Question

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?

Answer 1

`"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, `"CO"_2`, and water, `"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 `"C"_x"H"_y`. You know that one mole of this hydrocarbon would produce

• `xcolor(white)(a)"moles of C"`
• `ycolor(white)(a)"moles of H"`

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

`(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