Question

Combustion analysis of fluorene, a polycyclic aromatic hydrocarbon used to make dyes, plastics, and pesticides, produced 11.44 g CO2 and 1.80 g H2O. Calculate the empirical formula for fluorene.

Answer 1

No of moles of C = 11.44g times 1 mol CO_2/44.01g times 1 mol C/1 mol CO_2 = 0.260 mol C No of moles of H = 1.80 H_2 O times 1 mol H_2O/18.02g H_2O times 2 mol H/1 mol H_2O = 0.200 mol H Among the values of relative no of atoms of 4 elements, choose the least one (0.200) and divided the other values by this least one to get the simplest ratio of atoms C = 0.260/0.200 = 1.3 times 10 = 13 H = 0.200/0.200 = 1 times 10 = 10 Place these simplest ratio values as the lower subscript to the concerned element to get the empirical formula. C_13 H_10