Compound X contains only carbon, hydrogen, nitrogen, and...

Compound $X$ contains only carbon, hydrogen, nitrogen, and chlorine. When 1.00 $\mathrm{g}$ of $\mathrm{X}$ is dissolved in water and allowed to react with excess silver nitrate, AgNO $_{3},$ all the chlorine in $\mathrm{X}$ reacts and 1.95 $\mathrm{g}$ of solid AgCl is formed. When 1.00 $\mathrm{g}$ of $\mathrm{X}$ undergoes complete combustion, 0.900 $\mathrm{g}$ of $\mathrm{CO}_{2}$ and 0.735 $\mathrm{g}$ of $\mathrm{H}_{2} \mathrm{O}$ are formed. What is the empirical formula of $\mathrm{X} ?$

Key Concepts

Empirical Formula Determination

This concept involves deducing the simplest whole-number ratio of atoms in a compound based on its elemental composition. Experimental data, such as masses of combustion products or precipitates, are used to compute the number of moles of each element, which are then normalized to yield the empirical formula.

Combustion Analysis

In combustion analysis, an organic compound is burned in oxygen, producing CO? and H?O. The amount of CO? produced is directly related to the carbon content, while the water corresponds to the hydrogen content. By converting the masses of these products to moles, one can determine the relative amount of carbon and hydrogen present in the original compound.

Precipitation Reaction Analysis

This concept involves using a precipitation reaction—in this case, the reaction of a halide with silver nitrate to form a silver halide—to determine the amount of a specific element in a compound. By measuring the mass of the precipitate, one can calculate the number of moles of the halogen present, which contributes to the compound's overall composition.

Stoichiometric Calculations

Stoichiometry uses the relationships of reactants and products in a chemical reaction to convert between mass, moles, and number of atoms. It is vital in both combustion analysis and precipitation reactions, as it enables the determination of individual elemental contributions from measured masses, forming the basis for establishing the empirical formula.

Transcript

00:01 We know that x contains the following elements and in each of the experiments that they gave us, x has the total mass of 1 gram.

00:09 We're going to figure out what the masses of the masses and the number of moles for each element.

00:18 So let's start with the fact that 1 gram of x combines with silver nitrate to produce silver chloride.

00:30 It doesn't really matter what else is formed because i know that the silver chloride contains all the chlorine that came from x because there was no chlorine in the other reactors.

00:45 So if we know that 1 .95 grams of silver chloride are formed, those 1 .95 grams contain all the chlorine.

00:56 So what i'll do is i'll divide by the molar mass of chlorine, which is a hundred and 43 .32 grams for every mole.

01:04 This is equal to 0 .0136 moles of silver chloride.

01:11 That doesn't help me find chlorine, but i'm pretty close, because for every one mole of solar chloride, there's one more of chlorine.

01:21 So i have this many moles of chlorine.

01:28 Remember that i'm going to need the mass of chlorine as well later.

01:33 So i'll go ahead and multiply the moles of chlorine by the molar mass of chlorine, which is 35 .45 grams of coco for every one mole of coco.

02:03 This is equal to, we start pink colors here, 0 .482 grams of chlorine.

02:14 And they had this many moles of chlorine to start with.

02:17 To organize information, i'm going to start writing down.

02:26 0 .482 grams of chlorine as well at 0 .136 moles of that element.

02:41 Okay.

02:42 Then proceeding, we're also told that 1 gram of x produces water and carbon dioxide of combustion.

02:52 So this will form 0 .90 grams of carbon dioxide.

03:02 I'm going to use that to find the number of moles of carbon by dividing by the molar mass of carbon dioxide and then realizing that there is one mole of carbon in every one mole of co2.

03:21 So i find that i have 0 .0205 moles.

03:31 5 moles of carbon.

03:35 And you should also find the mass of carbon.

03:39 So i will multiply by 12 .01 grams for every mole of carbon.

03:52 And this compound contains 0 .2 .46 grams of carbon.

04:00 This is because if x contains those four elements, all of the carbon in x, it gets turned to carbon dioxide...

Please give Ace some feedback