What is the molecular formula of a compound that has a molecular mass of 54 and the empirical formula C_2H_3?

1 Answer
May 14, 2017

${C}_{4} {H}_{6}$.............

Explanation:

The $\text{molecular formula}$ is always a whole number multiple of the $\text{empirical formula}$.

i.e. "molecular formula"=nxx("empirical formula")

$54 \cdot g \cdot m o {l}^{-} 1 = n \times \left(2 \times 12.011 + 3 \times 1.00794\right) \cdot g \cdot m o {l}^{-} 1$

$= n \times \left(27.04\right) \cdot g \cdot m o {l}^{-} 1$

Clearly, $n = \frac{54.0 \cdot g \cdot m o {l}^{-} 1}{27.04 \cdot g \cdot m o {l}^{-} 1} = 2$

$\text{Molecular formula} = \left({C}_{2} {H}_{3}\right) \times 2 = {C}_{4} {H}_{6}$ as required.

This compound has two degrees of unsaturation, so it is a butadiene, or a cylocbutene, or an allene derivative, or an alkyne.........