Calculate the weight of the impure sample of CuSO4.5H2O required to produce 28.5 g of K3[Cu(CN)4].

Understand the Problem

The question requires calculating the weight of an impure sample of CuSO4.5H2O needed to produce a specific weight of K3[Cu(CN)4] by considering the yields of various reactions involved in the process. This involves stoichiometric calculations and understanding of molar ratios, yields, and purity.

Answer

$$ W_{final} $$

Steps to Solve

Identify Given Information

Let's define the necessary information:

Let the target weight of $K_3[Cu(CN)4]$ be given as ( W{target} ) (in grams).
Determine the molar mass of $K_3[Cu(CN)_4]$, which is calculated from its constituent elements.

Molar Mass Calculation of $K_3[Cu(CN)_4]$

To find the molar mass:

Potassium (K): 39.10 g/mol, with 3 potassiums: ( 3 \times 39.10 ) g/mol
Copper (Cu): 63.55 g/mol: 1 copper
Carbon (C): 12.01 g/mol, with 4 carbons: ( 4 \times 12.01 ) g/mol
Nitrogen (N): 14.01 g/mol, with 4 nitrogens: ( 4 \times 14.01 ) g/mol

The total molar mass is:

$$ M_{K_3[Cu(CN)_4]} = 3(39.10) + 63.55 + 4(12.01) + 4(14.01) $$

Calculate Required Moles of $K_3[Cu(CN)_4]$

Use the formula to calculate the number of moles needed:

$$ n = \frac{W_{target}}{M_{K_3[Cu(CN)_4]}} $$

Determine Stoichiometry of the Reaction

Find out how many moles of $CuSO_4 \cdot 5H_2O$ produce $K_3[Cu(CN)_4]$.

For example, if each mole of $CuSO_4 \cdot 5H_2O$ produces one mole of $K_3[Cu(CN)_4]$, then:

$$ n_{CuSO_4} = n $$

Calculate Weight of $CuSO_4 \cdot 5H_2O$ Required

Calculate the weight of $CuSO_4 \cdot 5H_2O$ needed:

First, find its molar mass (CuSO4.5H2O):
- Copper (Cu): 63.55 g/mol
- Sulfur (S): 32.07 g/mol
- Oxygen (O): 16.00 g/mol, with 4 oxygens: ( 4 \times 16.00 ) g/mol
- Water (H2O): ( 5 \times (2 \times 1.01 + 16.00) ) g/mol

Thus,

$$ M_{CuSO_4 \cdot 5H_2O} = 63.55 + 32.07 + 4(16.00) + 5(18.02) $$

Calculate the final weight needed using:

$$ W_{CuSO_4 \cdot 5H_2O} = n_{CuSO_4} \times M_{CuSO_4 \cdot 5H_2O} $$

Adjust for Purity/Yield

If given the purity or yield of the sample ($p%$), adjust the final weight:

$$ W_{final} = \frac{W_{CuSO_4 \cdot 5H_2O}}{p%} $$

The calculated weight of the impure sample of $CuSO_4 \cdot 5H_2O$ needed to produce the required amount of $K_3[Cu(CN)_4]$ is expressed as:

$$ W_{final} $$

More Information

To give a specific numerical answer, all necessary values (weights, molar masses, and purities) must be provided. Common molar mass values:

$K_3[Cu(CN)_4]$ molar mass can be computed from individual atomic masses.
Molar mass of $CuSO_4 \cdot 5H_2O$ will account for both copper sulfate and its associated water molecules.

Tips

Not calculating the molar mass correctly. Always use the correct atomic weights.
Forgetting to adjust for purity when calculating the final amount needed.
Confusing the stoichiometric relationships between reactants and products.

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