Question

A compound contains elements \(X\) 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) \(\mathrm{XY}_{4}\) (d) \(\mathrm{X}_{4} \mathrm{Y}\)

Short answer

Answer: (a) XY₂

Step-by-step solution

Find moles of X and Y in the compound

We can assume a certain mass of the compound and then calculate the masses of \(X\) and \(Y\). Let's assume that the mass of the compound is 20g. Then, the mass of \(X\) is 20g \(\cdot \frac{1}{4+1} = 4\)g, and the mass of \(Y\) is 20g \(\cdot \frac{4}{4+1} =16\)g. Let's assume the atomic mass of \(X\) as \(x\). Then, the atomic mass of \(Y\) is \(2x\). Next, we'll calculate the moles of \(X\) and \(Y\) in the compound: Moles of \(X = \frac{\text{mass of }X}{\text{ atomic mass of }X} = \frac{4\ \text{g}}{x}\) Moles of \(Y = \frac{\text{mass of }Y}{\text{ atomic mass of }Y} = \frac{16\ \text{g}}{2x} = \frac{8\ \text{g}}{x}\)

Calculate the mole ratio of X and Y

Now we'll calculate the mole ratio of elements \(X\) and \(Y\): Mole ratio \(= \frac{\text{moles of }X}{\text{moles of }Y} = \frac{\frac{4\ \text{g}}{x}}{\frac{8\ \text{g}}{x}} = \frac{4}{8} = \frac{1}{2}\)

Determine the empirical formula

We now know the mole ratio is \(\frac{1}{2}\), which means there are twice as many moles of \(Y\) as there are of \(X\). Therefore, the empirical formula is \(\mathrm{XY}_{2}\). The correct answer is (a) \(\mathrm{XY}_{2}\).

Key concepts

Mole Ratio

Understanding the mole ratio is crucial when dealing with chemical compounds. The mole ratio indicates the proportions of atoms of each element in a compound. It is derived from the comparison of moles of the different elements involved in the compound. To calculate this, you need to know the number of moles of each element, which can be found using their respective masses and atomic masses.

A common mistake is to confuse mass ratio with mole ratio. The mass ratio simply tells us how much one element weighs compared to another in a sample of a compound. However, to find the mole ratio, which is needed to write the empirical formula, we must convert these masses into moles. For example, if we have a 1:4 mass ratio for elements X and Y, this does not directly translate into a 1:4 mole ratio, as seen in the given exercise.

By dividing the mass of each element by its atomic mass, we get the number of moles and can calculate the mole ratio. This ratio will determine the subscripts in the empirical formula of the compound. In the example provided, the mole ratio was calculated to be 1:2, which led to finding that there are twice as many moles of Y as there are of X, giving us an empirical formula of XY2.

Atomic Mass

The atomic mass of an element is another critical factor in determining a compound's empirical formula. Atomic mass is the mass of an atom of a chemical element expressed in atomic mass units (u). The Periodic Table lists the average atomic mass of each element, which accounts for the occurrence of isotopes and their abundances.

To find the moles of an element in a compound, the given mass of that element must be divided by its atomic mass. In the original problem, the atomic mass ratio of X to Y was given as 1:2. Assuming the atomic mass of X is x, then the atomic mass of Y is 2x. This allows for converting the mass of each element to moles, which is pivotal for deriving the mole ratio.

Remember, a common point of confusion can arise if students overlook the difference between the atomic mass unit and grams. It's important to consistently use the same unit when working through such problems. Misunderstanding the concept of atomic mass can lead to incorrect empirical formulas. The exercise tackled this by calculating moles with the assumption of an atomic mass unit x, keeping calculations unspecific but universally applicable.

Compound Composition

The compound composition tells us what elements are present in the compound and in what quantities. It is usually expressed in terms of the empirical formula, which gives the simplest whole-number ratio of atoms of each element in the compound. Establishing the compound composition involves both qualitative and quantitative analysis; knowing what the elements are and how much of each is present.

While working through these problems, assume a certain mass for the entire compound to simplify the calculation of the composition. As illustrated in the sample problem, by assuming a 20g compound, it is straightforward to calculate the mass of each element present based on their known mass ratio. Once the masses are converted into moles, referring to their atomic masses, the empirical formula is determined.

In this context, compound composition does not simply mean how much of an element is present by mass but rather by the number of atoms, which is represented by moles. Thus, the mass contribution of each element is intrinsically connected to its moles through atomic mass, resulting in the most simplified form of the compound, the empirical formula XY2 for the given exercise.