Fumaric acid, which occurs in many plants contains, by mass, 41.4 % carbon, 3.47 % hydrogen, and 55.1 % oxygen. A 0.050 mol sample of this compound weighs 5.80 g. What is the molecular formula of this compound? (C:12 g/mol, H:1 g/mol, O:16g/mol)

Answer 1

41.4 % carbon, 3.47 % hydrogen, and 55.1 % oxygen. A 0.050 mol sample of this compound weighs 5.80 g
Take a 100 g sample which gives you 41.4 g C, 3.47 g H, 55.1 g O.
mols C = 41.4/12 = 3.45
mols H = 3.47/1 = 3.47
mols O = 55.1/16 = 3.44
So the ratio (rounded) is C 3.5; H 3.5; O 3.5. or C1H1O1 and the empirical mass is 12 + 1 + 16 = 29
mols = g/molar mass or molar mass = g/mol = 5.80/0.05 = 116
empirical mass x some whole number = molar mass
29 x ? = 116 and ? = 116/29 = 4 so the molecular formula is (C1H1O1)4 or C4H4O4.