Final answer:
The percent composition of water in copper(II) sulfate pentahydrate (CuSO4·5H2O) is calculated by dividing the molar mass of water by the total molar mass of the compound and multiplying by 100, resulting in 36.07% to three significant figures.
Step-by-step explanation:
Calculating Percent Composition of Water in CuSO4·5H2O
To find the percent composition of water in copper(II) sulfate pentahydrate (CuSO4·5H2O), you must first calculate the molar mass of the entire compound and the molar mass of the water within it.
The molar mass of anhydrous CuSO4 is 159.62 g/mol, and the molar mass of water (H2O) is 18.015 g/mol.
Since there are five moles of water for each mole of CuSO4,
the total molar mass of the water in the hydrate is 5 × 18.015 g/mol = 90.075 g/mol.
Therefore, the molar mass of CuSO4·5H2O is 159.62 g/mol (CuSO4) + 90.075 g/mol (5H2O) = 249.695 g/mol.
To find the percent water in CuSO4·5H2O, divide the molar mass of water by the total molar mass of the compound and multiply by 100:
- (90.075 g/mol / 249.695 g/mol) × 100 = 36.07%
The percent of water in CuSO4·5H2O to three significant figures is 36.07%.