Question

a compound has a molecular weight of 112.124 atomic mass units and the empirical formula c₃h₄o. what is the molecular formula of the compound? use the periodic table to help you.
a. c₆h₈o
b. c₉h₁₂o₃
c. c₈h₄o₂
d. c₄h₈o₂
e. c₆h₈o₂

Explanation:

Step1: Calculate empirical - formula weight

The atomic masses of $C = 12.01$ amu, $H = 1.008$ amu, and $O = 16.00$ amu.
For $C_{3}H_{4}O$, the empirical - formula weight $M_{e}=3\times12.01 + 4\times1.008+1\times16.00=36.03 + 4.032 + 16.00 = 56.062$ amu.

Step2: Find the ratio $n$

$n=\frac{\text{Molecular weight}}{\text{Empirical - formula weight}}=\frac{112.124}{56.062}=2$.

Step3: Determine the molecular formula

Multiply the sub - scripts in the empirical formula by $n$.
The molecular formula is $(C_{3}H_{4}O)\times2 = C_{6}H_{8}O_{2}$.

Answer:

E. $C_{6}H_{8}O_{2}$