mol cl=35g@ 33.9% =22.374    22.374/7.59     = 2.95 
   1mol f=19g@54.6%   =36.036    36.036/7.59      =4.75
  Total:66g

[macman104]

Try again, you need to take the mass, so for carbon 11.5g, and then find the moles of that.

So:

11.5 / 12.01
33.9 / 35
54.6 / 19

[ahawk1]

Try again, you need to take the mass, so for carbon 11.5g, and then find the moles of that.

So:

11.5 / 12.01
33.9 / 35
54.6 / 19

oh alright i completely understand how to do this now thanks a lot!!!!