Final answer:
To obtain 12.5g of copper (II) sulfate, one needs to calculate the moles of CuSO4 needed and convert that to moles of copper (II) sulfate pentahydrate using the molar mass of both compounds, then multiply by the molar mass of CuSO4·5H2O to find the required mass.
Step-by-step explanation:
To determine how much copper (II) sulfate pentahydrate, CuSO4·5H2O, is required to get 12.5g of copper (II) sulfate, we need to take into account the molar masses of both compounds. Copper (II) sulfate pentahydrate includes not only the copper sulfate but also the mass of five water molecules per formula unit.
The molar mass of CuSO4·5H2O is the sum of the molar mass of CuSO4 (159.62 g/mol) and the molar mass of 5 water molecules (5 × 18.015 g/mol). The molar mass of CuSO4·5H2O is therefore 159.62 + (5 × 18.015) = 249.70 g/mol.
To find the mass of copper (II) sulfate pentahydrate needed for 12.5g of CuSO4, we use the ratio of their molar masses:
- Calculate the moles of CuSO4 needed by dividing 12.5g by its molar mass. So, moles of CuSO4 = 12.5 g / 159.62 g/mol.
- Then, convert moles of CuSO4 to moles of CuSO4·5H2O, which is a 1:1 ratio since the hydrate has one formula unit of CuSO4.
- Multiply the moles of hydrated CuSO4 by its molar mass to find the mass of CuSO4·5H2O required.