Question

Many transition metal salts are hydrates: they contain a fixed number of water molecules bound per formula unit of the salt. For example, copper(II) sulfate most commonly exists as the pentahydrate, \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O} .\) If \(5.00 \mathrm{g}\) of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) is heated strongly so as to drive off all of the waters of hydration as water vapor, what volume will this water vapor occupy at \(350 .^{\circ}\) C and a pressure of 1.04 atm?

Short answer

The volume of water vapor produced when 5.00 g of copper(II) sulfate pentahydrate is heated strongly at 350°C and a pressure of 1.04 atm is approximately \(5.105 \ \mathrm{L}\).

Step-by-step solution

Calculate the moles of water in the hydrate

To find the moles of water, we first need to determine the moles of copper(II) sulfate pentahydrate by using the molecular weight and the given mass. Then, we can find the moles of water by multiplying the moles of the pentahydrate by 5, since there are 5 water molecules per formula unit of the salt. The molecular weight of \(\mathrm{CuSO}_4 \cdot 5\mathrm{H}_2\mathrm{O}\): Cu: 63.55 g/mol S: 32.07 g/mol O: 16.00 g/mol (9 oxygen atoms in total) H: 1.01 g/mol (10 hydrogen atoms in total) \(\mathrm{M}_{\mathrm{CuSO}_4 \cdot 5\mathrm{H}_2\mathrm{O}} = 63.55 + 32.07 + 9(16.00) + 10(1.01) = 249.73 \ \mathrm{g/mol}\) Moles of \(\mathrm{CuSO}_4 \cdot 5\mathrm{H}_2\mathrm{O}\) = \(\frac{5.00 \ \mathrm{g}}{249.73 \ \mathrm{g/mol}} = 0.02003 \ \mathrm{mol}\) Now, let's find the moles of water: Moles of water = moles of \(\mathrm{CuSO}_4 \cdot 5\mathrm{H}_2\mathrm{O} \cdot 5 = 0.02003 \ \mathrm{mol} \cdot 5 = 0.10015 \ \mathrm{mol}\)

Convert the temperature to Kelvin

In order to use the Ideal Gas Law, the temperature must be in Kelvin (K). To convert the temperature from Celsius to Kelvin, we add 273.15 K. \(T(^\circ C)= \ 350\),\ \(T(K) = 350 + 273.15 = 623.15\) K

Apply the Ideal Gas Law

Now that we have the moles of water vapor, the temperature in Kelvin, and the pressure, we can apply the Ideal Gas Law: \(PV=nRT\). Here, \(P = 1.04 \ \mathrm{atm}\) \(V = \mathrm{Volume \ of \ water \ vapor}\) \(n = 0.10015 \ \mathrm{mol}\) \(R = 0.08206 \ \frac{\mathrm{L \ atm}}{\mathrm{mol \ K}}\) (Ideal Gas Constant) \(T = 623.15 \ \mathrm{K}\) We need to solve for V (the volume of water vapor): \(V=\frac{nRT}{P} = \frac{0.10015 \ \mathrm{mol} \cdot 0.08206 \ \frac{\mathrm{L \ atm}}{\mathrm{mol \ K}} \cdot 623.15 \ \mathrm{K}}{1.04 \ \mathrm{atm}}\)

Calculate the volume of the water vapor

Now, we can calculate the volume of the water vapor: \(V = \frac{0.10015 \cdot 0.08206 \cdot 623.15}{1.04} = 5.105 \ \mathrm{L}\) So, the volume of the water vapor produced when 5.00 g of copper(II) sulfate pentahydrate is heated strongly at 350°C and a pressure of 1.04 atm is approximately 5.105 liters.

Key concepts

Transition Metal Salts

Transition metal salts are compounds formed by transition metals and nonmetals. In chemistry, these salts often appear as hydrates, which means they have water molecules integrated into their crystal structure. For instance, copper(II) sulfate can exist as a pentahydrate, symbolized as \(\mathrm{CuSO}_{4} \cdot 5\mathrm{H}_{2}\mathrm{O}\), which indicates there are five water molecules for each unit of copper(II) sulfate.

When heated, these hydrated salts can lose their water, a process known as dehydration, forming anhydrous salts. The water is released as vapor, and this property is useful for many applications such as in drying agents or in certain chemical syntheses. Understanding the behavior of such hydrates when subjected to heat is an essential concept, particularly when dealing with transition metal compounds in laboratory settings.

Molecular Weight Calculation

Calculating molecular weight, also referred to as molar mass, is a key step in many chemical calculations, including those involving hydrates. The molecular weight is the sum of the atomic masses of all atoms in a molecule. It's typically expressed in grams per mole (g/mol).

For hydrates, it's important to include the water molecules in this calculation. As in the exercise, the molecular weight of copper(II) sulfate pentahydrate is determined by adding the atomic masses of copper, sulfur, oxygen in both the \(\mathrm{CuSO}_{4}\) and the water molecules, and hydrogen from the water molecules. By knowing this, one can determine how many moles of the substance are in a given mass, which is crucial for conversions and chemical reaction calculations.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry and is expressed as \(PV = nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles of gas, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin. This law relates the four variables of the state of an ideal gas and allows chemists to predict how a gas will behave under different conditions.

Using the Ideal Gas Law, one can calculate any one of the variables if the others are known. In the given exercise, after the moles of water vapor were determined and the temperature converted to Kelvin, the Ideal Gas Law was used to calculate the volume that the water vapor would occupy under the specified conditions.

Stoichiometry

Stoichiometry involves the quantitative relationships between the reactants and products in a chemical reaction. For hydrates, stoichiometry can be used to relate the moles of hydrate to the moles of water, since the hydrate contains a stoichiometrically fixed number of water molecules. The process often involves balancing chemical equations and using molar ratios derived from those balanced equations to perform conversions between mass, moles, and molecules.

In the dehydration of a hydrate, stoichiometry helps us predict the amount of water produced. For example, in the copper(II) sulfate pentahydrate scenario, there is a 1:5 molar ratio between the salt and the number of water molecules. Knowing this, we can find out exactly how much water will be released when a known mass of the hydrate is heated. Such calculations are not only essential in laboratory experiments but also in industrial processes where precise amounts of materials are necessary.