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?
2.5
weel if u crry the 2.5 then do a backflip then it does it all
To determine how much MgSO4 anhydrous is needed to make the buffer with the same molar concentration, we need to first calculate the molar concentration of MgSO4.7H2O in the given recipe.
Step 1: Calculate the molar mass of MgSO4.7H2O
Molar mass of MgSO4 = 24.3 g/mol (Mg) + 32.1 g/mol (S) + (4 × 16.0 g/mol) (4 O) = 120.4 g/mol
Molar mass of H2O = 2 × (1.0 g/mol) (2 H) + 16.0 g/mol (O) = 18.0 g/mol
Molar mass of MgSO4.7H2O = 120.4 g/mol + (7 × 18.0 g/mol) = 246.5 g/mol
Step 2: Determine the amount of MgSO4.7H2O required in the buffer
Given that the buffer needs to be made with a concentration of 5% (w/v) MgSO4.7H2O, this means that 5g of MgSO4.7H2O is present in every 100 mL (or 0.1 L) of solution.
Step 3: Convert the amount of MgSO4.7H2O to moles
Since the molar mass of MgSO4.7H2O is 246.5 g/mol, we can calculate the number of moles of MgSO4.7H2O in 5g:
moles = mass / molar mass
moles = 5g / 246.5 g/mol = 0.0203 mol
Step 4: Determine the amount of MgSO4 anhydrous required
Since the molar ratio between MgSO4.7H2O and MgSO4 is 1:1, we can use the same number of moles calculated in Step 3 to determine the amount of MgSO4 anhydrous.
moles of MgSO4 = 0.0203 mol
To calculate the mass of MgSO4 anhydrous:
mass = moles × molar mass
mass = 0.0203 mol × 120.4 g/mol = 2.46 g
Therefore, you would need 2.46 grams of MgSO4 anhydrous to make 0.5L of the buffer with the same molar concentration as in the recipe.