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.

To determine the empirical and molecular formula of acetic acid, we start with its percentage composition and molecular mass.<br /><br />Given:<br />- Carbon (C): 39.9%<br />- Hydrogen (H): 6.7%<br />- Oxygen (O): 53.4%<br />- Molecular mass: 60 amu<br /><br />First, convert the percentages to grams assuming a 100 g sample:<br />- C: 39.9 g<br />- H: 6.7 g<br />- O: 53.4 g<br /><br />Next, convert these masses to moles by dividing by their respective atomic masses:<br />- Moles of C: \( \frac{39.9 \text{ g}}{12.01 \text{ g/mol}} = 3.32 \text{ mol} \)<br />- Moles of H: \( \frac{6.7 \text{ g}}{1.008 \text{ g/mol}} = 6.65 \text{ mol} \)<br />- Moles of O: \( \frac{53.4 \text{ g}}{16.00 \text{ g/mol}} = 3.34 \text{ mol} \)<br /><br />Now, find the simplest whole number ratio by dividing each by the smallest number of moles:<br />- Ratio of C: \( \frac{3.32}{3.32} = 1 \)<br />- Ratio of H: \( \frac{6.65}{3.32} = 2 \)<br />- Ratio of O: \( \frac{3.34}{3.32} = 1 \)<br /><br />The empirical formula is \( CH_2O \).<br /><br />To find the molecular formula, compare the empirical formula mass to the given molecular mass:<br />- Empirical formula mass of \( CH_2O \) = \( 12.01 + (2 \times 1.008) + 16.00 = 30.03 \text{ amu} \)<br /><br />Determine the ratio of the molecular mass to the empirical formula mass:<br />- \( \frac{60 \text{ amu}}{30.03 \text{ amu}} \approx 2 \)<br /><br />Multiply the subscripts in the empirical formula by this ratio to get the molecular formula:<br />- Molecular formula = \( (CH_2O)_2 = C_2H_4O_2 \)<br /><br />Thus, the empirical formula of acetic acid is \( CH_2O \), and the molecular formula is \( C_2H_4O_2 \).