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23. The Percentage Composition of Acetic Acid Is Found to Be 39.9% C,6.7% H,and 53.4% and Its Molecular Mass Was Determined by

Question

23. The percentage composition of acetic acid is found to be 39.9% C,6.7% H,and 53.4% and its molecular mass was determined by experiment to be 60 amu. Calculate the empirical and molecular formula of acetic acid.

Solution

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Rodrigo Especialista · Tutor por 3 anos

Resposta

To determine the empirical and molecular formula of acetic acid, we start with its percentage composition and molecular mass.Given:- Carbon (C): 39.9%- Hydrogen (H): 6.7%- Oxygen (O): 53.4%- Molecular mass: 60 amuFirst, convert the percentages to grams assuming a 100 g sample:- C: 39.9 g- H: 6.7 g- O: 53.4 gNext, convert these masses to moles by dividing by their respective atomic masses:- Moles of C: - Moles of H: - Moles of O: Now, find the simplest whole number ratio by dividing each by the smallest number of moles:- Ratio of C: - Ratio of H: - Ratio of O: The empirical formula is .To find the molecular formula, compare the empirical formula mass to the given molecular mass:- Empirical formula mass of = \( 12.01 + (2 \times 1.008) + 16.00 = 30.03 \text{ amu} \)Determine the ratio of the molecular mass to the empirical formula mass:- Multiply the subscripts in the empirical formula by this ratio to get the molecular formula:- Molecular formula = \( (CH_2O)_2 = C_2H_4O_2 \)Thus, the empirical formula of acetic acid is , and the molecular formula is .