[sundrops]

A particular carborane has the following mass percentages:
71.94% B
12.07% H
15.99% C

What is the empirical formula of this compound?

Now what I did is the following,
71.94g B * (1mol B / 100g B) = 0.7194mol B
12.07g H * (1mol H / 100g H) = 0.1207mol H
15.99g C * (1mol C / 100g C) = 0.1599mol C

then I wdivided by the lowest denominator:

B 0.7194/0.1207 = 5.96
H 0.1207 / 0.1207 = 1
C 0.1599/0.1207 = 1.32

but I'm having trouble writing out an empirical formula because don't have whole numbers. please help.

[AWK]

Calculate moles of B,C and H in 100 grams of compound, ie;
moles B=71.94/10.811
and so on
Then all moles divide by a smallest number of moles, and multiply evrentually by whole nambers, 2 3 to obtain numbers of moles close to whole numbers.

What you did ?
71.94g B * (1mol B / 100g B) = 0.7194mol B
12.07g H * (1mol H / 100g H) = 0.1207mol H
15.99g C * (1mol C / 100g C) = 0.1599mol C
You calculated mass of elements (not moles of elements) in 1 gram of carborane.

The rest of calculations are useless for this problem

[sundrops]

Thanks - that worked!

My answer was correct!