Calculate the empirical formula for a compound containing 18.4% aluminum, 32.6% sulfur and 49.0% Oxygen?

This is what i have so far:(18.4g Al)(1 mol Al/26.98) .682/.6821.000(32.6 g S)(1 mol S/32.07) 1.02/.682 1.4(49.0 g O)(1 mol O/16.00)3.06/.6824.49idk what to do after that? I think I might have done the sig figs wrong or maybe rounded? idk please help

Answer:

Yesnewspapers, plastic, aluminum, glass cardboardNo we do not have a compost pileYard too small for this need.

the air I breatheI inhale oxygen and exhale carbon dioxideI do it for the plants but mostly for my own selfish needsand paper :D

Divide each % by the respective atomic mass: Al 18.4/26.98 0.68199 S 32.6/32.066 1.0166 O 49/15.999 3.0627 Divide by smallest: Al 0.68199/0.68199 1.0 S 1.0166/0.68199 1.491 1.5 O 3.0627/0.68199 4.49 4.5 Multiply y 2: Al 2 S 3 O 9 Empirical formula: Al2S3O9 I would say that this is aluminium sulphite: Al2(SO3)3 Do not round too early during the calculations: