Question

Consider a hypothetical compound composed of elements \(\mathrm{X}, \mathrm{Y},\) and \(\mathrm{Z}\) with the empirical formula \(\mathrm{X}_{2} \mathrm{YZ}_{3}\). Given that the atomic masses of \(\mathrm{X}, \mathrm{Y},\) and \(\mathrm{Z}\) are 41.2,57.7 and \(63.9,\) respectively, calculate the percentage composition by mass of the compound. If the molecular formula of the compound is found by molar mass determination to be actually \(\mathrm{X}_{4} \mathrm{Y}_{2} \mathrm{Z}_{6},\) what is the percentage of each element present? Explain your results.

Short answer

The mass percentage composition of the compound in empirical formula X2YZ3 is approximately 24.8% X, 17.4% Y, and 57.7% Z. Similarly, the mass percentage composition of the compound in molecular formula X4Y2Z6 is also approximately 24.8% X, 17.4% Y, and 57.7% Z. This is because the molecular formula is simply a multiple of the empirical formula, resulting in the same ratio of elements and, therefore, the same mass percentage composition for each element.

Step-by-step solution

Determine the molar mass of the empirical formula X2YZ3

To find the mass percentages of elements in the compound, we need to determine the molar mass of the empirical formula, X2YZ3. We can do this by using the given atomic masses of X, Y, and Z: Molar mass of X2YZ3 = 2 * (Mass of X) + 1 * (Mass of Y) + 3 * (Mass of Z) = 2 * 41.2 + 1 * 57.7 + 3 * 63.9 = 82.4 + 57.7 + 191.7 = 331.8 g/mol

Calculate mass percentage for each element in the empirical formula X2YZ3

Next, we will calculate the mass percentage of each element by dividing the mass contribution of each element by the total molar mass of the empirical formula: % mass of X = (2 * 41.2) / 331.8 * 100 = (82.4 / 331.8) * 100 ≈ 24.8% % mass of Y = 57.7 / 331.8 * 100 = 17.4% % mass of Z = (3 * 63.9) / 331.8 * 100 = (191.7 / 331.8) * 100 ≈ 57.7% So, the mass percentage composition of the compound in empirical formula X2YZ3 is approximately 24.8% X, 17.4% Y, and 57.7% Z.

Determine the molar mass of the molecular formula X4Y2Z6

Now that we have the mass percentage composition of the empirical formula, we will do the same for the molecular formula X4Y2Z6: Molar mass of X4Y2Z6 = 4 * (Mass of X) + 2 * (Mass of Y) + 6 * (Mass of Z) = 4 * 41.2 + 2 * 57.7 + 6 * 63.9 = 164.8 + 115.4 + 383.4 = 663.6 g/mol

Calculate mass percentage for each element in the molecular formula X4Y2Z6

We will now calculate the mass percentage of each element in the molecular formula, similar to what was done for the empirical formula: % mass of X = (4 * 41.2) / 663.6 * 100 = (164.8 / 663.6) * 100 ≈ 24.8% % mass of Y = (2 * 57.7) / 663.6 * 100 = (115.4 / 663.6) * 100 ≈ 17.4% % mass of Z = (6 * 63.9) / 663.6 * 100 = (383.4 / 663.6) * 100 ≈ 57.7% Hence, the mass percentage composition of the compound in molecular formula X4Y2Z6 is approximately 24.8% X, 17.4% Y, and 57.7% Z.

Compare the results and explain

We can see that the mass percentage composition of elements in both the empirical formula X2YZ3 and the molecular formula X4Y2Z6 are the same. This is because the molecular formula is simply a multiple of the empirical formula, meaning that it has the same ratio of elements and therefore the same percentage of each element by mass.

Key concepts

Empirical Formula

The empirical formula represents the simplest whole-number ratio of elements in a compound. It does not give us the exact number of atoms in the molecule but tells us the ratio in which the elements combine. For instance, the empirical formula of water, H₂O, indicates that hydrogen and oxygen combine in a 2:1 ratio. If we had a hypothetical compound with an empirical formula of X₂YZ₃, it would indicate that for every two atoms of X, there is one atom of Y, and three atoms of Z present.

Knowing the empirical formula is the first step in understanding the compound's composition. By using the atomic masses of the elements involved, we can also calculate the relative mass contributions of each element, leading to the percentage composition by mass. With this information, scientists can then deduce the actual molecular formula of a substance, which may be a multiple of the empirical formula, as was the case in our textbook exercise where the compound’s empirical formula was found to be X₂YZ₃.

Molecular Formula

While the empirical formula provides the simplest ratio of the elements within a compound, the molecular formula gives the actual number of each type of atom in a molecule of the compound. For instance, the molecular formula of the compound from our sample exercise is X₄Y₂Z₆.

It's important to remember that the molecular formula can be the same as the empirical formula, or it can be its multiple. In this case, the molecular formula is a multiple of the empirical formula - the number of each atom type has been doubled. The molecular formula is particularly useful because it gives us a clearer picture of the compound's structure, which is critical for understanding its properties and behaviors.

Molar Mass Calculation

Molar mass is a fundamental concept in chemistry that represents the mass of one mole of a substance (usually in grams per mole). It is the sum of the atomic masses of all the atoms in a formula. In our exercise, the molar mass of the empirical formula, X₂YZ₃, and the molecular formula, X₄Y₂Z₆, was calculated by multiplying the atomic masses of X, Y, and Z with their respective number of atoms in each formula, and then adding these values together.

Careful calculation of the molar mass is critical in a wide range of chemical calculations, including stoichiometry, preparation of solutions, and conversion between moles and grams. In the applied problem, this allowed us to convert mass percentages into moles, providing a bridge between the physical weight of chemicals and their mole concept, which is more meaningful in chemical reactions.

Stoichiometry

Stoichiometry is the section of chemistry that deals with the quantitative relationships that govern the combination of elements in chemical reactions. It is based on the conservation of mass, where the total mass of the reactants equals the total mass of the products in a chemical reaction.

In stoichiometry, we use the mole concept, atomic masses, and molar mass as tools to understand and predict the outcomes of chemical reactions. For example, using stoichiometric calculations, we can determine the amount of reactants needed to produce a certain amount of product, or, as seen in our exercise, calculate the percentage composition of elements in a compound. Stoichiometry bridges the gap between the macroscopic world that we can measure and the atomic world that we express in chemical formulas.