chemistry
posted by Kenry on .
6.10g of boron reacted completely with 22.9g of chlorine to give 29.0g of the metallic chloride. Calculate the empirical formula of the chloride.

Use the following method:
Mass
Formula Mass
Moles
Divide by smallest ratio
Ratio 
I've gone through this time and time again and something isn't clicking. Here is the way to do it BUT the answers are not reasonable.
%B = (6.1/29)*100 = 21.03
%Cl = (22.9/29)*100 = 78.97
Take 100 g sample which then gives you 21.03 g B and 78.97 g Cl.
mols B = 21.03/10.81 = 1.95
mols Cl = 78.97/35.45 = 2.23
B = 1.95/1.95 = 1
Cl = 2.23/1.95 = 1.14
Then you multiply both values by a whole number; i.e. by 2, 3, 4, 5, 6, etc until you come up with numbers that can be rounded to whole numbers. Multiplying by 7 gives 1*7 = 7 for B
1.14*7 = 7.98 which rounds to 8.0 and that gives B7Cl8. As far as I know there is no such compound. The person making up the problem either made a typo, didn't calculate it right, OR just made up numbers that resulted in no such compound. Some profs do that but I never did. I think it's unfair to make up fake numbers to give fake compounds.