Problem 119 - General Chemistry

Question

A sample of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) was heated to \(110^{\circ} \mathrm{C}\), where it lost water and gave another hydrate of copper(II) ion that contains \(32.50 \%\) Cu. A \(98.77\) -mg sample of this new hydrate gave \(116.66 \mathrm{mg}\) of barium sulfate precipitate when treated with a barium nitrate solution. What is the formula of the new hydrate?

Step-by-Step Solution

Answer

The formula of the new hydrate is CuSO4·2H2O.

Step 1: Determine Moles of Precipitated BaSO4

To find the amount of barium sulfate precipitated, we first need to know the molar mass of BaSO4. It is calculated as:- Ba: 137.33 g/mol- S: 32.07 g/mol- O4: 4 × 16.00 = 64.00 g/molAdding these gives the molar mass of BaSO4 as 233.40 g/mol.Find moles of BaSO4:\[\text{moles of BaSO}_4 = \frac{116.66\, \text{mg}}{233.40\, \text{g/mol}} = \frac{116.66\times10^{-3}\, \text{g}}{233.40\, \text{g/mol}}\approx 0.000500\, \text{mol}\]

Step 2: Relate Moles of BaSO4 to Moles of Cu in Hydrate

Assuming complete reaction, each molecule of Cu^2+ will react with SO4^2- from the CuSO4 to form one molecule of BaSO4. Thus, the moles of Cu in the sample is equal to the moles of BaSO4:\[\text{moles of Cu} = 0.000500\, \text{mol}\]

Step 3: Calculate Mass of Cu from Moles

Determine the mass of Cu corresponding to the number of moles calculated:\[\text{mass of Cu} = 0.000500\, \text{mol} \times 63.55\, \text{g/mol} = 0.03178\, \text{g} = 31.78\, \text{mg}\]

Step 4: Determine Total Mass Percent Composition

Using the mass of Cu, calculate the mass percent composition of Cu in the hydrate:\[\text{mass percent of Cu} = \left(\frac{31.78\, \text{mg}}{98.77\, \text{mg}}\right) \times 100\% = 32.14\% \approx 32.50\%\]This is now consistent with the given composition.

Step 5: Solve for n in New Hydrate Formula CuSO4·nH2O

Let the new hydrate be CuSO4·nH2O. Calculate the total molar mass:Given 32.5% Cu in new hydrate, find the empirical formula considering:\[\text{molar mass of Cu} = 63.55\, \text{g/mol}\]Thus, from 32.5%, the rest (67.5%) is for the compound (CuSO4·nH2O minus Cu).For 1 mol Cu:- Cu contribution: 63.55g, - Total contribution: \(M\ = \frac{63.55}{0.325}\ = 195.54\, \text{g/mol}\).Water contribution to 32.5% copper is calculated from:\[195.54\,\text{g/mol} = 63.55\,\text{g/mol} + 96.07\,\text{g/mol} + 18.02\,\text{n} \\Rightarrow 195.54 = 159.62 + 18.02n \]Solve for \(n\):\[195.54 - 159.62 = 18.02n \implies 35.92 = 18.02n \implies n \approx 1.99 \approx 2\]

Step 6: Finalize Formula of New Hydrate

The hydration value \(n\) approximates to 2. Thus, the formula for the new hydrate is CuSO4·2H2O.

Key Concepts

Copper Sulfate Pentahydrate

Copper sulfate pentahydrate is a common chemical compound with the formula \( ext{CuSO}_4 imes 5 ext{H}_2 ext{O}\). It is known for its vibrant blue color and is often used in chemistry laboratories as a primary standard for the calibration of analyte solutions and in various other applications.
- The 'pentahydrate' means it has five water molecules for every copper sulfate molecule.- It is a hydrate, which is a compound that contains water molecules within its crystal structure.
When heated, copper sulfate pentahydrate loses these water molecules and forms an anhydrous version (without water) that is white or grayish. By heating to specific temperatures, it can lose different amounts of water and transform into various hydrates. The original exercise mentions heating to form a compound containing 32.50% copper, leading to formula determination based on the copper content.

Barium Sulfate Precipitation

In chemical reactions, precipitation occurs when substances in solution form a solid which then settles out of the solution. In this case, barium sulfate (\( ext{BaSO}_4\)) forms in a reaction involving barium nitrate and copper sulfate.
- Barium sulfate is highly insoluble in water, meaning it does not dissolve easily.- During the exercise, when barium ions (\( ext{Ba}^{2+}\)) react with sulfate ions \(( ext{SO}_4^{2-})\), barium sulfate precipitates as a solid due to its low solubility.
This process is useful for determining the amount of sulfate ions in a solution since the precipitate can be filtered, dried, and weighed. It is a key step in finding the copper amount in the new hydrate, as calculating moles of this precipitate directly relates to the copper present.

Molar Mass Calculation

Molar mass is critical in chemistry for converting between moles and grams. It is the mass of one mole of a particular substance in grams per mole, which allows chemists to quantify the amount of substance.
- To calculate the molar mass of a compound like barium sulfate \(( ext{BaSO}_4)\), sum the atomic masses of all its atoms. - Ba: 137.33 g/mol - S: 32.07 g/mol - O: 4 \times 16.00 g/mol = 64.00 g/mol
- The total molar mass of barium sulfate combines these values, which is 233.40 g/mol.
With this information, you can convert between grams and moles, important for stoichiometric calculations in reactions. This concept is necessary for determining the amount of precipitated barium sulfate which links back to how much copper was initially in the hydrate.

Mass Percent Composition

Mass percent composition is a way to express the concentration of an element within a compound. It indicates the percentage of the compound's mass that comes from a specific element.
- It is calculated by taking the mass of the element in one mole of the compound, dividing it by the molar mass of the compound, and then multiplying by 100 to get a percentage.
For example, in the exercise:

• The mass of copper in the new hydrate is 31.78 mg, part of the total mass of 98.77 mg.
• Mass percent copper is \(rac{31.78 \, \text{mg}}{98.77 \, \text{mg}} \times 100\% = 32.14\%\).

This percentage indicates copper's share in the total mass of the compound. This concept helps ensure consistency with the given composition and aids in the calculation of the new hydrate's formula.