Question

determine the percent composition of acetic acid CH3COOH

Answer 1

Answer:

See below

Explanation:

Using mole weights of  C =12   O = 16  H = 1

  the weight of acetic acid is  12 + 3 + 12 + 16 + 16  + 1 = 60  

C is then  (12+12 ) / 60 = 40 percent

O           (16 + 16 ) / 60  = 53.3 %

H               4 / 60          = 6 .67 %