A #22.35g# mass of an unknown compound consisting of potassium, chlorine, and oxygen, contained #7.13g# of potassium, #6.47*g# of chlorine. What is its empirical formula?

Question

1689 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-19T18:31:31

#KClO_3#

As always, we divide each constitiuent mass thru by the ATOMIC mass of each constituent atom.

#"Moles of potassium"# #=# #(7.13*g)/(39.10*g*mol^-1)# #=# #0.182*mol#.

#"Moles of chlorine"# #=# #(6.47*g)/(35.45*g*mol^-1)# #=# #0.182*mol#.

#"Moles of oxygen"# #=# #((22.35-7.13-6.47)*g)/(15.999*g*mol^-1)# #=# #0.546*mol#.

And now we divide thru by the LOWEST molar quantity, that of potassium/chlorine, to give an empirical formula of #KClO_3#.

How did I know that that there were #(22.35-7.13-6.47)*g# of #O#? The mass of oxygen was not given in the problem.

A #22.35*g# mass of an unknown compound consisting of potassium, chlorine, and oxygen, contained #7.13*g# of potassium, #6.47*g# of chlorine. What is its empirical formula?