Hi. I am trying to solve a couple of chemistry problems and am struggling with it. Can someone show me how they would do it. Here they are: A hydrocarbon mixture consists of 88.89% carbon and 11.11% hydrogen by mass. What is the empirical formula of the hydrocarbon? If the molecular weight of the hydrocarbon is approximately 54 g/mol, what is the molecular formula? Thanks!
Step 1: Divide through by the atomic masses (C: 7.4075; H: 11.11) Step 2: Divide through by the smallest number (C: 1; H: 1.4998) Step 3: Mulitple by the appropriate factor to get whole numbers (Or near enough) and round appropriately. This give you an emperical formula. (C: 2; H: 3) Step 4: Determine the Molecular Wieght of the emperical formula. ( 12*2 + 1*3 = 27) Step 5: Multiply the Emperical formula of the unknown by the Molecular mass of the unknown divided by the molecular mass of the emprical formula. \(C_2H_3\)*54/27 = \(C_4H_6\) Then realize that we pretty much have to be talking about either Butadiene, or Cyclobutene (I don't think Cyclobutene actually exists though, the strain on the SP2 hybridized bonds would be huge, and itw ouldn't take much to break it at all).
Cyclobutene exists. Surprisingly, so does cyclopropene, against all expectations of how much an sp2 bond can distort. Our chemical could also be methyl-cyclopropene or ethylacetylene.
Actually I would be interested to read an analysis of what's happening with the double bond in cyclopropene. It really doesn't seem like it should be able to exist. I wonder if it's really an sp2 hybrid, or if there's something more complicated going on.
My guess at this point (I'mn at work) would be that yes, it's a genuine SP2 hybrid, but, it has a stronger P charachter then S character - if that makes sense. But then again, the sigma network is what's actually hybridized, the Pi bonding takes place through a 'straight' P orbital.
It also occurs to me that the 3-4 Sigma bond is going to be longer than the 1-2 Pi bond, meaning the angles in Cyclobutene won't be exactly 90, it'll be a trapezoid, rather than a square :facepalm:
