## A compound is 54.53% C, 9.15% H, and 36.32% O by mass. What is its empirical formula?

The ratio of no of atoms of C : H : O= \( (\frac{54.53}{12}) : (\frac{9.15}{1}) : (\frac{36.32}{16}) \)

=> \( 4.54 : 9.15 : 2.27 \)

=> \( (\frac{4.54}{2.27}) : (\frac{9.15}{2.27}) : (\frac{2.27}{2.27}) \)

=> \( 2 : 4 : 1 \)

=> \( C_2 H_4 O_1 \)

=> \( C_2 H_4 O \)

Empirical Formula = \( C_2 H_4 O \)