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Empirical Formula:

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, $Al_2O_3$

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