What is the empirical formula of an #"aluminum sulfide"#, for which a mass of #2.03*g# contains a #0.73*g# mass of #"aluminum"#?

1 Answer
Jul 6, 2017

Answer:

We interrogate the empirical formula, and get #Al_2S_3#....

Explanation:

And so we work out moles of aluminum and moles of sulfur.

#"Moles of Al"=(0.73*g)/(26.98*g*mol^-1)=0.0271*mol#

#"Moles of S"=(2.03*g-0.73*g)/(32.06*g*mol^-1)=0.0406*mol#

We divide thru by the smallest molar quantity...........

And get #AlS_(1.498)#. But the #"empirical formula"# is specified to be the #"simplest WHOLE number ratio"#.........and thus we get an empirical formula of #Al_2S_3#.........

How did I know that there were #1.30*g# of sulfur?