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?

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Answer 1 · 2016-09-03T16:48:23

We need to find a molecular mass, and the best way of doing this is the Ideal Gas Equation: $PV=nRT$ .

We solve for $n$ and $n="Mass"/"Molar mass"$ $=$ $(PV)/(RT)$

And, $"Molar mass"$ $=$ $(RTxx"mass")/(PV)$

$=$ $(RT)/Pxx"mass"/V$

$=$ $(rhoRT)/P$ because $"mass"/V=rho="density"$

And thus,

$"Molar mass"$ $=$ $(rhoRT)/P$

$(1.88*g*cancel(L^-1)xx0.0821*cancel(L)*cancel(atm)*cancel(K^-1)*mol^-1xx273.15*cancel(K))/(1.00*cancel(atm))$

Well, this gives me an answer in $g*mol^-1$ , so I might be doing something right.

I get (finally!), $"Molar mass"$ $=$ $42.2*g*mol^-1$

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 $nxx(2xx1.00794 +12.01)*g*mol^-1=42.2*g*mol^-1$ .

Clearly, $n=3$ , and molecular formula $=$ $C_3H_6$ . We have $"propylene"$ or $"cyclopropene"$ .

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

A hydrocarbon gas with an empirical formula CH_2CH_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?