Maleic acid is an organic compound composed of \(41.39 \% \mathrm{C},\) 3.47$\% \mathrm{H}$ , and the rest oxygen. If 0.129 mole of maleic acid has a mass of 15.0 \(\mathrm{g}\) , what are the empirical and molecular formulas of maleic acid?

To convert the given percentage composition (by mass) of each element to moles, we'll first assume that we have 100 grams of the compound. This makes the mass of each element equal to its percentage composition. Then, we'll use the molar masses of each element to convert mass to moles. For Carbon (C): Percentage of C = 41.39% Mass of C in 100g of the compound = 41.39 g Moles of C = \(\frac{41.39\,\text{g}}{12.01\,\text{g/mol}} = \text{3.445}\,\text{moles}\) For Hydrogen (H): Percentage of H = 3.47% Mass of H in 100g of the compound = 3.47 g Moles of H = \(\frac{3.47\,\text{g}}{1.01\,\text{g/mol}} = \text{3.436}\,\text{moles}\) The remaining mass will be oxygen, as it is the only other element in the compound. For Oxygen (O): Percentage of O = 100% - (41.39% + 3.47%) = 55.14% Mass of O in 100g of the compound = 55.14 g Moles of O = \(\frac{55.14\,\text{g}}{16.00\,\text{g/mol}} = \text{3.446}\,\text{moles}\)

To find the empirical formula, we'll divide the moles of each element by the lowest number of moles to get the molar ratio. Lowest moles = 3.436 moles (H) Molar Ratio of C = \(\frac{3.445\,\text{moles}}{3.436\,\text{moles}} \approx 1\) Molar Ratio of H = \(\frac{3.436\,\text{moles}}{3.436\,\text{moles}} = 1\) Molar Ratio of O = \(\frac{3.446\,\text{moles}}{3.436\,\text{moles}} \approx 1\) The empirical formula is \(\text{CHO}\).

First, find the molar mass of the empirical formula. Molar mass of CHO = 12.01 g/mol (C) + 1.01 g/mol (H) + 16.00 g/mol (O) = 29.02 g/mol Next, we'll use the mass of 0.129 moles of the compound to calculate its molar mass. Molar mass of the compound = \(\frac{15.0\,\text{g}}{0.129\,\text{mole}} = \text{116.28}\,\text{g/mol}\) Divide the molar mass of the compound by the molar mass of the empirical formula to find the multiplier. Multiplier = \(\frac{116.28\,\text{g/mol}}{29.02\,\text{g/mol}} \approx 4\) Finally, multiply the empirical formula by the multiplier to get the molecular formula. Molecular Formula = CHO x 4 = C4H4O4