Question

Question #607c5

Answer 1

Indeed, that answer makes sense.

Explanation:

The idea here is that you need to convert the `1:1` mass ratio of hydrogen and oxygen that exists in the molecule, to a `x:y` mole ratio.

Here `x` represents the number of moles of hydrogen and `y` to the number of moles of oxygen.

Now, I have a feeling that this is a multiple-choice question, because no mention of carbon was made.

This is the case because you don't need to know anything about the actual percent composition of camphor In terms of carbon.

So let's assume that you have a sample of camphor that contains `m` grams of hydrogen. Since oxygen and hydrogen have the same percentage by mass in the sample, it follows that the mass of oxygen in the sample will also be equal to `m` grams.

How many moles of hydrogen would you get in `m` grams of hydrogen? Use hydrogen's molar mass to find

`mcolor(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794color(red)(cancel(color(black)("g")))) = m/1.00794"moles H"`

How about moles of oxygen?

`mcolor(red)(cancel(color(black)("g"))) * "1 mole O"/(16.0color(red)(cancel(color(black)("g")))) = m/16.0"moles O"`

Now look what happens when you take the mole ratio of the two elements

`overbrace(color(red)(cancel(color(black)("m")))/1.00794)^(color(blue)("moles of H")) * overbrace(16.0/color(red)(cancel(color(black)("m"))))^(color(blue)(["moles of O"]^(-1))) = 16/1.00794 = 15.87 ~~ 16`

The result comes out cleaner if you take the molar mass of hydrogen to be approximately `"1 g/mol"`.

This means that you have 16 moles of hydrogen for every one mole of oxygen in the sample.

If this is indeed a multiple-choice question, then the answer `"C"_10"H"_16"O"` makes sense, since it satisfies that `16:1` mole ratio between hydrogen and oxygen.