When "0.1 mols" of an unknown hydrocarbon "C"_x"H"_y was combusted, it produced "0.3 mols" of "CO"_2 and "0.4 mols" of "H"_2"O". What is the molecular formula of this hydrocarbon?
1 Answer
Explanation:
The trick here is to recognize the fact that you're dealing with a hydrocarbon, which means that it will only contain carbon and hydrogen.
Now, the complete combustion of a hydrocarbon produces carbon dioxide,
In your case, you know that when
Now, another important thing to realize here is that all of the carbon that was initially a part of the hydrocarbon is now a part of the carbon dioxide, and all of the hydrogen that was a part of the hydrocarbon is now a part of the water.
Since one mole of carbon dioxide contains
0.3color(red)(cancel(color(black)("moles CO"_2))) * "1 mole C"/(1color(red)(cancel(color(black)("mole CO"_2)))) = "0.3 moles C"
and
0.4color(red)(cancel(color(black)("moles H"_2"O"))) * "2 moles H"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "0.8 moles H"
Now, let's say that your hydrocarbon has a molecular formula
xcolor(white)(a)"moles of C" ycolor(white)(a)"moles of H"
If this is the case, you can say that
(0.1 xx x)color(white)(a)"moles of C"
(0.1 xx y)color(white)(a)"moles of H"
But you already know how many moles of carbon and hydrogen were produced by the reaction, so you can say that
0.1 xx x = 0.3 implies x = 0.3/0.1 = 3
and
0.1 xx y = 0.8 implies y = 0.8/0.1 = 8
Therefore, the molecular formula of the hydrocarbon is
color(green)("C"_3"H"_8) -> propane
![http://www.turbosquid.com/3d-models/propane-molecule-structure-3d-max/445795]()