Many chemical salts come as hydrated crystals. That is, there is a specific ratio of water molecules associated with the salt crystals e.g. MgSO4.7H2O (called Magnesium sulphate heptahydrate FW = 246.5) has 7 water molecules to every Mg2+ and SO42- ion pair.
But it is also possible to get MgSO4 as an anhydrous powder (without water) FW =120.4. Once dissolved in water these salts have the same chemical properties but in the solid state they obviously have a different FW.
You are given a recipe for a buffer that contains 5% (w/v) MgSO4.7H2O. However, you only have MgSO4 anhydrous available to you. How much MgSO4 anhydrous do you need to make 0.5L of the buffer with the same molar concentration as in the recipe? Show your calculations.
(can not do any of it pls help)
Sure, I'd be happy to help!
To determine how much MgSO4 anhydrous you need to make the buffer solution with the same molar concentration, we need to consider the molar masses and stoichiometry of the compounds involved.
1. First, we need to find the molar mass of MgSO4.7H2O and MgSO4. The molar mass of MgSO4.7H2O can be calculated as follows:
MgSO4.7H2O = (MgSO4) + (7H2O)
= (24.31 g/mol + 32.06 g/mol + (4 * 16.00 g/mol)) + (7 * (2 * 1.01 g/mol + 16.00 g/mol))
= 120.4 g/mol
2. Next, we need to find the molar mass of MgSO4:
MgSO4 = (24.31 g/mol + 32.06 g/mol + (4 * 16.00 g/mol))
= 120.4 g/mol
3. The molar mass of the water is 18.02 g/mol.
4. Now, we can calculate the molar mass of water in MgSO4.7H2O:
Water in MgSO4.7H2O = 7 * (2 * 1.01 g/mol + 16.00 g/mol)
= 14.14 g/mol
5. The remaining mass in MgSO4.7H2O comes from the MgSO4:
Mass of MgSO4 in MgSO4.7H2O = MgSO4.7H2O - (Water in MgSO4.7H2O)
= 120.4 g/mol - 14.14 g/mol
= 106.26 g/mol
6. Now we need to determine the molar concentration of MgSO4.7H2O in the recipe. We are given that the recipe contains 5% (w/v) MgSO4.7H2O, meaning that there are 5 grams of MgSO4.7H2O in every 100 grams of the solution. Since we want to make 0.5L of the buffer, we need to calculate the number of moles of MgSO4.7H2O required:
Moles of MgSO4.7H2O in recipe = (5 g / 100 g) / MgSO4.7H2O molar mass
= (0.05) / 120.4 mol
= 0.000415 mol
7. Finally, we can calculate how many grams of MgSO4 anhydrous are needed to provide the same number of moles of MgSO4:
Mass of MgSO4 anhydrous = Moles of MgSO4 * MgSO4 molar mass
= 0.000415 mol * 120.4 g/mol
= 0.0499 g (rounded to 3 significant digits)
So, you would need approximately 0.0499 grams of MgSO4 anhydrous to make 0.5L of the buffer with the same molar concentration as in the recipe.
Please note that this is a rough estimate and may not account for any impurities or variations in the purity of the MgSO4 anhydrous used.
To calculate the amount of MgSO4 anhydrous needed to make the buffer with the same molar concentration, we need to consider the difference in molecular weight between MgSO4.7H2O and MgSO4.
1. Calculate the molar mass of MgSO4.7H2O:
Mg: 24.31 g/mol
S: 32.07 g/mol
O: 16.00 g/mol (x4)
H: 1.01 g/mol (x14)
Water: 18.02 g/mol (x7)
Molar mass of MgSO4.7H2O = (24.31 + 32.07 + (16.00x4) + (1.01x14) + (18.02x7)) g/mol
= 246.52 g/mol
2. Calculate the molar mass of MgSO4:
Mg: 24.31 g/mol
S: 32.07 g/mol
O: 16.00 g/mol (x4)
Molar mass of MgSO4 = (24.31 + 32.07 + (16.00x4)) g/mol
= 120.37 g/mol
3. Calculate the amount of MgSO4 anhydrous needed:
Let's assume the buffer requires a molar concentration of X.
Molar concentration of MgSO4.7H2O = (5% w/v) / (molar mass of MgSO4.7H2O)
Molar concentration of MgSO4 = (molar concentration of MgSO4.7H2O) * (molar mass of MgSO4) / (molar mass of MgSO4.7H2O)
Volume of buffer needed = 0.5 L
Mass of MgSO4 anhydrous needed = (molar concentration of MgSO4) * (molar mass of MgSO4 anhydrous) * (volume of buffer needed)
Substituting the values into the equation:
Mass of MgSO4 anhydrous needed = (X / 246.52) * 120.37 * 0.5
Simplifying the equation gives the final answer in grams.
You need to provide the desired molar concentration (X) in order for us to calculate the exact amount of MgSO4 anhydrous needed.
To calculate the amount of MgSO4 anhydrous needed to make the buffer with the same molar concentration as in the recipe, you need to convert the recipe's percentage and molar concentration into a usable form.
First, let's convert the given percentage of MgSO4.7H2O (5% w/v) into a molar concentration.
Step 1: Calculate the mass of MgSO4.7H2O required for 0.5L of the buffer.
Given:
Percentage of MgSO4.7H2O = 5% (w/v)
Volume of buffer = 0.5L
To convert the percentage (w/v) to grams, we need to know the density of the solution. Since the density is not provided, we will assume the density is 1 g/mL for simplicity.
Step 2: Calculate the molar mass of MgSO4.7H2O (Magnesium sulphate heptahydrate).
The molar mass of MgSO4.7H2O is provided as 246.5 g/mol.
Step 3: Calculate the molar concentration (M) of MgSO4.7H2O in the recipe.
Molar concentration is calculated using the formula:
Molarity (M) = (Mass in grams / Molar mass) / Volume in liters
Given:
Mass of MgSO4.7H2O = 5% (w/v) * 0.5L = 0.025L
Molar concentration (M) = (0.025L / 246.5 g/mol) / 0.5L
Step 4: Calculate the amount of MgSO4 anhydrous needed to achieve the same molar concentration.
Now that we have the molar concentration of MgSO4.7H2O, we can convert it to the molar concentration of MgSO4 anhydrous.
The molar mass of MgSO4 anhydrous is provided as 120.4 g/mol.
To calculate the amount of MgSO4 anhydrous needed, we use the formula:
(Molar concentration * Molar mass) = Mass of MgSO4 anhydrous
Mass of MgSO4 anhydrous = (Molar concentration * Molar mass) * Volume in liters
Mass of MgSO4 anhydrous = (Molar concentration * 120.4 g/mol) * 0.5L
Now, you can plug in the calculated values to get the amount of MgSO4 anhydrous needed in grams.