I went through the math and ended up withy 52.04 for the molar mass of the hydrocarbon; I expect that small difference is in what we used for the units and I wouldn't worry about that. How to do the remainder which is to determine the empirical formula.
CxHy ==> xCO2 + yH2O
g CO2 = 2.4167
g H2O = 0.4967
You want to convert the 2.4167 g CO2 to mols C and the 0.4967 g H2O to mols H atoms. The easy way to do that is this way. I'll estimate; you should do these more accurately. mols = g/molar mass.
(2.4167/44) = 0.0549 mols CO2 = mols C.
(0.4967/18) = 0.0276 mols H2O = 0.0551 mol H atoms.
Now we find the whole number relation between C and H with no number being less than 1.00. The easy way to do that is to divide the smaller number by itself and divide the other number by the small number too. Since both numbers are essentially the same we know this is a 1:1 realtionship so the empirical formula is CH and the empirical mass is (12 for C and 1 for H for a total of 13).
So the molar mass is 52; the empirical mass is 13 so we ask ourselves the question of how many of the empirical units we have in the molar unit? That answer is (52/13) = 4 which means we have 4 units of CH strung together and the molecular formula is C4H4. The best structure I can think of is a four member ring with a double bond at the top and one at the bottom. I don't know that I can draw it on this forum.
H
|
C=C-H
|.|
C=C-H
|
H
Hope that looks ok. Ignore the . in the middle; I had to put it in to "HOPE" the spacing would come out ok.
A .7178 g sample of a hydrocarbon occupies a volume of 390.7 mL at 65C and 99.2kpa. When the sample is burned in excess oxygen, 2.4267 g CO2 and .4967g H2O are obtained. What is the molecular formula of the hydrocarbon? Write a plausible structural formula for the molecule.
This is what I have so far. Using PV=m/MRT, i calculated the molar mass to be 52.02g/mol. Now, I'm stuck. Could someone explain where I go from here?
3 answers
Let me try that again. We just can't draw structures on this forum but here's another try.
H
|
C=C-H
|...|
C=C-H
|
H
Again, ignore the three periods I've put in to try and make that bond between the two right most carbon atoms.
H
|
C=C-H
|...|
C=C-H
|
H
Again, ignore the three periods I've put in to try and make that bond between the two right most carbon atoms.
Third time is the charm.
H
|
C=C-H
|....|
C=C-H
|
H
Again, ignore the four periods I've put in to try and make that bond between the two right most carbon atoms.
H
|
C=C-H
|....|
C=C-H
|
H
Again, ignore the four periods I've put in to try and make that bond between the two right most carbon atoms.