Question

Combustion analysis of toluene, a common organic solvent, produces 2.93 mg of `CO_2` and .685 g of `H_2O`. The molecular mass is 46 g/mol. If the compound contains only carbon and hydrogen, what is the molecular formula?

Answer 1

I'm quite sure there's a typo here. There's no way that you made only `"2.93 mg"` of `"CO"_2`. I think it's in `"g"`. Furthermore, Toluene is NOT `"46 g"`. It's at least twice as heavy.

First I would see how many moles of each compound were made so we can construct the right side of the complete combustion reaction. Let Toluene be labeled as `"C"_i"H"_j`.

`kC_iH_j(l) + l"O"_2(g) -> m"CO"_2(g) + n"H"_2"O"(g)`

where we need to determine `k`, `l`, `m`, `n`, `i`, and `j`.

`2.93 cancel("g CO"_2) xx ("1 mol CO"_2)/(44.01 cancel("g CO"_2)) = "0.06658 mol CO"_2`

`0.685 cancel("g H"_2"O") xx ("1 mol H"_2"O")/(18.015 cancel("g CO"_2)) = "0.03802 mol H"_2"O"`

Then, let's normalize this so that the `"mol"`s of `"CO"_2` becomes `1`. After all, you can only have at minimum `1` equivalent of an atom within a molecule.

`0.06658/0.06658 = "1 mol CO"_2`

`0.03802/0.06658 = "0.5710 mol H"_2"O"`

We still need to get both of these to whole numbers, so what I would do is keep adding `0.5710` to itself until you get a whole number, and keep track of how many you used to determine the integer to multiply by for `"CO"_2`. I got `7xx0.5710 ~~ 4`. Thus, so far we have:

`k"C"_i"H"_j (l) + l"O"_2(g) -> 7"CO"_2(g) + 4"H"_2"O"(g)`

At this point we can start balancing the equation:

`"C"_i"H"_j (l) + 9"O"_2(g) -> 7"CO"_2(g) + 4"H"_2"O"(g)`

Then, we have `"MM"_"Tol" ~~ "46 g"`, and we know that we need a multiple of `7` carbons since we have `7xx"CO"_2`.

But we have a problem. Your molecular mass is not correct! Toluene is much more than `"46 g"`... In fact it is higher than `84`.

Looking it up, I see it is actually `\mathbf"92 g"`, so there are at minimum `7` carbons because `7xx12 = 84`. We can then add the remaining `8` hydrogens to get `"C"_7"H"_8`.

Therefore the final balanced equation is:

`color(blue)("C"_7"H"_8 (l) + 9"O"_2(g) -> 7"CO"_2(g) + 4"H"_2"O"(g))`