Coniine, a toxic substance isolated from poison hemlock,...

Coniine, a toxic substance isolated from poison hemlock, contains only carbon, hydrogen, and nitrogen. Combustion analysis of a $5.024 \mathrm{mg}$ sample yields $13.90 \mathrm{mg}$ of $\mathrm{CO}_{2}$ and $6.048 \mathrm{mg}$ of $\mathrm{H}_{2} \mathrm{O}$. What is the empirical formula of coniine?

Key Concepts

Combustion Analysis

This is an experimental technique in which a compound is completely burned in oxygen to produce carbon dioxide and water. By measuring the masses of these products, one can infer the amounts of carbon and hydrogen in the original compound. This method is particularly useful for determining the empirical formula of a compound that contains only carbon, hydrogen, and other known elements.

Empirical Formula Determination

This concept refers to finding the simplest whole-number ratio of the atoms in a compound. In combustion analysis, once the masses of the constituent elements are determined by converting the data from combustion products into element masses, they are compared after converting to moles, which reveals the ratio that constitutes the empirical formula.

Conversion of Mass to Moles

A critical step in many chemical calculations, this involves converting the measured mass of a substance to the number of moles using its molar mass. This conversion allows for the direct comparison of atom ratios in the compound, which is necessary for determining its empirical formula.

Mass Balance and Elemental Accounting

This involves using the conservation of mass principle to account for all atoms in a sample. When a compound is known to consist only of specific elements, determining the mass of one or more elements via experimental data allows the mass of the remaining element(s) to be calculated by subtracting from the total mass, ensuring that all components are considered.

Transcript

00:01 This is a combustion analysis problem, and it's pretty simple, but it takes a little time to do.

00:06 So let me give you the information.

00:08 We have a compound that contains c's, hs, and ns.

00:14 That's all it contains.

00:21 Our sample size is 5 .04 grams, milligrams.

00:30 It undergoes combustion, and it yields 6 .048 milligrams of carbon dioxide, or excuse me, water.

00:48 13 .90 milligrams of carbon dioxide.

01:00 So our first order of business will be able to obtain the masses of hydrogen carbon.

01:04 Let's do this one first.

01:06 So we've got 6 .048 milligrams of h2.

01:16 All of this is going to be h2o until i write something else.

01:20 And i'm not going to convert this since i want to, since i'm given milligrams here, i could convert it both ways, but i'm just going to ignore that.

01:28 Conversion and go straight to 18 .02 and this would be per millimole but i'm just going to go grams and grams because i'm going to convert back to grams.

01:45 I've got a two to one mole ratio of hydrogen to h2o and my molar mass of water is or h is 1 .008 grams per mole.

02:00 When i'm done with this i'll have 0 .6766 milligrams hydrogen.

02:15 That's how many milligrams of hydrogen i have.

02:19 My second calculation will be 13 .90 milligrams of co2.

02:34 Use the molar mass of co2.

02:38 This will have a 1 to 1 mole ratio for carbon dioxide and carbon and 12 .01 for the molar mass of carbon.

02:59 And this will equal 1 to 3 .793 milligrams of carbon.

03:19 Next, to find my mass of nitrogen, we're going to take our molar, our mass of our sample, which is 5 .024, and we're going to subtract from that the sum of those two.

03:48 Very good...

Please give Ace some feedback