Question

A percent composition analysis yields 52% carbon (C), 13% hydrogen (H), and 35% oxygen (O). What is the empirical formula for the compound?

Answer 1

c2h6o

Step-by-step explanation:

Answer 2

`C_2H_3O`

Step-by-step 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:

`n_C=(52g)/(12.01g/mol) =4.33\\\\n_H=(13g)/(2.02g/mol)=6.44\\\\n_O=(35g)/(16.00g/mol)=2.19`

Now we divide all the moles by those of oxygen as the fewest ones in order to obtain their subscripts in the empirical formula:

`C=(4.33)/(2.19) =2.0\\\\H=(6.44)/(2.19) =3.0\\\\O=(2.19)/(2.19) =1`

Thus, the empirical formula is:

`C_2H_3O`

Best regards!