You take three compounds, each consisting of two elements...

You take three compounds, each consisting of two elements (X, Y, and/or Z), and decompose them to their respective elements. To determine the relative masses of X, Y, and Z, you collect and weigh the elements, obtaining the following data: a. What are the assumptions needed to solve this problem? b. What are the relative masses of $\mathrm{X}, \mathrm{Y}$, and $\mathrm{Z}$ ? c. What are the chemical formulas of the three compounds? d. If you decompose $21 \mathrm{~g}$ of compound $\mathrm{XY}$, how much of each element is present?

You take three compounds, each consisting of two elements (X, Y, and/or Z), and decompose them to their respective elements. To determine the relative masses of X, Y, and Z, you collect and weigh the elements, obtaining the following data:
a. What are the assumptions needed to solve this problem?
b. What are the relative masses of $\mathrm{X}, \mathrm{Y}$, and $\mathrm{Z}$ ?
c. What are the chemical formulas of the three compounds?
d. If you decompose $21 \mathrm{~g}$ of compound $\mathrm{XY}$, how much of each element is present?

Key Concepts

Law of Definite Proportions

This principle states that a chemical compound always contains the same elements in the same fixed ratio by mass. It is fundamental in ensuring that the mass ratios observed in a compound are consistent, which is necessary for determining a compound’s empirical formula from experimental data.

Law of Multiple Proportions

This law explains that when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in ratios of small whole numbers. This concept is key to comparing different compounds containing the same elements and helps in deducing the relative masses of the elements involved.

Stoichiometry

Stoichiometry involves the calculation of relative quantities of reactants and products in a chemical reaction using the mole concept. It is used to relate the masses of different elements within compounds, which is essential for solving problems that require the determination of relative atomic masses and chemical formulas.

Empirical Formula Determination

The empirical formula represents the simplest whole-number ratio of atoms in a compound. Determining the empirical formula from experimental mass data is a central aspect of chemical analysis, as it provides information about the composition of the compound and the relative masses of its constituent elements.

Conservation of Mass

This concept asserts that the total mass in a closed system remains constant during any chemical reaction. It is a fundamental assumption in chemical calculations, ensuring that the sum of the masses of the elements in the products equals the mass of the original compound, which validates the stoichiometric relationships used in the problem.

Transcript

00:01 So in this baby, we're gonna talk about question 128 from doctor to which says you take three compounds each consisting in two elements x, y and r z, and decomposed them into their respective elements.

00:13 To determine the relative masses of x, y and z, you collect and weigh the elements, obtaining the following data.

00:20 So we have a table and in one column were given which elements are in that compound.

00:25 So our first compound is made of x and wire second, his native y and z, and the third is made of x and y.

00:31 And then we're given the masses of the elements after these compounds were decomposed into their elements.

00:38 Um so which questions every asked an aide were asked, what are the assumptions needed to solve this problem? so remember, we're trying to get the relative masses of these three elements.

00:49 Um, and the way we get that is, we have to know on the mass composition like we were given, and the number of each kind of adam would give us the relative masses.

01:00 But unfortunately, we don't know the chemical formulas of these three compounds.

01:06 So in order to determine the relative mass is we're going to have to assume a chemical formula.

01:13 Um so and be were asked, what are the relative masses? so we're going to have to go ahead and solve this problem.

01:19 So one thing we can do is we know that compound one in calm down.

01:23 Three in our table were both made of x and y, so we can determine their formulas by using the law of multiple proportions, which tells us that when you have, um, two elements that can combined for multiple different compounds, the ratio of the second element that combines with one gram of the first element in each compound should reduce to a whole number arm.

01:51 So you do that in this case, eh? so we want to know how much why will combine with one gram of x.

01:58 So in order to normalize this and one, i have 10.4 grams of aks and and two i have two grams of back, so those aren't the same.

02:04 We want them to both the one grams s o i divided.

02:08 Why? by x 4.2 grams by 0.4 grams and i get 10.5 grams and compound one compound to i take my seven grams of why and divide it by my two grams of x.

02:18 And that tells us i have 3.5 grams of bly for every one gram of x eso we have 10.5 grams in 3.5 grams for every gram of x.

02:28 And so we divide that by the smaller number.

02:30 So they went dark.

02:31 3.5 gives us one, um, and 10.5 divided by 3.5 gives us three.

02:37 So that's our simple whole number in this case is three.

02:40 And that tells us we have three times as much of our element.

02:44 Why, and compound wine as we do in compound to and come on three.

02:49 Um, so i'm gonna go ahead and assume the formula x y for compound three and assume the formula x y three for compound one in order to be consistent with that.

03:02 Um, so let's go ahead and do this for the first compound...

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