You perform a combustion analysis on a 255 mg sample of a substance that contains only C, H, and

You perform a combustion analysis on a 255 mg sample of a...

You perform a combustion analysis on a $255 \mathrm{mg}$ sample of a substance that contains only $\mathrm{C}, \mathrm{H},$ and $\mathrm{O},$ and you find that $561 \mathrm{mg} \mathrm{CO}_{2}$ is produced, along with $306 \mathrm{mg}$ $\mathrm{H}_{2} \mathrm{O} .$ If the substance contains only $\mathrm{C}, \mathrm{H},$ and $\mathrm{O},$ what is the empirical formula? If the molar mass of the compound is $180 \mathrm{~g} / \mathrm{mol}$, what is the molecular formula of the compound?

You perform a combustion analysis on a $255 \mathrm{mg}$ sample of a substance that contains only $\mathrm{C}, \mathrm{H},$ and $\mathrm{O},$ and you find that $561 \mathrm{mg} \mathrm{CO}_{2}$ is produced, along with $306 \mathrm{mg}$ $\mathrm{H}_{2} \mathrm{O} .$
If the substance contains only $\mathrm{C}, \mathrm{H},$ and $\mathrm{O},$ what is the empirical formula? If the molar mass of the compound is $180 \mathrm{~g} / \mathrm{mol}$, what is the molecular formula of the compound?

Key Concepts

Empirical Formula Determination

Empirical formula determination entails finding the simplest whole-number ratio of the elements in a compound based on experimental data. From the moles of individual elements (derived using stoichiometry from combustion products), the elemental ratios are calculated and reduced to the simplest form, yielding the empirical formula.

Molecular Formula Determination

Molecular formula determination uses the empirical formula along with the compound's molar mass to ascertain the actual number of atoms of each element in a molecule. The molar mass is divided by the mass of the empirical formula to find a multiplying factor that scales the empirical formula to the molecular formula.

Combustion Analysis

Combustion analysis is an experimental technique in which a compound, typically containing C, H, and sometimes O, is burned in excess oxygen to produce CO2 and H2O. Measuring the amounts of these products allows for the determination of the amounts of carbon and hydrogen present in the original compound, forming the basis for further compositional analysis.

Stoichiometry

Stoichiometry involves converting measured masses into moles using the molar masses of substances. In combustion analysis, it is applied to relate the mass of produced CO2 and H2O to the number of moles of carbon and hydrogen atoms, respectively, which is essential for determining elemental ratios.

Transcript

00:01 In this problem, we're going to find the molecular formula for the compound that we have analyzed using combustion analysis.

00:12 So in order to do this, we're going to determine the percent composition of each element in this compound.

00:20 So we've obtained two products.

00:21 We have obtained carbon dioxide and water.

00:24 And from these products, we can obtain the mass of the carbon and the hydrogen in the sample originally.

00:31 So we're going to take the mass of the carbon dioxide, convert it to grams.

00:37 In order to do that, we know there are 1 ,000 milligrams and 1 gram.

00:41 So divide the number of milligrams by 1 ,000.

00:45 Move a decimal place, three places to the left, and we get 0 .561 grams of carbon dioxide.

00:52 Then we are going to convert to moles of carbon dioxide using the molar mass of carbon dioxide.

01:00 When we have the same units on the top and bottom, they will cancel out.

01:04 And then we will get moles of carbon dioxide.

01:07 Then convert to moles of carbon, knowing that in one mole of carbon dioxide, we have one mole of just carbon, because the subscript on the carbon is one.

01:18 And then we need to convert from moles of carbon to the mass of carbon using the atomic mass of carbon.

01:26 Then we're able to get the mass of carbon obtained.

01:31 So that's the mass of carbon in the sample.

01:34 And then from the water we obtained, we do the same exact thing, except now we're just going to use the molar mass of the water.

01:42 And then in one formula unit of h2o, we have two moles of hydrogen.

01:51 So we're going to use that molar ratio instead.

01:56 And then we get the mass of the hydrogen.

01:59 So now we can find the percent composition by mass of each element compared to the original sample of the compound.

02:07 So we take the mass of the carbon, divided by the mass of the sample, making sure our units are both in grams, multiplied by it 100, and we get 60 % carbon.

02:18 Then we do the same for the hydrogen, and we get 13 .32 % hydrogen.

02:23 Then after that, since the only other element in this compound is oxygen, we can find the percent composition by mass of oxygen in the compound...

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