A compound is found to contain 53.70% iron and 46.30% sulfur. What is its empirical formula?

1 Answer
anor277
Feb 13, 2016

#Fe_2S_3#

Explanation:

In 100 g of the compound, there are 53.7 g iron, and 46.3 g of sulfur.

We divide thru by the ATOMIC masses of each constituent:

#Fe:##(53.7*cancelg)/(55.85*cancelg*mol^-1)# #=# #0.961# #mol#.

#S:##(46.3*cancelg)/(32.06*cancelg*mol^-1)# #=# #1.44# #mol#.

Now we divide thru by the LOWEST molar quantity to give...

#Fe:S# #=# #1:1.50#.

Now, the empirical formula is, by definition, the simplest #"WHOLE number ratio"# that defines constituent atoms in a species. To get whole numbers, clearly we have to multiply the given ratio by 2, to give (FINALLY):

#Fe_2S_3#

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