Chapter 3: Problem 130
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?
Short Answer
Expert verified
The percent by mass of \( \mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O} \) in the mixture is approximately 36.50%.
Step by step solution
Over 22 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Mass Composition
Understanding mass composition is crucial when dealing with hydrated compounds like \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\). Mass composition refers to the percentage by mass of each component in a compound.
For hydrated compounds, their composition includes both the anhydrous salt and the water of crystallization. The anhydrous salt is what remains after water is removed.
To determine the mass composition in mixtures, you must consider the individual contributions of each component. Start by identifying the total mass of the mixture. Then, partition it into the mass of each hydrated compound, which involves a detailed understanding of their chemical formulas and respective water content. By calculating the percentage of each component in the total mass, you can further identify the mass composition of the compound or mixture.
This process involves setting up equations based on known molecular masses and solving for components like the mass of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) in a mixture.
For hydrated compounds, their composition includes both the anhydrous salt and the water of crystallization. The anhydrous salt is what remains after water is removed.
To determine the mass composition in mixtures, you must consider the individual contributions of each component. Start by identifying the total mass of the mixture. Then, partition it into the mass of each hydrated compound, which involves a detailed understanding of their chemical formulas and respective water content. By calculating the percentage of each component in the total mass, you can further identify the mass composition of the compound or mixture.
This process involves setting up equations based on known molecular masses and solving for components like the mass of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) in a mixture.
Molar Mass Calculation
To work with hydrated compounds, understanding molar mass calculation is essential. The molar mass represents the mass of one mole of a compound, and calculating it involves summing the atomic masses of all atoms in the molecule.
For a compound like \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\), the molar mass includes the mass of \(\mathrm{CuSO}_{4}\) plus five moles of water (H₂O). The molecular weights of Cu, S, O, and each component of water (H and O) are obtained from the periodic table.
Similarly, calculate for \(\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\).
This helps in determining the amount of water lost and the molecular weight before and after heating. Mastery of molar mass calculations allows for accurate determination of components in chemical reactions.
For a compound like \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\), the molar mass includes the mass of \(\mathrm{CuSO}_{4}\) plus five moles of water (H₂O). The molecular weights of Cu, S, O, and each component of water (H and O) are obtained from the periodic table.
- Calculate \(\mathrm{CuSO}_{4}\)'s molar mass by adding the atomic weights: Cu (63.546), S (32.066), \(4 \times \text{O (15.999)}\)
- Add to it the water components: \(5 \times \text{H}_2\text{O (2 \times 1.008 for H + 15.999 for O)} = 90.075 \text{ g/mol}\)
Similarly, calculate for \(\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\).
This helps in determining the amount of water lost and the molecular weight before and after heating. Mastery of molar mass calculations allows for accurate determination of components in chemical reactions.
Water Loss in Chemical Reactions
In many chemical reactions, especially those involving hydrated compounds, water loss is an important aspect. As compounds like \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) are heated, they lose water molecules resulting in an anhydrous form of the salt.
The water loss can be quantified by calculating the difference in mass before and after the reaction. The initial mass minus the final mass equals the mass of the water that was removed during heating.
For example, if a total mixture of hydrated salts weighs \(5.020 \text{ g}\) initially and weighs \(2.988 \text{ g}\) after heating, the water loss can be calculated as: \[ \text{Water lost} = 5.020 \text{ g} - 2.988 \text{ g} = 2.032 \text{ g} \]
This reflects the cumulative mass of water molecules released during the reaction: \(5 \times \text{mole of } \mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) and \(7 \times \text{mole of } \mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\).
Knowing how to measure water loss helps in the calculation of how much of each component remains as the anhydrous substance, and this can lead to determining the percentage composition or other metrics important for analyzing chemical reactions.
The water loss can be quantified by calculating the difference in mass before and after the reaction. The initial mass minus the final mass equals the mass of the water that was removed during heating.
For example, if a total mixture of hydrated salts weighs \(5.020 \text{ g}\) initially and weighs \(2.988 \text{ g}\) after heating, the water loss can be calculated as: \[ \text{Water lost} = 5.020 \text{ g} - 2.988 \text{ g} = 2.032 \text{ g} \]
This reflects the cumulative mass of water molecules released during the reaction: \(5 \times \text{mole of } \mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) and \(7 \times \text{mole of } \mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\).
Knowing how to measure water loss helps in the calculation of how much of each component remains as the anhydrous substance, and this can lead to determining the percentage composition or other metrics important for analyzing chemical reactions.
