Answer:
[tex]C_2H_3O[/tex]
Explanation:
Hello!
In this case, since the percent composition analysis provide the by-mass percent of each atom in the molecule, we can assume we have 52 g of carbon, 13 g of hydrogen and 35 g of oxygen, so we are able to compute the moles based on their atomic masses:
[tex]n_C=\frac{52g}{12.01g/mol} =4.33\\\\n_H=\frac{13g}{2.02g/mol}=6.44\\\\n_O=\frac{35g}{16.00g/mol}=2.19[/tex]
Now we divide all the moles by those of oxygen as the fewest ones in order to obtain their subscripts in the empirical formula:
[tex]C=\frac{4.33}{2.19} =2.0\\\\H=\frac{6.44}{2.19} =3.0\\\\O=\frac{2.19}{2.19} =1[/tex]
Thus, the empirical formula is:
[tex]C_2H_3O[/tex]
Best regards!
