Let x, y, and z be the number of carbon (C), hydrogen (H) and oxygen (O) of terephthalic acid. The combustion reaction of the given compound can be expressed as,
CxHyOz + O2 --> CO2 + H2O
We are given that 0.250 mol of the acid is equal to 41.5 g. We calculate for the molar mass by dividing the mass by the number of moles.
M = (41.5 g)/ (0.250 mol) = 166 g/mol
Then, we calculate the number of moles of the given,
15.85 mg T acid / (41.5 g/0.250 mol) = 317/3320 mmol
Solving for the number of mols of CO2 in order to determine the number of mols of carbon,
(33.58 mg CO2) / (44 mg/1 mmol) = 0.763 mmol of CO2 (0.763 mmol of C)
For the number of mols of hydrogen from the produced water,
(5.16 mg H2O) / (18 mg H2O/1 mmol) = 0.287 mmol H2O (0.573 mmol of H)
Solving for the amount of oxygen from the terephthalic acid,
15.85 mg - (0.763 mmol C)(12 mg/1 mmol C) - (0.573 mmol H)(1 mg H/1 mmol H)
= 6.121 mg O
number of mols of O = (6.121 mg O)(1 mmol O/16 mg O)
= 0.383 mmol O
Thus, the empirical formula is,
C0.763H0.573O0.383
Dividing the numbers by 0.383
C2H(3/2)O
Eliminating the fraction,
C₈H₆O₄
