## Chemistry and Chemical Reactivity (9th Edition)

Empirical formula $C_8H_8O_3$ Molecular formula: $C_8H_8O_3$
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: $C_8H_8O_3$