Problem 125 - General Chemistry: Principles and Modern Applications

Question

A formula unit of the compound \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{SO}_{4}\) has nearly equal masses of (a) \(\mathrm{S}\) and \(\mathrm{O} ;\) (b) \(\mathrm{N}\) and \(\mathrm{O}\) (c) \(\mathrm{H}\) and \(\mathrm{N} ;\) (d) \(\mathrm{Cu}\) and \(\mathrm{O}\)

Step-by-Step Solution

Answer

For none of the pairs a, b, c, d, the masses are nearly equal in the given compound.

Step 1: List the atomic masses

Look up and list the atomic masses of the involved elements. The numbers are approximate: S (sulfur) = 32 g/mol, O (oxygen) = 16 g/mol, N (nitrogen) = 14 g/mol, H (hydrogen) = 1 g/mol, and Cu (copper) = 63.5 g/mol.

Step 2: Calculate the masses of each element in the formula

Using the chemical formula \([\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{SO}_{4}\), calculate the total mass of each constituent in one formula unit. For S, it's 32 g/mol. For O, it occurs five times (four in SO4 and one in Cu), so its mass is 5*16 = 80 g/mol. For N, it is in 4 NH3 groups, so its mass is 4*14 = 56 g/mol. H is in the 4 NH3 groups, its mass is 4*3*1 = 12 g/mol (each NH3 has 3 H atoms). Cu has a mass of 63.5 g/mol.

Step 3: Compare the masses

Compare the masses pairwise, as stated in the exercise. For a, S = 32 g/mol, O = 80 g/mol, not equal. For b, N = 56 g/mol, O = 80 g/mol, not equal. For c, H = 12 g/mol, N = 56 g/mol, not equal. For d, Cu = 63.5 g/mol, O = 80 g/mol, not equal.

Key Concepts

Atomic Mass Calculation

Understanding atomic mass is essential in chemistry, especially when dealing with coordination compounds. Atomic mass is the weighted average mass of an element's isotopes, measured in atomic mass units (amu). Here's how to calculate the atomic mass for a compound:

Start by looking up the atomic masses of the elements involved. These values are generally found on the periodic table and are given in grams per mole (g/mol).

• For example, sulfur (S) has an atomic mass of approximately 32 g/mol.
• Oxygen (O) has an atomic mass of 16 g/mol.
• Nitrogen (N) is around 14 g/mol.
• Hydrogen (H) is about 1 g/mol.
• Copper (Cu) is roughly 63.5 g/mol.

When you're working with complex compounds like \([\text{Cu(NH}_3\text{)}_4]\text{SO}_4\), knowing these basic atomic masses helps calculate the total mass of each element in the formula. By multiplying the atomic mass by the number of atoms in the compound, you can determine the total mass contribution of each element.

This is crucial for comparing and analyzing the relative masses of different components within the formula.

Chemical Formula Interpretation

Interpreting chemical formulas involves understanding the number and types of atoms present in a compound. In coordination chemistry, this can be particularly intriguing.

Take, for example, the compound \([\text{Cu(NH}_3\text{)}_4]\text{SO}_4\). This formula indicates:

• One copper atom (Cu).
• Four ammonia molecules (NH₃), which equates to 4 nitrogen atoms ({N}) and 12 hydrogen atoms (H).
• One sulfate ion (SO₄), consisting of one sulfur atom (S) and four oxygen atoms (O).

This formula provides a structured way to understand the compound's composition. By breaking down the formulas into individual atoms and molecules, you can better grasp how they interact and contribute to the compound's properties.

Chemical formula interpretation is a key skill that lets you visualize and calculate the compositions needed in various chemical reactions and processes.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It plays a vital role in chemistry, allowing us to calculate how much of each substance is involved.

In the context of \([\text{Cu(NH}_3\text{)}_4]\text{SO}_4\), stoichiometry helps us determine the exact proportions of each element in the formula. Here's how it works:

• First, calculate the total atomic mass for each component using the atomic masses from the periodic table.
• Apply these calculations to find the mass of the entire compound.
• Use these mass values to compare the masses of different combinations, as asked in problems like the original exercise.

With stoichiometry, you're not just counting atoms; you're understanding their proportional relationships, which can help predict the outcomes of chemical reactions. This is essential for balancing equations, determining yields, and optimizing reactions for desired products.

Understanding stoichiometry provides a deeper insight into how compounds form and interact, both in a lab and in real-world applications.