Question

A substance \(\mathrm{X}_{2} \mathrm{Z}\) has the composition (by mass) of 40.0\% X and 60.0\(\% \mathrm{Z}\) . What is the composition (by mass) of the compound \(\mathrm{XZ}_{2} ?\)

Short answer

The composition (by mass) of the compound XZ₂ is 14.29% X and 85.71% Z.

Step-by-step solution

Determine the mass of each element

Given the compound X₂Z, we have a mass composition of 40% X and 60% Z. We can assume a 100 g sample to simplify calculations. In this case, we have 40 g of element X and 60 g of element Z.

Calculate the moles of each element

To determine the composition of the compound XZ₂, we need to find the moles of each element in the 100 g sample. We do not know the atomic masses of elements X and Z, so let's assume the atomic mass of X is Mₓ and that of Z is Mz. Now we can determine the moles of each element using their mass and atomic mass: Moles of X = mass of X / atomic mass of X = 40 g / Mₓ Moles of Z = mass of Z / atomic mass of Z = 60 g / Mz

Find the mole ratio of X and Z in compound XZ₂

We know the compound X₂Z has two moles of element X and one mole of element Z. Based on the moles of the elements, we can write the mole ratios: Moles of X : Moles of Z in XZ₂ To obtain the moles of X and Z in compound XZ₂, we divide the moles of X in X₂Z by 2 since XZ₂ contains only one mole of X: Moles of X in XZ₂ = (40 g / Mₓ) / 2 = 20 g / Mₓ For the moles of Z, the compound XZ₂ contains two moles of Z, which is double the amount in X₂Z. To find the moles of Z in XZ₂, we multiply the moles of Z in X₂Z by 2: Moles of Z in XZ₂ = 2 × (60 g / Mz) = 120 g / Mz

Find the mass composition of the compound XZ₂

We have obtained the moles of X and Z in compound XZ₂. Now, we can find the mass of each element in XZ₂: Mass of X in XZ₂ = Moles of X in XZ₂ × Mₓ = (20 g / Mₓ) × Mₓ = 20 g Mass of Z in XZ₂ = Moles of Z in XZ₂ × Mz = (120 g / Mz) × Mz = 120 g Now, we'll sum the masses of X and Z in XZ₂ to find the total mass: Total mass of XZ₂ = mass of X in XZ₂ + mass of Z in XZ₂ = 20 g + 120 g = 140 g Finally, we'll determine the mass composition by finding the percentage contribution of each element in XZ₂: Mass percent of X in XZ₂ = (mass of X in XZ₂ / total mass of XZ₂) × 100% = (20 g / 140 g) × 100% = 14.2857% Mass percent of Z in XZ₂ = (mass of Z in XZ₂ / total mass of XZ₂) × 100% = (120 g / 140 g) ×100% = 85.7143% Thus, the composition (by mass) of the compound XZ₂ is 14.29% X and 85.71% Z.

Key concepts

Chemical Formulas

Chemical formulas are essential in chemistry for representing compounds using symbols for elements and numerical subscripts. These subscripts show the number of each type of atom in the compound. For example, the formula \(\mathrm{X}_2\mathrm{Z}\) indicates that there are two atoms of element \(X\) and one atom of element \(Z\) in a molecule of this compound.
Similarly, the compound \(\mathrm{XZ}_2\) signifies one atom of \(X\) paired with two atoms of \(Z\). By knowing the chemical formula, one can determine the specific ratio of atoms in the compound, making it an essential tool for understanding the compound's composition and reactions.

• Components: Symbols for the elements (\(X\) and \(Z\) in the examples).
• Subscripts: Numerical values indicating the number of each atom.

Understanding these symbols and what they represent helps in calculations concerning the compound's mass and ratio.

Mole Ratio

The mole ratio is a concept used in chemistry to convey the proportion between different elements in a compound. This proportion is dictated by the subscripts in a chemical formula. In the case of \(\mathrm{X}_2\mathrm{Z}\), the mole ratio of \(X\) to \(Z\) is 2:1, meaning there are two moles of \(X\) for every mole of \(Z\).
When the structural formula changes to \(\mathrm{XZ}_2\), the ratio becomes 1:2, indicating one mole of \(X\) reacts with or is paired with two moles of \(Z\). The mole ratio is crucial for calculations involving stoichiometry, allowing chemists to predict how much of one substance will react with a given amount of another.

• Why it's important: Helps in balancing chemical equations.
• How it's used: Determines proportions in reactions and mixtures.

This concept enables scientists to understand not just the composition, but also the interactions between different elements in compounds.

Moles Calculation

Calculating the number of moles involved in a substance is foundational in chemistry. Moles provide a bridge between the atomic scale and the macroscopic scale that allows for comparison and calculation. To find the number of moles of an element, you divide its mass by its atomic or molar mass.
In the problem, to calculate the moles of \(X\) in \(\mathrm{X}_2\mathrm{Z}\), the formula used was:\[\text{Moles of X} = \frac{\text{mass of X}}{\text{atomic mass of X}} = \frac{40 \text{ g}}{M_x}\]Similarly, the moles of \(Z\) are calculated with:\[\text{Moles of Z} = \frac{\text{mass of Z}}{\text{atomic mass of Z}} = \frac{60 \text{ g}}{M_z}\]

• Calculation Steps: Identify mass and divide by atomic mass.
• Purpose: Necessary for determining the mole ratio.

These calculations provide a quantitative foundation for understanding the chemical processes and reactions involving the substance.

Mass Percent

Mass percent is a way of expressing a component's concentration within a mixture or a chemical compound. In other words, it shows how much of the total mass is made up by a specific element. The mass percent is calculated by dividing the mass of the element by the total mass of the compound, and then multiplying by 100 to convert it to a percentage.
For the compound \(\mathrm{XZ}_2\), the mass percent of \(X\) and \(Z\) was calculated using:\[\text{Mass percent of X} = \left(\frac{\text{mass of X in } \mathrm{XZ}_2}{\text{total mass of } \mathrm{XZ}_2}\right) \times 100\%\]\[\text{Mass percent of Z} = \left(\frac{\text{mass of Z in } \mathrm{XZ}_2}{\text{total mass of } \mathrm{XZ}_2}\right) \times 100\%\]This results in \(14.29\%\) for \(X\) and \(85.71\%\) for \(Z\).

• Relevance: Helpful in identifying concentration and purity.
• Use Cases: Useful in commercial products and formulations.

This straightforward calculation is fundamental in fields ranging from industrial chemistry to analytical labs.