Menthol, a flavoring agent obtained from peppermint oil,...

Menthol, a flavoring agent obtained from peppermint oil, contains carbon, hydrogen, and oxygen. On combustion analysis, 1.00 g of menthol yields 1.161 g of $\mathrm{H}_{2} \mathrm{O}$ and 2.818 $\mathrm{g}$ of $\mathrm{CO}_{2}$ . What is the empirical formula of menthol?

Key Concepts

Empirical Formula Determination

The empirical formula shows the simplest whole-number ratio of atoms in a compound. By converting the measured masses from combustion into moles, and comparing the relative amounts of each element, one can deduce the most reduced formula that represents the compound's composition.

Stoichiometric Conversions

Stoichiometry involves converting between mass, moles, and molecules using molar masses and balanced chemical equations. In a combustion analysis, stoichiometric calculations are used to convert the mass of carbon dioxide and water produced into the moles of carbon and hydrogen, which are essential for determining the empirical formula.

Combustion Analysis

This is a quantitative analytical technique in which an organic compound is burned in excess oxygen to produce carbon dioxide and water. The masses of these combustion products are measured and used to determine the amounts of carbon and hydrogen present in the original compound, forming the basis for compositional analysis.

Transcript

00:01 To find the empirical formula, we're going to use the fact that in this combustion analysis, menthol, this oxygen gas, makes carbon dioxide, and water.

00:16 This happens in the following amount.

00:18 Menthol has, which contains carbon, hydrogen, and oxygen, is a 1 .20 gram sample, and we isolate this much carbon dioxide and water.

00:32 According to the information of the problem.

00:35 So our first step is to determine the number of moles of carbon and of hydrogen.

00:49 So to do that, we're going to take all the mass of carbon dioxide and divide it by the molar mass of co2.

01:11 And this step might be silly, but we'll include it for completion.

01:15 There's one mole of carbon in every one mole of co2.

01:22 Notice how nicely our units cancel.

01:26 And we find that this is equal to 0 .0 .6403 moles of carbon.

01:46 And then for hydrogen, all of the hydrogen that was in the menthol ends up in the water.

01:51 So we're going to use the mass of the water and essentially do the same thing, divide by the molar mass, and then multiply by two, because there are two moles of hydrogen in every one mole of water and this gives us 0 .1289 moles of hydrogen.

02:29 Okay, we're going to use these numbers later, but more immediately we first need to find the number of grams for each one.

02:39 So first we're going to find, next we're going to find the mass of carbon and hydrogen, and we're going to do that by multiplying each one of these by their respective molar masses.

02:54 So 12 grams of carbon in every mole of carbon and 1 .01 grams of hydrogen in every mole of hydrogen.

03:05 Notice these units cancel.

03:09 And we get 0 .7690 grams of carbon and 0 .1301 grams of hydrogen.

03:33 Now, the only reason we need to do this step right now is because we were told that this was an oxygen containing compound...

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