Question

In the combustion of 5.00 grams of an alcohol the chemist produced 11.89 grams of `"CO"_2` and 6.09 grams of `"H"_2"O"`. What is the identity of the alcohol?

Alcohols burn rapidly in oxygen. A chemist is burning an alcohol in an attempt to determine its identity. She knows the possibilities are: `"CH"_3"OH"`, `"C"_3"H"_5"OH"`, `"C"_4"H"_9"OH"`, or `"C"_5"H"_11"OH"`.

Answer 1

The alcohol is `"C"_4"H"_9"OH"`.

Explanation:

Let's first write the chemical equation for this combustion reaction:

`"alcohol"color(white)(l) + "O"_2(g) rarr "CO"_2(g) + "H"_2"O"(g)`

What we can do first is use the molar masses of `"CO"_2` and `"H"_2"O"` to convert the given masses of these products to moles:

`bb("CO"_2:`

`11.89cancel("g CO"_2)((1color(white)(l)"mol CO"_2)/(44.01cancel("g CO"_2))) = ul(0.270color(white)(l)"mol CO"_2`

`bb("H"_2"O":`

`6.09cancel("g H"_2"O")((1color(white)(l)"mol H"_2"O")/(18.015cancel("g H"_2"O"))) = ul(0.338color(white)(l)"mol H"_2"O"`

We can get an idea of the coefficients of the chemical equation by finding the mole ratio of `"H"_2"O"` to `"CO"_2`:

`(0.338color(white)(l)"mol H"_2"O")/(0.270color(white)(l)"mol CO"_2) = color(red)(1.25`

So a realistic mole ratio of `"H"_2"O": "CO"_2` is `color(red)(ul(5:4` (`1.25 = 5/4` as a fraction)

We can now update our chemical equation:

`"alcohol"color(white)(l) + "O"_2(g) rarr 4"CO"_2(g) + 5"H"_2"O"(g)`

What we now realize is that in order to balance the equation, the alcohol must have `color(red)(4` carbon atoms in it (or a multiple of 4), so we can see that the answer is `color(blue)("C"_4"H"_9"OH"`.

But what if you weren't given the options?

Well, we can also realize that the alcohol should have a hydrogen count as a multiple of `5xx2=ul(10`.

Furthermore, alcohols have one `"OH"` group, so there should be `1` (or a multiple of `1`) oxygen atom.

So our alcohol can be condensed to

`"C"_4"H"_10"O"`

Keep in mind it can be any multiple of this; this is the empirical formula of the alcohol, so we have to look now at the `5.00` `"g alcohol"` used.

Let's calculate the mass of `"CO"_2` that forms if `5.00` `"g"` of this alcohol

`5.00cancel("g C"_4"H"_10"O")((1cancel("mol C"_4"H"_10"O"))/(74.12cancel("g C"_4"H"_10"O")))((4cancel("mol CO"_2))/(1cancel("mol C"_4"H"_10"O")))((44.01color(white)(l)"g CO"_2)/(1cancel("mol CO"_2)))`

`= color(green)(ul(11.9color(white)(l)"g CO"_2`

It works, since we were given `11.9` `"g CO"_2` in the question, so the alcohol is indeed

`color(blue)(ulbar(|stackrel(" ")(" ""C"_4"H"_9"OH")" "|)`