Determine the mass of CuSO4 • 5H2O needed to prepare 100.00 mL of 0.65000 M copper sulfate solution.
From this post I assume you want to prepare CuSO4 solution and not CuSO4.5H2O.
How many mols CuSO4 do you need? That's M x L = 0.65000 x 0.100 = 0.06500
grams CuSO4 = mols x molar mass CuSO4 = ?
But you want to weigh CuSO4.5H2O so how do you handle that?
Use the mass shown by ? x (molar mass CuSO4.5H2O/molar mass CuSO4) = this you weigh out, place it in a 100 mL volumetric flask, add a a little distilled water, swirl until all of the CuSO4.5H2O is dissolve, then add distilled to the mark on the volumetric flask, stopper, mix thoroughly, label. Done.
Would it be 16.229g?
That looks good to me.
To determine the mass of CuSO4 • 5H2O needed to prepare the given copper sulfate solution, we can follow the steps below:
Step 1: Determine the molar mass of CuSO4 • 5H2O.
The molar mass of copper sulfate pentahydrate (CuSO4 • 5H2O) can be calculated by summing the molar masses of its individual elements. Referencing the periodic table, we find:
- Molar mass of Cu = 63.55 g/mol
- Molar mass of S = 32.07 g/mol
- Molar mass of O = 16.00 g/mol
The molar mass of H2O is calculated by adding the molar masses of hydrogen (H) and oxygen (O):
- Molar mass of H = 1.01 g/mol
- Molar mass of O = 16.00 g/mol
Therefore, the molar mass of CuSO4 • 5H2O is:
(63.55 g/mol) + (32.07 g/mol) + (4 × (16.00 g/mol)) + (5 × ((2 × (1.01 g/mol)) + (16.00 g/mol))) = 249.68 g/mol
Step 2: Calculate the number of moles of CuSO4 required.
Given that the solution's concentration is 0.65000 M and the volume is 100.00 mL (0.10000 L), we can use the formula:
moles = concentration (M) × volume (L)
moles = 0.65000 M × 0.10000 L = 0.06500 mol
Step 3: Calculate the mass of CuSO4 • 5H2O.
The mass of CuSO4 • 5H2O needed is obtained by multiplying the number of moles by the molar mass:
mass = moles × molar mass
mass = 0.06500 mol × 249.68 g/mol ≈ 16.230 g
Therefore, approximately 16.230 grams of CuSO4 • 5H2O are needed to prepare 100.00 mL of 0.65000 M copper sulfate solution.