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.

Answer 1

Use the following method:
Mass
Formula Mass
Moles
Divide by smallest ratio
Ratio

Answer 2

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.

Answer 3

I AGRÉE WITH YOU

Answer 4

agreed

Answer 5

That's true, thanks

Answer 6

I like the explicandum, it is very understandable.

Answer 7

Coming to the last part, its not just 7 that can round it up. Check again. Numbers b4 7. Just My observation

Answer 8

I know there is only BCl3
B8Cl8
B9Cl9

Let's look into these instead