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.