if 125g of magnesium sulfate heptahydrate is completely dehydrated, how many grams of anhydrous salt will remain?
i would really appreciate anyone's help
MgSO4.7H2O ==> MgSO4 + 7H2O
mols MgSO4.7H2O = grams/molar mass = about 0.5 mol but that's approximate.
Therefore, you should obtain approximately 0.5 mol MgSO4.
g = mols x molar mas = ?
thank you!!
To determine the mass of anhydrous salt that will remain after completely dehydrating 125g of magnesium sulfate heptahydrate, we need to consider the water content in the compound.
Magnesium sulfate heptahydrate contains 7 water molecules per formula unit. The molar mass of magnesium sulfate heptahydrate is calculated as follows:
Magnesium (Mg): 24.31 g/mol
Sulfur (S): 32.07 g/mol
Oxygen (O): 16.00 g/mol
Mass of heptahydrate = (Mg + S + 7O) x 7 + 7H2O = (24.31 + 32.07 + (7 x 16.00)) x 7 + (7 x 18.02) = 246.47 g/mol
Now, we can set up a ratio to find the remaining mass of anhydrous salt using the molar masses of the compounds involved:
(125 g heptahydrate) / (246.47 g/mol heptahydrate) = (x g anhydrous salt) / (120.37 g/mol anhydrous salt)
Solving for x:
x = (125 g heptahydrate) x (120.37 g/mol anhydrous salt) / (246.47 g/mol heptahydrate)
x = 60.891 g
Therefore, approximately 60.891 grams of anhydrous salt will remain after completely dehydrating 125 grams of magnesium sulfate heptahydrate.
To find the number of grams of anhydrous salt that will remain after completely dehydrating 125g of magnesium sulfate heptahydrate, we need to first understand the compound's chemical formula and the process of dehydration.
The chemical formula of magnesium sulfate heptahydrate is MgSO4 * 7H2O, which indicates that each molecule of magnesium sulfate is associated with seven water molecules. When this compound is dehydrated, the water molecules are removed, leaving behind the anhydrous (without water) form of the salt.
To determine the amount of anhydrous salt remaining, we need to calculate the molar mass of both the heptahydrate and the anhydrous magnesium sulfate.
The molar mass of magnesium sulfate heptahydrate can be calculated as follows:
MgSO4: 24.31 g/mol (molar mass of magnesium) + 32.06 g/mol (molar mass of sulfur) + (4 x 16.00 g/mol) (4 oxygen atoms) = 120.37 g/mol
7H2O: (2 x 1.01 g/mol) (14 hydrogen atoms) + (16.00 g/mol) (7 oxygen atoms) = 126.14 g/mol
Summing the molar masses of the salt and water, we get:
120.37 g/mol (salt) + 126.14 g/mol (water) = 246.51 g/mol
Therefore, one mole of magnesium sulfate heptahydrate weighs 246.51 grams.
To determine the number of moles of magnesium sulfate heptahydrate in 125g, we divide the given mass by the molar mass:
125 g / 246.51 g/mol = 0.507 moles
Next, we need to consider the molar ratio between the heptahydrate and the anhydrous salt. From the formula MgSO4 * 7H2O, we can determine that for every one mole of MgSO4 * 7H2O, one mole of anhydrous MgSO4 is obtained.
Therefore, the number of moles of anhydrous salt remaining is also 0.507 moles.
Finally, we calculate the mass of the anhydrous salt by multiplying the number of moles by its molar mass:
0.507 moles x (24.31 g/mol + 32.06 g/mol + (4 x 16.00 g/mol)) = 41.08 g
Hence, if 125g of magnesium sulfate heptahydrate is completely dehydrated, approximately 41.08 grams of anhydrous salt will remain.