Quinone, which is used in the dye industry and in chemical...

Quinone, which is used in the dye industry and in chemical photography, is an organic compound containing only $\mathrm{C}, \mathrm{H},$ and $\mathrm{O} .$ Determine the empirical formula if $0.105 \mathrm{~g}$ of the compound gives $0.257 \mathrm{~g} \mathrm{CO}_{2}$ and $0.0350 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}$ when burned completely.

Key Concepts

Combustion Analysis

A technique in analytical chemistry where a compound is burned completely in oxygen to produce combustion products, typically CO2 and H2O, which are then analyzed to determine the amounts of carbon and hydrogen originally present in the compound.

Empirical Formula

The simplest whole-number ratio of the elements in a chemical compound. It is derived by converting the amounts of elements determined from experiments (such as combustion analysis) into moles and then reducing these ratios to the simplest integral values.

Stoichiometry

The branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. In combustion analysis, stoichiometry is used to relate the mass of CO2 and H2O produced back to the moles of carbon and hydrogen in the original compound.

Mole Concept

A foundational principle in chemistry that involves converting between the mass of a substance and the number of particles it contains, using the substance's molar mass. This concept is essential when interpreting combustion analysis data to determine the number of moles of each element present in the compound.

Transcript

00:01 So let's look at the combustion of quinone with oxygen to create carbon dioxide and water.

00:08 So let's say that we start out with 0 .105 grams of the quinone.

00:19 And after we combust it, we find that we get 0 .257 grams of carbon dioxide and 0 .035 grams of water.

00:33 Let's try to figure out what the formula is of the quinone, the empirical formula, the lowest whole number ratio of atoms in that formula.

00:49 And we only have carbon, hydrogen, and oxygen there.

00:54 The carbon in the quinone goes to form co2.

00:57 The hydrogen in there goes to form water.

01:00 But the oxygen in there goes into both of our products.

01:06 That's going to be a little harder to figure out.

01:08 So we'll leave that for last.

01:10 So let's first start with the carbon dioxide and the card, finding out the carbon.

01:16 So we have 0 .257 grams of co2.

01:23 We can convert that into moles.

01:27 44 .01 grams is the molar mass.

01:32 Now, we know that in carbon dioxide, for every mole of carbon dioxide, there's one mole of carbon in it.

01:40 So that'll get us the amount of carbon that there is.

01:45 And that is 5 .84 times 10 to the minus 3 moles.

01:59 And we can do the same thing with our hydrogen, which came from our water.

02:05 So let's look for water.

02:06 We have 0 .0350 grams of water, molar mass, 18 .015 grams.

02:21 Now for water, there's two moles of h in every mole of water.

02:30 And so we had 3 .89 times 10 to the minus 3 moles of hydrogen.

02:41 Now we're going to use the value of moles in a little while because the ratio of moles will tell us the empirical formula, but we still need to figure out what's going on with oxygen.

02:54 And to be able to do that, we need the grams of carbon and the grams of hydrogen that there are...

Please give Ace some feedback