what is the empirical formula of A compound is found to contain 39.12 % carbon, 8.772 % hydrogen, and 52.11 % oxygen by mass.

C₃H₈O₃

Explanation:

Let's assume we have 100 g of said compound. Then we would have:

39.12 g of C8.772 g of H52.11 g of O

Now we convert those masses into moles, using their respective atomic weights:

C ⇒ 39.12 g ÷ 12 g/mol = 3.26 mol CH ⇒ 8.722 g ÷ 1 g/mol = 8.722 mol HO ⇒ 52.11 g ÷ 16 g/mol = 3.26 mol O

Then we divide those moles by the smallest number among them:

C ⇒ 3.26 mol C / 3.26 = 1H ⇒ 8.722 mol H / 3.26 = 2.68O ⇒ 3.26 mol O / 3.26 = 1

Finally we multiply those numbers by 3, so as to convert the 2.68 of H into an integer:

C ⇒ 1 * 3 = 3H ⇒ 2.68 * 3 = 8O ⇒ 1 * 3 = 3

Thus the empirical formula is C₃H₈O₃

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