Final answer:
Upon heating 17.0g of MgSO4 · 7H2O, the mass of anhydrous magnesium sulfate (MgSO4) remaining would be approximately 8.20g after removing all the water of hydration.
Step-by-step explanation:
If 17.0g of MgSO4 · 7H2O (magnesium sulfate heptahydrate) is thoroughly heated and all the water is removed, you would be left with the mass of anhydrous magnesium sulfate (MgSO4). The molar mass of MgSO4 is approximately 120.37 g/mol and that of water (H2O) is 18.015 g/mol. Since the original compound contains 7 moles of water for every mole of MgSO4, you can calculate the mass of the water in the sample and subtract it from the original mass to find the mass of anhydrous MgSO4 left over.
To calculate, you would use the percentage by mass of MgSO4 in the hydrate and apply it to the initial mass:
Percentage by mass of MgSO4 = (Molar mass of MgSO4) / (Molar mass of compound) × 100%
Percentage by mass of MgSO4 = (120.37 g/mol) / (120.37 g/mol + 7 × 18.015 g/mol) × 100%
This results in a percentage of about 48.24%
Therefore, the mass of anhydrous MgSO4 remaining would be:
Mass of MgSO4 = 17.0 g × 48.24% = 8.20g (to two decimal places)