Problem 77 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

A compound contains elements \(\mathrm{X}\) and \(\mathrm{Y}\) in \(1: 4\) mass ratio. If the atomic masses of \(X\) and \(Y\) are in \(1: 2\) ratio, the empirical formula of compound should be (a) \(\mathrm{XY}_{2}\) (b) \(\mathrm{X}_{2} \mathrm{Y}\) (c) XY \(_{4}\) (d) \(\mathrm{X}_{4} \mathrm{Y}\)

Step-by-Step Solution

Answer

The empirical formula of the compound is \textbf{(a) XY2}.

Step 1: Understand Mass Ratio of Elements

The problem gives a mass ratio of elements X to Y in the compound as 1:4. This means that for every unit mass of X, there are 4 unit masses of Y.

Step 2: Understand Atomic Mass Ratio of Elements

The atomic masses of elements X and Y are given in a 1:2 ratio. This means if the atomic mass of X is m, then the atomic mass of Y is 2m.

Step 3: Calculate Moles of Elements from Masses

Since the mass ratio is 1:4 and the atomic mass ratio is 1:2, we can calculate the mole ratio. For element X, there is 1 unit of mass/m atomic mass = 1/m moles. For element Y, there are 4 units of mass/2m atomic mass = 2/m moles.

Step 4: Determine Simplest Whole Number Ratio

To find the empirical formula, divide the moles of each element by the smallest number of moles to get a whole number ratio. The moles of X is 1/m and for Y is 2/m. Dividing by 1/m (the smallest) we get a mole ratio X:Y as 1:2.

Step 5: Write the Empirical Formula

Using the simplest whole number mole ratio of X to Y (1:2), the empirical formula of the compound becomes XY2.

Key Concepts

Mass Ratio in Compounds

Understanding the mass ratio in compounds is crucial when it comes to determining their empirical formulas. An empirical formula represents the simplest whole number ratio of elements in a compound. It gives us information about the relative proportions of each element within a molecule without detailing how they are bonded.

For instance, if a compound is said to have elements X and Y in a 1:4 mass ratio, it means that for every 1 gram of element X, there are 4 grams of element Y. This ratio does not necessarily reflect the atoms' actual count in the molecule but the mass comparison between the two constituents. To further clarify, if you have 10 grams of X, you would need 40 grams of Y to maintain the specified mass ratio.

Remember, the mass ratio is distinct from the mole ratio, which is the number of moles of one substance compared to the number of moles of another substance within the compound. The challenge in empirical formula calculation comes from translating this mass ratio into an atomic or mole ratio, as that leads us to the simplest formula representing the compound.

Atomic Mass Ratio

The atomic mass ratio complements the mass ratio by providing a deeper insight into the makeup of a compound. To understand it, we should first grasp that the atomic mass of an element is measured in atomic mass units (u), which gives the mass of a single atom of that element.

For example, if elements X and Y have an atomic mass ratio of 1:2, and we consider that the atomic mass of X is m u, then the atomic mass of Y would be 2m u. It's important to highlight that this is a ratio between the masses of single atoms, not the total amount of substance.

To distinguish, the mass ratio tells us how much more one element weighs compared to another in bulk, while the atomic mass ratio tells us the same but on an atomic level. The atomic mass ratio becomes particularly helpful when we aim to convert the mass ratios into mole ratios since one mole of any element has a mass in grams equal to its atomic mass in atomic mass units.

Mole Concept

The mole concept is a bridge connecting the mass of substances to the number of particles they include, such as atoms or molecules. One mole is defined as the amount of any substance that contains the same number of particles as there are atoms in exactly 12 grams of carbon-12. This number is Avogadro's number, which is approximately equal to 6.022 x 10^23 particles.

When you calculate moles from mass, as seen in the exercise, you take the mass of the element and divide it by its atomic mass. In the given problem, the moles of X would calculate as 1 unit of mass divided by m (the atomic mass of X), which equals 1/m moles. Likewise, for element Y, with a mass four times that of X and an atomic mass twice as much, the calculation would be 4 units of mass divided by 2m, resulting in 2/m moles.

With the mole concept, the process of finding an empirical formula involves converting mass ratios into mole ratios and then expressing these ratios with the smallest whole numbers. This step is what allows us to write a concise and accurate representation of the compound's composition, known as its empirical formula.