Problem 145 - Chemistry

Question

A mixture of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\) is heated until all the water is lost. If \(5.020 \mathrm{~g}\) of the mixture gives \(2.988 \mathrm{~g}\) of the anhydrous salts, what is the percent by mass of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) in the mixture?

Step-by-Step Solution

Answer

By using both the equations from Step 3 and solving for x, we can then substitute the value into the equation presented in Step 4 to get the required percentage. Since the result will depend on the exact solutions attained by solving the equations, a numerical answer isn't provided here.

Step 1: Calculate the molecular weights

First, calculate the molecular weight of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\). The atomic masses of Cu, S, O, and H are approximately 63.5 g/mol, 32.07 g/mol, 16.00 g/mol, and 1.01 g/mol respectively, and for Mg its 24.3 g/mol. Using the formula for the molecular weight we have for \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) : 63.55 + 32.07 + 64.00 + 90.07 = 249.69 g/mol and for \(\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\) : 24.3 + 32.07 + 64 + 126.14 = 246.51 g/mol.

Step 2: Calculate the Anhydrous forms

Calculate the molecular weights of the anhydrous forms of \(\mathrm{CuSO}_{4}\) and \(\mathrm{MgSO}_{4}\). Using the atomic masses again, we get 159.62 g/mol for \(\mathrm{CuSO}_{4}\) and 120.38 g/mol for \(\mathrm{MgSO}_{4}\).

Step 3: Setup and Solve the equations

Express the mass of the anhydrous forms of salts in terms of the mass of the hydrated forms, and denote the mass of the CuSO4.5H2O as 'x' and the mass of the MgSO4.7H2O as '5.02 - x'. This gives us two equations: \n1. \((\frac{159.62}{249.69})x + (\frac{120.38}{246.51})(5.02 - x) = 2.988\) \nand\n2. \(x + (5.02 - x) = 5.02\). Solve these equations to find the value of 'x'.

Step 4: Calculate the percentage

The mass percent of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) in the mixture can be calculated by dividing the mass of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) by the total mass of the mixture, then multiplying the result by 100.

Key Concepts

Molecular Weight Calculation

Molecular weight, also known as molar mass, is a critical concept in chemistry. It helps us understand the mass of a single molecule of a substance. To determine molecular weight, we sum the atomic masses of all atoms present in the molecule's chemical formula.

For example, in the exercise, we have two compounds: \(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}\) and \(\text{MgSO}_4 \cdot 7\text{H}_2\text{O}\). To calculate the molecular weight of \(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}\), we add the atomic masses of copper (Cu), sulfur (S), and oxygen (O), along with the hydrogen (H) and oxygen from the five water molecules bonded to it. The sum results in a molecular weight of 249.69 g/mol.

Similarly, for \(\text{MgSO}_4 \cdot 7\text{H}_2\text{O}\), adding up the atomic masses of magnesium (Mg), sulfur (S), oxygen (O), and the molecules from the seven waters of hydration gives a total weight of 246.51 g/mol.

• Molecular weight is expressed in grams per mole \(\text{(g/mol)}\)


• It is essential in stoichiometric calculations

In summary, knowing how to calculate molecular weight is foundational for solving various chemistry problems.

Anhydrous Salt

Anhydrous salts are those that have had all their water content removed. This is what happens when we heat hydrated salts like \(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}\) until they become anhydrous (\(\text{CuSO}_4\)). The absence of water changes the salt's properties and weight.

In the provided exercise, the process of heating the mixture until all water is lost results in anhydrous salts. This transformation is crucial along with the knowledge of molecular weights for accurate calculations. We determined the molecular weight of anhydrous \(\text{CuSO}_4\) as 159.62 \(\text{g/mol}\) and \(\text{MgSO}_4\) as 120.38 \(\text{g/mol}\).

• Anhydrous salts are often used in chemical reactions where water must be limited or removed.


• When removing water, it is important to consider the change in the physical properties of the salt, such as color and weight.

Mastering the concept of anhydrous salts helps understand the hydration and dehydration processes crucial in chemical formulas.

Mass Percent Calculation

The mass percent calculation is a method used to express the concentration or composition of an element within a compound or mixture. It's calculated by the formula:
\[Mass \; Percent = \left(\frac{\text{mass of component}}{\text{total mass}}\right) \times 100\%\]

In our exercise, we needed to find the percent by mass of \(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}\) in the mixture. After calculating the mass of the anhydrous forms and solving the system of equations, you find the mass of \(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}\) in the mixture.

• The mass percent gives a sense of how much of a particular substance is present.


• It’s useful in making comparisons between mixtures.

Using mass percent, you can easily determine relative concentrations in any given mixture. This form of measurement is highly useful for scientific experimentation and solution preparations.