Question

when 6.25 grams of pure iron are allowed to react with oxygen,a black oxide forms.if the product weighs 8.15g,what is the empirical formula of the oxide?

Answer 1

By the law of conservation of mass, and assuming this reaction goes to completion, if 6.25 g of Fe reacted, 8.15 - 6.25 = 1.9 g O2 reacted with it. Now find moles:

For Fe: 6.25/56 = 0.11
For O2: 1.9/32 = 0.06

Now divide both by the lower number to find the molar ratio of the product (i.e. the empirical formula):

0.06/0.06 = 1
0.11/0.06 ~ 2

They give you actual masses so this results in the actual molecular formula. We see that 2 moles of Fe (2 moles of Fe atoms) reacts with 1 mole of O2 (2 moles of O atoms). This means your oxide is Fe2O2 which becomes FeO for the empirical formula. Your reaction is:

\[2Fe (s)+O_{2}(g) \rightarrow Fe_{2}O_{2}(s)\]