Answer:
Empirical Formula: [tex] Al_2O_3[/tex]
Explanation:
Since total mass is 306, we have:
Mass of $Al = \frac{52.9}{100}\times 306=161.87 $
Mass of $O= \frac{47.1}{100}\times 306=144.126$
Hence,
Moles of $Al=\frac{Mass}{molar mass}=\frac{161.87}{27}=6$
Moles of $O= \frac{Mass}{molar mass}=\frac{144.126}{16}=9$
Hence the formula is $Al_6O_9$ , or, $Al_2O_3$
