On a réuni, à propos d'un composé organique formé de...

On a réuni, à propos d'un composé organique formé de carbone, d'hydrogène et d'azote, les informations suivantes : - Un échantillon de $1,12 \mathrm{~g}$, vaporisé à $80^{\circ} \mathrm{C}$, occupe un volume de $725 \mathrm{~mL}$, sous une pression de 1002,6 mbar. - La combustion d'un échantillon de $128 \mathrm{mg}$ donne $250 \mathrm{mg}$ de dioxyde de carbone $\mathrm{CO}_2$. - À la suite d'un traitement qui transforme tout l'azote contenu dans le composé en diazote gazeux, appliqué à un échantillon de $1,225 \mathrm{~g}$, on a recueilli sur une cuve à eau $335 \mathrm{~mL}$ de diazote, à $18^{\circ} \mathrm{C}$ et sous une pression atmosphérique de 997,2 mbar. Quelle est la formule moléculaire de ce composé ?

On a réuni, à propos d'un composé organique formé de carbone, d'hydrogène et d'azote, les informations suivantes :
- Un échantillon de $1,12 \mathrm{~g}$, vaporisé à $80^{\circ} \mathrm{C}$, occupe un volume de $725 \mathrm{~mL}$, sous une pression de 1002,6 mbar.
- La combustion d'un échantillon de $128 \mathrm{mg}$ donne $250 \mathrm{mg}$ de dioxyde de carbone $\mathrm{CO}_2$.
- À la suite d'un traitement qui transforme tout l'azote contenu dans le composé en diazote gazeux, appliqué à un échantillon de $1,225 \mathrm{~g}$, on a recueilli sur une cuve à eau $335 \mathrm{~mL}$ de diazote, à $18^{\circ} \mathrm{C}$ et sous une pression atmosphérique de 997,2 mbar.
Quelle est la formule moléculaire de ce composé ?

Key Concepts

Gas Stoichiometry

The application of stoichiometric principles to chemical reactions involving gases, utilizing conversions between volume, pressure, temperature, and moles. It is vital for solving problems where gas measurements are tied to the quantities of reactants or products.

Empirical Formula Determination

The process of finding the simplest whole-number ratio of atoms in a compound from experimental data. It involves converting mass percentages or measured masses of elements into moles and then establishing the ratio among them.

Molecular Formula Determination

This involves using the empirical formula along with the compound’s molar mass to determine the actual number of atoms of each element in a molecule. The molecular formula is a multiple of the empirical formula.

Ideal Gas Law

This law establishes the relationship among pressure, volume, temperature, and the number of moles of a gas (PV = nRT). It is fundamental in converting volume measurements under specific conditions to moles, which is essential in quantitative gas analysis.

Combustion Analysis

A method used to determine the elemental composition of a compound by burning it in excess oxygen and analyzing the amounts of resulting products like CO2 and H2O. This technique allows for calculation of the amounts of carbon and hydrogen (and sometimes other elements) in the original compound.

Transcript

00:01 So in this question, we're going to calculate the molecular formula of this compound.

00:08 So for the first part, we're giving the combustion gives us a certain amount of carbon dioxide and water.

00:16 So all of the carbon dioxide is now in the carbon dioxide and all of the hydrogen is now in the water.

00:21 So we can calculate the percentage of carbon and hydrogen.

00:24 So we're going to convert the mass into grams and then convert to moles using.

00:31 The molar mass of the compound.

00:35 So in this case, for carbon dioxide, divide by molar mass of carbon dioxide.

00:39 In one mole of carbon dioxide, we have one mole of carbon.

00:43 So convert to molds of carbon and multiply by the atomic mass of carbon, divide by the total mass of the compound and multiply by 100.

00:51 This will give you the percentage of carbon in the compound.

00:54 We do the same thing with water.

00:56 So in one mole of hydrogen, we have 1 .008, or in one mole of water, we have two moles of hydrogen because of the subscript on the hydrogen.

01:09 The rest of the calculation would be the same except we use the molar mass of water and hydrogen.

01:17 And then we're going to get the percentage of hydrogen in this compound.

01:22 Subtract this sum from 100 because the only other element is nitrogen, but the total mass of the comma v100 grams.

01:28 The percentages are equal to the mass of the elements.

01:35 So we're going to then divide by the atomic masses of the elements to get the moles of each element in the compound.

01:40 Divide by the smallest number of moles of each element to get the subscript on the elements in the compound...

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