A compound has the empirical formula CHCl. A 256 mL flask, at 373 K and 750 torr, contains 0.800 g of the compound, what is the molecular formula?

Nov 14, 2015

Assuming ideality, the molecular formula is ${C}_{2} {H}_{2} C {l}_{2}$

Explanation:

From the ideal gas law, $M M = \frac{m R T}{P V}$; where $M M$ $=$ molecular mass; $m$ $=$ mass; $P$ $=$ pressure in atmospheres; $V =$ volume in litres; $R$ $=$ gas constant with appropriate units.

So, $\frac{0.800 \cdot g \times 0.0821 \cdot \cancel{L} \cdot \cancel{a t m} \cdot {\cancel{K}}^{- 1} \cdot m o {l}^{- 1} \times 373 \cdot \cancel{K}}{0.256 \cdot \cancel{L} \times 0.987 \cdot \cancel{a t m}}$ $=$ $97.0$ $g \cdot m o {l}^{- 1}$.

$n \times \left(12.01 + 1.01 + 2 \times 35.45\right) \cdot g \cdot m o {l}^{- 1}$ $=$ $97.0 \cdot g \cdot m o {l}^{- 1}$.

Clearly, $n$ $=$ $1$. And molecular formula $=$ ${C}_{2} {H}_{2} C {l}_{2}$.

I seem to recall (but can't be bothered to look up) that vinylidene chloride, ${H}_{2} C = C {\left(C l\right)}_{2}$ is a low boiling point gas, whereas the 1,2 dichloro species is a volatile liquid. At any rate we have supplied the molecular formula as required.