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Determine the percent composition of acetic acid CH3COOH
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Determine the percent composition of acetic acid CH3COOH

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Answer:

See below

Step-by-step 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 %

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