Ethanol=C2H5OH

C2= 2*12

H5=5*1

O=16

H=1

24+5+16+1= 46g/mole

Carbon= (24/46)*100= 52.2%

Hydrogen= 5+1=6 (6/46)*100= 13.0%

Oxygen= (16/46)*100= 34.8%

To determine the percent composition of ethanol, we need to determine the weight of each element from ethanol.

We'll write the chemical formula of ethanol: C2H5OH.

To calculate the weight of each element, we need to know how many moles of each element ethanol has. This thing could be easy determined just looking at the subscripts of each constituent element. Therefore ethanol has : 2 mols of carbon, 1 mole of oxygen and 6 mols of hydrogen.

We'll calculate the mass of these elements:

Mass of carbon: 2 mols* 12.01g/mol = 24.02 grams

Mass of oxygen: 1mol*16g/mol = 16 grams

Mass of hydrogen: 6 moles*1.01 g/mol = 6.06 grams

Now, we'll add the results to determine the mass of 1 mol of ethanol:

Mass of C2H5OH = 24.02 + 16 + 6.06 = 46.08 g

Now, we'll determine the percent composition of each element:

mass percent of C: (24.02/46.08)*100% = 52.14%

mass percent of O: (16/46.08)*100% = 34.77%

mass percent of H: (6.06/46.08)*100% = 13.15%

**The requested mass percents are: 52.14% ; 34.77% ; 13.15%.**