8.1g of XSO4.7H2O crystal when heated at 90degree for 15 minutes gave 6.9g of the salt,which gave 4.9g when heated further at 120degree. If X is 72. Calculate the amount of water of crystallization in the intermediate and final salts.

[X=72,S=32,O=16,H=1]

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To calculate the amount of water of crystallization in the intermediate and final salts, we need to determine the mass lost at each temperature and then compare it to the molar mass of water (H2O) to find the number of moles of water lost.

Let's start by calculating the number of moles of XSO4.7H2O in the initial sample:

Molar mass of XSO4.7H2O = (1 * 72) + (1 * 32) + (4 * 16) + (7 * (2 * 1 + 16)) = 158 g/mol

Number of moles of XSO4.7H2O = mass / molar mass = 8.1 g / 158 g/mol = 0.05127 mol

Next, let's calculate the mass lost at 90 degrees Celsius:

Mass lost at 90 degrees = initial mass - mass at 90 degrees = 8.1 g - 6.9 g = 1.2 g

Now, let's calculate the number of moles of water lost at 90 degrees Celsius:

Molar mass of H2O = (2 * 1) + 16 = 18 g/mol

Number of moles of water lost = mass lost / molar mass = 1.2 g / 18 g/mol = 0.06667 mol

After heating further at 120 degrees Celsius, the mass lost is calculated as:

Mass lost at 120 degrees = mass at 90 degrees - mass at 120 degrees = 6.9 g - 4.9 g = 2.0 g

Now, let's calculate the number of moles of water lost at 120 degrees Celsius:

Number of moles of water lost = mass lost / molar mass = 2.0 g / 18 g/mol = 0.11111 mol

Finally, we can determine the amount of water of crystallization in the intermediate and final salts by subtracting the moles of water lost at each stage from the initial formula:
- Intermediate Salt: XSO4.7H2O - 0.06667 mol H2O = XSO4.6.93333H2O
- Final Salt: XSO4.6.93333H2O - 0.11111 mol H2O = XSO4.6.82222H2O

So, the amount of water of crystallization in the intermediate salt is approximately 6.93333 moles of water, and in the final salt, it is approximately 6.82222 moles of water.

If I work this out I find that the water of crystallization(s) are not whole numbers. There are compounds where the hydration number is not a whole number. Do you have any indication that this is the case for this problem.