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?

Feb 14, 2016

${\text{C"_3"H}}_{8}$

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, ${\text{CO}}_{2}$, and water, $\text{H"_2"O}$.

In your case, you know that when $0.1$ moles of this hydrocarbon underwent combustion, the reaction produced $0.3$ moles of carbon dioxide and $0.4$ moles of water.

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 $1$ mole of carbon, and one mole of water contains $2$ moles of hydrogen, it follows that the hydrocarbon contained

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 ${\text{C"_x"H}}_{y}$. You know that one mole of this hydrocarbon would produce

• $x \textcolor{w h i t e}{a} \text{moles of C}$
• $y \textcolor{w h i t e}{a} \text{moles of H}$

If this is the case, you can say that $0.1$ moles of this hydrocarbon will produce

$\left(0.1 \times x\right) \textcolor{w h i t e}{a} \text{moles of C}$
$\left(0.1 \times y\right) \textcolor{w h i t e}{a} \text{moles of H}$

But you already know how many moles of carbon and hydrogen were produced by the reaction, so you can say that

$0.1 \times x = 0.3 \implies x = \frac{0.3}{0.1} = 3$

and

$0.1 \times y = 0.8 \implies y = \frac{0.8}{0.1} = 8$

Therefore, the molecular formula of the hydrocarbon is

$\textcolor{g r e e n}{{\text{C"_3"H}}_{8}} \to$ propane 