Answer
Empirical formula $C_3H_4$
Molecular formula: $C_9H_{12}$
Work Step by Step
Atomic weights (g/mol): $C: 12.011,\ H: 1.008$
In 100.0 g:
C: $89.94\ g \div 12.011 = 7.488\ mol$
H: $10.06\ g\div 1.008 = 9.980\ mol$
Normalizing by the smallest amount of moles:
C: $7.488/7.488= 1.0$
H: $9.980/7.488 = 1.333$
To get integer numbers on both proportions, we multiply each by 3 which yields:
Empirical formula $C_3H_4$
Empirical formula's molar mass: $M_u=40.06\ g/mol$
$M/M_u=3.0$
Molecular formula: $C_9H_{12}$
