How much CuSO4.5H20 would you need to prepare 6 grams of copper ammine complex, [Cu(NH3)]SO4.H20?
There is only 1 ammonia molecule in the copper ammine complex.
please answer asap, thanks!
6 g Cu[NH3]SO4.H2O x (molar mass CuSO4.5H2O/molar mass Cu[NH3]SO4.H2O) = ?
To determine the amount of CuSO4·5H2O needed to prepare 6 grams of the copper ammine complex, [Cu(NH3)]SO4·H2O, we need to consider the molar mass and stoichiometry of these compounds.
1. Calculate the molar mass of the copper ammine complex, [Cu(NH3)]SO4·H2O:
- Atomic mass of Cu = 63.55 g/mol
- Atomic mass of N = 14.01 g/mol
- Atomic mass of H = 1.01 g/mol
- Atomic mass of S = 32.07 g/mol
- Atomic mass of O = 16.00 g/mol
- Atomic mass of H2O = 18.02 g/mol
Molar mass of [Cu(NH3)]SO4·H2O = (63.55 + 14.01 + (3 × 1.01) + 32.07 + (4 × 16.00) + 18.02) g/mol
= 248.77 g/mol
2. Determine the molar mass of CuSO4·5H2O:
- Atomic mass of Cu = 63.55 g/mol
- Atomic mass of S = 32.07 g/mol
- Atomic mass of O = 16.00 g/mol
- Atomic mass of H2O = 18.02 g/mol
Molar mass of CuSO4·5H2O = (63.55 + 32.07 + (4 × 16.00) + (5 × 18.02)) g/mol
= 249.68 g/mol
3. Determine the moles of the copper ammine complex:
Moles = Mass / Molar mass
= 6 g / 248.77 g/mol
= 0.0241 mol
4. Determine the moles of CuSO4·5H2O needed:
The copper ammine complex, [Cu(NH3)]SO4·H2O, has a 1:1 stoichiometric ratio with CuSO4·5H2O. Therefore, the number of moles of CuSO4·5H2O required is also 0.0241 mol.
5. Determine the mass of CuSO4·5H2O needed:
Mass = Moles × Molar mass
= 0.0241 mol × 249.68 g/mol
= 6.03 g
Therefore, you would need approximately 6.03 grams of CuSO4·5H2O to prepare 6 grams of the copper ammine complex, [Cu(NH3)]SO4·H2O.
To determine the amount of CuSO4.5H2O needed to prepare 6 grams of the copper ammine complex [Cu(NH3)]SO4.H2O, we need to calculate the molar mass of both compounds and use stoichiometry.
First, let's calculate the molar mass of each compound:
1. Copper ammine complex ([Cu(NH3)]SO4.H2O):
- Molar mass of Cu: 63.55 g/mol
- Molar mass of NH3: 17.03 g/mol (since there is only one ammonia molecule in the complex)
- Molar mass of SO4: 32.06 g/mol
- Molar mass of H2O: 18.02 g/mol
Summing these up, the molar mass of the complex is:
63.55 g/mol + 17.03 g/mol + 32.06 g/mol + 18.02 g/mol = 130.66 g/mol
2. CuSO4.5H2O (Copper(II) sulfate pentahydrate):
- Molar mass of Cu: 63.55 g/mol
- Molar mass of S: 32.06 g/mol
- Molar mass of O: 16.00 g/mol
- Molar mass of H2O: 18.02 g/mol (since there are five water molecules in the formula)
Summing these up, the molar mass of CuSO4.5H2O is:
63.55 g/mol + 32.06 g/mol + (16.00 g/mol x 4) + (18.02 g/mol x 5) = 249.68 g/mol
Next, we can use stoichiometry to calculate the required amount of CuSO4.5H2O.
Let's assume x grams of CuSO4.5H2O is needed to produce 6 grams of the copper ammine complex.
Using the molar mass, we can set up a proportion based on their molar masses:
(6 g copper ammine complex) / (130.66 g/mol) = (x g CuSO4.5H2O) / (249.68 g/mol)
Cross multiplying, we get:
(6 g copper ammine complex) * (249.68 g/mol) = (130.66 g/mol) * (x g CuSO4.5H2O)
Simplifying, we find:
x g CuSO4.5H2O = (6 g copper ammine complex) * (249.68 g/mol) / (130.66 g/mol)
Calculating, we get:
x g CuSO4.5H2O ≈ 11.46 g
Therefore, you would need approximately 11.46 grams of CuSO4.5H2O to prepare 6 grams of the copper ammine complex [Cu(NH3)]SO4.H2O.