Question

Determine the empirical formula of a compound with the following composition by

mass: 60.0 % C, 8.0 % H and 32.0 % O. If 100 mg of this compound is neutralized

with exactly 10.0 mL of 0.0500 M NaOH in a reaction with a 1:1 stoichiometry,

what is the molecular formula of this compound?

Answer 1

Empirical Formula =$C_7H_6O_3$

Molecular Formula =$C_{14}H_{12}O_{6}$