# A hydrocarbon gas with an empirical formula CH_2 has a density of 1.88 grams per liter at 0°C and 1.00 atmosphere. What is a possible formula for the hydrocarbon?

Sep 3, 2016

We need to find a molecular mass, and the best way of doing this is the Ideal Gas Equation: $P V = n R T$.

#### Explanation:

We solve for $n$ and $n = \text{Mass"/"Molar mass}$ $=$ $\frac{P V}{R T}$

And, $\text{Molar mass}$ $=$ $\frac{R T \times \text{mass}}{P V}$

$=$ $\frac{R T}{P} \times \frac{\text{mass}}{V}$

$=$ $\frac{\rho R T}{P}$ because $\text{mass"/V=rho="density}$

And thus,

$\text{Molar mass}$ $=$ $\frac{\rho R T}{P}$

$\frac{1.88 \cdot g \cdot \cancel{{L}^{-} 1} \times 0.0821 \cdot \cancel{L} \cdot \cancel{a t m} \cdot \cancel{{K}^{-} 1} \cdot m o {l}^{-} 1 \times 273.15 \cdot \cancel{K}}{1.00 \cdot \cancel{a t m}}$

Well, this gives me an answer in $g \cdot m o {l}^{-} 1$, so I might be doing something right.

I get (finally!), $\text{Molar mass}$ $=$ $42.2 \cdot g \cdot m o {l}^{-} 1$

Now the molecular formula is always a mulitiple of the empirical formula, of course, the multiple might be $1$.

("Empirical formula")xxn="Molecular formula"

And $n \times \left(2 \times 1.00794 + 12.01\right) \cdot g \cdot m o {l}^{-} 1 = 42.2 \cdot g \cdot m o {l}^{-} 1$.

Clearly, $n = 3$, and molecular formula $=$ ${C}_{3} {H}_{6}$. We have $\text{propylene}$ or $\text{cyclopropene}$.

This is a first year problem, I take it. I ask because it gives me an idea as to how to bowl.