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?

We know that maleic acid has \(41.39\% \mathrm{C}\), \(3.47\% \mathrm{H}\), and the rest is oxygen. So, we can write these percentages as grams of each element in 100 g of maleic acid: Carbon: 41.39 g Hydrogen: 3.47 g Oxygen: 100 - 41.39 - 3.47 = 55.14 g Now, divide the grams of each element by its molar mass to get the number of moles: moles of Carbon: \(\frac{41.39}{12.01} = 3.448\) moles of Hydrogen: \(\frac{3.47}{1.008} = 3.444\) moles of Oxygen: \(\frac{55.14}{16.00} = 3.446\)

Divide each mole value by the smallest mole value to get the ratio of each element in the empirical formula: Carbon: \(\frac{3.448}{3.444} = 1\) Hydrogen: \(\frac{3.444}{3.444} = 1\) Oxygen: \(\frac{3.446}{3.444} = 1\) The whole-number ratio is 1:1:1. Thus, the empirical formula is CHO.

First, we will find the molar mass of the empirical formula: Empirical formula molar mass: \(12.01\mathrm{C} + 1.008\mathrm{H} + 16.00\mathrm{O} = 29.018\, \mathrm{g/mol}\) Given that 0.129 moles of maleic acid have a mass of 15.0 g, we can calculate the molar mass of maleic acid: Molar mass of maleic acid = \(\frac{15.0\, \mathrm{g}}{0.129\, \mathrm{mol}} = 116.28\, \mathrm{g/mol}\) Now, divide the molar mass of maleic acid by the molar mass of its empirical formula to find the whole-number multiple: Multiple: \(\frac{116.28}{29.018} \approx 4\) Finally, multiply the empirical formula by the whole-number multiple to get the molecular formula: C(1 x 4)H(1 x 4)O(1 x 4) = C4H4O4 Therefore, the molecular formula of maleic acid is C4H4O4.