# What is the molecular formula of a compound with the empirical formula CH and a molar mass of 39.01 grams per mole?

May 26, 2016

$\text{Molecular formula}$ $=$ $\left(\text{Empirical formula}\right) \times n$

#### Explanation:

And thus the molecular mass must be a whole number multiple of the empirical formula mass.

Here, clearly, $39.01 \cdot g \cdot m o {l}^{-} 1$ $=$ $3 \times \left(12.011 + 1.008\right) \cdot g \cdot m o {l}^{-} 1$

Thus the molecular formula $=$ ${C}_{3} {H}_{3}$.

As far as I know, this result make no sense at all, inasmuch as there is no molecule with a formula of ${C}_{3} {H}_{3}$. (If I am not seeing something bleeding obvious, then someone will tell me!)

There is a cyclopropenyl cation, ${C}_{3} {H}_{3}^{+}$, but this would have a different molecular mass; why? Allene, ${C}_{3} {H}_{4}$ does not fit this description.