Answer
Empirical formula $C_8H_8O_3$
Molecular formula: $C_8H_8O_3$
Work Step by Step
Atomic weights (g/mol): $C: 12.011,\ H: 1.008,\ O: 15.999$
In 100.0 g:
C: $63.15\ g \div 12.011 = 5.258\ mol$
H: $5.30\ g\div 1.008 = 5.258\ mol$
O: $31.55\ g\div15.999=1.972\ mol$
Normalizing by the smallest amount of moles:
C,H: $5.258/1.972= 2.666$
O: $1.972/1.972= 1.0$
To get integer numbers on both proportions, we multiply each by 3 which yields:
Empirical formula $C_8H_8O_3$
Empirical formula's molar mass: $M_u=152.15\ g/mol$
$M/M_u=1.0$
Molecular formula: $C_8H_8O_3$
