Question

Question #7a4c6

Answer 1

`"C"_4"H"_6`

Explanation:

To obtain the molecular formula of this unknown compound, we first need to know the mole ratio between carbon and hydrogen.

We are only given the mass percentages of carbon and hydrogen. We don't know how many grams how much of the compound we actually have, so let's assume that we have `"100 g"`.

Carbon: `88.8%` of `"100 g = 88.8 g"`

Hydrogen: `11.2%` of `"100 g = 11.2 g"`

We have `"88.8 g C"` and `"11.2 g H"`. Let's convert these to moles.

The Periodic Table of Elements indicates that the molar mass of carbon is about `"12.0 g/mol"` and that of hydrogen is about `"1.0 g/mol"`. We can use these as conversion factors to convert from grams to moles.

`"88.8" cancel("g C") xx "1 mol C" / ("12.0" cancel("g C")) = "7.40 mol C"`

`"11.2" cancel("g H") xx "1 mol H" / ("1.0" cancel("g H")) = "11.2 mol H"`

We now know that there are `7.40` moles of carbon for `11.2` moles of hydrogen. This is equivalent to saying there are `7.40` atoms of carbon for every `11.2` atoms of hydrogen, so we can write

`"C"_7.40"H"_11.2`

But this doesn't look like a formula at all--the subscripts need to be integers. Usually you have to play around with the numbers until you get nice integers.

Let's divide both subscripts by `7.40` so that carbon has a subscript of `1`.

`"C"_1"H"_1.51`

Okay, we're close! `1.51 ~~ 1.50`, so we can multiply both numbers by `2`.

`"C"_2"H"_3`

So, the empirical formula of this compound is `"C"_2"H"_3`; this is the simplest whole number ratio of atoms of `"C"` to atoms of `"H"`.

Now we need to find the molecular formula. The molar mass of `"C"_2"H"_3` is `(2 xx 12.0) + (3 xx 1.0) = "27 g/mol"`. However, the problem states that the unknown compound has a molar mass of `"54.1 g/mol"`.

Since `54.1 / 27 ~~ 2`, we multiply each subscript by `2`. The molecular formula is thus `color(blue)("C"_4"H"_6)`.