Question

The deep blue compound \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{SO}_{4}\) is made by the reaction of copper(II) sulfate and ammonia. \(\mathrm{CuSO}_{4}(\mathrm{aq})+4 \mathrm{NH}_{3}(\mathrm{aq}) \longrightarrow \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{SO}_{4}(\mathrm{aq})\) (a) If you use \(10.0 \mathrm{g}\) of \(\mathrm{CuSO}_{4}\) and excess \(\mathrm{NH}_{3}\), what is the theoretical yield of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{SO}_{4} ?\) (b) If you isolate \(12.6 \mathrm{g}\) of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{SO}_{4},\) what is the percent yield of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{SO}_{4} ?\)

Short answer

(a) 15.4 g; (b) 81.8%

Step-by-step solution

Calculate Molar Mass of CuSO4

First, calculate the molar mass of \( \mathrm{CuSO}_4 \). Use the atomic masses: Cu = 63.55 g/mol, S = 32.07 g/mol, O = 16.00 g/mol. \(\mathrm{CuSO}_4\) has 1 Cu, 1 S, and 4 O atoms. Add them up: \(63.55 + 32.07 + 4 \times 16.00 = 159.62 \ \mathrm{g/mol}\).

Calculate Moles of CuSO4

Using the mass of \(\mathrm{CuSO}_4\) provided (10.0 g), calculate the moles by dividing by its molar mass: \( \text{moles} = \frac{10.0 \ \mathrm{g}}{159.62 \ \mathrm{g/mol}} \approx 0.0627 \ \mathrm{mol} \).

Determine Moles of Product

From the balanced chemical equation, \( 1 \ \mathrm{mol} \ of \ \mathrm{CuSO}_4 \) reacts to form \( 1 \ \mathrm{mol} \) of \( \mathrm{Cu(NH}_3\mathrm{)}_4\mathrm{SO}_4 \). Thus, \( 0.0627 \ \mathrm{mol} \) of \( \mathrm{CuSO}_4 \) will produce \( 0.0627 \ \mathrm{mol} \) of \( \mathrm{Cu(NH}_3\mathrm{)}_4\mathrm{SO}_4 \).

Calculate Molar Mass of Cu(NH3)4SO4

Calculate the molar mass of \( \mathrm{Cu(NH}_3\mathrm{)}_4\mathrm{SO}_4 \): Cu = 63.55 g/mol, S = 32.07 g/mol, O = 16.00 g/mol, N = 14.01 g/mol, H = 1.01 g/mol. The compound has 1 Cu, 1 S, 4 O, 4 N, and 12 H atoms. Add the masses: \(63.55 + 4(14.01 + 3 \times 1.01) + 32.07 + 4 \times 16.00 = 245.7 \ \mathrm{g/mol}\).

Calculate Theoretical Yield

Multiply the moles of product by its molar mass to get the theoretical yield: \( 0.0627 \ \mathrm{mol} \times 245.7 \ \mathrm{g/mol} \approx 15.4 \ \mathrm{g} \).

Calculate Percent Yield

Use the formula: \( \text{Percent Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\% \). Substitute the values: \( \left(\frac{12.6 \ \mathrm{g}}{15.4 \ \mathrm{g}} \right) \times 100\% \approx 81.8\% \).

Key concepts

Chemical Reactions

Understanding chemical reactions is fundamental in chemistry. These reactions involve breaking and forming chemical bonds, resulting in substances transforming into new products. In the reaction given, copper(II) sulfate (\(\mathrm{CuSO}_{4}\)) reacts with ammonia (\(\mathrm{NH}_{3}\)) to form a complex compound, \(\mathrm{Cu(NH}_{3}\mathrm{)}_{4}\mathrm{SO}_4\). This balanced chemical equation tells us the proportions of reactants and products.

• The original materials (reactants) present are copper(II) sulfate and ammonia.
• The product formed is a deep blue compound that shows the result of the reaction.
• The equation reflects the conservation of mass, meaning the quantity of each type of atom is the same on both sides of the arrow.

Here, for each mole of \(\mathrm{CuSO}_{4}\), one mole of \(\mathrm{Cu(NH}_{3}\mathrm{)}_{4}\mathrm{SO}_4\) is produced, which helps in calculations for molar amounts.

Molar Mass Calculations

Molar mass calculations are key to converting between mass and moles, which is crucial in stoichiometry. The molar mass is the mass of one mole of a substance. To calculate it, sum the atomic masses of all atoms in a compound.
In the exercise, the molar mass of \(\mathrm{CuSO}_{4}\) was calculated as follows:

• The atomic mass of copper is 63.55, sulfur is 32.07, and oxygen is 16.00 g/mol.
• Add them up: 63.55 + 32.07 + 4(16.00) = 159.62 g/mol.

Similarly, for the compound \(\mathrm{Cu(NH}_{3}\mathrm{)}_{4}\mathrm{SO}_4\):

• Add copper, nitrogen, hydrogen, sulfur, and oxygen masses: 63.55 + 4(14.01 + 3 \times 1.01) + 32.07 + 4 \times 16.00 = 245.7 g/mol.

Using these calculations, you can convert grams to moles, which helps in finding theoretical yields and understanding reaction scales.

Percent Yield Calculations

Percent yield calculations tell us how efficient a chemical reaction is by comparing the actual yield (real amount produced) to the theoretical yield (amount predicted by calculations). It is expressed as:\[\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\%\]This simple formula shows the efficiency of the reaction process.
In the given problem:

• We found the theoretical yield of \(\mathrm{Cu(NH}_{3}\mathrm{)}_{4}\mathrm{SO}_4\) as 15.4 g.
• The actual yield obtained was 12.6 g.
• Plug these values into the formula: \(\left( \frac{12.6}{15.4} \right) \times 100\% \approx 81.8\%\)

A percent yield of 81.8% indicates that the reaction was not perfectly efficient, possibly due to side reactions or incomplete conversion of reactants.