What would be the mass of a bag of anhydrous magnesium sulfate, MgSO4, if it contained the same amount of magnesium as a 1.00 kg bag of Epsom salts, MgSO4 * 7H2O? Give your answer in grams.
There are long ways and short ways to do this but the best way is to use something called the chemical factor. That isn't taught in chemistry courses anymore. The chemical factor has gone the way of normality of solutions; that isn't taught anymore either. Shame. mm will stand for molar mass. am stands for atomic mass.
1,000g MgSO4.7H2O x (am Mg/MgSO4.7H2O) x (mm MgSO4/am Mg) = ?
You see the first factor convert from grams MgSO4.7H2O to grams Mg while the second factor converts from Mg to MgSO4. Obviously the atomic mass of Mg cancels so you need only one factor which is
1,000 g MgSO4.7H2O x (mm MgSO4/mm MgSO4.7H2O) = ? g MgSO4.
Most schools now teach this type problem by calculating the percent MgSO4 in MgSO4.7H2O and changing that to 1000 g sample instead of a 100 g sample. Post your work if you get stuck.
Well, you're really taking the "mass" in "mass-terpiece" to a whole new level with this question! Let's dive into the magnesium-filled world of chemistry.
MgSO4 * 7H2O may look like a mouthful, but it's just good old Epsom salts with seven water molecules hanging around, having a hydration party. Now, if we strip away those pesky water molecules, we're left with anhydrous magnesium sulfate, which is just the solo version.
Given that a 1.00 kg bag of Epsom salts has the same amount of magnesium as our magnesium sulfate party guest, it means they're basically twins separated at birth. The only difference is those extra water molecules. So you want to know the mass of just the magnesium sulfate.
Since the mass of the Epsom salts is 1.00 kg, we need to subtract the mass of the water molecules from the equation. The molar mass of H2O is approximately 18 g/mol, and the molar mass of MgSO4 is around 120 g/mol. So, there are 7 mol of water in the Epsom salts.
By multiplying the number of moles of water (7 mol) by the molar mass of water (18 g/mol), we find that the total mass of water is 126 g.
Now, we subtract that 126 g from the total mass of the Epsom salts (1.00 kg or 1000 g), and we get the mass of the anhydrous magnesium sulfate. I hope you're ready for this punchline...
*drumroll*
The mass of the anhydrous magnesium sulfate is 874 grams! Ta-da!
So, humor aside, the anhydrous magnesium sulfate would have a mass of 874 grams.
To determine the mass of the anhydrous magnesium sulfate (MgSO4), we need to compare it to the hydrated form of magnesium sulfate (MgSO4 * 7H2O).
First, we'll find the molar mass of magnesium sulfate (MgSO4):
Molar mass of Mg = 24.31 g/mol
Molar mass of S = 32.07 g/mol
Molar mass of O = 16.00 g/mol
Therefore, molar mass of MgSO4 = (1 * Mg) + (1 * S) + (4 * O)
= (1 * 24.31) + (1 * 32.07) + (4 * 16.00)
= 24.31 + 32.07 + 64.00
= 120.38 g/mol
Since there is 1.00 kg (1000 g) of Epsom salts, we can calculate the number of moles present in that amount using the molar mass of the hydrated compound:
Number of moles in 1.00 kg of MgSO4 * 7H2O = (mass of Epsom salts) / (molar mass of MgSO4 * 7H2O)
= 1000 g / (246.47 g/mol)
= 4.06 mol
Now, we need to consider that 1 mole of MgSO4 * 7H2O contains 1 mole of MgSO4. Therefore, the number of moles of MgSO4 is also 4.06 mol.
To determine the mass of the anhydrous MgSO4, we can use the equation:
Mass = Number of moles * Molar mass
Mass of anhydrous MgSO4 = 4.06 mol * 120.38 g/mol
= 488.63 g
Therefore, the mass of the anhydrous magnesium sulfate (MgSO4) would be 488.63 grams.
To determine the mass of a bag of anhydrous magnesium sulfate (MgSO4), if it contained the same amount of magnesium as a 1.00 kg bag of Epsom salts (MgSO4 * 7H2O), we need to consider the molar ratio between the two compounds.
1. First, calculate the molar mass of MgSO4 and MgSO4 * 7H2O:
- MgSO4:
Molar mass of magnesium (Mg) = 24.31 g/mol
Molar mass of sulfur (S) = 32.07 g/mol
Molar mass of oxygen (O) = 16.00 g/mol (4 oxygen atoms in the formula)
Therefore, the molar mass of MgSO4 is:
(24.31 g/mol) + (32.07 g/mol) + (16.00 g/mol x 4) = 120.37 g/mol
- MgSO4 * 7H2O:
Molar mass of water (H2O) = 18.02 g/mol (2 hydrogen atoms and 1 oxygen atom per molecule)
Therefore, the molar mass of MgSO4 * 7H2O is:
(24.31 g/mol) + (32.07 g/mol) + (16.00 g/mol x 4) + (18.02 g/mol x 14) = 246.47 g/mol
2. Next, we need to calculate the molar ratio of magnesium (Mg) in the two compounds:
In MgSO4, there is one magnesium atom per molecule.
In MgSO4 * 7H2O, there is also one magnesium atom per molecule.
Therefore, the molar ratios of magnesium are the same (1:1).
3. Now, we know that the 1.00 kg bag of Epsom salts (MgSO4 * 7H2O) contains the same amount of magnesium as the bag of anhydrous magnesium sulfate.
To find the mass of the anhydrous MgSO4, we can use the following equation:
(1.00 kg) * (1000 g/kg) = x g (mass of MgSO4)
The mass of anhydrous MgSO4 would be equal to 1000 grams.
Therefore, the mass of the bag of anhydrous magnesium sulfate (MgSO4) would be 1000 grams.
Remember to double-check the calculations to ensure accuracy.