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 you do the buffer part.
To convert from g MgSO4.7H2O to MgSO4, it's grams MgSO4.7H2O needed x (molar mass MgSO4/molar mass MgSO4.7H2O) = ?
2.44g
To calculate the amount of MgSO4 anhydrous needed to make 0.5L of the buffer with the same molar concentration as the recipe (5% w/v), we can follow these steps:
Step 1: Calculate the molar mass of MgSO4•7H2O:
Molar mass of MgSO4 = (24.31 g/mol for Mg) + (32.06 g/mol for S) + (4 * 16.00 g/mol for O) = 120.37 g/mol
Molar mass of H2O = (2 * 1.01 g/mol for H) + (16.00 g/mol for O) = 18.02 g/mol
Molar mass of MgSO4•7H2O = (120.37 g/mol for MgSO4) + (7 * 18.02 g/mol for H2O) = 246.48 g/mol
Step 2: Calculate the amount of MgSO4 anhydrous needed using the molar concentration and the molar mass.
Given that the molar concentration is 5% w/v, we can convert it to grams/L:
5% w/v = 5g/100ml = 50g/L
As we want to make 0.5L of the buffer, the required mass of MgSO4•7H2O would be:
Mass = 0.5L * 50g/L = 25g
Step 3: Determine the amount of MgSO4 anhydrous needed.
Since the molar mass of MgSO4•7H2O is 246.48 g/mol and we need 25g, we can use the formula:
Amount of MgSO4 in moles = Mass (in grams) / Molar mass
Amount of MgSO4 in moles = 25g / 246.48 g/mol ≈ 0.101 mol
However, MgSO4 anhydrous has a molar mass of 120.37 g/mol. Therefore, we divide the amount of MgSO4 in moles by the ratio of their molar masses:
Amount of MgSO4 anhydrous in moles = Amount of MgSO4•7H2O in moles / (Molar mass of MgSO4 / Molar mass of MgSO4•7H2O)
Amount of MgSO4 anhydrous in moles = 0.101 mol / (120.37 g/mol / 246.48 g/mol) ≈ 0.203 mol
Step 4: Calculate the mass of MgSO4 anhydrous needed.
Mass of MgSO4 anhydrous = Amount of MgSO4 anhydrous in moles * Molar mass of MgSO4 anhydrous
Mass of MgSO4 anhydrous = 0.203 mol * 120.37 g/mol ≈ 24.49 g
Therefore, to make 0.5L of the buffer with the same molar concentration as in the recipe, you will need approximately 24.49 grams of MgSO4 anhydrous.