Problem 14 - Chemistry: Principles and Reactions

Question

What is the mass percent of chromium in the chloride salt of \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{OH})\right]^{2+} ?\)

Step-by-Step Solution

Answer

Answer: The mass percent of chromium in the chloride salt of the complex ion \(\left[\mathrm{Cr}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}(\mathrm{OH})\right]^{2+}\) is approximately 24.41%.

Step 1: Find the molecular weight of the complex ion and the chloride salt

First, we'll find the molecular weight of the complex ion. In the given ion \(\left[\mathrm{Cr}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}(\mathrm{OH})\right]^{2+}\), we have: - 1 chromium atom (Cr) - 5 water molecules (H2O) - 1 hydroxide ion (OH) Let's calculate the molecular weight of these individual components: Cr = 51.996 g/mol H2O = (2 × 1.008) + 16 = 18.016 g/mol OH = 15.999 + 1.008 = 17.007 g/mol The molecular weight of the complex ion is the sum of the molecular weight of each component: Molecular weight of complex ion = 1 × Cr + 5 × H2O + 1 × OH = 51.996 + 5 × 18.016 + 17.007 = 142.096 g/mol Now, let's find the molecular weight of the chloride salt. Since the complex ion has a charge of 2+, it needs 2 chloride ions (Cl-) to form the salt: 2 × Cl = 2 × 35.453 = 70.906 g/mol The molecular weight of the chloride salt = Molecular weight of complex ion + 2 × Cl = 142.096 + 70.906 = 213.002 g/mol

Step 2: Calculate the mass percent of chromium

Now, we can calculate the mass percent of chromium in the chloride salt. The mass percent is given by the ratio the mass of chromium to the mass of the chloride salt multiplied by 100%: Mass percent of chromium = \(\frac{\text{mass of chromium}}{\text{mass of chloride salt}} \times 100\% = \frac{51.996 \text{ g/mol}}{213.002 \text{ g/mol}}\times 100\% = 24.407\%\) So, the mass percent of chromium in the chloride salt of \(\left[\mathrm{Cr}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}(\mathrm{OH})\right]^{2+}\) is approximately 24.41%.

Key Concepts

Molecular Weight

Understanding molecular weight is crucial in chemistry, as it allows us to quantify the mass of a molecule in atomic mass units. It is calculated by summing the atomic weights of all the atoms in a molecule. In our example, atomic weights for chromium (Cr), hydrogen (H), oxygen (O), and chlorine (Cl) are taken from the periodic table.

To calculate the molecular weight of the complex ion \(\left[\mathrm{Cr}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}(\mathrm{OH})\right]^{2+}\), we determined the sum of the weights of one chromium atom, five water molecules (each consisting of two hydrogen atoms and one oxygen atom), and one hydroxide ion (composed of one oxygen atom and one hydrogen atom). For the chloride salt, we then added the weight of two chlorine atoms to account for the negative charges that balance the 2+ charge of the complex ion. These summed weights give the entire molecule’s mass.

Chemical Formula

The chemical formula provides a concise representation of the composition and proportions of atoms in a compound. In our exercise, the complex ion's formula indicates one chromium atom, five water molecules, and one hydroxide ion, which are combined with two chloride ions to form the neutral chloride salt.

To accurately determine the chemical formula, one must understand the subscripts used, representing the quantity of each type of atom or molecule within the compound, and the charges, signifying how ions combine to form neutral compounds. In this context, the chemical formula helped us identify the atoms and calculate their respective contributions to the molecule's molecular weight.

Stoichiometry

Stoichiometry is the section of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It is based on the conservation of mass and the concept of moles. In the case of our problem, we used stoichiometry to find the mass percent of chromium in the chloride salt. By dividing the mass of chromium by the total mass of the compound and multiplying by 100, we obtained its mass percent.

This calculation is fundamental in stoichiometry, as it can predict the amount of substances required or produced in a given reaction. The result is crucial for numerous applications, such as determining the purity of a sample or formulating mixtures in the correct proportions.