Final answer:
To find the value of n in the hydrate formula Al(NO3)3·nH2O, the mass of the water lost during heating (6.549 grams) is converted to moles of water and divided by the moles of anhydrous Al(NO3)3. The result is approximately 9, indicating that n is 9 in the formula for the hydrated compound.
Step-by-step explanation:
To calculate the value of n in the formula Al(NO3)3·nH2O(s), we first determine the mass of water that has been driven off during the heating process. The initial mass of the hydrate was 15.152 grams and after heating, the mass of anhydrous Al(NO3)3 is 8.603 grams. Therefore, the mass of water lost is:
15.152 grams (initial) - 8.603 grams (final) = 6.549 grams
Next, we convert the mass of water to moles by using the molar mass of water, which is approximately 18.015 g/mol:
6.549 grams / 18.015 g/mol = 0.363 moles of water
The molecular mass of Al(NO3)3 is approximately 213.00 g/mol. Using the final mass of the anhydrous compound, we calculate the moles of Al(NO3)3:
8.603 grams / 213.00 g/mol = 0.0404 moles of Al(NO3)3
To find the number of moles of water per mole of Al(NO3)3, we divide the moles of water by the moles of aluminum nitrate:
0.363 moles of water / 0.0404 moles of Al(NO3)3 ≈ 9 moles of water
Since the closest whole number is 9, we conclude that n is approximately equal to 9, which means the hydrated formula would be Al(NO3)3·9H2O.