Empirical formula for carbon oxygen and chlorine?

1 Answer
Mar 26, 2018

Where are the DATA?

Explanation:

You should list percentage by mass of carbon, oxygen, and chlorine... A typical question SHOULD read...

#"you are given a chemical containing carbon, and chloride,"# #"and oxygen, whose percentage by mass is the following."#

#%C=18.92%#

#%Cl=55.9286%#

#"What is the empirical formula?"#

#%O# WOULD NOT be quoted...but of course you don't need it...why not? We assume an #100*g# mass of stuff, and we determine the empirical formula, the simplest whole number ratio representing constituent atoms in a species, by dividing thru by the atomic masses of each constituent element...

And so we get a trial empirical formula...

#C_((18.92*g)/(12.011*g*mol^-1))Cl_((55.93*g)/(35.45*g*mol^-1))O_((100*g-55.93*g-18.92*g)/(16.0*g*mol^-1))-=C_(1.575)Cl_(1.57)O_1.57#

#-="C""Cl""O"#

And then we would be given an estimate of the molecular mass, here #126.9*g*mol^-1#....and since we know that #"molecular formula"-=nxx"empirical formula"#...

#126.9*g*mol^-1=nxx(12.011+35.45+16.00)*g*mol^-1#

#n=2#, and our molecular formula is #C_2Cl_2O_2#...i.e. #Cl(O=)C-C(=O)Cl#.