If 17.0g of MgSO4*7H2O is thoroughly heated,what mass of anhydrous magnesium sulfate will remain ?
I don't even know where to begin on this one
Change 17.0 g to mols of the hydrate, then multiply by 1 mol MgSO4/1 mol MgSO4*7H2O and that times molar mass of the anhydrous material. It's almost the same as the SiO2 problem. Check my thinking.
To solve this problem, you need to understand how to calculate the mass of water lost when a hydrate is heated.
First, we need to determine the molar mass of MgSO4•7H2O (magnesium sulfate heptahydrate).
The molar mass of magnesium (Mg) is 24.31 g/mol, the molar mass of sulfur (S) is 32.07 g/mol, and the molar mass of oxygen (O) is 16.00 g/mol.
For hydrogen (H), we multiply its atomic weight (1.01 g/mol) by the number of hydrogen atoms in a water molecule (2) and multiply this by seven since there are seven water molecules in MgSO4•7H2O.
Calculating the molar mass:
(24.31 g/mol Mg) + (32.07 g/mol S) + (4 * 16.00 g/mol O) + (7 * 2 * 1.01 g/mol H) = 246.47 g/mol
Now, we can use the molar mass to determine how much water is in 17.0 g of MgSO4•7H2O.
Dividing the given mass by the molar mass:
(17.0 g MgSO4•7H2O) / (246.47 g/mol MgSO4•7H2O) = 0.069 mol MgSO4•7H2O
Since the molar ratio between anhydrous magnesium sulfate (MgSO4) and water is 1:7, there are seven times more moles of water than moles of anhydrous magnesium sulfate.
Multiply the number of moles of MgSO4•7H2O by the ratio 7:1 to find the number of moles of water:
0.069 mol MgSO4•7H2O * (7 mol H2O / 1 mol MgSO4•7H2O) = 0.483 mol H2O
Now, we need to calculate the mass of this quantity of water:
Mass of water = (0.483 mol H2O) * (18.02 g/mol H2O) = 8.70 g H2O
Finally, subtract the mass of water lost from the initial mass of MgSO4•7H2O to find the mass of anhydrous magnesium sulfate that remains:
Mass of anhydrous MgSO4 = Initial mass - Mass of water lost
Mass of anhydrous MgSO4 = 17.0 g - 8.70 g = 8.30 g
Therefore, after thoroughly heating 17.0 g of MgSO4•7H2O, 8.30 g of anhydrous magnesium sulfate will remain.
Let me know if you need help with anything else!
To solve this problem, we need to understand the concept of hydrates and anhydrous compounds. A hydrate is a compound that contains water molecules, while an anhydrous compound is the same compound but without the water molecules. In this case, we are given 17.0 g of hydrated magnesium sulfate (MgSO4*7H2O) and we need to find the mass of anhydrous magnesium sulfate that remains after heating.
To solve this problem, we can follow these steps:
Step 1: Determine the molar mass of hydrated magnesium sulfate (MgSO4*7H2O).
- The molar mass of MgSO4 is:
- Mg: 24.31 g/mol
- S: 32.07 g/mol
- O: 16.00 g/mol (since there are four oxygen atoms in MgSO4)
- Summing these values: (24.31 + 32.07 + (16.00*4)) = 120.37 g/mol
- The molar mass of 7H2O is:
- H: 1.01 g/mol (since there are 14 hydrogen atoms in water)
- O: 16.00 g/mol (since there are 7 oxygen atoms in 7H2O)
- Summing these values: (1.01*14) + (16.00*7) = 126.14 g/mol
- To get the molar mass of the whole hydrate, sum the molar masses of the MgSO4 and 7H2O:
- 120.37 + 126.14 = 246.51 g/mol
Step 2: Calculate the number of moles of the hydrated magnesium sulfate.
- Divide the given mass (17.0 g) by the molar mass of the hydrate:
- 17.0 g / 246.51 g/mol = 0.069 moles
Step 3: Determine the molar mass of anhydrous magnesium sulfate (MgSO4).
- Same as mentioned before, the molar mass of MgSO4 is 120.37 g/mol.
Step 4: Calculate the mass of anhydrous magnesium sulfate that remains after heating.
- Multiply the number of moles of hydrated magnesium sulfate by the molar mass of anhydrous magnesium sulfate.
- 0.069 moles * 120.37 g/mol = 8.30 g
Therefore, if 17.0 g of MgSO4*7H2O is completely heated, approximately 8.30 g of anhydrous magnesium sulfate will remain.