# 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: $\text{CH"_3"OH}$, $\text{C"_3"H"_5"OH}$, $\text{C"_4"H"_9"OH}$, or $\text{C"_5"H"_11"OH}$.

Aug 2, 2017

The alcohol is $\text{C"_4"H"_9"OH}$.

#### Explanation:

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

$\text{alcohol"color(white)(l) + "O"_2(g) rarr "CO"_2(g) + "H"_2"O} \left(g\right)$

What we can do first is use the molar masses of ${\text{CO}}_{2}$ and $\text{H"_2"O}$ 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 $\text{H"_2"O}$ to ${\text{CO}}_{2}$:

(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 ${\text{H"_2"O": "CO}}_{2}$ is color(red)(ul(5:4 ($1.25 = \frac{5}{4}$ as a fraction)

We can now update our chemical equation:

$\text{alcohol"color(white)(l) + "O"_2(g) rarr 4"CO"_2(g) + 5"H"_2"O} \left(g\right)$

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 $\text{OH}$ group, so there should be $1$ (or a multiple of $1$) oxygen atom.

So our alcohol can be condensed to

$\text{C"_4"H"_10"O}$

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 $5.00$ $\text{g alcohol}$ used.

Let's calculate the mass of ${\text{CO}}_{2}$ that forms if $5.00$ $\text{g}$ 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 $11.9$ ${\text{g CO}}_{2}$ in the question, so the alcohol is indeed

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