What is the empirical formula for a compound that is 60.9% arsenic and 39.1% sulfur?

1 Answer
Apr 2, 2016

Answer:

#As_2S_3#

Explanation:

How did we get this formula?

We assume #100.0# #g# of compound, and there are #60.9# #g# of arsenic, and #39.1# #g# of sulfur.

We divide each mass thru by the molar mass of each element:

#As# #=# #(60.9*g)/(74.92*g*mol^-1)# #=# #0.813# #mol#.

#S# #=# #(39.1*g)/(32.06*g*mol^-1)# #=# #1.22# #mol#.

Now, we divide thru by the LOWEST molar quantity, #0.813# #mol#:

We get #AsS_(1.500)#. But we want the whole number ratio, so we get #As_2S_3# as required.