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"#.

1 Answer
Aug 2, 2017

Answer:

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")" "|)#