The aluminum sulfate hydrate [Al2(SO4)3
.
xH2O] contains8.10 percent Al by mass. Calculate x, that
is, the number of water molecules associated with each Al2(SO4)3 unit.
decimal percent Al= (2*27)/(2*27+3*32+12*16 + x*18)
so the decimal percent was given to be .0810, solve for x.
To calculate the value of x, we need to determine the ratio of moles of water to moles of aluminum sulfate.
1. Find the molar mass of Al2(SO4)3:
- Aluminum (Al): 2 * atomic mass = 2 * 26.98 g/mol = 53.96 g/mol
- Sulfur (S): 3 * atomic mass = 3 * 32.07 g/mol = 96.21 g/mol
- Oxygen (O): 12 * atomic mass = 12 * 16.00 g/mol = 192.00 g/mol
Total molar mass = 53.96 g/mol + 96.21 g/mol + 192.00 g/mol = 342.17 g/mol
2. Calculate the mass of Al in one mole of Al2(SO4)3:
- Percentage of Al in Al2(SO4)3 = 8.10%
- Mass of Al in one mole of Al2(SO4)3 = 8.10% * 342.17 g/mol = 27.74 g
3. Calculate the number of moles of Al:
- Moles of Al = Mass of Al / Molar mass of Al = 27.74 g / 26.98 g/mol = 1.03 mol
4. Since there are two Al atoms in one mole of Al2(SO4)3, there must be 2.06 moles of Al2(SO4)3.
5. The molar ratio between water and Al2(SO4)3 is:
- Moles of water / Moles of Al2(SO4)3 = x
To find x, we need to convert grams of water to moles using the molar mass of water:
- Molar mass of water (H2O) = 2 * 1.01 g/mol + 16.00 g/mol = 18.02 g/mol
The mass of water in one mole of Al2(SO4)3 can be calculated as:
- Mass of water = Mass of Al2(SO4)3 - Mass of Al = 342.17 g/mol - 27.74 g = 314.43 g
Converting the mass of water to moles:
- Moles of water = Mass of water / Molar mass of water = 314.43 g / 18.02 g/mol = 17.44 mol
Therefore, the molar ratio is:
- x = Moles of water / Moles of Al2(SO4)3 = 17.44 mol / 2.06 mol = 8.47
Rounding the value to the nearest whole number, we find that x ≈ 8. Therefore, there are approximately 8 water molecules associated with each Al2(SO4)3 unit.
To calculate the number of water molecules (x) associated with each Al2(SO4)3 unit in the aluminum sulfate hydrate, we need to use the given information about the mass percentage of aluminum (Al) in the compound.
Here's how you can calculate x:
Step 1: Determine the molar mass of Al2(SO4)3:
- The molar mass of Al is 26.98 g/mol.
- The molar mass of S is 32.07 g/mol.
- The molar mass of O is 16.00 g/mol.
Therefore, the molar mass of Al2(SO4)3 is:
2(26.98 g/mol) + 3(32.07 g/mol) + 12(16.00 g/mol) = 342.14 g/mol.
Step 2: Calculate the molar mass of Al in Al2(SO4)3:
- The molar mass of Al is 26.98 g/mol.
- Therefore, the mass of Al in Al2(SO4)3 is:
(2)(26.98 g/mol) = 53.96 g/mol.
Step 3: Calculate the mass of water (H2O) in Al2(SO4)3:
- The mass percentage of Al is given as 8.10 percent by mass.
- Therefore, the mass of Al in a given amount of Al2(SO4)3 is 8.10 percent of the mass of Al2(SO4)3.
- Assuming the mass of Al2(SO4)3 is 100 grams, the mass of Al is:
(8.10/100) × 100 g = 8.10 g.
Step 4: Determine the number of moles of Al:
- The number of moles of Al is calculated by dividing the mass of Al by its molar mass:
8.10 g / 26.98 g/mol = 0.300 mol.
Step 5: Calculate the number of moles of water (H2O):
- Since the molecular formula of water is H2O, the molar mass is 18.02 g/mol.
- The mass of water in Al2(SO4)3 is equal to the difference between the mass of Al2(SO4)3 and the mass of Al:
100 g - 8.10 g = 91.9 g.
- The number of moles of water is calculated by dividing the mass of water by its molar mass:
91.9 g / 18.02 g/mol = 5.10 mol.
Step 6: Determine the ratio of moles of water to moles of Al:
- Dividing the moles of water by the moles of Al gives us the ratio:
5.10 mol / 0.300 mol = 17 mol.
Step 7: Determine the value of x:
- Since the formula of the hydrate is Al2(SO4)3 · xH2O, the value of x represents the number of water molecules associated with each Al2(SO4)3 unit.
- We have determined that there are 17 moles of water for every 0.300 moles of Al2(SO4)3.
- Therefore, the value of x is equal to 17.
So, the number of water molecules associated with each Al2(SO4)3 unit in the aluminum sulfate hydrate is 17.